gate 2021...gate examination has been emerging as one of the most prestigious competitive exam for...

TM

Electronics & Communication Engineering

(Volume - I)

GATE 2021

TOPIC WISE GATE SOLUTIONS1987-2020

GATE examination has been emerging as one of the most prestigious competitive exam for engineers. Earlier it was considered to be an exam just for eligibility for pursuing PG courses, but now GATE exam has gained a lot of attention of students as this exam open an ocean of possibilities like : 1. Admission into IISc, IITs, IIITs, NITs A good GATE score is helpful for getting admission into IISc, IITs, IIITs, NITs and many other

renowned institutions for M.Tech./M.E./M.S. An M.Tech graduate has a number of career opportunities in research fields and education industries. Students get ` 12,400 per month as stipend during their course.

2. Selection in various Public Sector Undertakings (PSUs) A good GATE score is helpful for getting job in government-owned corporations termed

as Public Sector Undertakings (PSUs) in India like IOCL, BHEL, NTPC, BARC, ONGC, PGCIL, DVC, HPCL, GAIL, SAIL & many more.

3. Direct recruitment to Group A level posts in Central government, i.e., Senior Field Officer (Tele), Senior Research Officer (Crypto) and Senior Research Officer (S&T) in Cabinet Secretariat, Government of India, is now being carried out on the basis of GATE score.

4. Foreign universities through GATE GATE has crossed the boundaries to become an international level test for entry into

postgraduate engineering programmes in abroad. Some institutes in two countries Singapore and Germany are known to accept GATE score for admission to their PG engineering programmes.

5. National Institute of Industrial Engg. (NITIE) • NITIE offers PGDIE / PGDMM / PGDPM on the basis of GATE scores. The shortlisted

candidates are then called for group Discussion and Personal Interview rounds. • NITIE offers a Doctoral Level Fellowship Programme recognized by Ministry of HRD

(MHRD) as equivalent to PhD of any Indian University. • Regular full time candidates those who will qualify for the financial assistance will receive

` 25,000 during 1st and 2nd year of the Fellowship programme and ` 28,000 during 3rd, 4th and 5th year of the Fellowship programme as per MHRD guidelines.

6. Ph.D. in IISc/ IITs • IISc and IITs take admissions for Ph.D. on the basis of GATE score. • Earn a Ph.D. degree directly after Bachelor’s degree through integrated programme. • A fulltime residential researcher (RR) programme.

7. Fellowship Program in management (FPM) • Enrolment through GATE score card • Stipend of ` 22,000 – 30,000 per month + HRA • It is a fellowship program • Application form is generally available in month of sept. and oct.

Note : In near future, hopefully GATE exam will become a mandatory exit test for all engineering students, so take this exam seriously. Best of LUCK !

IMPORTANCE of GATE

Section Question No. No. of Questions Marks Per Question Total Marks

General Aptitude

1 to 5 5 1 5 6 to 10 5 2 10

Technical +

Engineering Mathematics

1 to 25 25 1 25

26 to 55 30 2 60

Total Duration : 3 hours Total Questions : 65 Total Marks : 100 Note : 40 to 45 marks will be allotted to Numerical Answer Type Questions

Pattern of Questions : (i) Multiple Choice Questions (MCQ) carrying 1 or 2 marks each in all the papers and

sections. These questions are objective in nature, and each will have a choice of four answers, out of which the candidate has to select (mark) the correct answer.

Negative Marking for Wrong Answers : For a wrong answer chosen in a MCQ, there will be negative marking. For 1-mark MCQ, 1/3 mark will be deducted for a wrong answer. Likewise for, 2-marks MCQ, 2/3 mark will be deducted for a wrong answer.

(ii) Numerical Answer Type (NAT) Questions carrying 1 or 2 marks each in all the papers and sections. For these questions, the answer is a signed real number, which needs to be entered by the candidate using the virtual numeric keypad on the monitor (Keyboard of the computer will be disabled). No choices will be shown for these type of questions. The answer can be a number such as 10 or – 10 (an integer only). The answer may be in decimals as well, for example, 10.1 (one decimal) or 10.01 (two decimal) or –10.001 (three decimal). These questions will be mentioned with, up to which decimal places, the candidates need to make an answer. Also, an appropriate range will be considered while evaluating the numerical answer type questions so that the candidate is not penalized due to the usual round-off errors. Wherever required and possible, it is better to give NAT answer up to a maximum of three decimal places.

Note : There is NO negative marking for a wrong answer in NAT questions.

GATE Exam Pattern

GATE ACADEMY Team took several years’ to come up with the solutions of GATE examination. It is because we strongly believe in quality. We have significantly prepared each and every solution of the questions appeared in GATE, and many individuals from the community have taken time out to proof read and improve the quality of solutions, so that it becomes very lucid for the readers. Some of the key features of this book are as under : This book gives complete analysis of questions chapter wise as well as year wise. Video Solution of important conceptual questions has been given in the form of QR code and

by scanning QR code one can see the video solution of the given question. Solutions has been presented in lucid and understandable language for an average student. In addition to the GATE syllabus, the book includes the nomenclature of chapters according to

text books for easy reference. Last but not the least, author's 10 years experience and devotion in preparation of these

solutions. Steps to Open Video solution through mobile.

Note : For recent updates regarding minor changes in this book, visit -

www.gateacademy.co.in. We are always ready to appreciate and help you.

(1) Search for QR Code scanner

in Google Play / App Store.

(2) Download & Install any

QR Code Scanner App.

(3) Scan the given QR Code

for par�cular ques�on.

(4) Visit the link generated & you’ll

be redirect to the video solu�on.

What is special about this book ?

It is our pleasure, that we insist on presenting “GATE 2021 Electronics & Communication Engineering (Volume - I)” authored for Electronics & Communication Engineering to all of the aspirants and career seekers. The prime objective of this book is to respond to tremendous amount of ever growing demand for error free, flawless and succinct but conceptually empowered solutions to all the question over the period 1987 - 2020. This book serves to the best supplement the texts for Electronics & Communication Engineering but shall be useful to a larger extent for Electrical Engineering and Instrumentation Engineering as well. Simultaneously having its salient feature the book comprises : Step by step solution to all questions. Complete analysis of questions, i.e. chapter wise as well as year wise. Detailed explanation of all the questions. Solutions are presented in simple and easily understandable language. Video solutions for good questions. It covers all GATE questions from 1987 to 2020 (34 years). The authors do not sense any deficit in believing that this title will in many aspects, be different from the similar titles within the search of student. We would like to express our sincere appreciation to Mrs. Sakshi Dhande Mam (Co-Director, GATE ACADEMY Learning Pvt. Ltd.) for her constant support and constructive suggestions and comments in reviewing the script. In particular, we wish to thank GATE ACADEMY expert team members for their hard work and consistency while designing the script. The final manuscript has been prepared with utmost care. However, going a line that, there is always room for improvement in anything done, we would welcome and greatly appreciate the suggestions and corrections for further improvement.

Umesh Dhande

(Director, GATE ACADEMY Learning Pvt. Ltd.)

PREFACE

S. No. Topics Page No.

1. Network Theory 1. Basic Concepts of Networks 2. Network Theorems 3. Two port Networks 4. Transient Analysis 5. Sinusoidal Steady State Analysis 6. Phasor & Locus Diagram 7. Complex Power 8. Resonance 9. Magnetic Coupling 10. Network Functions & Filters 11. Graph Theory

2. Signals & Systems 1. Basics of Signals 2. Classification of Systems 3. Laplace Transform 4. Continuous Time Convolution 5. Continuous Time Fourier Transform 6. Continuous & Discrete Time Fourier Series 7. Z - Transform 8. Discrete Time Convolution 9. DTFT and DFT

3. Digital Electronics 1. Number Systems 2. Boolean Algebra 3. Logic Gates

CONTENTS

4. Combinational Circuits 5. Sequential Circuits 6. Logic Families 7. Semiconductor Memories 8. ADC and DAC 9. Microprocessor

4. Control Systems 1. Basics of Control Systems 2. Block Diagrams & Signal Flow Graph 3. Time Response Analysis 4. Routh’s Stability Criterion 5. Root Locus 6. Polar Plot & Nyquist Stability Criterion 7. Bode Plot 8. Frequency Response of Second Order Systems 9. Controllers & Compensators 10. State Space Analysis

5. Engineering Mathematics 1. Linear Algebra 2. Differential Equations 3. Integral & Differential Calculus 4. Vector Calculus 5. Maxima & Minima 6. Mean Value Theorem 7. Complex Variables 8. Limit & Series Expansion 9. Probability & Statistics 10. Numerical Methods

1.1 A network contains linear resistors and

ideal voltage source. If values of all the resistors are doubled, then the voltage across each resistor is

(A) halved (B) doubled (C) increased by four times (D) not changed 1.2 The two electrical sub networks

1 2andN N are connected through three resistors as shown in figure. The voltage across 5 ohm resistor and 1 ohm resistor are given to be 10 V and 5 V, respectively. Then voltage across 15 ohm resistor is

(A) – 105 V (B) + 105 V (C) – 15 V (D) + 15 V

1.3 A DC voltage source is connected across

a series RLC circuit. Under steady state conditions, the applied DC voltage drops entirely across the

(A) R only (B) L only (C) C only (D) R and L combination 1.4 The current, i(t) through a 10 Ω resistor

in series with an inductance, is given by 0( ) 3 4sin (100 45 )i t t= + +

04sin (300 60 )t+ + Amperes. The RMS value of the current and the

power dissipated in the circuit are

(A) 41A, 410 W, respectively

(B) 35 A, 350 W, respectively (C) 5 A, 250 W, respectively (D) 11 A, 1210 W, respectively

1.5 The voltages

1 2 3, andC C CV V V across the

capacitors in the circuit in figure, under steady state, are respectively

(A) 80 V, 32 V, 48 V (B) 80 V, 48 V, 32 V (C) 20 V, 8 V, 12 V (D) 20 V, 12 V, 8 V

1Basic Concepts

of Networks

15 �

5 �

1 �

5V –�

10 V –�1N 2N

1 H 2 F10 k� 25 k�2 H

3 F

�

�40 k�

�

1CV100 V 1 F

2CV

3CV

1993 IIT Bombay

1995 IIT Kanpur

1996 IISc Bangalore

1.2 Topic Wise GATE Solutions [EC] GATE ACADEMY ®

1.6 The current 4i in the circuit of figure is

equal to

(A) 12 A (B) –12 A (C) 4 A (D) None of these 1.7 The voltage V in figure is equal to

(A) 3 V (B) – 3 V (C) 5 V (D) None of these 1.8 The voltage V in figure is always equal to

(A) 9 V (B) 5 V (C) 1 V (D) None of these 1.9 The voltage V in figure is

(A) 10 V (B) 15 V (C) 5 V (D) None of these

1.10 In the circuit of figure the equivalent impedance seen across terminals AB is

(A) (16/3) Ω (B) (8/3) Ω (C) (8/3 12 )j+ Ω (D) None of these

1.11 The nodal method of circuit analysis is

based on (A) KVL and Ohm’s law. (B) KCL and Ohm’s law. (C) KCL and KVL. (D) KCL, KVL and Ohm’s law. 1.12 The voltage across the terminals a and b

in figure is

(A) 0.5 V (B) 3.0 V (C) 3.5 V (D) 4.0 V

1.13 A delta-connected network with its star

equivalent is shown in figure. The resistances 1 2,R R and 3R (in ohms) are respectively

(A) 1.5, 3 and 9 (B) 3, 9 and 1.5 (C) 9, 3 and 1.5 (D) 3, 1.5 and 9

2 3 Ai �1 5 Ai �

0 7 Ai �

4 ?i � 3 4 Ai �

I

+–

5 V

4 V

+–

4 V

V

2 �

3 �

2 �

�

�

5 VV

2 A

3 �

V 10 V 5 V

3j �

4j� �

2 � 4 �

A

B

eqZ

4 �2 �

3 A2 �

2 � 1 �

1 V

a

b

a

b c

5 � 30 �

15 �

1R

a

b c

3R2R

1997 IIT Madras

1998 IIT Delhi

1999 IIT Bombay

1.3GATE ACADEMY ® Network Theory : Basic Concepts of Networks

1.14 In the given circuit, the voltage ( )v t is

(A) at bte e− (B) at bte e+

(C) at btae be− (D) at btae be+

1.15 The value of the voltage source E is

(A) – 16 V (B) 4 V (C) – 6 V (D) 16 V

1.16 The voltage 0e in figure is

(A) 2 V (B) 4 V3

(C) 4 V (D) 8 V

1.17 If each branch of a Delta circuit has impedance 3 ,Z then each branch of the equivalent Y circuit has impedance

(A) 3

Z (B) 3Z

(C) 3 3 Z (D) 3Z

1.18 The voltage 0e in the figure is

(A) 48 V (B) 24 V (C) 36 V (D) 28 V

1.19 The dependent current source shown in

the figure

(A) delivers 80 W (B) absorbs 80 W (C) delivers 40 W (D) absorbs 40 W

1.20 Twelve 1 Ω resistances are used as edges

to form a cube. The resistance between two diagonally opposite corners of the cube is

(A) 56

Ω (B) 1 Ω

(C) 65

Ω (D) 32

Ω

1.21 An ideal sawtooth voltage waveform of frequency 500 Hz and amplitude 3 V is generated by charging a capacitor of 2 Fμ in every cycle. The charging requires

(A) constant voltage source of 3 V for 1 ms.

(B) constant voltage source of 3 V for 2 ms.

(C) constant current source of 3 mA for 1 ms.

(D) constant current source of 3 mA for 2 ms.

1 � 1 �

ate ( )v t 1 H bte

1 V2 V

4 V 5 V

1v

E

10 V

2v 0 V

4 �

0e12 V

2 �

2 �4 �

2 �

10 � 12 �8 A

6 �

16 V

0e

5 �

5 �

1 20 VV �1 A

5

V

2000 IIT Kharagpur

2001 IIT Kanpur

2002 IISc Bangalore

2003 IIT Madras


1.22 The rms value of the periodic waveform

given in figure is

(A) 2 6 A (B) 6 2 A

(C) 4 / 3 A (D) 1.5 A

1.23 If 1 2 4R R R R= = = and 3 1.1R R= in the

bridge circuit shown in the figure, then the reading in the ideal voltmeter connected between a and b is

(A) 0.238 V (B) 0.138 V (C) 0.238 V− (D) 1.0 V

1.24 In the interconnection of ideal sources

shown in the below figure, it is known that the 60 V source is absorbing power. Which of the following can be the value of the current source I ?

(A) 10 A (B) 13 A (C) 15 A (D) 18 A

1.25 A fully charged mobile phone with a 12 V battery is good for a 10 minute talk-time. Assume that, during the talk-time, the battery delivers a constant current of 2 A and its voltage drop linearly varies from 12 V to 10 V as shown in the figure. How much energy does the battery deliver during this talk-time?

(A) 200 J (B) 12 kJ (C) 13.2 kJ (D) 14.4 kJ

1.26 In the circuit shown, the power supplied

by the voltage source is

(A) 0 W (B) 5 W (C) 10 W (D) 100 W

1.27 If 6VA BV V− = , then C DV V− is

(A) – 5 V (B) 2 V (C) 3 V (D) 6 V

i t( )

6 A

t0

– 6 A

T/2 T

10 V

1R

2R 3R

4R

a bV

I 60 V

12 A

20 V

10 min0t

10 V

12 V

v t( )

1 �

1 �1 �

1 �

1 �

1 A

2 A

10 V

RR

R R RR R

R

BV

DVCV

AV

5 V

10 V

2 �

1 �

2 A

2004 IIT Delhi

2005 IIT Bombay

2009 IIT Roorkee

2010 IIT Guwahati

2012 IIT Delhi


1.28 Consider a delta connection of resistors

and its equivalent star connection is shown below. If all elements of the delta connection are scaled by a factor k, k > 0, the elements of the corresponding star equivalent will be scaled by a factor of

(A) 2k (B) k

(C) 1/k (D) k

1.29 Three capacitors 1 2,C C and 3C whose values are 10 F, 5 Fμ μ and 2 Fμ respectively, have breakdown voltages of 10 V, 5 V and 2 V respectively. For the interconnection shown below, the maximum safe voltage in Volts that can be applied across the combination, and the corresponding total charge in Cμ stored in the effective capacitance across the terminals are respectively,

(A) 2.8 and 36 (B) 7 and 119 (C) 2.8 and 32 (D) 7 and 80

Consider the following figure.

1.30 The current sI in Amps in the voltage

source, and voltage sV in Volts across the current source respectively, are

(A) 13, 20− (B) 8, 10−

(C) 8,20− (D) 13,20−

1.31 The current in the 1 Ω resistor in Amps is

(A) 2 (B) 3.33 (C) 10 (D) 12

1.32 Consider the configuration shown in the

figure which is a portion of a larger electrical network. [Set - 01]

For 1R = Ω and currents 1 2 A,i =

4 1A,i = − 5 4 A,i = − which one of the following is TRUE?

(A) 6 5 Ai =

(B) 3 4 Ai = −

(C) Data is sufficient to conclude that the supposed currents are impossible.

(D) Data is insufficient to identify the currents 2,i 3i and 6i .

1.33 A Y-network has resistance of 10 Ω each in two of its arms, while the third arm has a resistance of 11 Ω . In the equivalent

-Δ network, the lowest value (in Ω ) among the three resistance is ________. [Set - 01]

aR

cRbR

CR BR

AR

2C 3C

1C

1 �2 �

5 �

SV 2 A

SI10 V

RR

R

1i

2i

3i

4i

5i

6i

2013 IIT Bombay

Common Data for Questions 1.30 & 1.31

2014 IIT Kharagpur


1.34 A periodic variable x is shown in the figure as a function of time. The root-mean-square (rms) value of x is ________. [Set - 01]

1.35 In the figure shown, the value of current

I (in Amperes) is _______. [Set - 03]

1.36 For the Y-network shown in the figure,

the value of 1R (in Ω ) in the equivalent Δ -network is _________. [Set - 03]

1.37 The circuit shown in the figure represents

a [Set - 04]

(A) voltage controlled voltage source. (B) voltage controlled current source. (C) current controlled current source. (D) current controlled voltage source.

1.38 The magnitude of current (in mA) through the resistor 2R in the figure shown is _______. [Set - 04]

1.39 The equivalent resistance in the infinite

ladder network shown in the figure, is eR . [Set - 04]

The value of eRR

is_________.

1.40 In the network shown in the figure, all

resistors are identical with 300R = Ω . The resistance abR (in Ω ) of the

network is __________. [Set - 01]

1.41 In the given circuit, the values of 1V and

2V respectively are [Set - 01]

(A) 5 V, 25 V (B) 10 V, 30 V (C) 15 V, 35 V (D) 0 V, 20 V

/2T/2T

t

( )x t

1

0

5 �

10 �5 V 1 A

I5 �

1R

5 �3 �

7.5 �

iI 1 iA I R

2 mA10 mA 2 k� 4 k�

1 k�

3 k�

1R

2R

3R

4R

R R R

R R R R

R

eR

R

RR R

RR

R

RR

a

b

R

RabR

R

R

R

R

1V2V 4� 4� 2 I

I

4�

��

��

5A

2015 IIT Kanpur


1.42 In the circuit shown, the voltage xV (in Volts) is ______. [Set - 03]

1.43 In the given circuit, each resistor has a

value equal to 1 Ω .

What is the equivalent resistance across

the terminals a and b ? [Set - 02] (A) 1 / 6 Ω (B) 1 / 3 Ω (C) 9 / 20 Ω (D) 8 /15 Ω 1.44 In the circuit shown in the figure, the

magnitude of the current (in amperes) through 2R is ______. [Set - 02]

1.45 In the figure shown, the current i (in

ampere) is __________. [Set - 03]

1.46 A connection is made consisting of

resistance A in series with a parallel combination of resistances B and C. Three resistors of value 10 , 5 , 2 Ω Ω Ω are provided. Consider all possible permutations of the given resistors into the positions A, B, C and identify the configurations with maximum possible overall resistance, and also the ones with minimum possible overall resistance. The ratio of maximum to minimum values of the resistances (up to second decimal place) is _____. [Set - 02]

1.47 Consider the network shown below with

1 1 ,R = Ω 2 2R = Ω and 3 3R = Ω . The network is connected to a constant voltage source of 11V.

The magnitude of the current (in

amperes, accurate to two decimal places) through the source is _______.

