1. the eigen values of a skew-symmetric matrix are (b ...€¦ · the trigonometric fourier series...
TRANSCRIPT
GATE-2010
1. 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
Soln.1 A skew symmetric matrix is a square matrix and has property
that aij = -aji for all
11 12 1
21 22 2
1 2
...
...... ... ... ...
...
n
n
n n nn
a a a
a a aA
a a a
=
Diagonal entries are zero
i.e. aii = o
It has the properties
(i) AT = -A where AT is matrix transpose. (ii) All Eigen values of skew symmetric matrices are imaginary or zero.
Option (C) is the correct
2. The trigonometric Fourier series for the waveform f(t) shown below contains
(A) Only cosine terms and zero value for the dc component
(B) Only cosine terms and a positive value for the dc component
(C) Only cosine terms and a negative value for the dc component
(D) Only sine terms and a negative value for the dc component
Soln.2 The function f (t) is an even function and is symmetric to y- axis so it contains only cosine terms.
GATE-2010
Note that the positive part of waveform has amplitude A while the negative part has amplitude 2A.
DC value is the average value i.e. the area of the waveform lying above and below this line will be equal. So the line for average value will be shifted to negative value, so DC value is negative.
Option (C) is correct
3. A function n(x) satisfied the differential equation d2n(x)/dx2 -n(x)/L2 = 0 where L is a constant. The boundary conditions are: n(0)=K and n(∞) =0. The solution to this equation is
(A) n(x) = K exp(x/L) (B) n(x) = K exp(-x/√L )
(C) n(x) = K2 exp(-x/L) (D) n(x) = K exp(-x/L)
Soln.3 Given differential eqn. is
− = 0
Let =
− = 0
− = 0
= ±
So, the solution is
= +
Applying the boundary conditions
n(0)=K i.e. n(0)=C1+C2=K
n(∞)=0 then n(∞)=0 C1=0
So C2=K
Thus n(x) =K exp(-x/L)
Option (D) is correct.
GATE-2010
4. For the two-port network shown below, the short-circuit admittance parameter matrix is
Soln. (4)
V1 V2
1 2
1’
2’
0.5Ω0.5Ω
0.5Ω I2I2
The governing eqn. for the admittance matrix is
"##$ = "% %% %$ "&&$ # = %& + %&
# = %& + %&
% = '()( *)+, = -./ǁ-./ = -./ = 01
% = '() *)(+, = -./ = 2 V 0.5
0.5I
% = ')( *)+, = -./ = 2 V20.5
0.5I0.5
% = ') *)(+, = -./ǁ-./ = -./ = 01
GATE-2010
5. For parallel RLC circuit, which one of the following statements is NOT correct?
(A) The bandwidth of the circuit deceases if R is increased
(B) The bandwidth of the circuit remains same if L is increased
(C) At resonance, input impedance is a real quantity
(D) At resonance, the magnitude of input impedance attains its minimum value.
Soln. (5) Parallel RLC circuit has the following characteristic equation.
1 + 34 1 + 4 = 0
Where 34 is the bandwidth
From the above eqn. we can draw the following conclusions:
• Bandwidth RLC circuits independent of L • At resonance, imaginary part of input impedance is zero, i.e. It is real. • At resonance the admittance is minimum thus input impedance is maximum.
Thus the option (D) is correct
6. At room temperature, a possible value for the mobility of electrons in the inversion layer of a silicon n-channel MOSFET is
(A) 450 cm2/V-s
(B) 1350 cm2/V-s
(C) 1800 cm2/V-s
(D) 3600 cm2/V-s
Soln.(6) Option (B) is correct
7. Thin gate oxide in a CMOS process in preferably grown using
(A) wet oxidation (B) dry oxidation
(C) epitaxial deposition (D) ion implantation
Soln.(7) Dry oxidation alone can provide uniform film with good dielectric properties for thin gate oxide in a CMOS
Option (B) is correct
GATE-2010
8. In the silicon BJT circuit shown below, assume that the emitter area of transistor Q1 is half that of transistor Q2.
The value of current I0 is approximately
(A) 0.5 mA (B) 2mA
(C) 9.3 mA (D) 15mA
Soln. (8)
-10V
Q1 Q2
+
-
+
-i1 R=9.3KΩ
β1 =700 β2 =715
I0
0.7V
KVL to circuit with Q1
5.6 + 0.7 = 0
= --.89.: = 9.:9.: = ;<
The emitter area of Q1 is half that of transfer Q2
So Ii = I0/2
Therefore I0 = 2mA
So option (B) is correct
GATE-2010
9. The amplifier circuit shown below uses a silicon transistor. The capacitors CC and CE can be assumed to be short at signal frequency and the effect of output resistance r0 can be ignored. If CE is disconnected from the circuit, which one of the following statements is TRUE?