1.48 In the circuit shown below, the Thevenin

voltage thV is

5 A xV 8 � 0.25 xV

10 �

0.5 xV

20 �

a

b

60 V

1R

5 �

2R

3 �

5 �0.04 xv xv

i

5 �

1 � 1 �

1 �

8 V

8 V1 A

1 �

+–

11V

1R1R

2R

3R

3R

2R

1R1R

1R1R

2016 IISc Bangalore

2017 IIT Roorkee

2018 IIT Guwahati

2020 IIT Delhi


(A) 4.5 V (B) 3.6 V (C) 2.4 V (D) 2.8 V 1.49 The current I in the given network is

(A) 02.38 96.37 A∠ − (B) 0 A (C) 02.38 23.63 A∠ −

(D) 02.38 143.63 A∠

1.1 D 1.2 A 1.3 C 1.4 C 1.5 B

1.6 B 1.7 A 1.8 D 1.9 A 1.10 B

1.11 B 1.12 C 1.13 D 1.14 D 1.15 A

1.16 C 1.17 A 1.18 D 1.19 A 1.20 A

1.21 D 1.22 A 1.23 C 1.24 A 1.25 C

1.26 A 1.27 A 1.28 B 1.29 C 1.30 D

1.31 C 1.32 A 1.33 29 1.34 0.408 1.35 0.5

1.36 10 1.37 C 1.38 2.8 1.39 2.62 1.40 100

1.41 A 1.42 8 1.43 D 1.44 5 1.45 –1

1.46 2.14 1.47 8 1.48 B 1.49 D

Concepts of Absorbed and Delivered Power

Concepts of Voltmeter and Ammeter

Initially : Let network contains three resistors of same value.

Voltage across each resistor by using voltage division rule is given by,

3 3RR VV VR

= × =

1 �1A 2A 2 �

2V2 � 4 �

thV

�

�

+–

Z

Z

Z

0120 90 V∠ −

(80 35)j− Ω

0120 30 V∠ −

I

Scan for Video

Explanation

Scan for Video

Explanation

R R R

V

I

Answers Basic Concepts of Networks

Explanations Basic Concepts of Networks

1.1 (D)


Later : ' 2R R=

Voltage across each resistor by using voltage division rule is given by,

''3 'RRV VR

= ×

2'6R

RV VR

= ×

'3 3RR VV VR

= × =

If resistors are doubled then current will decrease by same factor and voltage across each resistor remains constant. Hence, voltage across each resistor is not changed. Hence, the correct option is (D).

Given circuit is shown below,

From the figure,

210 2 A5

i = = , 35 5 A1

i = =

Applying KCL at 1N , 1 2 3 0i i i+ + = 1 2 3( ) 7 Ai i i= − + = − Voltage across 15 Ω resistor is, 15 1 15 15 ( 7) 105 VV iΩ = × = × − = − Hence, the correct option is (A). Key Point Passive sign convention :

1.

2.

3.


In given circuit, DC voltage is applied. Hence, in steady state inductor behaves as a short circuit and capacitor behaves as an open circuit.

Applying KVL in the loop shown, CV V= Hence, the applied DC voltage drops entirely across the capacitor. Hence, the correct option is (C). Key Point For DC source, 0 0fω = = Hence, 0LX L= ω = (Short circuit)

and 1CX

C= = ∞

ω (Open circuit)

Given : 0( ) 3 4sin(100 45 )i t t= + +

04sin(300 60 )t+ + RMS value of i(t) is given by,

2 2

2 4 4(3)2 2rmsI = + +

9 8 8 5 ArmsI = + + =

'R

V

'R 'R'I

15 �

1 �

5V –�

–V�

1N 2N

1i

2i

3i

5 �

10 V –�

v

i Rv iR�

v

i L

div L

dt�

v

iC

dvi C

dt�

R L C

V

R L

V

C

CV0 AI �

1.2 (A)

1.3 (C)

1.4 (C)


Power dissipated in 10 Ω resistor is given by,

2 2(5) 10rmsP I R= = × 250 W=

Hence, the correct option is (C).


In given circuit, DC voltage is applied. Hence, in steady state inductor behaves as a short circuit and capacitor behaves as an open circuit.

Hence, modified circuit is shown below,

1

40 100 80 V40 10CV = × =

+ [By VDR]

As, current through 25 kΩ is 0 A. So, entire 80 V drops across 2C and 3C .

2

3 80 48 V2 3CV = × =

+ [By VDR]

3

2 80 32 V2 3CV = × =

+ [By VDR]

Hence, the correct option is (B).

Key Point 1. Voltage division rule for resistor :

2. Voltage division rule for inductor :

3. Voltage division rule for capacitor :

Concept of Voltage division rule in capacitor :

1 H 2 F10 k� 25 k�2 H

3 F

�

�40 k�

�

1CV100 V 1 F

2CV

3CV

10 k� 25 k�

40 k�1 F 3 F

2 F

�

1CV100 V

2CV

�

3CV

S.C. S.C.

2CV

�

3CV3 F

2 F

80 V

�

�

1R

V

1RV

2R2RV

1

1

1 2

R

RV

V

R R

��

�1

RV

R1

1

1

1 2

R

RV

V

R R

��

�2

RV

R2

1L

V

1LV

2L2LV

1

1

1 2

R

RV

V

L L

��

�1

LV

L1

1

1

1 2

R

RV

V

L L

��

�2

LV

L2

1C

V

1CV

2C2CV

1

1

1 2

R

RV

V

C C

��

�1

CV

C2

1

1

1 2

R

RV

V

C C

��

�2

CV

C1

1.5 (B)


Capacitor 1C and 2C are connected in series. So, same current will flow through them and hence, same charge will store in both the capacitor. Let Q will be the charge stored in capacitor 1C and 2C then,

1 21 2C C eqQ CV C V C V= = =

where, 1 2

1 2eq

C CCC C

=+

By using above equations,

11 C eqCV C V=

1

1 21

1 2C

C CC V VC C

=+

1

2

1 2C

CV VC C

=+

Similarly, 22 C eqC V C V=

2

1 22

1 2C

C CC V VC C

=+

2

1

1 2C

CV VC C

=+


Applying KCL at node a, 1 0 5 7 12 AI i i= + = + = From the figure, 4i I= − 12 A= − Hence, the correct option is (B).


Applying KVL in the loop shown, 5 4 4 0V− − + + = 3 VV = Hence, the correct option is (A).


Applying KVL in the loop shown, 2 2 5 0V v− + + × + = 9V v= + In above expression, voltage across the current source (v) is unknown. Hence, exact voltage cannot be determined. Hence, the correct option is (D).


Voltage V is taken across the constant voltage source of 10 Volts. Hence, voltage V = 10 Volts 3 Ω resistor and 5 V voltage source has no effect on the voltage V. Hence, the correct option is (A).

2 3 Ai �1 5 Ai �

4 ?i � 3 4 Ai �

I

a

i0 =

7A

+–

5 V

4 V

+–

4 V

V

3 �

2 �

2 �

�

�

5 VV

2 A

� �v

3 �

V 10 V 5 V

1.6 (B)

1.7 (A)

1.8 (D)

1.9 (A)



From figure, 4 4,

2 2P R

Q S

Z ZZ Z

= =

Thus, P R

Q S

Z ZZ Z

=

Hence, given circuit is a balanced bridge.

Hence, (2 2) || (4 4) 4 || 8eqZ = + + =

4 8 84 8 3eqZ ×= = Ω

+

Hence, the correct option is (B). Key Point Condition for Balanced Bridge :

P R

Q S

Z ZZ Z

=

For balanced bridge, current through load terminal i.e. LZ will be zero.

Nodal method = KCL + Ohm’s law In nodal method, we apply KCL first, and to find the branch current, we use Ohm’s law.

Hence, it is a combination of KCL and Ohm’s law. Hence, the correct option is (B).


Let voltage across terminal ab is V. Applying KCL at node a,

1 32 2

V V− + =

1 32

V − =

3.5 VV = Hence, the correct option is (C).

Given : 5 , 15 ,ab bcR R= Ω = Ω 30caR = Ω

Applying Δ to Y conversion,

15 30 3

50ab ca

aab bc ca

R RR RR R R

× ×= = = = Ω+ +

215 5 1.5

50ab bc

bab bc ca

R RR RR R R

× ×= = = = Ω+ +

330 15 9

50bc ca

cab bc ca

R RR RR R R

× ×= = = = Ω+ +

Hence, the correct option is (D).


3j �

4j� �

2QZ � �A

B

eqZ

4PZ � �

4RZ � �2SZ � �

2 � 4 �

O.C.

0 A

A

B

eqZ

4 �2 �

PZ

A

B

QZ

RZSZ

LZ

3 A2 �

2 � 1 �

1 V

a

b

�

�

V

a

b cbcR

1aR R�

a

b c

3cR R�2bR R�

caRabR

1.10 (B)

1.11 (B)

1.12 (C)

1.13 (D)

1.14 (D)


Applying KCL at node a, 0at btI e e− + + = at btI e e= + …(i) Voltage across inductor is given by,

( ) dIv t Ldt

=

Put the value of I from equation (i),

( ) 1 ( )at btdv t e edt

= +

( ) at btv t ae be= + Hence, the correct option is (D).


Applying KVL in branch shown in the figure, 0 1 5 10 0E+ + + + = 16 VE = − Hence, the correct option is (A).


Applying KCL at node a,

0 0 012 04 4 4

e e e− + + =

03 34e = 0 4 Ve =


Given : 3ab bc caZ Z Z Z= = = Δ to Y conversion is given by,

ab ca

aab bc ca

Z ZZZ Z Z

×=+ +

23 3 3

3 3 3 3 3 3aZ Z Z ZZ

Z Z Z Z×= = =

+ +

3

ab bcb

ab bc ca

Z Z ZZZ Z Z

×= =+ +

3

bc cac

ab bc ca

Z Z ZZZ Z Z

×= =+ +

Hence, the correct option is (A). Key Point (i) If all the elements of Δ network are

same, then Δ to Y conversion decreases the impedance by a factor of 3.

(ii) If all the elements of Y network are

same, then Y to Δ conversion increases the impedance by a factor of 3.

1 � 1 �

ate ( )v t 1 H bte

I

a

1 V2 V

4 V 5 V

1v

E

10 V

2v0 V

4 �

0e12 V

2 �

2 �4 �

a

a

b c

bcZ

aZ

a

b c

cZbZ

caZabZ

a

b c

/ 3Z

a

b c

ZZ

/ 3Z/ 3Z

Z

a

b c3Z

Z

a

b c

3Z3Z

ZZ

1.15 (A)

1.16 (C)

1.17 (A)



Using source transformation :

Applying KCL at node 0e ,

0 0 080 16 012 12 6

e e e− −+ + =

04 112e =

0112 28 V

4e = =



Applying KCL at node V,

20 4 05 5

V V− + − =

2 85V = 20 VoltV =

Power across dependent current source is given by, deliver 20 4 80 WP = × = Dependent current source is delivering power of 80 W to the circuit. Hence, the correct option is (A).

Key Point Concept of absorbed and delivered power : (i) Power is absorbed by the element if

current enters into the positive terminal. (ii) Power is delivered by the element if

current leaves from the positive terminal. E.g. 1 :

E.g. 2 :

E.g. 3 :

E.g. 4 :

Cube circuit is shown below,

2 �

10 � 12 �8 A

6 �

16 V

0e

2 �

12 �80 V6 �

16 V

0e10 �

0e

5 �

5 �

1 20 VV �1 20

A A5 5

4 A

�

�

V

V

10 V

2 A

10 V absorb 10 2 20 WP P� � � �

10 V deliver 20 WP P� � �

10 V

2 A10 V deliver 10 2 20 WP P� � � �

10 V absorb 20 WP P� � �

10 V

2 A�

10 V absorb 10 ( 2) 20 WP P� � � � � �

10 V deliver 20 WP P� �

10 V�

2 A10 V deliver 10 2 20 WP P

��

10 V absorb 20 WP P�

� �

1 �1 �

1 �

v

/3i

/3i

/6i

/3i

/3i

/3i

/6i

/6i/6i

/6i

/6i

1 �

1 �

1 �

1 �

1 �

1 �

1 �

1 �

1 �

i

/3i

1.18 (D)

1.19 (A)

1.20 (A)


Applying KVL in the loop shown,

1 1 13 6 3i i iv = × + × + ×

56eq

vRi

= = Ω

Hence, the correct option is (A). Key Point (i) If twelve 1 H inductors are used as edges

to form a cube. The inductance between two diagonally opposite corners of the cube will be,

5 H6eqL =

(ii) If twelve 1 F capacitors are used as edges to form a cube. The capacitance between two diagonally opposite corners of the cube will be,

6 F5eqC =

(iii) The resistance of a wire cube :

(a) Equivalent resistance across diagonal

of cube is given by,

1856

R R=

(b) Equivalent resistance across edge of cube is given by,

137

12R R=

(c) Equivalent resistance across diagonal of face of cube is given by,

1434

R R=

(iv) Equivalent Resistance across Diagonal of Cube :

(v) Equivalent Resistance across Edge of

Cube :

(vi) Equivalent Resistance across Diagonal

of Face of Cube :

Given : 500 Hz, 2 Ff C= = μ

1 1 2 msec500

Tf

= = =

Amplitude = 3 V . Method 1 :

33( ) 1500

2 10CV t t t−= =×

[ 0 2 msect< < ] Current through capacitor is given by,

( )CdI C V tdt

=

62 10 (1500 )dI tdt

−= × × ×

33 10 3 mAI −= × = [ 0 2 msect< < ] Hence, the correct option is (D). . Method 2 : Voltage across capacitor is given by,

R

R

R

R

R

R

R

R

R

R

R

R

1

2

3

4

6 5

87

Scan for Video

Explanation

Scan for Video

Explanation

Scan for Video

Explanation

( )CV t

3 V

2 msec

t

1.21 (D)


1C CV I dt

C=

C CCV I t= 62 10 3CI t −= × × 66 10CI t −= ×

Only option (D) satisfies above relation. Hence, the correct option is (D).


12 ; 02( )

6;2

t TtTi t

T t T

− < <= < <

RMS value is given by,

2

0

1 ( )T

rmsI i t dtT

=

2/2

2

0 /2

1 12 6T T

rmsT

I t dtT T = − +

/2

2 22

0 /2

1 144 6T T

rmsT

I t dtT T = +

( )/23

2 /20

1 144 363

TT

rms T

tI tT T

= +

3

21 144 18

24rmsTI T

T T

= +

[ ]1 6 18 2 6 ArmsI T TT

= + =

Hence, the correct option is (A).

Given : 1 2 4R R R R= = = , 3 1.1R R=


Ideal voltmeter is connected between node a and b. Hence, replace it with its internal resistance

inR = ∞ , i.e. current through ideal voltmeter will be zero.

Note : (i) Ideal voltmeter measures open circuit

voltage. Hence, inR = ∞ .

(ii) Ideal ammeter measures the short circuit current. Hence, 0inR = .

10 5 VaRV

R R= × =

+ [By VDR]

1.1 10 5.238 V1.1b

RVR R

= × =+

[By VDR]

Here the polarity is not mentioned assuming ab a bV V V= − i.e. a bV V> .

Hence, 5 5.238 0.238 Vab a bV V V= − = − = −


i t( )

6 A

t0

– 6 A

T/2 T

10 V

1R

2R 3R

4R

a bV

10 V

R

1.1R

R

a b0 A

O.C.

R

1.22 (A)

1.23 (C)


Key Point


It is given that 60 V voltage source is absorbing power i.e. current must enter into the positive terminal of voltage source. Applying KCL at node a, ' 12 0I I+ − = 12 'I I= − Current I will be less than 12 A. Hence, the correct option is (A).

Given : Current supplied by the battery = 2 A. Total time T = 10 min = 600 sec

. Method 1 :

Equation of ( )v t is given by,

10 12( ) 12 1210 60 300

tv t t− − = + = + ×

where, t is in second Power delivered is given by, ( ) ( ) ( )p t v t i t= × ; where, ( ) 2 Ai t =

( ) 12 2300

tp t − = +

( ) 24 Watts150

tp t − = +

Total energy delivered by the battery in 10 minutes i.e. 600 sec is given by,

600

0

( )E p t dt=

600

0

24150

tE dt− = +

6002

6000

0

1 24[ ]150 2

tE t −= +

1 600 600 24 600150 2

E −= × × + ××

1200 14400 13200 JE = − + = 13.2 kJE = Hence, the correct option is (C). . Method 2 :

The power delivered by the battery is given by,

I 60 V

12 A

20 V

'I

a

Mobile

v t( )

10 min0t

10 V

12 V

v t( )

10 min (600 sec)0t

10 V

12 V

v t( )

A1

A2

1.24 (A)

1.25 (C)


( ) ( ) ( )p t v t i t= × where, ( ) 2 Ai t = The energy delivered during time T (600 sec) is given by,

600

0 0

( ) ( ) ( )T

E p t dt v t i t dt= = ×

600

0

2 ( )E v t dt=

2E = [Area of v(t) from 0 to 600 sec] [ ]1 22E A A= +

12 2 600 10 600 13200 J2

E = × × + × =

13.2 kJE = Hence, the correct option is (C).

. Method 1 : Nodal Analysis .

Assume C is the reference node, hence 0CV = .

Apply KCL at node A,

10 1 02 2

A AV V− + + =

10 2 0A AV V− + + + =

4 VAV = Apply KCL at node B,

101 2 01

BV −− − + =

13 VBV =

Apply KCL at node D,

10 102 1

A BV Vi − −= +

10 4 10 132 1

i − −= +

3 3 0 Ai = − = Hence, power supplied by 10 V voltage source 10 10 0 0 Watti= × = × = Hence, the correct option is (A). . Method 2 : Superposition theorem .

(i) Consider only 1 A current source :

Applying KVL in outer loop, 1 1 1( 1) ( 1) 2 0i i i+ + + + =

1 114 2 A2

i i= − = −

(ii) Consider only 2 A current source :

Scan for

Video Solution

1 �

1 �1 �

1 �

1 �

1 A

2 A

10 V

D i

A B

C

1 �

1 �1 �

1 �

1 �

1 A

2 A

10 V

i

1 �

1 �1 �

1 �

1 �

1 A

1( 1)i �

1i

1A

1i

O.C.

S.C.

1 �

1 �1 �

1 �

1 �

2 A

2( 2)i �

2i

O.C.S.C.

1.26 (A)


Applying KVL in outer loop, 2 2 2 2( 2) ( 2) ( 2) ( 2) 0i i i i+ + + + + + + = 2 24 8 0 2 Ai i+ = = − (iii) Consider only 10 V voltage source :

Apply KVL in the outer loop,

310 2.5 A4

i = =

According to superposition theorem, the current i is given by, 1 2 3i i i i= + +

1 ( 2) 2.5 0 A2

i = − + − + =

So, power supplied by 10 V voltage source is, 10 0 0 WP V i= × = × = Hence, the correct option is (A). . Method 3 :

Applying KVL in outer loop,

10 (3 ) (3 ) (2 ) (2 ) 0i i i i− + + + + + + + + = 10 10 4i= + 0 Ai = Hence, power supplied by the voltage source is given by, 10 0 0 WP Vi= = × =


Given : 6 VA BV V− =

The circuit is shown below,

From figure, 6 A 3 A2 2

A BV Vi −= = =

The same current will flow through branch between node C and D. Using source transformation :

From above figure,

3 1 2 5 VDCV = × + =

5 VCD C DV V V= − = −

Without using source transformation :

5 1 5 VDCV = × =

1 �

1 �1 �

1 �

1 �

3i

4 0i �

O.C.

4 0i �

+–

10 V4 0i �

1 �

1 �1 �

1 �

1 �

1 A

2 A

10 V

3 i� i

3A

2 i�

3 i�

2 i� i

Scan for

Video Solution

RR

R R RR R

R

BV

DVCV

AV

5 V

10 V

2 �

1 �

2 A

i

i

RR

R R RR R

R

BV

DVCV

AV

5 V

10 V

2 �

i

2 V 1 �

3A

3A

RR

R R RR R

R

BV

DVCV

AV

5 V

10 V

2 �

1 �

2 A

3 Ai �

3 A

5 A

1.27 (A)


5 VCDV = − Hence, the correct option is (A).

Key Point

For a particular one port network, Incoming current = Outgoing current Hence,

Given delta to star conversion is shown below,

Before scaling :

b cA

a b c

R RRR R R

=+ +

If ' , ' , 'a a b b c cR kR R kR R kR= = =

After scaling :

' ''' ' '

b cA

a b c

R RRR R R

=+ +

2( ) ( )'

( )b c b c

Aa b c a b c

kR kR k R RRkR kR kR k R R R

= =+ + + +

'A AR kR= Hence, the correct option is (B).

Given : 1 1010 F, 10 VC V= μ =

2 205 F, 5 VC V= μ =

3 302 F, 2 VC V= μ =

. Method 1 :

Charge in a capacitor is given by, Q CV=

Capacitor

Breakdown voltage or maximum operating

voltage

Maximum charge stored in capacitor

1 10 FC = μ 10 V 1(max) 100 CQ = μ

2 5 FC = μ 5 V 2(max) 25 CQ = μ

3 2 FC = μ 2 V 3(max) 4 CQ = μ

Since, 2C and 3C are in series. So, Charge on both capacitor will be same and equal to the min ( 2 3,Q Q )

23 4 CQ = μ

Equivalent capacitance of 2C and 3C is,

2 323

2 3

5 2 10 F5 2 7

C CCC C

×= = = μ+ +

Equivalent voltage is,

2323

23

4 C 2.8 V(10 / 7) Feq

QV VC

μ= = = =μ

In parallel, voltage will be same, hence 2.8 V will appear across 1C also. Charge stored in 1C is given by, 1 1 eqQ CV=

Scan for

Video Solution

1N 2N

2 �BVAV

DVCV

1N

AV

CV

3 A

3 A BV

DV

3 A

3 A

2N

aR

cRbR

CR BR

AR

2C 3C

1C

1.28 (B)

1.29 (C)


1 10 F 2.8 V 28 CQ = μ × = μ In parallel, total charge is given by, 1 23TQ Q Q= + (28 4) C 32 CTQ = + μ = μ Hence, the correct option is (C). . Method 2 :

Capacitor 1C 2C 3C Value (in Fμ ) 10 5 2

Breakdown Voltage (Volts) 10 5 2

From above figure, 0 2 3V V V= +

3 02 0 0

3 2

2 25 2 7

C VV V VC C

= = =+ +

2 03 0 0

2 3

5 52 5 7

C VV V VC C

= = =+ +

Now, 0 210 V, 5 VV V≤ ≤ 3and 2 VV ≤ (i) If 0 10 VV = ,

Then 22 10 20 2.85 V

7 7V ×= = =

So, 2 5 VV < [No Breakdown]

35 10 50 7.13 V

7 7V ×= = =

So, 3 2 VV > [Breakdown] Hence, 0V must be less than 10 V. (ii) If 2 5 VV =

07 355 17.5 V2 2

V = × = =

0 10 VV > [Breakdown] Hence, 2V must be less than 5 V. (iii) If 3 2 VV =

07 2 2.8 10 V5

V = × = <

[No Breakdown]

Also, 02

2 0.8 V 5 V7VV = = <

[No Breakdown] None of them are in breakdown. Hence, minimum value of voltage across the combination can be 0 2.8 VV = .

Total charge 0eqQ C V=

where, 2 31

2 3eq

C CC CC C

= ++

2 5 8010 F2 5 7eqC ×= + = μ

+

80 2.8 32 C7

Q = × = μ

Hence, the correct option is (C). . Method 3 :

Here, capacitors 3C has minimum breakdown voltage and hence the breakdown voltage of capacitor 3C will decide the maximum safe

voltage SV applied across the circuit as shown in figure.

23

2 3S

CV VC C

= ×+

[By VDR]

522 5 SV= ×

+

2.8 VSV =

2C 3C

1C2V 3V

0V

2C 3C

1C2V 3V

V

2C 3C

1C2V 2 V

SV


Also, total charge is given by, eq SQ C V=

where, 2 31

2 3eq

C CC CC C

= ++

2 5 8010 F2 5 7eqC ×= + = μ

+

80 2.8 32 C7

Q = × = μ



Circuit can be redrawn as shown below,

110 A 5 A2

I = =

210 A 10 A1

I = =

Applying KCL at node a, 1 2 2 0sI I I+ + − =

10 5 2sI + + =

13 AsI = −

Applying KVL in loop (1), 5 2 10 0sV− + × + =

20 VsV =


In parallel connection, voltage will be same. So, voltage across 2 Ω and 1 Ω is same.

Current in the 1 Ω resistor,

210 10 A1

I = =


Given : 1 41 , 2 A, 1 A,R i i= Ω = = − 5 4 Ai = −


Applying KCL at node (1), 1 4 2 0i i i− − + =

2 1 4i i i= +

2 2 1 1 Ai = − =

Applying KCL at node (2), 2 5 3 0i i i− − + =

3 2 5i i i= +

3 1 4 3 Ai = − = −

Applying KCL at node (3), 1 3 6 0i i i− − =

6 1 3i i i= −

6 2 ( 3) 5 Ai = − − =


According to question,

1 �2 �

5 �

SV 2 A

SI10 V

1 �2 �

5 �

SV 2 A

SI 1I 2I

10 V

a

1

Scan for

Video Solution

RR

R1i

3i

4i

5i

1i6i

3i2i

2i

2

31

1.30 (D)

1.31 (C)

1.32 (A)

1.33 29


. Method 1 : Equivalent Δ network is shown below,

where, a b b c c aab

c

R R R R R RRR

+ +=

10 10 11 10 10 1110abR × + × + ×= 32= Ω

a b b c a cbc

a

R R R R R RRR

+ +=

10 10 11 10 10 11 3210bcR × + × + ×= = Ω

a b b c a cac

b

R R R R R RRR

+ +=

10 10 11 10 10 11 2911acR × + × + ×= = Ω

Lowest resistance 29acR = Ω . Hence, the lowest value among the three resistance is 29 Ω . . Method 2 :

a b b c a cab

c

R R R R R RRR

+ +=

a b b c a cbc

a

R R R R R RRR

+ +=

a b b c a cac

b

R R R R R RRR

+ +=

Since, numerator values are same. So minimum value depends on denominator. b a cR R R> = So, ac bc abR R R< = 29acR = Ω Hence, the lowest value among the three resistance is 29 Ω .

Given waveform is shown below,

For the given signal,

2 , 02( )

0,2

Tt tTx t

T t T

< ≤= < ≤

RMS value of periodic waveform is given by,

RMS value 2

0

1 ( )T

x t dtT

=

2/2

2

0 /2

1 2 (0)T T

rmsT

x t dt dtT T = +

/2/2 32

2 20 0

1 4 1 43

TT

rmstx t dt

T T T T

= = × ×

3

34 1 0.408

24 6rmsTx

T= × = =

Hence, the root-mean-square (rms) value of x is 0.408.


10 �

a c

b

10 �

( )cR( )aR

11 � ( )bR

N

acR

bcRabR

a c

b/2T/2T

t

( )x t

1

0

5 �

10 �5 V 1 A

I5 �V

1.34 0.408

1.35 0.5


. Method 1 : Nodal Analysis . Applying KCL at node V,

5 1 05 15

V V− + − =

4 215V = 15 V

2V =

Hence, 1 A 0.5 A15 2VI = = =

Hence, the value of current I is 0.5 A. . Method 2 : Mesh Analysis .

Applying KVL in loop shown in figure, 5 5(1 ) 15 0I I− − − + = 5 5 5 15 0I I− − + + = 20 10I =

1 A 0.5 A2

I = =

Hence, the value of current I is 0.5 A. . Method 3 : Superposition Theorem . (i) Consider only 5 V voltage source :

15 5 1 A

5 5 10 20 4I = = =

+ +

(ii) Consider only 1 A current source :

25 1

5 5 10I = ×

+ + [By CDR]

25 1 A

20 4I = =

Hence, current I is given by,

1 21 1 1 A4 4 2

I I I= + = + =

Hence, the value of current I is 0.5 A.


Applying Y to Δ conversion,

1a b b c c a

b

R R R R R RRR

+ +=

Given : 5 , 7.5a bR R= Ω = Ω , 3cR = Ω

Hence, 15 7.5 7.5 3 3 5

7.5R × + × + ×=

1 10R = Ω

Hence, the value of 1R in the equivalent Δ -network is 10 Ω .


From figure, 0 i iI A I= (output current depends on input current)

5 �

10 �5 V 1 A

I

5 �

(1 )I�

5 V+–

5 � 5 �

10 �

I1

O.C.

5 � 5 �

1 A 10 �

I2

S.C.

1R

aR cR

bR

5 � 3 �

7.5 �

iI

0 1 iI A I�

R

1.36 10

1.37 (C)


It is a current source, controlled by current iI .

So, it is current controlled current source. Hence, the correct option is (C).


. Method 1 : Mesh Analysis .

Applying KVL in loop, 1 ( 2) 4 3 2(10 ) 0I I I I× + − × + × − − = 10 8 20 0I − − = 2.8 mAI = Hence, the magnitude of current through the resistor 2R is 2.8 mA.

. Method 2 : Superposition theorem . 1. Consider 10 mA current source :

Current 1I is given by,

12 10 2 mA

8 2I = × =

+ [By CDR]

2. Consider 2 mA current source :

Current 2I is given by,

24 2 0.8 mA

6 4I = × =

+ [By CDR]

According to superposition theorem, total current I due to both source is given by, 1 2I I I= + 2 0.8 2.8 mAI = + = Hence, the magnitude of current through the resistor 2R is 2.8 mA.

Given ladder network is shown below,

Above circuit can be replaced by circuit shown below,

eqeq

eq

R RR R

R R×

= ++

2 mA10 mA 2 k� 4 k�

1 k�

3 k�

1R

2R

3R

4R

2 mA10 mA 2 k� 4 k�

1 k�

3 k�

10 mA I

(10 )I� ( 2)I �

2 mA

I

10 mA 2 k� 4 k�

1 k�

3 k�

1I

2 mA2 k� 4 k�

1 k�

3 k�

2I

Scan for

Video Solution

R R

R R

R

eqR

R �

R

eR

R

R

eqR

eqR

R

eR

1.38 2.8

1.39 2.62


2

( )eq eq

eqeq

R RR RRR

R R+ +

=+

2( )eq eq eq eqR R R R RR RR+ = + +

2 2 2eq eq eqRR R R RR+ = +

2 2 0eq eqR RR R− − =

2 24 5

2 2eqR R R R RR ± + ±= =

eqR cannot be negative. So,

1 5 1.6182eqR R R

+= =

Therefore, e eqR R R= +

1.618eR R R= + 2.618R=

2.618 2.62eRR

= =

Hence, the value of eRR

is 2.62.


The given circuit has three balanced bridges. Hence, modify the given circuit as discussed below,

. Method 1 : Using concept of open circuit :

1 1 1 1 1

2 2abR R R R R= + + +

1 3abR R

=

300 1003 3abRR = = = Ω

Hence, the resistance abR is 100 Ω . . Method 2 : Using concept of short circuit : In case of resistor, the current through resistor is 0 A only when potential difference across the resistor is 0 V. i.e. resistor behaves as a short circuit.

R

RR R

RR

R

RR

a

b

R

RabR

R

R

R

R

RR R

RR

R

R

a

b

R

RabR

R

RR

O.C. O.C. O.C.

R

a

b

R

R

RabR

R R

R R

R R

R R

2R

a

b

abR

R R 2R

RR R

RR

R

R

a

b

R

RabR

R

RR

S.C. S.C. S.C.

1.40 100


26 6 6abR R RR = + =

2 300 1006abR ×= = Ω

Hence, the resistance abR is 100 Ω .


Applying KCL at node 1V , 5 2I I I= + + 5 4I=

5 A4

I =

Voltage is same for parallel branches i.e. 1 4V I=

1 5 VoltV = Applying KVL in first loop, 2 15 4V V= × +

2 20 5 25 VoltV = + = Hence, the correct option is (A).


Applying KCL at node (1),

0.255 0.5 020 10

x x xx

V V V V−− + + + =

0.75 0.5520 10 1

x x xV V V= + +

1.5 10520

x x xV V V+ +=

12.5 100xV =

8 VoltxV =

Hence, the voltage xV is 8 V.


Fig. (a)

Applying Δ to Y conversion,

R

a

b

R

R

RabR

R R

R R

R R

R R

/ 6R

/ 6R

a

b

abR

1V2V 4� 4� 2 I

I

4�

��

��

5A

1V 1V

I

5 A xV 8 � 0.25 xV

10 �

0.5 xV

20 �

1

Scan for

Video Solution

a

b

1 �

1 �

1 �

1 �

1 �

1 �

1 �1 �

1 �

1.41 (A)

1.42 8

1.43 (D)


Fig. (b)

Modified figure is shown below,

Fig. (c)

Applying Y to Δ conversion,

Fig. (d)


Fig. (e)

4 4 4 85 5 5 15abR = + = Ω



. Method 1 : Mesh Analysis .

Applying KVL in the outer loop,

60 5(0.16 ) 0.2 3 0x x xv v v− + + × + =

25 Vxv =

Current through 2R is,

225 5 A

5 5xvI = = =

Hence, the magnitude of the current through 2R is 5 A.


a

b

1 �

1 �1 �

1 � 1 �

1 �

1/3 �

1/3 �

1/3 �

a

b

1 �

1 �

1 �

4/3 �

4/3 �

4/3 �

a

b

1 �

1 �

1 �

4 �

4 �

4 �

a

b

4

5�

4

5�

4

5�

abR

Scan for

Video Solution

60 V

1R

5 �

2R

3 �

5 �0.04 xv xv

60 V

5 � 3 �

5 �0.04 xv xv

0.16 xv 0.2 xv

2I

60 V

5 � 3 �

5 �0.04 xv xv

V

2I

1.44 5


Applying KCL at node V,

60 0.04 05 8x

V Vv− − + =

13 1.6 480xV v− = …(i)

Current through 3 Ω and 5 Ω is same.

So, 2 1.68 5

xx

V vI V v= = = …(ii)

From equation (i) and (ii), 25 Voltxv =

and 40 VoltV =

240 5 A

8 8VI = = =

Hence, the magnitude of the current through 2R is 5 A.



Since, node D, node E, and node F are reference nodes, hence 0D E FV V V= = =

From above figure, 8 VC FV V− = Apply KCL at node ,A

1 1 18 01

A F A E A C

A D

V V V V V V

V V

− − −+ +

− −+ =

8 8 01 1 1 1

A A A AV V V V− −+ + + =

4 16AV = 4 VoltAV =

Hence, 18 4 4 A

1i −= =

3 4 A1 1

A F AV V Vi −= = =

Apply KCL at node C, 2 11 0i i− + + = 2 1 4 5 Ai = + = Apply KCL at node F, 3 2 0i i i− + + = 3 2 (4 5) Ai i i= − = − 1 A= − Hence, the current i is –1 A. . Method 2 : Thevenin’s Theorem : Calculation of :THR (when dependent sources are not present) Replace all independent source by their internal resistance, i.e. short circuit all the independent voltage source ( 0inR = ) and open circuit all the independent current source ( )inR = ∞ .

1 11 ||1 1 ||1

2 2THR = + = +

1THR = Ω

i

5 �

1 � 1 �

1 �

8 V

8 V1 A

1 �

5 �

1 � 1 �

1 �

1 A

C

A

1 �

1i 2i

3iF

i

E

D

18 VSV �

28 VSV �

5 �

1 � 1 �

1 �

1 �

THR

S.C.

S.C.

O.C.

C

F

D

EA

1.45 – 1


Calculation of :THV

Apply KVL in loop I,

1 18 1 0I I− + − + =

192

I =

Apply KVL in loop II,

2 2(1 ) 8 0I I+ + − =

22 7I =

272

I =

Apply KVL in loop III,

1 2 0THV I I− − + =

2 1THV I I= −

7 9 1 V2 2THV = − = −

Thevnin’s equivalent circuit :

1 1 A1

i −= = −

Hence, the current i is –1 A.

. Method 3 : Superposition Theorem :

(i) Consider only 1 A current source :

Fig. (a)

Above circuit can be simplified as given below,

Fig. (b)

Since, there is a short circuit across branch CD, hence we can remove branch CD. The modified figure is shown below,

5 �

1 �

1 A1 A

1 �

THV

1 �

8 V

8 V

1 A

1 �

21 I�

21 I� �

2I 1I 0 A

0 A

LoopI

LoopII Loop

III

C

F

D

EA

1 1I �

1THR � �

1 VTHV � � i

F

D

5 �

1 �

1 �

1 �

1 A1 �

i

18 VSV �

28 VSV �

A

C

D

FE

5 �

1 �

1 �

1 �

1 A1 �

1i

S.C.

A

C

D

FE

S.C.

5 �

1 A0.5 �

1i

C

D

A

0.5 �


Fig. (c)

From figure (c), 1 1 Ai = − (ii) Consider only 8 V voltage source :

Fig. (d)

Above circuit can be simplified as shown below,

Fig. (e)

From figure (e), current 0'i is given by,

08 8' 6 A11 (1 || 0.5) 1

3

i = = =+ +

2 01 1' 6 4 A

1 0.5 1.5i i= − × = − × = −

+

[By CDR]

(iii) Consider only 8 V voltage source :

Fig. (f)

Above circuit can be simplified as given below,

Fig. (g)

From figure (g),

08 8 6 A11 (1|| 0.5) 1

3

= = =+ +

i

3 01 1 6 4 A

1 0.5 1.5= × = × =

+i i

[By CDR] According to superposition theorem, total current i is given by, 1 2 3 1 4 4 1 A= + + = − − + = −i i i i

Hence, the current i is – 1 A.

According to the question,

5 �

1 A

1i

C

D

5 �

1 �

1 �

1 �

1 �

2i

18 VSV �

A

C

D

FE

S.C.

O.C.

0'i

1 �

0.5 �

1 �

2i

18 VSV �

A

C

D

F

0'i

5 �

1 �

1 �

1 �

1 �

3i2

8 VSV �

A

C

D

FE

O.C.

S.C.

0i

1 �

0.5 �

1 �

3i28 VSV �

A

D

FE

0i

Scan for

Video Solution

1.46 2.14


B C

eq AB C

R RR RR R

= ++

eqR will be maximum when 10AR = Ω .

(max)5 2 8010

7 7eqR ×= + = Ω

eqR will be minimum when 2AR = Ω

(min)10 5 802

15 15eqR ×= + = Ω

(max)

(min)

15 2.147

eq

eq

RR

= =

Hence, the ratio of maximum to minimum values of the resistances is 2.14. Key Point (i) Parallel combination of resistors

decreases the equivalent resistance of the combination.

(ii) Series combination of resistors increases the equivalent resistance of the combination.


Given : 1 1 ,R = Ω 2 2R = Ω and 3 3R = Ω

From the symmetry of given circuit, we can conclude that point B and C, D and E are equipotential points since the resistive distance of both the points from terminals A as well as from F are equal. We can remove the resistance connected between two equipotential points as current through it will always be zero and the simplified circuit is shown below,


Current supplied by voltage source is given by,

eq

VIR

=

where, 1 1 1 3||2 2 2 2eqR = + +

BR

CR

AR

P Q

+–

11V

1R1R

2R

3R

3R

2R

1R1R

1R1R

A

B C

D E

F

+– 11V

3 �

I

1 �

A

,D E

,B C

1 �

1 �1 �

1 �1 �

3 �

F

+– 11 V

11||1

2� �

I

11||1

2� �

11||1

2� �

33 || 3

2� �

1.47 8


1 32 21 1 32 2

eqR×

= ++

3 / 412

= +

118eqR = Ω

Therefore, 1111/8eq

VIR

= = 8 A=

Hence, the magnitude of the current through the source is 8 A.

Given circuit is shown in figure,

We have to find open circuit voltage thV .