(A) The input resistance Ri increases and the magnitude of voltage gain AV decreases
(B) The input resistance Ri decreases and the magnitude of voltage gain AV decreases
(C) Both input resistance Ri and the magnitude of voltage gain AV decrease
(D) Both input resistance Ri and the magnitude of voltage gain AV increase
Soln. (9) By removing CE
• Input impedance increases by (1+= RE) • Voltage gain decreases by the same factor
Option (A) is correct
10. Assuming the OP-AMP to be ideal, the voltage gain of the amplifier shown below is
(A) -R2/R1 (B) -R3/R1
(C) - (R2 || R3)/R1 (D) - (R2+R3)/R1
Soln. (10) Since the OP-AMP is ideal, voltage at point A will be zero. The above circuit can be simplified to
GATE-2010
DC
Vi
R1
R3
V0
A
R2
+
-
writing the current eqn.
)>-3( = -),3**3?
),)> = − 3**3?3(
<@ = − 3**3?3(
The option (C) is correct
11. Match the logic gates in Column A with their equivalents in Column B.
(A) P–2, Q-4, R-1, S-3 (B) P-4, Q-2, R-1, S-3
(C) P–2, Q-4, R-3, S-1 (D) P-4, Q-2, R-3, S-1
Soln. (11) P gate A = < + B = <. B
So P matches with 4 (P-4)
Q gate A = <. B = < + B
Matches with (Q-2)
R gate Exor ie. A = <B + <B
Now see gate 3 A = <. B + <C. B = <B + <B ie. is EXOR gate
R matches with 3
Now gate 1
GATE-2010
<. B + <BD = < B + <B ie. Ex Nor ie. S
So S-1
Option D is correct
12. For the output F to be 1 in the logic circuit shown, the input combination should be
(A) A = 1, B= 1. C = 0 (B) A = 1, B= 0,C = 0
(C) A = 0, B= 1. C = 0 (D) A = 0, B= 0, C = 1
Soln. (12) Best way is to try the various inputs
(A) A=1, B=1,C=0 A=1, B=1 output of gate 1 is 0
output of gate 2 is 1
c=0 output of gate 3 is 0
with these inputs F = 0
similarly try options B & C both give F = 0
only option D ie
A = 0, B = 0, C = 1
Gate 1 output 0
Gate 2 output 1
C is 1
F = 1 Thus proved
Option (D) is correct
13. In the circuit shown, the device connected to Y5 can have address in the range
GATE-2010
(A) 2000 - 20FF (B) 2D00 – 2DFF
(C) 2E00 – 2EFF (D) FD00 – FDFF
Soln. (13) To connect to Y5, inputs CBA in 101
A15 A14 A13 A12 A11 A10 A9 A8 A7 A6 A5 A4 A3 A2 A1 A0
0 0 1 0 1 1 0 1 0 0 0 0 0 0 0 0
2 D 0 0 (Min.)
0 0 1 0 1 1 0 1 1 1 1 1 1 1 1 1 (Max.)