..Method 1.. Using source transformation,

Applying KVL in loop 1, 1 (1) 2 2 2 4 0I I I− + − − − = 1 2 4 5I+ − =

1 Amp5

I = −

14 (2) 4 25thV I = + = + − ×

4 0.4 3.6VthV = − =

..Method 2.. By nodal method,

Applying KCL at node xV ,

( 2)1 01 2

x x thV V V+ −− + + =

1 A 12 2

x thx

V VV = + + −

32 2

x thV V=

3th

xVV = …. (i)

Applying KCL at node thV ,

( 2)2 A2 2

th x thV V V− += +

4 2th x thV V V= − − +

4 2 2th xV V= − −

2 6th xV V− =

Putting the value of xV from equation (i),

2 63th

thVV − =

6 63

th thV V− =

5 63

thV =

6 3 18 3.6 V5 5thV ×= = =


Given circuit is shown in below figure,

Scan for

Video Solution

1 �1A 2A 2 �

2V2 � 4 �

thV

�

�

+–

1V2 Ω

2 V 2 Ω 4 Ω

thV

+

−

1 Ω+–I

+– 4 V

+

–

2 0I =

Loop 1

1 �1A 2A 2 �

2V2 � 4 �

thV

�

�

+–xV thV2xV �

1.48 (B)

1.49 (D)


080 35 87.32 23.63LZ j= − = ∠ −

From figure, 1 1 0LV I Z− + =

0

11 0

120 9087.32 23.63L

VIZ

∠ −= =∠ −

01 1.374 66.37I = ∠ −

Similarly, 2 2 0LV I Z− + =

0

22 0

120 3087.32 23.68L

VIZ

− ∠ −= =∠ −

01.374 6.32= − ∠ − From circuit, 1 2I I I+ =

2 1I I I= − 0 0( 1.374 6.32 ) (1.374 66.37 )I = − ∠ − − ∠ −

( 1.36 0.341 ) (0.55 1.258 )I j j= − + − −

1.91 1.408I j= − +

02.373 143.60 AI = ∠


LZ

LZ

LZ

1V1I

2II

+

−2V

0120 90∠ −

0120 30∠ −

(80 35)j= − Ω

+

−

1.1 Which of the following signals is/are

periodic? (A) ( ) cos2 cos3 cos5s t t t t= + +

(B) ( ) exp( 8 )s t j t= π

(C) ( ) exp( 7 )sin10s t t t= − π

(D) ( ) cos2 cos4s t t t=

1.2 Let ( )tδ denote the delta function. The

value of the integral 3(t)cos2t dt

∞

− ∞

δ is

(A) 1 (B) –1

(C) 0 (D) 2π

1.3 If a signal ( )f t has energy E, the energy of the signal (2 )f t is equal to

(A) E (B) 2E

(C) 2E (D) 4E

1.4 Consider the sequence

[ ] [ ]4 5 1 2 4x n j j= − − +

↑

The conjugate antisymmetric part of the sequence is

(A) [ ]4 2.5 2 4 2.5j j j− − −

(B) [ ]2.5 1 2.5j j−

(C) [ ]5 2 0j j−

(D) [ ]4 1 4−

1.5 The function x(t) is shown in the figure.

Even and odd parts of a unit-step function u(t) are respectively,

(A) 1 1, ( )2 2

x t (B) 1 1, ( )2 2

x t−

(C) 1 1, ( )2 2

x t− (D) 1 1, ( )2 2

x t− −

1.6 The power of the signal

( ) 8cos 20 4sin(15 )2

s t t tπ = π − + π

is

(A) 40 (B) 41

(C) 42 (D) 82

1 Basics of Signals

( )x t

t0

1�

1

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2001 IIT Kanpur

2004 IIT Delhi

2005 IIT Bombay


1.7 The derivative of the symmetric function drawn in given figure will look like

(A)

(B)

(C)

(D)

1.8 The Dirac delta function ( )tδ is defined

as

(A) 1 0

( )0 otherwise

tt

=δ =

(B) 0

( )0 otherwise

tt

∞ =δ =

(C)1 0

( )0 otherwise

tt

=δ =

& ( ) 1t dt

∞

−∞

δ =

(D)0

( )0 otherwise

tt

∞ =δ =

& ( ) 1t dt

∞

−∞

δ =

1.9 Three functions 1 2( ), ( )f t f t and 3( )f t ,

which are zero outside the interval [ ]0, T are shown in the figure. Which of the following statements is correct?

(A) 1 2( ) and ( )f t f t are orthogonal

(B) 1 3( ) and ( )f t f t are orthogonal

(C) 2 3( ) and ( )f t f t are orthogonal

(D) 1 2( ) and ( )f t f t are orthonormal

1.10 A function is given by 2( ) sinf t t=

cos2t+ . Which of the following is true? (A) f has frequency components at 0 and

1 / 2 Hzπ . (B) f has frequency components at 0 and

1 / Hzπ . (C) f has frequency components at 1 / 2π

and 1 / Hzπ . (D) f has frequency components at 0,

1 / 2 and 1 / Hzπ π .

1( )f t

2

0

2�

/3T

2 /3T Tt

2 ( )f t

2

1�

t

1

3 ( )f t

1

0

3�

t

0 T

T

2006 IIT Kharagpur

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2009 IIT Roorkee

2.3GATE ACADEMY ® Signals & Systems : Basics of Signals

1.11 For a periodic signal

( ) 30sin100 10cos300v t t t= + 6sin(500 / 4)t+ + π

The fundamental frequency in rad/s is (A) 100 (B) 300 (C) 500 (D) 1500

1.12 A discrete-time signal 2[ ] sin ( )x n n= π , n

being an integer, is [Set - 01] (A) Periodic with period π . (B) Periodic with period 2π . (C) Periodic with period / 2π . (D) Not periodic.

1.13 The waveform of a periodic signal ( )x t

is shown in the figure.

A signal ( )g t is defined by 1( ) .2

tg t x − =

The average power of ( )g t is ________. [Set - 01]

1.14 Two sequences 1[ ]x n and 2[ ]x n have the

same energy. Suppose 1[ ] 0.5 [ ]nx n u n= α where α is a positive real number and

[ ]u n is the unit step sequence. Assume

21.5; for 0,1[ ]0; other wise.

nx n ==

Then the value of α is _____. [Set - 03]

1.15 Consider the plot of ( )f x versus x as

shown below. [Set - 01]

Suppose 5

( ) ( ) .x

F x f y dy−

= Which one

of the following is a graph of ( )F x ? (A)

(B)

(C)

(D)

( )x t

t

3

3�

1�

2�

4�

1

2 3

40

( )f x

2�

5�

x5�

2�

0

( )F x

5�

x5�

0

( )F x

5�

x5�

0

( )F x

5�

x5�

0

( )F x

5�x

5�

0

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2014 IIT Kharagpur

2015 IIT Kanpur

2016 IISc Bangalore


1.16 A continuous-time function x(t) is

periodic with period T. The function is

sampled uniformly with a sampling

period sT . In which one of the following

cases is the sampled signal periodic?

[Set - 01]

(A) 2 sT T= (B) 1.2 sT T=

(C) Always (D) Never

1.17 The energy of the signal sin(4 )( )4

tx ttπ=

π

is _________. [Set - 02]

1.18 Consider the signal

2( ) 1 2cos(2 ) 3sin3

f t t tπ = + π +

4cos2 4

tπ π + +

where t is in seconds. Its fundamental time period, in seconds, is _______.

1.19 The output [ ]y n of a discrete - time

system for an input [ ]x n is

[ ] max [ ]k n

y n x k−∞≤ ≤

=

The unit impulse response of the system is

(A) unit step signal [ ]u n .

(B) 0 for all n.

(C) unit impulse signal [ ]nδ .

(D) 1 for all n.

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2020 IIT Delhi


1.1 A, B, D 1.2 A 1.3 B 1.4 A 1.5 A

1.6 A 1.7 C 1.8 D 1.9 B 1.10 B

1.11 A 1.12 D 1.13 2 1.14 1.5 1.15 C

1.16 B 1.17 0.25 1.18 12 1.19 A

Given : Option (A) : ( ) cos2 cos3 cos5s t t t t= + + The combination of periodic signals will be periodic, if the ratio of any two frequencies (or any two time periods) is a rational number, i.e.

ratio can be represented in the form of pq

, where

p and q are integers.

Here, 1 2 rad/secω = , 2 3rad/secω = ,

3 5rad/secω =

1

2

23

ω =ω

, 1

3

25

ω =ω

, 2

3

35

ω =ω

Hence, it is a periodic signal. Option (B) : ( ) exp( 8 )s t j t= π ( ) cos(8 ) sin(8 )s t t j t= π + π

Here, 1 8ω = π , 2 8ω = π

1

2

8 18

ω π= =ω π

Hence, it is periodic. Option (C) : ( ) exp( 7 )sin10s t t t= − π This is a decreasing exponential signal. Hence, obviously it is non periodic. Option (D) : ( ) cos2 cos4s t t t=

Since, 2cos cos cos( ) cos( )A B A B A B= + + −

So, 1cos2 cos4 [cos2 cos6 ]2

t t t t= +

Here, 1 2 rad/secω = , 2 6 rad/secω =

1

2

26

ω =ω

Thus, it is also a periodic signal. Hence, the correct options are (A), (B) and (D).

Impulse function satisfies the following property,

0

( ) ( ) ( ) (0)t

x t t dt x t x∞

=− ∞

δ = =

Hence, 0

3 3( )cos cos2 2 t

t tt dt∞

− ∞ =

δ =

03 0cos cos0 12× = = =


Given : Energy of the signal ( )f t is E.

Thus, 2( )E f t dt∞

− ∞

=

Let ( ) (2 )y t f t= and energy of ( )y t is 1E

Scan for

Video Solution

Scan for

Video Solution

Answers Basics of Signals

Explanations Basics of Signals

1.1 (A), (B) and (D)

1.2 (A)

1.3 (B)


So, 21 ( )E y t dt

∞

−∞

=

21 (2 )E f t dt

∞

−∞

=

Let 2t z= , 2dt dz= , 12

dt dz=

21 ( )

2dzE f z

∞

−∞

=

1 2EE =

Hence, the correct option is (B). Key Point If energy of any signal ( )x t is E, then energy

of signal ( )x at will be Ea

.

Given :

[ ] [ ]4 5 1 2 4x n j j= − − +

↑

For any signal [ ]x n , conjugate antisymmetric part is given by,

[ ] [ ]*[ ]

2CAS

x n x nx n

− −=

[ ] [ ]* 4 5 1 2 4x n j j= − + −

↑

[ ] [ ]* 4 1 2 4 5x n j j− = − − +

↑

Hence, [ ] [ ]4 5 1 2 4 4 1 2 4 51[ ]

2CAS

j j j jx n

− − + − − − + = ↑ ↑

[ ]8 5 4 8 51[ ]2CAS

j j jx n

− − − = ↑

[ ][ ] 4 2.5 2 4 2.5CASx n j j j= − − −

↑


Given :

For any signal ( )y t , even part and odd part is given by ( )ey t and 0( )y t .

where, ( ) ( )( )2e

y t y ty t + −=

and 0( ) ( )( )

2y t y ty t − −=

According to the question, ( ) ( )y t u t=

For even part, ( ) ( )( )2e

u t u tu t + −=

( )x t

t0

1�

1

( )u t

t0

1

( )u t�

t0

1

( ) ( )u t u t� �

t0

1

( ) ( )( )

2e

u t u tu t

� ��

t0

1/ 2

1.4 (A)

1.5 (A)


Thus, ( ) ( ) 1( )2 2e

u t u tu t + −= =

For odd part 0( ) ( )( )

2u t u tu t − −=

Therefore, 0( ) ( ) 1( ) ( )

2 2u t u tu t x t− −= =


Given :

( ) 8cos 20 4sin(15 )2

s t t tπ = π − + π

( ) 8sin(20 ) 4sin(15 )s t t t= π + π Let 1 2( ) ( ) ( )s t x t x t= +

where, 1( ) 8sin(20 )x t t= π

and 2( ) 4sin(15 )x t t= π

Let power of 1( )x t is 1P and power of 2 ( )x t is

2P . Signals with different frequencies are

orthogonal to each other.

Thus, 1( )x t and 2 ( )x t are orthogonal to each

other, therefore total power is given by addition of 1P and 2P .

i.e. 1 2P P P= +

where, 2 21

18 32 Watt

2 2AP = = =

2 22

24 8 Watt

2 2AP = = =

(32 8) 40 WattP = + =


Key Point

Given :

. Method 1 :

Let the given waveform is of function ( ).f x

Now,

ve, when 0

( )ve, when 0

xd f xxdx

+ <= − >

( ) ( )u t u t� �

t0

1

1�

0

( ) ( )( )

2

u t u tu t

� ��

t0

1/ 2

1/ 2�

1( )

2x t

t0

1/ 2

1/ 2�

Scan for

Video Solution

B

B

�� 0 �

1.6 (A)

1.7 (C)


Key Point

[ ]increasing function Positiveddx

=

and [ ]decreasing function Negativeddx

=

Thus, function ( )f x must be increasing for 0x < and decreasing for 0x > .

From figure,

( )f x is an even symmetric function.

Thus, derivative of any even symmetric function will always be an odd function. Key Point

[ ]even function Odd functionddx

=

[ ]odd function Even functionddx

=

From option (A) :

It is an even symmetric function with both of its side as positive. Hence, it does not satisfy the above criteria. From option (B) :

It is an even symmetric function with both of its side as positive. Hence, it does not satisfy the above criteria.

From option (C) :

The above figure is an odd symmetric function. For 0x > , ( )f x is decreasing function thus

'( )f x should have negative sides and for 0, ( )x f x< is increasing function thus '( )f x

should have positive sides. Therefore, it clearly satisfies the above criteria. From option (D) :

It is an odd symmetric function but from the above criteria

[ ]decreasing function Negativeddx

=

and [ ]increasing function Positiveddx

=

For 0x > , ( )f x is decreasing function thus '( )f x should have negative sides and for

0, ( )x f x< is increasing function thus '( )f x should have positive sides. Hence, option (D) does not satisfies the above criteria. Hence, the correct option is (C). . Method 2 :

It is a Gaussian function 2

( )xe− .

Let 2

( ) xf x e−=

Now, 2

'( ) ( ) ( )xd df x f x edx dx

−= =

2

'( ) 2 xf x xe−= −

( )f x

Increasing Decreasing

'( )f x

+ ve + ve

x

'( )f x

x+ ve + ve

+ ve

– ve

'( )f x

x

+ ve

– ve

'( )f x

x


x '( )f x 0 0 1 0.735− 1− 0.735

∞ 0 − ∞ 0

The waveform for '( )f x will be,


Definition of dirac delta function ( )tδ is given by,

, 0( )

0, otherwiset

t∞ =

δ =

and ( ) 1t dt∞

−∞

δ =


Two functions ( )u t and ( )v t are said to be orthogonal in interval ( , )a b ,

( ) *( ) 0b

a

u t v t dt =

The function 1 2( ), ( )f t f t and 3( )f t are real functions and defined only in the interval (0, T). (i) To check 1 3( ) and ( ) :f t f t Since, 3( )f t is a real function, Therefore, *

3 3( ) ( )f t f t= (Real function)

*1 3 1 3

0 0

( ) ( ) ( ) ( )T T

I f t f t dt f t f t dt= =

I = [Area under the product of 1 3f f from 0 to T]

22 2 03 3T TI T = × − × − =

Hence, 1( )f t and 3( )f t are orthogonal. (ii) To check 2 3( ) and ( ) :f t f t

1 3 2 23 3 3T T TI T = × + × + × =

Hence, 2 ( )f t and 3( )f t are not orthogonal. (iii) To check 1 2( ) and ( ) :f t f t

22 43 3 3T T TI = × − × = −

Hence, 1( )f t and 2 ( )f t are not orthogonal.

[ ]21

0

( )T

f t dt = Energy of 1( )f t

[ ]21

0

8( ) 4 43 3 3

T T T Tf t dt = × + × =

0

'( )f x

x

( )t�

t0

1 3( ) ( )f t f t

2

0

2�

/3T

2 /3T Tt

2 3( ) ( )f t f t

2

0 /3T 2 /3T Tt

3

1

1 2( ) ( )f t f t

2

0

4�

/3T

2 /3T Tt

1.8 (D)

1.9 (B)


[ ]22

0

( )T

f t dt = Energy of 2 ( )f t

[ ]22

0

( ) 1 1 4 23 3 3

T T T Tf t dt T= × + × + × =

[ ]23

0

( )T

f t dt = Energy of 3( )f t

[ ]23

0

11( ) 1 9 13 3 3 3

T T T T Tf t dt = × + × + × =

But [ ]21

0

( ) 1T

f t ≠ and [ ]23

0

( ) 1T

f t ≠

So, they are not orthonormal. Hence, the correct option is (B). Key Point

(i) If ( ) *( ) 0b

a

u t v t dt =

Then, ( )u t and ( )v t are said to be orthogonal.

(ii) If ( ) *( ) 0b

a

u t v t dt =

and 2 2( ) ( ) 1b b

a a

u t dt v t dt= =

Then, ( )u t and ( )v t are said to be orthonormal.

Given : 2( ) sin cos2f t t t= +

(1 cos2 )( ) cos22

tf t t−= +

1 1( ) cos22 2

f t t= +

Then, ( )f t contains only two frequency components. 0, 2 rad/secω =

10, Hzf =π


Given : ( ) 30sin100 10cos300v t t t= +

6sin 5004

t π + +

Here, 1 100 rad/secω = , 11

2 2100 50

T π π π= = =ω

2 300 rad/secω = , 22

2 2300 150

T π π π= = =ω

3 500 rad/secω = , 33

2 2500 250

T π π π= = =ω

Fundamental time period of 1 2 3, andT T T is calculated as,

1× 2× 3× 4× 5×

1 50T π=

50π

25π 3

50π 2

25π

10π

2 150T π=

150π

75π

50π 2

75π

30π

3 250T π=

250π

125π 3

250π 2

125π

50π

Fundamental time period, 0 sec50

T π=

Fundamental frequency,

00

2 2 100 rad/sec/ 50T

π πω = = =π


Key Point Fundamental frequency is given by, [ ]1 2 3HCF , ,ω = ω ω ω

[ ]HCF 100, 300, 500 100 rad/secω = =

[ ]1 2 3LCM , , ,.....T T T T=

Scan for

Video Solution

1.10 (B)

1.11 (A)


Given : [ ] 2sin( )x n n= π

where, 20ω = π

For a discrete time signal to be periodic,

0

0 2m pN q

ω= = =π

ratio of integers.

Here, 2

0 2 2mN

π π= =π

02N m = π

So, there is not any integer value of m, which makes the 0N integer.

Thus, the signal is not periodic. Hence, the correct option is (D).

. Method 1 : Given :

1 1( ) ( 1)2 2

tg t x x t− = = −

Using time scaling property :

/2( )2

t t tx t x= ⎯⎯⎯→

[Time expansion]

Using time shifting property :

1 12 2

t tt tx x= − − ⎯⎯⎯→

[Right shift]

[ ] [ ]5

2

1

1( ) ( )6avgP g t g t dt

−

=

[ ]23

1

1 6( ) (1 )6 4avgP g t t dt

−

= −

[ ]3

2

1

1 36( ) ( 2 1)6 16avgP g t t t dt

−

= × − +

[ ]33 2

1

3 2( )8 3 2avg

t tP g t t−

= − +

[ ] 3 1( ) 9 9 1 3 18 3avgP g t = + − + + +

[ ] 3 16( ) 2 W8 3avgP g t = =

Hence, the average power of ( )g t is 2 W.

. Method 2 : Power of any signal is only affected by amplitude scaling and is NOT affected by amplitude inversion, time inversion, time shifting and time scaling.

Scan for

Video Solution

( )x t

t

3

3�

1�

2�

4�

1

2 3

40

3T �

2

tx

� � � �

t

3

3�

2�

4� 2

4 6

80

1( )

2

tx g t

��

� �

t

3

3�

1�

3� 2

31 50

1.12 (D)

1.13 2


Thus, power of ( )g t = Power of ( )x t

i.e. [ ]1 ( )2

tP x P x t− = …(i)

[ ]/2

2

/2

1( ) ( )T

T

P x t x t dtT −

=

[ ]21 2

2

1 1

1 6( ) 03 2

P x t t dt dt−

− = +

[ ]11 3

2

1 1

1( ) 9 33 3

tP x t t dt− −

= =

[ ]( ) 1 1 2 WP x t = + =

From equation (i),

[ ]1 ( ) 2 W2

tP x P x t− = =

Hence, the average power of ( )g t is 2 W.

Given : [ ] [ ]1 0.5nx n u n= α

and [ ]21.5, for 0,1

0, otherwisenx n

==

Energy of [ ] [ ] 21 1

nx n x n

∞

=− ∞

=

[ ] 2 2 2

00.5 (0.5)n n

n nu n

∞ ∞

=− ∞ =

= α = α 2 2 2 2 4(0.5) (0.5)= α + α + α +

2 2 2

24

11 (0.5) 314

α α α= = =− −

…(i)

Energy of [ ] [ ]1 2

2 20n

x n x n=

=

2 2( 1.5) ( 1.5) 1.5 1.5= + = + = 3 …(ii)

From equation (i) and (ii),

Since, energy of [ ]1x n and [ ]2x n are equal.

Thus, 24 33

α =

2 94

α =

3 1.52

α = =

Hence, the value of α is 1.5.

Given : 5

( ) ( )x

F x f y dy−

= …(i)

From the given figure,

'( ) ( ) 0F x f x= < , when 0x < …(ii)

'( ) ( ) 0F x f x= > , when 0x > …(iii)

So, the function becomes,

( ) 0; 0

'( )( ) 0; 0

f x xF x

f x x> >

= < <

Key Point

ddx

[increasing function] = positive

and ddx

[decreasing function] = negative

Thus, our function ( )F x must be decreasing for 0x < and increasing for 0x > .

From figure (a),

which does not satisfy the above criteria.

Scan for

Video Solution

( )F x

5�

x5�

0decreasing increasingincreasing

increasingincreasing decreasing

1.14 1.5

1.15 (C)


From figure (b),

which does not satisfy the above criteria. From figure (c),

Therefore, it clearly satisfies the above criteria. From figure (d),

which does not satisfy the above criteria. Hence, the correct option is (C).

Key Point Given : ( )f x = odd signal

Option (A) : odd signal Option (B) : odd signal Option (C) : even signal Option (D) : even signal

( ) odd signal even signalf x dx = =

ramp signal parabolic signal=


Consider a signal,

( ) sin tx tT

2π=

which is periodic with time period T. For sampling, we put st nT=

[ ] 02sin sinsTx n n n

Tπ= = ω

Hence, 02 sT

Tπω =

So the time period N is,

0

2s

TNT

π= =ω

For signal to be periodic, there exists atleast one value of m such that

s

TN mT

= is an integer.

From option (A), 2 sT T=

Thus, 2N m= So there is no integer m such that N is integer. So 2 sT T= won’t result in periodic signal. From option (B), 1.2 sT T=

Therefore, 65

N m=

So, for 5, 10, .....m = N becomes integer. Thus, the two options (C) and (D) get automatically eliminated. Hence, the correct option is (B).

Given : sin(4 )( )4

tx ttπ=

πsinc(4 )t= …(i)

( )F x

5�

x5�

0decreasing

decreasing

increasingincreasing

increasingincreasing

( )F x

5�

x5�

0

decreasingdecreasing increasingincreasing

( )F x

5�x

5�

0

constant

constantconstant

constant

Scan for

Video Solution

Scan for

Video Solution

1.16 (B)

1.17 0.25


. Method 1 : Energy of sinc ( )t is always equal to 1.

Thus, energy of sinc( )kt is given by,

[ ] 1Energy sinc( )ktk

= …(ii)


[ ] 1Energy sinc(4 )4

t =

0.25 JE = Hence, the energy of the signal is 0.25 J. . Method 2 :

Fourier transform of rect tA τ

is given by,

. .rect sinc( )F TtA A f ←⎯⎯→ τ τ τ

By duality property,

. .sinc( ) rectF T fA t A − τ τ ←⎯⎯→ τ

. .sinc( ) rectF T fA t A τ τ ←⎯⎯→ τ …(iii)

From equation (i) and (iii), 4, 1Aτ = τ =

14

A =

. . 1sinc(4 ) rect ( )4 4

F T ft X f ←⎯⎯→ =

From Parseval’s theorem, energy is given by,

2 2( ) ( )E x t dt X f df∞ ∞

− ∞ − ∞

= =

22

2

2

1( )4

E X f df df∞

− ∞ −

= =

[ ]1 2 2 0.25 J16

E = × + =

Hence, the energy of the signal is 0.25 J.

Given :

2( ) 1 2cos 3sin 2cos3 2 4

x t t tπ π π = + π + + +

. Method 1 : Let 1 2 3( ) 1 ( ) ( ) ( )x t x t x t x t= + + +

For 1( ),x t 10ω = π

12 2T π= =π

For 2( ),x t 20

23πω =

22 3

2 / 3T π= =

π

For 3( ),x t 30 2

πω =

32 4/ 2

T π= =π

∴ Fundamental period of ( )x t , [ ]1 2 3LCM of , , T T T T=

[ ]LCM of 2, 3, 4T = ∴ 12T = Hence, the fundamental time period for given signal ( )f t is 12 seconds.

. Method 2 : Fundamental frequency of ( )x t ,

1 2 30 0 0 0GCD of , , ω = ω ω ω

02GCD of , ,

1 3 2π π π ω =

0 6πω =

f2� 2

1/ 4

1rect

4 4

f� � � �

0

1.18 12


0

2 2 12sec/ 6

T π π= = =ω π

Hence, the fundamental time period for given signal ( )f t is 12 seconds.

Given : [ ] max ( )

k ny n x k

−∞≤ ≤=

To find response of the system for impulse input, i.e. impulse response , taking [ ] [ ]x n n= δ then output

[ ] [ ] max ( )k n

y n h n k−∞≤ ≤

≅ = δ

[ ] 0h n = for 0n <

1 for 0n ≥ As impulse will occur at 0k = and from the given condition k n≤ , so impulse response will be 1 for all 0n ≥ , which is same as the definition of unit step signal [ ]u n .

Hence the correct option is (A).

1.19 (A)

1.1 The subtraction of a binary number Y

from another binary number X, done by adding the 2's complement of Y to X, results in a binary number without overflow. This implies that the result is

(A) negative and is in normal form. (B) negative and is in 2's complement

form. (C) positive and is in normal form. (D) positive and is in 2's complement

form.

1.2 2's complement representation of a 16-bit

number (one sign bit and 15 magnitude bits) is FFFF. Its magnitude in decimal representation is

(A) 0 (B) 1 (C) 32,767 (D) 65,535

1.3 A signed integer has been stored in a byte

using the 2's complement format. We wish to store the same integer in a 16 bit word. We should

(A) copy the original byte to the less significant byte of the word and fill the more significant byte with zeros.

(B) copy the original byte to the more significant byte of the word and fill the less significant byte with zeros.

(C) copy the original byte to the less significant byte of the word and make each bit of the more significant byte equal to the most significant bit of the original byte.

(D) copy the original byte to the less significant bytes well as the more significant byte of the word.

1.4 An equivalent 2's complement

representation of the 2's complement number 1101 is

(A) 110100 (B) 001101 (C) 110111 (D) 111101

1.5 The 2 's complement representation of

–17 is (A) 101110 (B) 101111 (C) 111110 (D) 110001

1.6 4-bit 2's complement representation of a

decimal number is 1000. The number is (A) +8 (B) 0 (C) – 7 (D) – 8

1 Number Systems

1987 IIT Bombay

1993 IIT Bombay

1997 IIT Madras

1998 IIT Delhi

2001 IIT Kanpur

2002 IISc Bangalore


1.7 The range of signed decimal numbers

that can be represented by 6-bit 1's complement number is

(A) – 31 to + 31 (B) – 63 to + 63 (C) – 64 to + 63 (D) – 32 to + 31 1.8 11001, 1001 and 111001 correspond to

the 2's complement representation of which one of the following sets of number?

(A) 25, 9 and 57 respectively. (B) – 6, – 6 and – 6 respectively. (C) – 7, – 7 and – 7 respectively. (D) – 25, – 9 and – 57 respectively.

1.9 Decimal 43 in Hexadecimal and BCD

number system is respectively (A) B2, 01000011 (B) 2B, 01000011 (C) 2B, 0011 0100 (D) B2, 01000100

1.10 A new Binary Coded Pentary (BCP)

number system is proposed in which every digit of a base-5 number is represented by its corresponding 3-bit binary code. For example, the base-5 number 24 will be represented by its BCP code 010100. In this numbering system, the BCP code 100010011001 corresponds to the following number in base-5 system

(A) 423 (B) 1324 (C) 2201 (D) 4231

1.11 X = 01110 and Y = 11001 are two 5-bit

binary numbers represented in two's complement format. The sum of X and Y represented in 2's complement format using 6 bits is

(A) 100111 (B) 001000 (C) 000111 (D) 101001

1.12 The two numbers represented in signed

2's complement form are P = 11101101 and Q = 11100110 If Q is subtracted from P, the value

obtained in signed 2's complement form is

(A) 100000111 (B) 00000111 (C) 11111001 (D) 111111001

1.13 The number of bytes required to

represent the decimal number 1856357 in packed BCD (Binary Coded Decimal) form is ________. [Set - 02]

1.14 P, Q and R are the decimal integers

corresponding to the 4-bit binary number 1100 considered in signed magnitude, 1's complement and 2 's complement representation, respectively. The 6-bit 2 's complement representation of (P + Q + R) is

(A) 111101 (B) 110010 (C) 111001 (D) 110101

2004 IIT Delhi

2005 IIT Bombay

2006 IIT Kharagpur

2007 IIT Kanpur

2008 IISc Bangalore

2014 IIT Kharagpur

2020 IIT Delhi

3.3GATE ACADEMY ® Digital Electronics : Number Systems

1.1 B 1.2 B 1.3 C 1.4 D 1.5 B

1.6 D 1.7 A 1.8 C 1.9 B 1.10 D

1.11 C 1.12 B 1.13 4 1.14 D

Representation of Signed Binary Numbers

Special case of 2’s complement representation

Subtraction of a binary number Y from another number X is given below, ( )X Y X Y− = + − 2 'sX= + complement of Y. Example : Let 6X = and 9Y = Now, Z X Y= − 2 'sX= + complement of Y. Binary of 9 1 0 0 1Y = = Sign magnitude representation of 9 1 1 0 0 1− = 1's complement representation of 9 1 0 1 1 0− = 2's complement representation of 9−

1 0 1 1 0 1 1 0 1 1 1= + = And, sign magnitude representation of 6 using 5 bits is 0 0 1 1 0 2 'sZ X= + complement of Y.

0 0 1 1 01 0 1 1 11 1 1 0 1

+

Note : Carry not generated. Since, MSB of Z is 1, so Z is a negative number and result Z is available in 2's complement form.

To determine actual result, we have to find 2's complement of answer. 1's complement of 1 1 1 0 1 = 1 0 0 1 0 2's complement of 1 1 1 0 1 = 1 0 0 1 0 + 1 = 1 0 0 1 1 So, actual number is – 3. We can conclude that if carry is not generated, result is negative and is in 2's complement form. Hence, the correct option is (B). Key Point 1. MSB of signed magnitude representation,

1's complement representation and 2 's complement representation of any number denote its sign.

(a) If MSB is 1, then number is – ve. (b) If MSB is 0, then number is + ve. 2. If MSB of 1's complement representation

and 2 's complement representation of any number is 1, then number will be – ve.

3. 1's complement representation, 2 's complement representation and sign magnitude representation is known as “signed representation”.

4. During conversion from one signed representation to another, we will not change MSB.

5. Signed magnitude representation, 1's complement representation and 2 's complement representation of any positive number will be same.

Example : Representation +6 Sign magnitude representation 0110 1's complement representation 0110 2's complement representation 0110

Scan for Video

Explanation

Scan for Video

Explanation

Answers Number Systems

Explanations Number Systems

1.1 (B)


Given : 2's complement representation of 16-bit number = FFFF = 2(1111 1111 1111 1111)

Actual number = 2's complement of given number 1's complement of number 2(1000 0000 0000 0000)=

2's Complement of number = 1000000000000000 + 1

1000000000000001

Sign representation of 1000000000000001 1.= − Thus, magnitude will be 1. Hence, the correct option is (B).

Consider an example of – 9 Binary representation of 9 = 1001 Sign magnitude representation of 9 in 8 bit format = 0000 1001 Sign magnitude representation of – 9 in 8 bit format = 1000 1001 1’s complement representation of – 9 in 8 bit format = 1111 0110 2’s complement representation of – 9 in 8 bit format = 1111 0111 Now, Sign magnitude representation of – 9 in 16 bit format = 1000 0000 0000 1001 1’s complement representation of – 9 in 16 bit format = 1111 1111 1111 0110 2’s complement representation of – 9 in 16 bit format = 1111 1111 1111 0111 Consider an example of +13 Sign magnitude representation of + 13 in 8 bit format = 0000 1101 Sign magnitude representation of + 13 in 16 bit format = 0000 0000 0000 1101

Conclusion :

So, option (C) is correct i.e., copy the original byte to the less significant byte of the word and make each bit of the more significant byte equal to the most significant bit of the original byte. Hence, the correct option is (C).

Given : Number = 1101 According to the sign bit extension the sign bit can be extended towards left. 2 's complement representation of ‘– 3’ = 1101 2 's complement representation of ‘– 3’ using 6 bit = 111101 2 's complement representation of ‘– 3’ using 8 bit = 11111101 Hence, the correct option is (D).

Key Point 2's complement representation of 2 's complement of any number is equal to that number itself. Example :

2 's complement 2 's complement3 11101 10011− ⎯⎯⎯⎯⎯→ ⎯⎯⎯⎯⎯→

↓

– 3 (original number)

Scan for

Video Solution

Scan for

Video Solution

1.2 (B)

1.3 (C) 1.4 (D)


Binary representation of 17 = 010001 Binary representation of –17 = 110001 1's Complement of –17 = 101110 + 1 2's Complement of –17 = 101111 Hence, the correct option is (B).

The given number 1000 comes under special case of 2’s complement which has single ‘1’ followed by zeros. Decimal equivalent number of 2’s complement representation with single ‘1’ followed by ‘n’ number of zeros (2)n= − . Given 2's complement number contain 3 zeros after ‘1’, so the decimal number is 3(2) 8− = − . Hence, the correct option is (D).

Key Point Some more examples of special case :

For 1's complement number, range of signed decimal number of n bit 1(2 1)n−= − − to 1(2 1)n−+ − Here 6n = Therefore, range of signed decimal numbers 6 1(2 1)−= − − to 6 1(2 1)−+ − 5(2 1)= − − to 5(2 1)+ − 31= − to + 31 Hence, the correct option is (A).

Key Point 1. For 2's complement representation, the

range of decimal number of n-bit 1(2)n−= − to 1(2 1)n−+ − 2. For 1's complement representation, the

range of decimal number of n-bit 1(2 1)n−= − − to 1(2 1)n−+ −

All numbers are in 2's complement form. 2's complement of (2's complement of a number) = Original number (i) 11001 : Magnitude sign bit (MSB) of 2’s

complement is 1, hence number is negative.

(ii) 1001 : MSB of 2’s complement is 1 hence,

number is negative

(iii) 111001 : MSB of 2’s complement is 1 hence,

number is negative.


Scan for

Video Solution

1's complement of 11001 = 10110

2's complement of 11001 = 1 0111

Original Number =

+ 1

– 7

1's complement of 1001 = 1110

2's complement of 1001 = 1 111

Original Number =

+ 1

– 7

1's complement of 111001 = 100110

2's complement of 111001 = 1 00111

Original Number =

+ 1

– 7

Scan for

Video Solution

1.5 (B)

1.6 (D)

1.7 (A)

1.8 (C)


(i) Decimal to Hexadecimal :

Hence, 10 16(43) (2B)→

(ii) Decimal to BCD : 10 BCD(43) (01000011)=


Given : 5 2(24) (010100)→

Since, every number is represented by its corresponding 3 bit.

54 2 3 1

100010011001 (4231) =


Given : 01110X = and 11001Y = MSB of both numbers represent sign bit. Since MSB of X is 0, hence it is positive number and since MSB of Y is 1, hence it is negative number.

Since carry is discarded in the addition of two number represented in 2's complement, hence 00111X Y+ = X Y+ , in 6 - bit representation = 000111 Hence, the correct option is (C).

Key Point 1. If a binary number is represented using

N-bits (for 2's complement format) and it has been asked to represent it using (N+M) bits, then just copy MSB M-times. Examples :

(i) 1011 represents –5 using 4 bits 11011 represents –5 using 5 bits 111111011 represents –5 using 9 bits (ii) 0011 represents +3 using 4 bits 00011 represents +3 using 5 bits 000000011 represents +3 using 9 bits 2. In the 1's complement addition and

subtraction, if there is a carry out then bring the carry around and add it to the LSB. This carry is called the end around carry.

3. In the 2's complement addition and subtraction, if there is a carry out then just ignore the carry.

Given : P = 11101101, Q = 11100110 . Method 1 : Both P and Q are in signed 2's complement form. The MSB denotes sign of given number i.e. if MSB is 0 then given number is positive and if MSB is 1 then given number is negative. P = 11101101 Since MSB of P is 1, hence it is a negative number. Original number P = 2's complement of P Thus, the 2's complement of P will be = 10010011 = – 19 Similarly, the 2's complement of Q will be

16 43

2 11 (B)

Scan for

Video Solution

Scan for

Video Solution

01110

1

Discard carry

X

Y�

Z

�

�

11001�

00111

= 10011001

1

10011010

= – 26

+

1.9 (B)

1.10 (D)

1.11 (C)

1.12 (B)


Thus, 19 ( 26) 7P Q− = − − − = +

Now, 7 00000111+ =

The signed 2’s complement of a positive number is identical to the sign magnitude form of the positive number. 7 00000111+ =

Hence, the correct option is (B). . Method 2 : Given : P = 11101101 Q = 11100110 ( )P Q P Q− = + −

2 'Q s− = complement of 00011010Q =


Key Point

Short trick to find 2’s complement : Move from right to left, keep bits till first ‘1’ as it is, then complement each bit. Example : To find 2’s complement of 101011100100

2’s complements = 010100011100

To convert a decimal number into BCD number, each decimal digit is represented in 4-bits, as shown below, 1 → 0001 3 → 0011 8 → 1000 5 → 0101 5 → 0101 7 → 0111 6 → 0110

minimum1 byte1byteis required

10(1856357) 0001 1000 0101→

1byte

1byte

0110 0011 0101 0111

Thus, total 4 byte are required. Hence, the number of required to represent the decimal number 1856357 in packed BCD form is 4. Key Point (i) 1 Nibble = 4 bit (ii) 1 Byte = 8 bit

Given : P, Q and R are the decimal integers corresponding to a 4-bit binary number 1100 in signed magnitude, 1’s complement and 2’s complement representation. If a number in signed magnitude form is 10( ) 1100P =

Then MSB = 1 signifies that number is negative and other bit represents the binary of magnitude of that number 10 2( ) (100) 4P = − = −

If a number in 1’s complement representation is 10( ) 1100Q =

Then MSB = 1 signifies no. is negative. To represent only negative number in 1’s

= 1 1 1 0 1 1 0 1P– = 0 0 0 1 1 0 1 0Q

0 0 0 0 0 1 1 11

Carry discarded(Rule of 2’scomplement addition)

Scan for

Video Solution

101011100100

First ‘1’

Move

010100011100

Same

Complemented

1.13 4

1.14 (D)


complement we write the positive number first and then take its 1’s complement as to represent -3 in 4-bits 3 0011+ = 3 1100− = Hence, to find binary equivalent of any number represented in 1’s complement with MSB=1. Take 1’s complement of bits other than MSB and place negative sign. 10( ) 1100Q =

1 → negative 100 → 1’s complement of actual binary 10 2( ) (011) 3Q = − = −

Similarly if a number with MSB = 1 is represented in 2’s complement, then take the 2’s complement of bits other than MSB and place negative sign. 10( ) 1100R =

1 → negative 100 → 2’s complement of actual number 10 2( ) (100) 4R = − = −

4 3 4 11P Q R+ + = − − − = − To represent – 11 in 2’s complement in 6 bits, writing +11 first 11 001011+ = Now to represent – 11, taking 2’s complement of +11 11 110101− = Hence, the correct option is (D).

1.1 The transfer function of a linear system

is the (A) ratio of the output 0( )v t and the

input ( )iv t .

(B) ratio of the derivatives of the output and the input.

(C) ratio of the Laplace transform of the output and that of the input with all initial conditions assumed to be zero.

(D) none of the above.

1.2 The unit impulse response of a linear

time invariant system is the unit step function u(t). For t > 0, the response of the system to an excitation ( ),ate u t−

0a > will be

(A) atae− (B) 1 (1 )atea

−−

(C) (1 )ata e−− (D) 1 ate−−

1.3 The transfer function of a tachometer is of the form

(A) Ks (B) Ks

(C) ( 1)

Ks +

(D) ( 1)

Ks s +

1.4 The open loop transfer function of an

unity feedback system is 2

22 6 5

( 1) ( 2)s s

s s+ +

+ +

The characteristic equation of the closed loop system is

(A) 22 6 5 0s s+ + = (B) 2( 1) ( 2) 0s s+ + = (C) 2 22 6 5 ( 1) ( 2) 0s s s s+ + + + + = (D) 2 22 6 5 ( 1) ( 2) 0s s s s+ + − + + =

1.5 The open-loop DC gain of a unity

negative feedback system with closed-

loop transfer function 24

7 13s

s s+

+ + is

(A) 413

(B) 49

(C) 4 (D) 13

1.6 Despite the presence of negative

feedback, control system still have problems of instability because the

(A) components used have non linearities.

(B) dynamic equations of the systems are not known exactly.

(C) mathematical analysis involves approximations.

(D) system has large negative phase angle at high frequencies.

1Basics of

Control Systems

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1.7 The transfer function of a plant is

25( )

( 5) ( 1)T s

s s s=

+ + +

The second-order approximation of T(s) using dominant pole concept is

(A) 1( 5) ( 1)s s+ +

(B) )1()5(

5++ ss

(C) 1

52 ++ ss

(D) 1

12 ++ ss

1.8 A system with transfer function

2( 9) ( 2)( )

( 1) ( 3) ( 4)s sG s

s s s+ +=

+ + +

is excited by sin ( )tω . The steady state output of the system is zero at

(A) 1 rad/secω = (B) 2 rad/secω = (C) 3 rad/secω = (D) 4 rad/secω =

1.9 The open-loop transfer function of a dc

motor is given as ( ) 10( ) 1 10a

sV s sω =

+. When

connected in feedback as shown below, the approximate value of aK that will reduce the time constant of the c losed loop system by one hundred times as compared to that of the open-loop system is

(A) 1 (B) 5 (C) 10 (D) 100

1.10 The input 23 ( ),te u t− where ( )u t is the

unit step function, is applied to a system

with transfer function 2 .3

ss

−+

If the

initial value of the output is 2,− then the value of the output at steady state is ______. [Set - 03]

1.11 Negative feedback in a closed-loop

control system DOES NOT [Set - 01] (A) reduce the overall gain. (B) reduce bandwidth. (C) improve disturbance rejection. (D) reduce sensitivity to parameter

variation.