2 D F F
Option (B) is correct
14. Consider the z-transform X(z) = 5z2 + 4z-1 + 3; 0<|z| < ∞ . The inverse z-transform x[n] is
(A) 5δ[n + 2] + 3δ[n] + 4δ[n – 1]
(B) 5δ[n - 2] + 3δ[n] + 4δ[n + 1]
(C) 5 u[n + 2] + 3 u[n] + 4 u[n – 1]
(D) 5 u[n - 2] + 3 u[n] + 4 u[n + 1]
Soln. (14) We know that
EF + -G ↔ I-
X(z)= 5z2 + 4z-1+3
0< *I* < ∞
x(n) = 5 E + 2 + 0E − + 6E
Option (A) is correct
15. Two discrete time systems with impulse responses h1[n] = δ[n -1] and h2[n] = δ[n– 2] are connected in cascade. The overall impulse response of the cascaded system is
(A) δ[n - 1] + δ[n - 2] (B) δ[n - 4]
(C) δ[n - 3] (D) δ[n - 1] δ[n - 2]
Soln. (15) We know
ℎFG = EF − G LI = I
ℎFG = EF − 2G LI = I
Overall impulse response of cascaded systems in Z- domain:
GATE-2010
H(z) = H1 (z) .H2(z)
= z-1 . z-2
= z-3
Overall impulse response in discrete time domain is
ℎFG = E − 6
Option (C) is correct
16. For an N-point FFT algorithm with N = 2m which one of the following statements is TRUE?
(A) It is not possible to construct a signal flow graph with both input and output in normal order
(B) The number of butterflies in the mth stage is N/m
(C) In-place computation requires storage of only 2N node data
(D) Computation of a butterfly requires only one complex multiplication
Soln. (16) Computation of a Butterfly requires only one complex multiplication for N point Fast Fourier Transform (FFT) with N=2m.
It requires only one complex multiplication to compute a Butterfly
Option (D) is correct
17. The transfer function Y(s)/R(s) of the system shown is
(A) 0 (B) 1/(s+1) (C) 2/(s+1) (D) 2/(s+3)
Soln. (17)
P(s)
From the above diagram.
GATE-2010
P (s) = R (s) – Q (s)
And M1 = NO PQ − NO OQ = 0
So P(s) = R(s) -0 = R(s)
So %1 = NO OQ = 3O OQ therefore RO 3O = OQ Option (B) is correct
18. A system with transfer function Y(s)/X(s) = s/(s+p) has an output y(t) = cos(2t-π/3) for the input signal x(t) = p cos(2t - π/2) Then, the system parameter ‘p’ is
(A) √3 (B) 2/√3 (C) 1 (D) √3/2
Soln. (18) From the given input and output signals, we can find the phase difference between the output and input signals i.e.
S = − T: − − T = TU = 60-
and ω=2rad/sec. (from the given equation)
Given RO VO = OOQW = XYXYQW
S = 50- − Z[\ ] ^
Which is also 300
Z[ W = _0- W = Z[_0- = `6 ^ = :
Option (B) is correct
19. For the asymptotic Bode magnitude plot shown below, the system transfer function can be
(A) (10s+1)/(0.1s+1) (B) (100s+1)/(0.1s+1)
GATE-2010
(C) (100s)/(10s+1) (D) (0.1s+1)/(10s+1)
Soln. (19) At ω=0.1, slope changes from odB to +20dB
Factor + -.
So, system transfer function T(s) = aQ b,.(Q b(, = a-OQ -.OQ
At ω=0.1, magnitude =0dB 20log k =0
or, k=1
thus T(s) = Q-OQ-.O Option (A) is correct
20. Suppose that the modulating signal is m(t) = 2cos (2πfmt) and the carrier signal is xc(t) = Ac cos(2πfct), which one of the following is a conventional AM signal without over-modulation?
(A) x(t) = Acm(t) cos(2πfct)
(B) x(t) = Ac[1 + m(t)]cos(2πfct)
(C) x(t) = Ac cos(2πfct) + Ac/4 m(t) cos(2πfct)
(D) x(t) = Ac cos(2πfmt) cos(2πfct) + Ac sin(2πfmt) sin(2πfct)
Soln. (20) The conventional AM signal is given as
x(t) = Ac [1+m(t)] cos 2πfct
= Ac cos (2πfct) + Ac m (t) cos 2πfct
Let us check option (B)
Modulation index =2/1=2
(since modulating signal has amplitude 2)
So, it is AM signal with over modulation.