1.12 A first-order low-pass filter of time

constant T is excited with different input signals (with zero initial conditions up to t = 0). Match the excitation signals X, Y, Z with the corresponding time responses for t ≥ 0 [Set - 01]

X : Impulse P : /1 t Te−− Y : Unit step Q :

/(1 )t Tt T e−− −

Z : Ramp R : /1 t TeT

−

(A) X → R, Y → Q, Z → P (B) X → Q, Y → P, Z → R (C) X → R, Y → P, Z → Q (D) X → P, Y → R, Z → Q 1.13 The response of the system

( 2)( )( 1)( 3)

sG ss s

−=+ +

to the unit step

input ( )u t is ( )y t . [Set - 02]

The value of dydt

at 0t += is ________.

10

1 10s�+–

( )R s ( )s�aK

( )a

V s

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4.3GATE ACADEMY ® Control Systems : Basics of Control Systems

1.1 C 1.2 B 1.3 A 1.4 C 1.5 B

1.6 A 1.7 D 1.8 C 1.9 C 1.10 0

1.11 B 1.12 C 1.13 1

The transfer function of a linear time invariant system is the ratio of the Laplace transform of the output and that of the input with all initial conditions assumed to be zero. Continuous time system :

Transform domain :

( )( )( )

Y sH sX s

= = Transfer function


Given : Impulse response is unit step function

i.e. 1( )H ss

=

Excitation or input is, ( ) ( ), 0atr t e u t a−= > Taking Laplace transform of ( )r t ,

1( )R ss a

=+

The Laplace transform of output is given by, ( ) ( ) ( )C s R s H s=

1 1 1 1( )( )

C ss s a a s s a

= = − + +

Taking inverse Laplace transform of ( )C s ,

1( ) [1 ]atc t ea

−= −


DC tachometers are used to convert rotational speed into proportional DC voltage. Let, ( )e t is proportional speed.

( )tω is rotational speed.

Hence, ( ) ( )e t t∝ ω

( ) ( )e t K t= ω

Since, ( ) dtdtθω =

Therefore, ( )( ) d te t Kdtθ=

Taking Laplace transform of ( )e t ,

( ) ( )E s Ks s= θ

( )( )

E s Kss

=θ


Given : The open loop transfer function of a unity feedback system is,

2

22 6 5( )

( 1) ( 2)s sG s

s s+ +=

+ +

The characteristic equation is given by, 1 ( ) 0G s+ =

2

22 6 51 0

( 1) ( 2)s s

s s+ ++ =

+ +

2 22 6 5 ( 1) ( 2) 0s s s s+ + + + + =


( )h t ( ) ( ) ( )y t x t h t� �( )x t

( )H s ( ) ( ) ( )Y s X s H s�( )X s

Answers Basics of Control Systems

Explanations Basics of Control Systems

1.1 (C)

1.2 (B)

1.3 (A)

1.4 (C)


Given : 24CLTF

7 13s

s s+=

+ +

For unity negative feedback system the closed loop transfer function is given by,

( )CLTF1 ( )

G sG s

=+

where, G(s) = open loop transfer function

2( ) 4

1 ( ) 7 13G s s

G s s s+=

+ + +

21 ( ) 7 13

( ) 4G s s s

G s s+ + +=

+

2 21 7 13 6 91

( ) 4 4s s s s

G s s s+ + + += − =

+ +

24( )

6 9sG s

s s+=

+ +

For DC gain, s = 0

0

4( )9sG s

==


This question is based on Lyapunov’s stability criterion, which is not in GATE syllabus. Consider a system, where only linear elements are present. Block diagram of such system is shown below,

where, A = Amplifier gain As amplifier is a linear element here. Hence, the output u is linearly related with input e. i.e. u e∝ u Ae=

The characteristics equation is given by,

1 0( 1)

As s

+ =+

2 0s s A+ + = System is stable for all positive values of A that can be proved by Routh-Hurwitz criteria.

2s 1 A1s 1 00s A 0

However, practically amplifier response is non-linear transistor or Op-Amp characteristics is non-linear.

Due to non-linearity of component used, now the system becomes non-linear so its stability can be determined using “Lyapunov’s stability criteria”. Hence, the correct option is (A).

Given : 25( )

( 5) ( 1)T s

s s s=

+ + +

rd3 order system

In dominant pole concept, DC gain should be same after approximation. DC gain before approximation is,

5( 0) 1.0(0 5) (0 0 1)

T s = = =+ + +

Scan for

Video Solution

1

( 1)s s �+–

( )r t ( )c tA

Amplifier

e u

Linear

u

e

u

e

1.5 (B)

1.6 (A)

1.7 (D)


Since, insignificant pole is more than five times of the dominant pole, hence insignificant pole can be neglected. Therefore, second order approximation can be written as given below,

2 25 1( )

( 5)( 1) ( 1)T s

s s s s s= ≈

+ + + + +

DC gain after approximation,

0

1( ) 11sT s

== =


Key Point (i) Time constant due to dominant pole is

higher than insignificant pole. (ii) Dominant pole gives slower response

compared to insignificant pole. (iii) For approximation of transfer function

we neglect insignificant poles. (iv) For existence of dominant pole, the ratio

of insignificant pole to the significant pole (dominant pole) should be greater than or equal to 5.

(v) In dominant pole concept, overall time constant of system is given by dominant pole.

Given : 2( 9) ( 2)( )

( 1) ( 3) ( 4)s sG s

s s s+ +=

+ + +

( ) sin ( )x t t= ω Taking Laplace transform of ( )x t ,

[ ] 2 2( ) sin ( )X s L ts

ω= ω =+ ω

( )( )( )

Y sG sX s

=

( ) ( ) ( )Y s G s X s=

2

2 2( 9) ( 2)( )

( ) ( 1) ( 3) ( 4)s sY s

s s s sω + +=

+ ω + + +

Using final value theorem, steady state output is given by,

0lim ( ) lim ( )t s

y t sY s→ ∞ →

=

2

2 20

( 9) ( 2)lim ( ) lim( ) ( 1) ( 3) ( 4)t s

s s sy ts s s s→ ∞ →

ω + +=+ ω + + +

Final value theorem is applicable only for absolute stable system. But 2 2( )s + ω term in denominator is making system as a marginal stable system because, 2 2 0s + ω = s j= ± ω [Imaginary axis poles]

For existence of final value theorem, 2 2( )s + ω

should be cancel out with 2( 9)s + . Hence

2 9ω = 3 rad/secω = At 3 rad/secω = , steady state output will be zero. Hence, the correct option is (C).

Given : closed loop open loop1

100τ = τ

0.5�

0.866j

j�

�

0.866j�

5�

Insignificantpole

Dominantpole

Scan for

Video Solution

1.8 (C)

1.9 (C)


where, τ represents time constant.

From figure, Open loop transfer function of DC motor

= ( ) 10( ) 1 10a

sV s sω =

+

Location of pole of open loop transfer function is shown below,

Time constant is defined as reciprocal of magnitude of negative real root. Hence, open loop 10 secτ =

Since, closed loop open loop1

100τ = τ

closed loop10 0.1 sec100

τ = = … (i)

Closed-loop transfer function for negative unity feedback is given by,

( )( )1 ( )

G sT sG s

=+

Here, 10( )1 10aG s K

s = +

10( ) 1 10( )

10( ) 11 10

a

a

Ks sT s

R s Ks

ω + = = + +

10( )1 10 10

a

a

KT ss K

=+ +

10( )10 (10 1)

a

a

KT ss K

=+ +

Location of pole of closed loop transfer function is shown below,

From above figure,

closed loop10

10 1aKτ =

+ … (ii)


10 0.1 sec10 1aK

=+

10 110 1 10aK

=+

10 1 100aK + =

10 99aK =

9.9 10aK = Hence, the correct option is (C).

Given : 2( ) 3 ( ),tr t e u t= −

2( )3

sH ss

−=+

and initial value of the output is 2.−

. Method 1 :

Taking Laplace transform of r(t),

10

1 10s�+–

( )R s ( )s�aK

( )a

V s

�

j�

1

10

�

�

j�

1 10

10

aK��

�

Scan for

Video Solution

2( ) 3 ( )tr t e u t� � 2

3

s

s

��

��

Steady stateoutput = ( )y t

1.10 0


3( )2

R ss−=−

Transfer function of the system is,

( )( )( )

Y sH sR s

=

( ) ( ) ( )Y s R s H s=

3 2( )2 3

sY ss s− − = − +

3( )3

Y ss−=+

Using final value theorem steady state output is given by,

0lim ( ) lim ( )ss t s

y y t sY s→∞ →

= =

0

3lim3ss s

y ss→

− = +

0

3 0lim 00 3ss s

y→

− × = = +

Hence, the value of the output at steady state is 0. . Method 2 :

3( )3

Y ss−=+

Taking inverse Laplace transform, 3( ) 3 ty t e−= − Steady state output is given by, lim ( )ss t

y y t→∞

=

3lim 3 tss t

y e−

→∞ = −

0ssy = Hence, the value of the output at steady state is 0.

Negative feedback in a closed loop control system, (i) increases bandwidth. (ii) reduces gain. (iii) improve disturbance rejection.

(iv) reduce sensitivity to parameter variation. Hence, the correct option is (B).

For first order system with time constant T,

1( ) , ( ) 1G s H ssT

= =

Block diagram of first order system is shown below,

Fig. First order system

Closed loop transfer function of negative unity feedback system is given by,

( ) ( )( ) 1 ( )

Y s G sR s G s

=+

( ) 1 /1( ) 1

Y s sTR s

sT

=+

1 /( ) ( )11

sTY s R s

sT

=+

1( ) ( )1

Y s R ssT

=+

For impulse response, ( ) 1R s =

/1( ) t Ty t eT

−= , for 0t ≥

Hence, X R→

For step response, 1( )R ss

=

1 (1 ) ( )( )(1 ) (1 )

sT sTY ss sT s sT

+ −= =+ +

1 1( )1(1 )

T TY ss sT s T s

T

= − = −+ +

/( ) (1 )t Ty t e−= − , for 0t ≥

1

sT( )R s ( )Y s

1.11 (B)

1.12 (C)


Hence, Y P→

For ramp response, 21( )R ss

=

2 21 1( ) 1(1 )

T TY ss sT s s s

T

= = − ++ +

/( ) (1 )t Ty t t T e−= − − , for 0t ≥ Hence, Z Q→ Hence, the correct option is (C). Key Point

ddt

(ramp response) = step response

ddt

(step response) = impulse response

Given : ( 2)( )( 1) ( 3)

sG ss s

−=+ +

and

( ) ( )r t u t=

. Method 1 :

1( )R ss

=

Laplace transform of the output is given by, ( ) ( ) ( )Y s G s R s=

1 ( 2)( ) ( )( 1) ( 3)

sY s G ss s s s

−= × =+ +

(0)y = Initial value Applying initial value theorem, (0) lim ( )

sy sY s

→∞=

( 2)(0) lim( 1) ( 3)s

sys s→∞

−=+ +

21(0) lim

1 31 1s

sys

s s→∞

− =

+ +

0=

( ) (0)dyL sY s ydt

= −

( 2)( )( 1) ( 3)

dy s sL sY sdt s s s

− = = + +

( 2)( 1) ( 3)

dy sLdt s s

− = + +

0

lims

t

dy dysLdt dt→∞

=

= ( 2)lim

( 1) ( 3)s

s ss s→∞

−=+ +

2

20

21lim

1 31 1s

t

sdy sdt s

s s→∞

=

− =

+ +

0

21lim 1

1 31 1s

t

dy sdt

s s→∞

=

− = =

+ +

Hence, the value of dydt

at 0t += is 1.

. Method 2 :

( ) ( ) ( )Y s R s G s=

( 2)( )( 1) ( 3)

sY ss s s

−=+ +

2 / 3 3 / 2 5 / 6( )1 3

Y ss s s

−= + −+ +

Taking inverse Laplace of ( )Y s ,

32 3 5( )3 2 6

t ty t e e− −−= + −

33 5( ) 02 2

t td y t e edt

− −= − +

0

3 5( ) 12 2t

d y tdt =

= − + =

Hence, the value of dydt

at 0t += is 1.

( )G s ( )y t( )r t

( )G s ( )Y s( )R s

Scan for

Video Solution

1.13 1

1.1 The rank of matrix

cannot be more than (A) m (B) n (C) mn (D) None 1.2 Solve the following system of equations 1 2 3 3x x x+ + =

1 2 0x x− =

1 2 3 1x x x− + =

(A) Unique solution (B) No solution (C) Infinite number of solutions (D) Only one solution

1.3 The eigen values of the matrix

0 11 0

A =

are

(A) 1, 1 (B) –1, –1, (C) j, – j (D) 1, – 1

1.4 The eigen values of the matrix

2 1 0 00 3 0 00 0 2 00 0 1 4

− − −

are

(A) 2, – 2, 1, – 1

(B) 2, 3, – 2, 4

(C) 2, 3, 1, 4

(D) None of the above

1.5 Given an orthogonal matrix

1 1 1 11 1 1 11 1 0 00 0 1 1

A

− − =

− −

then the value of

1[ ]TAA − is

(A)

1 0 0 04

10 0 04

10 0 02

10 0 02

(B)

1 0 0 02

10 0 02

10 0 02

10 0 02

1 Linear Algebra

( )m n× ( )m n<

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(C)

1 0 0 00 1 0 00 0 1 00 0 0 1

(D)

1 0 0 04

10 0 04

10 0 04

10 0 04

1.6 Given matrix , the eigen vector

is

(A) (B)

(C) (D)

1.7 Let 2 0.10 3

A− =

and 1

120

aA

b

− =

Then (a + b) is

(A) (B)

(C) (D)

1.8 The rank of the matrix is

(A) 0 (B) 1 (C) 2 (D) 3 1.9 The eigen values and the corresponding

eigen vectors of a matrix are given by

Eigen value Eigen vector

1

11

v =

2

11

v = −

The matrix is

(A) (B)

(C) (D)

1.10 For the matrix the eigen value

corresponding to the eigen vector

is (A) 2 (B) 4 (C) 6 (D) 8

1.11 It is given that 1 2, ,........ MX X X are M

non zero, orthogonal vectors. The dimension of the vector space spanned by the 2M vectors 1 2, ,........ MX X X ,

1 2, ,........ MX X X− − − is (A) 2 M (B) M + 1 (C) M (D) Dependent on the choice of

1 2, ,........ MX X X

1.12 All the four entries of the matrix

are non-zero and one of

its eigen values is zero. Which of the following statements is true ?

(A) (B)

4 24 3

−

32

43

21

−

12

−

720

320

1960

1120

1 1 11 1 01 1 1

−

2 2×

1 8λ =

2 4λ =

6 22 6

4 66 4

2 44 2

4 88 4

4 22 4

101101

2 2×11 12

21 22

P PP

P P

=

11 22 12 21 1P P P P− =

11 22 12 21 1P P P P− = −

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5.3GATE ACADEMY ® Engineering Mathematics : Linear Algebra

(C)

(D) 1.13 The system of linear equations 4x + 2y = 7 2x + y = 6 has (A) a unique solution. (B) no solution. (C) an infinite number of solutions. (D) exactly two distinct solutions.

1.14 The eigen values of the following matrix

are

1 3 53 1 60 0 3

− − −

(A) 3,3 5,6j j+ −

(B) 6 5, 3 ,3j j j− + + − (C) 3 ,3 ,5j j j+ − + (D) 3, 1 3, 1 3j j− + − −

1.15 The eigen values of a skew-symmetric

matrix are (A) always zero. (B) always pure imaginary. (C) either zero or pure imaginary. (D) always real.

1.16 The system of equations has NO solution for values of λ and μ

given by (A) (B) (C) (D)

1.17 Given that5 32 0

A− − =

and 1 00 1

I =

, the value of 3A is (A) 15A + 12I (B) 19A + 30I (C) 17A + 15I (D) 17A + 21I

1.18 The minimum eigen value of the

following matrix is

3 5 25 12 72 7 5

(A) 0 (B) 1 (C) 2 (D) 3 1.19 Let A be an matrix and B an

matrix. It is given that determinant ( )mI AB+ = determinant ( )nI BA+ , where kI is the identity matrix. Using the above property, the determinant of the matrix given below is

2 1 1 11 2 1 11 1 2 11 1 1 2

(A) 2 (B) 5 (C) 8 (D) 16

1.20 For matrices of same dimension M, N

and scalar c, which one of these properties DOES NOT ALWAYS hold? [Set - 01]

(A) ( )T TM M=

(B) ( ) ( )T TcM c M=

(C) ( )T T TM N M N+ = + (D) MN NM=

11 22 12 21 0P P P P− =

11 22 12 21 0P P P P+ =

6+ + =x y z4 6 20+ + =x y z4+ + λ = μx y z

6, 20λ = μ = 6, 20λ = μ ≠6, 20λ ≠ μ = 6, 20λ ≠ μ ≠

m n× n m×

×k k

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1.21 A real (4 4)× matrix A satisfies the equation 2A I= where I is the (4 4)× identity matrix. The positive eigen value of A is __________. [Set - 01]

1.22 Consider the matrix :

6

0 0 0 0 0 10 0 0 0 1 00 0 0 1 0 00 0 1 0 0 00 1 0 0 0 01 0 0 0 0 0

J =

which is obtained by reversing the order of the columns of the identity matrix 6.I

Let 6 6α ,P I J= + where is a non-negative real number. The value of for which det ( ) 0P = is _________.

[Set - 01] 1.23 The determinant of matrix A is 5 and the

determinant of matrix B is 40. The determinant of matrix AB is ______. [Set - 02]

1.24 The system of linear equations2 1 3 53 0 1 41 2 5 14

abc

= −

have [Set - 02]

(A) a unique solution. (B) infinitely many solutions. (C) no solution. (D) exactly two solutions. 1.25 The maximum value of the determinant

among all 2 2× real symmetric matrices with trace 14 is_____. [Set - 02]

1.26 Which one of the following statements is NOT true for a square matrix? [Set - 03]

(A) If A is upper triangular, the eigen values of A are the diagonal elements of it.

(B) If A is real symmetric, the eigen values of A are always real and positive.

(C) If A is real the eigen values of A and TA are always the same.

(D) If all the principal minors of A are positive, all the eigen values of A are also positive.

1.27 Consider a system of linear equations 2 3 1x y z− + = − , 3 4 1x y z− + = and 2 4 6x y z k− + − = The value of k for which the system has

infinitely many solution is _______. [Set - 01]

1.28 The value of p such that the vector 123

is an eigen vector of the matrix 4 1 2

2 114 4 10p

−

is [Set - 01]

1.29 Let ( ) az bf zcz d

+=+

. If 1 2( ) ( )f z f z= ,

for all 1 2, 2, 4z z a b≠ = = and 5c = , then d should be equal to ______.

[Set - 02] 1.30 The value of x for which all the eigen-

values of the matrix given below are real is [Set - 02]

10 5 4

20 24 2 10

jx

+

−

(A) 5 j+ (B) 5 j− (C) 1 5j− (D) 1 5j+

1.31 For 1 tan

,tan 1

xA

x = −

the

determinant of 1TA A− is [Set - 03] (A) 2sec x (B) cos4x (C) 1 (D) 0

αα

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1.32 Let 4M I= , (where I denotes the

identity matrix) and 2,M I M I≠ ≠ and 3M I≠ . Then, for any natural number

1,k M − equals [Set - 01]

(A) 4 1kM + (B) 4 2kM + (C) 4 3kM + (D) 4kM 1.33 A sequence x[n] is specified as,

[ ]

[ ]1 1 1

, for 21 1 0 0

nx nn

x n = ≥ −

The initial conditions are x[0] = 1, x[1] = 1, and x[n] = 0 for n < 0. The value of x [12] is ______ [Set - 01]

1.34 The value of x for which the matrix 3 2 49 7 136 4 9

Ax

= − − − +

has zero as an

eigen value is _____. [Set - 02]

1.35 The matrix

0 3 72 5 1 30 0 2 40 0 0

a

A

b

=

has det

( ) 100A = and trace ( ) 14A = . The value of a b− is _____.[Set - 02] 1.36 Consider a 2 × 2 square matrix

x

Aσ = ω σ

where, x is unknown. If the eigen values of the matrix A are ( )jσ + ω and ( )jσ − ω , then x is equal to [Set - 03]

(A) j+ ω (B) j− ω (C) + ω (D) −ω 1.37 If the vectors 1 2(1, 0, 2), (0, 1, 0)e e= =

and 3 ( 2, 0, 1)e = − form an orthogonal basis of the three dimensional real space

3R , then the vector 3(4, 3, 3)u R= − ∈ can be expressed as [Set - 03]

(A) 1 2 32 1135 5

u e e e= − − +

(B) 1 2 32 1135 5

u e e e= − − −

(C) 1 2 32 1135 5

u e e e= − + +

(D) 1 2 32 1135 5

u e e e= − + −

1.38 The rank of the matrix 5 10 101 0 23 6 6

M =

is [Set - 01] (A) 0 (B) 1 (C) 2 (D) 3 1.39 Consider the following statements about

the linear dependence of the real valued functions 1 21,y y x= = and 2

3y x= , over the field of real numbers.

I. 1 2 3, andy y y are linearly independent on 1 0x− ≤ ≤ .

II. 1 2 3, andy y y are linearly dependent on 0 1x≤ ≤ .

III. 1 2 3, andy y y are linearly independent on 0 1x≤ ≤ .

IV. 1 2 3, andy y y are linearly dependent on 1 0x− ≤ ≤ .

Which one among the following is correct? [Set - 01]

(A) Both I and II are true (B) Both I and III are true (C) Both II and IV are true (D) Both III and IV are true

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1.40 Consider the 5 5× matrix

1 2 3 4 55 1 2 3 44 5 1 2 33 4 5 1 22 3 4 5 1

A

=

It is given that A has only one real eigen value. Then the real eigen value of A is [Set - 01]

(A) 2.5− (B) 0 (C) 15 (D) 25 1.41 The rank of the matrix

1 1 0 0 00 0 1 1 00 1 1 0 01 0 0 0 10 0 0 1 1

− − − − −

is ________.

[Set - 02]

1.42 Let M be a real 4 4× matrix. Consider

the following statements : S1 : M has 4 linearly independent

eigenvectors. S2 : M has 4 distinct eigenvalues. S3 : M is non-singular (invertible)

matrix. Which one among the following is

TRUE? (A) S1 implies S2 (B) S1 implies S3 (C) S2 implies S1 (D) S3 implies S2

1.43 Consider matrix 2 2

2k kA

k k k

= − and

vector 1

2

xX

x

=

. The number of distinct

real values of k for which the equation 0AX = has infinitely many solutions is

________.

1.44 The number of distinct Eigen values of

the matrix

2 2 3 30 1 1 10 0 3 30 0 0 2

A

=

is equal to _______.

1.45 If 1v , 2v … 6v are six vectors in 4 ,

which one of the statements is FALSE? (A) Any four of these vectors form a

basis for 4 . (B) It is not necessary that these vectors

span 4 .

(C) If 1 3 5 6{v , v , v , v } spans 4 , then it

forms a basis of 4 . (D) These vectors are not linearly

independent. 1.46 Consider the following system of linear

equations. 1 2 12x x b+ = ; 1 2 22 4x x b+ = ;

1 2 33 7x x b+ = ; 1 2 43 9x x b+ =

Which one of the following conditions ensures that a solution exists for the above system?

(A) 2 1 1 3 42 and 3 6 0b b b b b= − + =

(B) 3 1 1 3 42 and 6 3 0b b b b b= − + =

(C) 2 1 1 3 42 and 6 3 0b b b b b= − + =

(D) 3 1 1 3 42 and 3 6 0b b b b b= − + =

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2019 IIT Madras

2020 IIT Delhi


1.1 A 1.2 A 1.3 D 1.4 B 1.5 A

1.6 C 1.7 A 1.8 C 1.9 A 1.10 C

1.11 C 1.12 C 1.13 B 1.14 D 1.15 C

1.16 B 1.17 B 1.18 A 1.19 B 1.20 D

1.21 1 1.22 1 1.23 200 1.24 B 1.25 49

1.26 B 1.27 2 1.28 17 1.29 10 1.30 B

1.31 C 1.32 C 1.33 233 1.34 1 1.35 3

1.36 D 1.37 D 1.38 C 1.39 B 1.40 C

1.41 4 1.42 C 1.43 2 1.44 3 1.45 A

1.46 C

Let A be a matrix with dimension of m n× . m n< A is non-square matrix, for non-square matrix rank is given as, ( ) min ( , )A m nρ ≤ ( )A mρ ≤ Thus, rank of matrix A cannot be more than m. Hence, the correct option is (A).

Given : The system of equations are, 1 2 3 3x x x+ + =

1 2 0x x− =

1 2 3 1x x x− + =

. Method 1 : It is in the form of a non-homogeneous equation is given by, AX B=

Where, 1 1 11 1 01 1 1

A = −

−

, 1

2

3

xB x

x

=

Number of unknowns ( ) 3n = .

The augmented matrix is given by,

[ ]1 1 1 : 3

: 1 1 0 : 01 1 1 : 1

A B = −

−

By elementary transformation

2 2 1R R R→ − and 3 3 1R R R→ − ,

[ ]1 1 1 : 3

: 0 2 1 : 30 2 0 : 2

A B = − − −

− −

3 3 2R R R→ − ,

[ ]1 1 1 : 3

: 0 2 1 : 30 0 1 : 1

A B = − − −

There exists three non-zero rows, So, ( ) 3, ( : ) 3A A Bρ = ρ =

( ) ( : ) 3A A B nρ = ρ = =

This shows that the given system of non-homogeneous equations has a unique solution. Hence, the correct option is (A).

Answers Linear Algebra

Explanations Linear Algebra

1.1 (A)

1.2 (A)


. Method 2 : Here we can see that the determinant of coefficient matrix is not equal to 0. The point is to be noted if 0A ≠ then we have unique solution. Hence, the correct option is (A). Key Point For a non-homogenous equation PX Q= : 1. If rank [ : ] rank[ ]P Q P≠ No solution exist. 2. If rank [ : ] rank[ ]P Q P= = Number of

variables Unique solution exist. 3. If rank [ : ] rank[ ]P Q P= < Number of

variables Infinite numbers of solutions exist.

Given : 0 11 0

A =

. Method 1 : The characteristic equation of the matrix A is given by, 0A I− λ = [where, λ = Eigen values]

1

01

−λ=

−λ

1λ = ± Thus, the eigen values of the matrix are 1 and – 1. Hence, the correct option is (D). . Method 2 : Let 1λ and 2λ be the eigen values of the matrix A . By the properties of eigen values, 1 2 0 1 1Aλ λ = = − = − …(i)

and 1 2 0 0 0λ + λ = + = …(ii) By checking all the options, only option (D) satisfies both equations. From option (D), 1( 1) 1− = − and 1 1 0− = Hence, the correct option is (D). Key Point For any square matrix : 1. Sum of eigen values is equal to the trace

of the matrix (sum of leading diagonal elements).

2. Product of eigen values is equal to the determinant of the matrix.

Given :

2 1 0 00 3 0 00 0 2 00 0 1 4

A

− =

− −

The characteristic equation is given by, 0A I− λ =

2 1 0 00 3 0 0

00 0 2 00 0 1 4

− λ −− λ

=− − λ

− − λ

(2 )(3 )( 2 )(4 ) 0− λ − λ − − λ − λ = 1 22, 3,λ = λ =

3 42, 4λ = − λ =

The roots of characteristics equation of a matrix gives its eigen values. Hence, the correct option is (B).

Given :

1 1 1 11 1 1 11 1 0 00 0 1 1

A

− − =

− −

2 1 0λ − =

1.3 (D)

1.4 (B)

1.5 (A)


. Method 1 :

1 1 1 11 1 1 11 1 0 00 0 1 1

A

− − =

− −

The transpose of matrix A is given by,

1 1 1 01 1 1 01 1 0 11 1 0 1

TA

− =

− − −

4 0 0 00 4 0 00 0 2 00 0 0 2

TA A

=

This is a diagonal matrix. So, the determinant is given by, 4 4 2 2 64TA A = × × × =

1 Adj( )T

TT

AAA AAA

− =

16 0 0 00 16 0 0

Adj( )0 0 32 00 0 0 32

TAA

=

1

16 0 0 00 16 0 01[ ]0 0 32 0640 0 0 32

TAA −

=

1

1 0 0 04

10 0 04[ ]

10 0 02

10 0 02

TAA −

=


. Method 2 : A is an orthogonal matrix. (Given) By the properties of orthogonal matrices,

TAA I= (Identity matrix)

1 0 0 00 1 0 00 0 1 00 0 0 1

TAA I

= =

Thus, [ ] 11( )TAA I I−− = =

1

1 0 0 00 1 0 0

( )0 0 1 00 0 0 1

TAA −

=

Note : Since, the matrix given in the question is not orthogonal, so the given information in the question framed, is wrong.

Given : 4 24 3

A− =

The characteristic equation is given by, λ 0A I− =

4 2

04 3

− − λ=

− λ

(4 )( 3) 8 0+ λ λ − − =

2 20 0λ + λ − = 1 24, 5λ = λ = − For any eigen vector [ ]X for a matrix [ ]A corresponding to eigen value λ , the following matrix equation satisfies, [ ][ ] 0A I X− λ =

For 1 4λ = ,

1

2

8 20

4 1xx

− = −

1 28 2 0x x− + =

1 28 2x x=

1.6 (C)


Let ,

Thus, 14

X =

is an eigen vector corresponding

to the eigen value

4λ = . For 2 5λ = − ,

1

2

1 20

4 8xx =

1 22 0x x+ = Let 1 2x K= ,

1

22

2 2x Kx − −= =

1

2

21

xK

x = −

Thus, 21

X = − is an eigen vector corresponding

to the eigen value

5λ = − . Hence, the correct option is (C).

Given : 2 0.10 3

A− =

and 1

120

aA

b

− =

Inverse of matrix A is given by,

1 Adj AA

A− =

3 0.1

Adj 0 2

A =

and 6A =

So, 113 0.112

0 26 0

aA

b

− = =

By comparing both sides,

0.1 1and6 3

a b= =

a + b


Given : 1 1 11 1 01 1 1

A = −

. Method 1 : By elementary transformation,

3 3 1R R R→ −

1 1 11 1 00 0 0

A = −

2 2 1R R R→ −

1 1 10 2 10 0 0

A = − −

There exist two non-zero rows. So, ( ) 2Aρ =

Hence, the correct option is (C). . Method 2 : The rank of a matrix is defined as the number of linearly independent rows or columns whichever is minimum present in the matrix. In the given matrix A, there exists linear relationship is given below, 1 2 32C C C+ = So, there are only two linearly independent columns 1C and 2C . ( ) 2Aρ = Hence, the correct option is (C).

Given : For a 2 2× matrix A (Let), Eigen values are 1 8λ = and 2 4λ =

Eigen vectors are 1

11

v =

and 2

11

v = − .

1x K=

2 14 4x x K= =

1

2

14

xK

x

=

1 22x x= −

1 160 3

= + 21 760 20

= =

Scan for

Video Solution

1.7 (A)

1.8 (C)

1.9 (A)


. Method 1 : By the property of eigen values, 1 2 8 4 12λ + λ = + = = Trace of matrix A

and 1 2 8 4 32 Aλ ⋅λ = × = =

By checking all options, only option (A) has sum of diagonal elements equal to 12. Hence, the correct option is (A). . Method 2 : Modal matrix for any square matrix is given by,

1 2[ ] [ ]M v v=1 11 1 = −

From the concept of diagonalization of a matrix, the matrix A is given by,

1 1

2

0[ ] [ ]

0A M M −λ

= λ

11 1 8 0 1 1

1 1 0 4 1 1A

− = − −

1 1 8 0 1 111 1 0 4 1 12

A− − = − −−

12 414 122

A− − −= − −

6 22 6 =

Hence, the correct option is (A). Key Point Concept of diagonalization : Any square matrix with all its Eigen values are distinct can be represented as, 1[ ] [ ] [ ]A M D M −= where, D is a diagonal matrix whose diagonal

elements are Eigen values of A. M is a non-singular matrix whose

columns are respective Eigen vectors of A.

Note : M is also referred to as Modal matrix.

Given : Matrix A =

Eigen vector 101101

X =

. Method 1 : For any eigen vector [ ]X for a matrix [ ]A corresponding to eigen value λ , the following equation satisfies,

On solving above matrix,

4 2 0− λ + = 6λ = Hence, the correct option is (C). . Method 2 : [ ][ ] 0A I X− λ =

0AX X− λ =

AX X= λ

4 2 101 1012 4 101 101 = λ

404 202 101202 404 101

+ = λ +

606 101606 101 = λ

By comparing both sides, Hence, the correct option is (C).

Given : M non-zero orthogonal vectors are

1 2, ,........ MX X X if there are two orthogonal vectors, we require two dimensions to define them (say X and Y). If there are three orthogonal vectors, we require three dimensions to define them (say X, Y, and Z).

4 22 4

[ ][ ] 0A I X− λ =

4 2 1010

2 4 101− λ

= − λ

(4 )(101) 2(101) 0− λ + =

6λ =

1.10 (C)

1.11 (C)


The dimension of the vector space that is spanned by 2M vectors, given by

1 2 3

1 2 3

, , ....., , ....

M

M

X X X XX X X X− − − −

There are basically M orthogonal vectors 1 2, ,........ MX X X with positive and negative

signs. Thus, only M dimensions are required to define them. Hence, the correct option is (C).

Given : 11 122 2

21 22

[ ]P P

PP P×

=

Let, 1λ and 2λ are two Eigen values of P.

By the property of square matrices,

1 2 Pλ λ =

According to question, 1 0λ = .

So, 0P = 11 22 12 21 0P P P P− =


Given : The system of linear equations, 4x + 2y = 7 2x + y = 6 It is in the form of non-homogenous equation given by, AX B=

where, 4 22 1

A =

and 76

B =

. Method 1 : The augmented matrix is given by,

[ ] 4 2 : 7:

2 1 : 6A B =

By elementary transformation,

12 2 2

RR R→ −

4 2 : 7

[ : ] 50 0 :2

A B =

So, ( ) 1Aρ = , ( : ) 2A Bρ = ( ) ( : )A A Bρ ≠ ρ Thus, the given system has no solution. Hence, the correct option is (B). . Method 2 : The system of linear equations, …(i) …(ii) Multiplying 2 in both sides of equation (ii), …(iii) By comparing equation (i) and (iii), its left hand side is equal. So, right hand side should also be equal. But . Thus, given system is inconsistent and no solution exists. Hence, the correct option is (B).

Given : 1 3 53 1 60 0 3

A− = − −

. Method 1 :

1 3 53 1 60 0 3

A− = − −


1 3 5 0 03 1 6 0 0 00 0 3 0 0

− λ− − − λ =

λ

4 2 7x y+ =2 6x y+ =

4 2 12x y+ =

7 12≠

1.12 (C)

1.13 (B) 1.14 (D)


1 3 5

3 1 6 00 0 3

− − λ− − − λ =

− λ

2( 3)( 2 10) 0λ − λ + λ + = 3λ =

2 4 402

− ± −λ =

2 6 1 32

j j− ±λ = = − ±

Thus, the eigen value of the matrix are 3, 1 3, 1 3j j+ − + − − respectively.

Hence, the correct option is (D). . Method 2 : The characteristic equation for a 3 3× matrix is given by, 3 2(trace of )Aλ − λ

11 22 33( ) 0M M M A+ + + λ − =

Here, ( 1)( 3 0) 3( 9 0) 5(0)A = − − − − − − +| | 30A =

Trace of ( 1) ( 1) 3 1A = − + − + =

Minor 11 ( 1 3 0 6) 3M = − × − × = −

22 ( 1 3 0 5) 3M = − × − × = −

[ ]33 ( 1 1) (3 3) 10M = − × − − × − =

So, 3 2 4 30 0λ − λ + λ − = 3, 1 3jλ = − ± Thus, the eigen value of the matrix are

3, 1 3, 1 3j j+ − + − − respectively. Hence, the correct option is (D). . Method 3 : Here, the point is to be noted that the complex roots occur always in conjugate pair, the eigen values are the roots of characteristics equation, here by checking the options we can see that option (D) is correct because two roots are in the form of complex conjugate pair and one is real. Hence, the correct option is (D).

By properties of skew symmetric matrices, eigen values of skew symmetric matrix is either zero or pure imaginary. Consider the example below,

Let 0 1 11 0 21 2 0

A− −

= −

Here, A is skew symmetric matrix.

[ ]1 1

1 21 2

A I−λ − − − λ = −λ −

−λ


[ ] [ ]2 4 1 2 1 2 0 −λ λ + + −λ + − + λ =

3 4 2 2 0−λ − λ − λ + − − λ = 3 6 0λ + λ = 2 6 0 λ λ + =

6 ,0jλ = ± Hence, the correct option is (C). Key Point (i) All the diagonal elements of skew

symmetric matrix are either zero or pure imaginary.

(ii) Element above and below the diagonal of skew symmetric are same but have opposite signs.

Given : The system of equations,

Scan for

Video Solution

6+ + =x y z4 6 20+ + =x y z4+ + λ = μx y z

1.15 (C)

1.16 (B)


It is in the form of a non-homogeneous equation given by, AX B=

where, 1 1 11 4 61 4

A =

λ

and 620B =

μ

The augmented matrix is given by,

1 1 1 : 6

[ : ] 1 4 6 : 201 4 :

A B =

λ μ


3 3 2R R R→ −

1 1 1 : 6

[ : ] 1 4 6 : 200 0 6 : 20

A B =

λ − μ −

If no solution exists then, ( ) [ : ]A A Bρ ≠ ρ

It is possible only if 6 0λ − = Hence, 6λ = ( ) 2Aρ =

and if 20μ ≠

Then [ : ] 3A Bρ =

( ) [ : ]A A Bρ ≠ ρ

Thus, for no solution 6λ = and 20μ ≠ .

In other case, if we take 20μ = and 6λ ≠ , then both the matrices A and AB will have same ranks of 3 giving a unique solution to the problem. Hence, the correct option is (B).

Given : 5 3 1 0

and2 0 0 1

A I− − = =

The characteristic equation of matrix A is given by, 0A I− λ =

5 3

02

− − λ −=

−λ

2 5 6 0λ + λ + = By Cayley Hamilton theorem, every square matrix satisfies its own characteristic equations. 2 5 6 0A A+ + =

2 5 6A A I= − − …(i) Multiplying by A on both sides, 3 25 6A A A= − − From equation (i),

{ }3 5( 5 6 ) 6 0A A I A= − − − − =

3 19 30A A I= +


Given : 3 5 25 12 72 7 5

A =

. Method 1 : Applying elementary transformation,

3 1 3R R R→ +

3 5 25 12 75 12 7

A =

Applying elementary transformation, 3 3 2R R R→ −

3 5 25 12 70 0 0

A =

Thus, A is a singular matrix i.e. . Now, product of eigen value = Determinant of matrix.

Scan for

Video Solution

0A =

1 2 3 0λ λ λ =

1.17 (B)

1.18 (A)


i.e. at least one of the eigen values of A is equal to zero. Since all given options are positive, minimum eigen value must be . Hence, the correct option is (A). . Method 2 :

3 5 25 12 72 7 5

A =


3 5 2

5 12 7 02 7 5

− λ− λ =

− λ

2( 20 33) 0λ λ − λ + = 0λ = , 18.18, 1.81 So, 0λ = is minimum eigen value. Hence, the correct option is (A).

Given : A is an m n× matrix, B is an n m× matrix and kI is k k× identity matrix

and m nI AB I BA+ = +

. Method 1 : Given matrix,

2 1 1 11 2 1 11 1 2 11 1 1 2

P

=

Using matrix algebra P can be represented as,

4 4

1 0 0 0 1 1 1 10 1 0 0 1 1 1 10 0 1 0 1 1 1 10 0 0 1 1 1 1 1m n

P

× ×

= +

4P I AB= + Here, 4m = Then 4A m n n→ × = × 4B n m n→ × = × Let 1n = , Then 4 1A → × and 1 4B → ×

4 1

1111

A ×

=

, [ ]1 4 1 1 1 1B × =

[ ]

11

1 1 1 111

AB

=

1 1 1 11 1 1 11 1 1 11 1 1 1

AB

=

According to question, m nP I AB I BA= + = +

4 1P I AB I BA= + = +

[ ]

11

1 1 1 1 411

BA

= =

Thus, [ ]nP I BA= +

1P I BA= +

1 4 5P = + = Hence, the correct option is (B). . Method 2 : Direct calculation of determinant of given matrix.

Let,

2 1 1 11 2 1 11 1 2 11 1 1 2

P

=

0λ =

Scan for

Video Solution

1.19 (B)


2 1 1 1 1 1

2 1 2 1 11 2 11 1 2 1 1 2

P = −

1 2 1 1 2 111 1 1 11 1 2

1 1 2 1 1 1+ −

2[2(4 1) 1(2 1) 1(1 2)]P = − − − + −

1[1(4 1) 1(2 1) 1(1 2)]− − − − + −

1[1(2 1) 2(2 1) 1(1 1)]+ − − − + −1[1(1 2) 2(1 2) 1(1 1)]− − − − + −

2(4) 1(1) 1( 1) 1(1)P = − + − −

8 1 1 1 5P = − − − =


Let any 2 2× matrix andM N which are respectively,

a b

Mc d =

and p q

Nr s =

Here, a b p q

MNc d r s =

ap br aq bs

MNcp dr cq ds

+ + = + +

and p q a b

NMr s c d =

pa qc pb qd

NMra sc rb sd

+ + = + +

Thus, MN NM≠ Hence, the correct option is (D). Objective approach : Properties in options A, B and C are general properties of any matrix. But the product of matrices is a conditional operation.

Given : A is a 4 4× real matrix such that, 2A I= 2 0A I− = Let, λ represent the eigen values of A. By Cayley-Hamilton theorem, 2 1 0λ − = 1 21, 1λ = λ = −

Positive value is 1 1λ = .

Hence, the positive eigen value of A is 1.

Consider,

(i) Let 2 2

1 0 0 10 1 1 0

P I J = + α = + α

1

1P

α = α

21P = − α

(ii) Let 4 4

1 0 00 1 00 1 0

0 0 1

P I J

α α = + α =

α α

1 0 0 1

(1) 1 0 ( ) 0 10 0 1 0 0

Pα α

= α − α αα

2 2(1 ) ( )[ (1 )]P = − α − α α − α

2 2(1 )P = − α

Similarly for 6 6P I J= + α ,

2 3(1 ) 0P = − α =

2 3(1 ) 0− α = 1α = ± Since, is a non-negative real number

1α = . Hence, the value of α is 1.

α

1.20 (D)

1.21 1

1.22 1


Given : 5, 40A B= =

By the property of determinant, (5) (40) 200AB A B= = =

Hence, the determinant of matrix AB is 200.

Given : The system of linear equations,

2 1 3 53 0 1 41 2 5 14

abc

= −

It is in the form of non-homogenous equation given by, AX B= .

where, 2 1 3 53 0 1 , 41 2 5 14

A B = = −

Number of unknowns (n) = 3. . Method 1 : The augmented matrix is given by,

2 1 3 : 5[ : ] 3 0 1 : 4

1 2 5 : 14A B

= −


2 2 12 3R R R→ − and 3 3 12R R R→ −

2 1 3 : 5[ : ] 0 3 7 : 23

0 3 7 : 23A B

= − − −

3 3 2R R R→ +

2 1 3 : 5

[ : ] 0 3 7 : 230 0 0 : 0

A B = − − −

So, ( ) 2Aρ = , ( : ) 2A Bρ =

( ) ( : )A A B nρ = ρ < Thus, equations have infinitely many solutions. Hence, the correct option is (B).

. Method 2 : The augmented matrix is given by,

2 1 3 : 5[ : ] 3 0 1 : 4

1 2 5 : 14A B

= −

Here, we can see that 3 1 22R R R= − , that means

3R is dependent, so we can vanish (zero) it completely, hence number of equations are less than number of variables it represents infinite number of solutions. Hence, the correct option is (B).

Given : A 2 2× real symmetric matrix. Trace of matrix = 14 . Method 1 : A symmetric matrix is a matrix whose transpose is equal to the matrix itself. TA A=

Let a b

Ab c

=

2A ac b= − …(i)


14a c+ = 14c a= − From equation (i), 2(14 )A a a b= − −

2 214A a a b= − − …(ii) This is a function in two variable a and b. First order partial derivatives are,

14 2Ap aa

∂= = −∂

2Aq b

b∂= = −∂

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Video Solution

1.23 200

1.24 (B)

1.25 49


The strationary points are given by,

0 7p a= =

0 0q b= =

Second order partial derivatives are,

2 2

2 2, 0,A A

r sa a b

∂ ∂= = − = =

∂ ∂ ∂

2

2 2A

tb

∂= = −

∂

In case r is a negative number, check the value of 2rt s− .

2 4 0 4rt s− = − =

Hence, 2 0rt s− > and 0r < (Maxima occurs)

Thus, maximum value of A occurs at stationary point (7, 0). From equation (ii),

2 2max 14(7) 7 0A = − −

max 49A =

Hence, the maximum value of the determinant is 49. . Method 2 :

Let a b

Ab c =

2A ac b= − …(i)

For maximum value of determinant A, b = 0 From equation (i),

A ac=


14a c+ = By the trail the posible values of a and c for example (10, 4), (9, 5), (8, 6), (7, 7) (11, 3) etc. (7, 7) gives maximum determinat which is 49. Hence, the maximum value of the determinant is 49.

. Method 1 : Option (A) : For any triangular matrix, the diagonal elements represents its eigen values. So, this option is True. Option (B) : From the properties of symmetric matrices, the eigen values of a symmetric matrix are always real. Thus, it is not certain whether the eigen values are positive or negative. So, this option is not true. Hence, the correct option is (B). . Method 2 : Let A is a square matrix, which is real and symmetric.

Hence, TA A= (symmetric) …(i) If matrix A is real, Then, complex conjugate of matrix A is,

A A= …(ii) Now, transpose of conjugate of matrix A

( )TA Aθ=

( )T TA A Aθ = = [From equation (ii)]

and A Aθ = [From equation (i)] Thus, A is a hermitian matrix. The eigen value of hermitian matrix is always real but not necessarily positive. Hence, the correct option is (B).

Given : The system of linear equations is, 2 3 1x y z− + = −

3 4 1x y z− + =

2 4 6x y z k− + − =

1.26 (B)

1.27 2


It is in the form of non-homogeneous equation given by, AX B=

where, 1 2 31 3 42 4 6

A−

= − − −

, 11Bk

− =

Number of unknowns ( ) 3n = . The augmented matrix is given by,

[ ]1 2 3 : 1

: 1 3 4 : 12 4 6 :

A Bk

− − = − − −


2 2 1 3 3 1and 2R R R R R R→ − → +

[ ]1 2 3 : 1

: 0 1 1 : 20 0 0 : 2

A Bk

− − = − +

−

So, ( ) 2Aρ = Condition to have infinite number of solutions is, [ : ] ( )A B A nρ = ρ <

So, ( ) ( : ) 2A A Bρ = ρ = If 2 0k − = As ( : ) 2A Bρ = , there must be two non-zero rows i.e. 2 0k − = 2k = Hence, the value of k for which the system has infinitely many solution is 2.

Given : 4 1 2 1

2 1 , 214 4 10 3

A p X = =

−

For any Eigen vector [ ]X for a matrix [ ]A corresponding to Eigen value λ , the following equation satisfies, [ ][ ] 0A I X− λ =

AX X= λ

3 3 3 1 3 1

4 1 2 1 12 1 2 2

14 4 10 3 3p

× × ×

= λ

−

3 1 3 1

127 2

36 3p

× ×

λ + = λ

λ

On comparing both sides, 12λ =

7 2p + = λ

So, eigen value 12λ = and 7 2 12p + = ×

17p =

Hence, the value of p is 17.

Given : ( ) az bf zcz d

+=+

…(i)

where, 2, 4a b= = and 5c =

According to question, for all 1 2z z≠ :

1 2( ) ( )f z f z=

From equation (i),

1 2

1 2

az b az bcz d cz d

+ +=+ +

1 2 2 1(2 4)(5 ) (2 4)(5 )z z d z z d+ + = + +

1 2 1 2 1 2 110 2 20 4 10 20z z dz z d z z z+ + + = +

22 4dz d+ +

1 2 1 22 ( ) 20( )d z z z z− = −

10d = Hence, the value of d is 10.

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Video Solution

1.28 17

1.29 10


Given : 10 5 4

20 24 2 10

jA x

+ =

−

For a Hermitian matrix, the eigen values are always real. Transpose conjugate of A is given by,

10 5 4

( ) 20 24 2 10

T

T

jA A xθ

− = =

−

10 4

5 20 24 2 10

xA jθ

= −

−

As eigen values of A are real, it means A is Hermitian matrix. A Aθ =

10 4 10 5 4

5 20 2 20 24 2 10 4 2 10

x jj x

+ − =

− −

5 and 5x j x j= + = − 5x j= − Hence, the correct option is (B).

Given : 1 tan

tan 1x

Ax

= −

The transpose of matrix A is given by,

1 tan

tan 1T x

Ax

− =

…(i)

2 21 tan secA x x= + =

The inverse of matrix A is given by,

1 Adj AAA

− =

12

1 tan1tan 1sec

xA

xx− − =

…(ii)

Multiplying equations (i) and (ii),

12

1 tan 1tan 1 sec

T xA A

x x− − =

1 tantan 1

xx

−

2

12 2

1 tan 2 tan1sec 2 tan 1 tan

T x xA A

x x x− − −

= −

Taking the determinant,

2

12 2


T x xA A

x x x− − −

= −

2 21

2 2


T x xA A

x x x− − − = −

1 2 2 24

1 [(1 tan ) 4 tan ]sec

TA A x xx

− = − +

1 4 24

1 [1 tan 2 tan ]sec

TA A x xx

− = + +

1 2 24

1 [1 tan ]sec

TA A xx

− = +

4

14

sec 1sec

T xA Ax

− = =


Given : 4M I=

2,M I M I≠ ≠ and 3M I≠ .

Then, 4k kM I=

4kM I= [ ]kI I=

Or 4( 1) 4kM I I+ = =

Multiplying 1M − on both sides,

1 4( 1) 1kM M M I− + −× = ×

4 3 1kM M+ −= Hence, the correct option is (C).

1.30 (B)

1.31 (C) 1.32 (C)


Given : A sequence is specified as,

[ ] 1 1 1

[ 1] 1 0 0

nx nx n = −

…(i)

Let, 1 11 0

A =

The characteristic equation is given by,

0A I− λ =

1 1

01− λ

=− λ

2 1 0−λ + λ − =

2 1 0λ − λ − = By Cayley Hamilton theorem, every square matrix satisfies its own characteristics equation.

2 0A A I− − =

2 ( )A A I= +

Squaring both sides,

4 2( ) ( ) 2A A I A I A A I= + + = + +

[ 2I I= ]

4 ( ) 2 3 2A A I A I A I= + + + = +

[ 2A A I= + ] Again squaring both sides,

8 2 2(3 2 ) 9 4 12A A I A I A= + = + +

[ 2I I= ]

8 9( ) 4 12 21 13A A I I A A I= + + + = +

[ 2A A I= + ]

Now, 12 4 8 (3 2 )(21 13 )A A A A I A I= = + +

12 1 1 1 0144 89 144 89

1 0 0 1A A I = + = +

12 233 144144 89

A =

Put n = 12 in equation (i),

[ ][ ]

1212 1 1 111 1 0 0

xx =

[ ][ ]12 233 144 111 144 89 0

xx =

[ ]12 233 0 233x = + =

Hence, the value of x [12] is 233.

Given : 3 2 49 7 136 4 9

Ax

= − − − +

One eigen value is zero. By the property of square matrices,

1 2 3A = λ λ λ

Let 1 0λ = ,

0A =

3 2 49 7 13 06 4 9 x

=− − − +

3( 63 7 52) 2( 81 9 78)x x− + + − − + +

4( 36 42) 0+ − + =

3(7 11) 2(9 3) 4(6) 0x x− − − + =

21 33 18 6 24 0x x− − + + = 3 3 0x − = 1x = Hence, the value of x is 1.

Key Point If any of the eigen values of a matrix is zero, then the matrix is a singular matrix.

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Video Solution

1.33 233

1.34 1


Given :

0 3 72 5 1 30 0 2 40 0 0

a

A

b

=

det ( ) 100A = and trace ( ) 14A = 5 2 14a b+ + + = [Sum of principal diagonal elements is equal to the trace of matrix] 7a b+ = …(i)

0 3 7 3 7

2 0 2 4 5 0 2 40 0 0 0

aA

b b= − +

0 7 0 31 0 0 4 3 0 0 2

0 0 0 0 0

a a

b− +

3 7

0 5 0 2 4 1 0 3 00 0

aA

b= + − × + ×

Upper Triangularmatrix

3 75 0 2 4

0 0

aA

b=

Let 1 aλ = , 2 2λ = , 3 bλ = . Product of eigen values = Determinant of the matrix 5 2 100a b× × × = 10ab = … (ii) From equation (i), 7a b= − Put in equation (ii), (7 ) 10b b− =

27 10b b− = 2 7 10 0b b− + =

( 5) ( 2) 0b b− − =

2, 5b =

Thus, 5, 2a =

Now, 5 2 3a b− = − =

Hence, the value of a b− is 3.

Given : x

Aσ = ω σ

The eigen values of the matrix A are, 1 jλ = σ + ω and 2 jλ = σ − ω

By the property of square matrices, Trace of the matrix = Sum of eigen values. ( ) ( )j jσ + σ = σ + ω + σ − ω

and A = product of Eigen values

2 2 2( ) ( )x j jσ − ω = σ + ω σ − ω = σ + ω

By comparing,

2x−ω = ω x = −ω


Given : Vectors are, 1 2(1, 0, 2), (0, 1, 0)e e= =

and 3 ( 2, 0, 1)e = −

Given 3u R∈ (a 3 dimensional space or system of axes which is defined by 1 2,e e and 3e ).

Using linear algebra in matrix form, 1 2 3u ae be ce= + +

Where u, 1 2,e e and 3e are all 3 1× vectors.

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Video Solution

1.35 3

1.36 (D)

1.37 (D)


Then, 4 1 0 23 0 1 03 2 0 1

a b c−

= + + −

On solving, 2 4a c− = …(i) 3b = and 2 3a c+ = − …(ii) From equation (i) and (ii),

2 11, 3,5 5

a b c− −= = =

So, 1 2 32 113

5 5u e e e−= + −


Given : 5 10 101 0 23 6 6

M =


2 2 15R R R→ − and 3 3 15 3R R R→ − .

5 10 100 10 00 0 0

M = −

There are two non-zero rows. So, ( ) 2Mρ = Hence, the correct option is (C).

Given : 2

1 2 31, ,y y x y x= = = From Wronskian’s matrix,

If ( ) ( ) ( )'( ) '( ) '( ) 0''( ) ''( ) ''( )

f x g x h xW f x g x h x

f x g x h x= =

then functions are linearly dependent.

Here, 2( ) 1, ( ) , ( )f x g x x h x x= = =

210 1 2 20 0 2

x xW x= =

0W ≠

So, functions are linearly independent. Thus, statement (I) & (III) are correct. Hence, the correct option is (B).

Given :

1 2 3 4 55 1 2 3 44 5 1 2 33 4 5 1 22 3 4 5 1

A

=

For any Eigen vector [ ]X for a matrix [ ]A corresponding to Eigen value λ , the following equation satisfies, [ ][ ] 0A I X− λ =

AX X= λ

Let [ ]

ab

X cde

=

Now,

1 2 3 4 55 1 2 3 44 5 1 2 33 4 5 1 22 3 4 5 1

a ab bc cd de e

= λ

2 3 4 5a b c d e a+ + + + = λ …(i) 5 2 3 4a b c d e b+ + + + = λ …(ii) 4 5 2 3a b c d e c+ + + + = λ …(iii) 3 4 5 2a b c d e d+ + + + = λ …(iv)

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Video Solution

1.38 (C)

1.39 (B)

1.40 (C)


2 3 4 5a b c d e e+ + + + = λ …(v)

By adding equations (i), (ii), (iii), (iv) and (v), 15( ) ( )a b c d e a b c d e+ + + + = λ + + + +

15λ =


Given :

1 1 0 0 00 0 1 1 00 1 1 0 01 0 0 0 10 0 0 1 1

A

− −

= − − −


4 4 5R R R→ +

1 1 0 0 00 0 1 1 00 1 1 0 01 0 0 1 00 0 0 1 1

A

− −

= − − −

3 3 4R R R→ +

1 1 0 0 00 0 1 1 01 1 1 1 01 0 0 1 00 0 0 1 1

A

− −

= − − − −

2 2 3R R R→ +

1 1 0 0 01 1 0 0 01 1 1 1 01 0 0 1 00 0 0 1 1

A

− −

= − − − −

1 1 2R R R→ +

0 0 0 0 01 1 0 0 01 1 1 1 01 0 0 1 00 0 0 1 1

A

−

= − − − −

There exist four non-zero rows in above matrix. So, ( ) 4Aρ = Hence, the rank of the matrix is 4.

If the eigen values of a matrix are distinct, then it will have linearly independent eigen vectors. Number of distinct eigen value = Number of independent eigen vectors. If the matrix is invertible then 0A ≠

Product of eigen values is equal to determinant of A, since 0A ≠

It implies that all eigen values are non zero. For a invertible matrix, eigen values can be repeated as 2, 2, 4, 4. Hence, the correct option is (C).

Given : 12 2

2

2 xk kA X

k k k x = = −

It is in the form of homogenous equation given by, 0AX = …(i) Homogenous linear equations are always consistent. If ( )Aρ = Number of variables, Then the system have unique solution when

0A ≠ .

( )Aρ < Number of variables then the system has infinite number of solution

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1.41 4 1.42 (C)

1.43 2


Here the system must have infinitely many solutions. Hence, ( )Aρ < Number of variables Which is possible only if 0A =

2 2

20

k kk k k

=−

3 22 ( ) 0k k k k− − =

3 3 22 2 0k k k− + = 2 32 0k k− = 2 (2 ) 0k k− = 0,0,2k =

Hence, number of distinct values of K for infinite number of solutions is 2.

Given : [ ]

2 2 3 30 1 1 10 0 3 30 0 0 2

A

=

Given matrix is an upper triangular matrix for which Eigen values are given as the principle diagonal element. Hence, Eigen values of [ ]A are

1 2λ = , 2 1λ = , 3 3λ = , 4 2λ =

Hence, number of distinct Eigen values is 3.

Given : 1v , 2v … 6v are six vectors in 4 .

Four vectors that form a basis for four dimensional space must be linearly independent to each other. Hence, the correct option is (A).

..Method 1..

Given : Linear equations are

1 2 12x x b+ =

1 2 22 4x x b+ =

1 2 33 7x x b+ =

1 2 43 9x x b+ =

Now, augmented matrix can be written as,

[ ]1

2

3

4

1 2 :2 4 :

:3 7 :3 9 :

bb

A Bbb

=

Applying 2 2 12R R R→ −

3 3 13R R R→ −

4 4 13R R R→ −

1

2 1

3 1

4 1

1 2 :0 0 : 20 1 : 30 3 : 3

bb bb bb b

− − −

Applying 4 4 33R R R→ −

1

2 1

3 1

4 1 3

1 2 :0 0 : 20 1 : 30 0 : 6 3

bb bb b

b b b

− − + −

For solution to exist

( ) ( : ) 2A A Bρ = ρ =

k = 0, 0, 2

same

distinct

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Video Solution

1.44 3

1.45 (A)

1.46 (C)


∴ 2 1 1 3 42 ; 6 3 0b b b b b= − + =

Hence the correct option is (C).

..Method 2..

Given : Linear equations are, 1 2 12x x b+ = …(i)

1 2 22 4x x b+ = …(ii)

1 2 33 7x x b+ = …(iii)

1 2 43 9x x b+ = …(iv)

From option (A) : 2 12b b= and 1 3 43 6 0b b b− + =

Multiplying equation (i) by 2, 1 2 12 4 2x x b+ = ...(v)

Comparing equation (ii) and equation (v), 2 12b b= (satisfies the first condition)

Now, justifying equation 1 3 43 6 0b b b− + =

1 2 1 2 1 23( 2 ) 6(3 7 ) 3 9 0x x x x x x+ − + + + =

1 1 1 2 1 23 6 18 42 3 9 0x x x x x x+ − − + + =

1 212 27 0x x− − ≠

Condition (2) is not satisfied. Hence, option (A) is wrong. From option (B) : 3 1 1 32 and 6 3 0ab b b b b= − + =

Multiplying equation (i) by 2, 1 2 12 4 2x x b+ = …(vi)

Comparing equation (iii) and equation (vi), 3 12b b≠

Hence, option (B) is wrong. From option (C) : 2 1 1 3 42 and 6 3 0b b b b b= − + =

Multiplying equation (i) by 2,

1 2 12 4 2x x b+ = …(vii)

Comparing equation (ii) and equation (vii), 2 12b b= (satisfied the first condition)

Now, justifying equation: 1 3 46 3 0b b b− + = …(viii)

Putting the value of 1 3 4, ,b b b in equation (viii),

1 2 1 2 1 26( 2 ) 3(3 7 ) 3 9 0x x x x x x+ − + + + =

1 2 1 2 1 26 12 9 21 3 9 0x x x x x x+ − − + + =

1 1 2 29 9 21 21 0x x x x− + − =

0 0= Hence, 1 3 46 3 0b b b− + =

Both conditions are satisfied, Hence, the correct option is (C)

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