Now check option (c)
X(t)=Accos(2 fct) +Ac/4 m(t)cos(2 fct)
Hence Modulation Index =2.Ac/4 =1/2
Thus, it is AM signal without over modulation
Option (C) is correct
21. Consider an angle modulated signal x(t) = 6cos[2πx106t+2sin(8000πt) +
4cos(8000πt)] V. The average power of x(t) is :
(A) 10W (B) 18W (C) 20W (D) 28W
GATE-2010
Soln. (21) Angle modulated signal as given is
x(t) = 6 cos [2π . 106 + 2 sin (8000πt)+4 cos (8000πt) V.
Here
ωc = 2π x 106
ωm= 800 π
so the wave is
x(t) = 6 cos [ωc t + 2 sinωmt+4 cosωmt]
Average power = ()c`)2 =( U)2 =36/2=18w
Option (B) is correct
22. If the scattering matrix [S] of a two port network is
then the network is
(A) lossless and reciprocal (B) lossless but not reciprocal
(C) not lossless but reciprocal (D) neither lossless nor reciprocal
Soln. (22) The given matrix is
= " 0.2 0.50.5 0.$ Since, ∡00=cos0+jsin0=1
∡900=cos900+jsin900
For reciprocal network Sij = Sji i.e. S12 = S21
In the present problem S12 =S21 = j 0.9
So the network is reciprocal.
Now it could be checked for the condition of lossless.
For the losses network s-matrix has to be unitary. It gives rise to two conditions
e *fg* = e fghi fgj = − − −
Or
GATE-2010
e fOfkjhi = 0l1 ≠ − − − 2
Applying condition (1) to
e *fg*
i.e. ef + f #On opq;
i.e. (0.2)2 + (j 0.9)2
or, 0.04 – 0.81 ≠ 1
so the network is not lossless
Option (C) is correct
23. A transmission line has a characteristic impedance of 50 Ω and a resistance of 0.1 Ω /m. if the line is distortion less, the attenuation constant (in Np/m) is
(A) 500 (B) 5 (C) 0.014 (D) 0.002
Soln. (23) For distortion less line attenuation constant and velocity of propagation should be independence of frequency.
Given
Z0 = 50Ω
R = 0.1 Ω/m
Since the line is distortion less it should satisfy the following:
LG = CR
i.e. 4 = 3r
Characteristic impedance
I- = s4 = s3r
and Attenuation constant t = `uv = `u`wx,. Or α = 3y-
For the present case
α = -./- =0.002 Np/m
Option (D) is correct
24. Consider the pulse shape s(t) as shown. The impulse response h(t) of the filter matched to this pulse is
GATE-2010
Impulse response of the matched filter is h(t) =s(T-t)
s(T+t) is left side shifted version of s(t) by T.
s(T-t) is the mirror image of s(T+t) on the Y-axis.
Option (C) is correct
25. The electric field component of a time harmonic plane EM wave traveling in a nonmagnetic lossless dielectric medium has an amplitude of 1 V/m. If the relative permittivity of the medium is 4, the magnitude of the time-average power density vector (in W/m2) is:
(A) 1/30π (B) 1/60π (C) 1/120π (D) 1/240π
Soln. (25) Given z = V/m
ϵ r = 4
Magnitude of the time average power density
GATE-2010
Pav = |
Where = s~ ∈ - = s,ϵ,
= s ,ϵ, = |, = -T
@ = (, = -T W/m2
Option C is correct
26. If ey = X1/x, then y has a
(A) maximum at x= e (B) minimum at x= e
(C) maximum at x= e-1 (D) minimum at x= e-1
Soln. (26) Given:
=
Taking log, p = p( = p
= . p Differentiating
= − . p + .
= − +
= 0
rr, −p + = 0
= 0p =
i.e. = ∞ = = =− ? − ? −p ?
For = =− ? − ? − ? = − ?
GATE-2010
Since 1 − Hence maximum at =
Option (A) is correct
27. A fair coin is tossed independently four times. The probability of the event “the number of time heads shown up is more than the number of times tails shown up” is : [2 marks]
(A) 1/16 (B) 1/8 (C) ¼ (D) 5/16
Soln. (27) There are only five events when the number of times heads show up is more than the number of times tails show up are the following:
H H H H
H H H T
H H T H
H T H H
T H H H
Since the fair coin in tossed independently four times the probability of each event is
= U
All the event are mutually exclusive so the total probability
= + + + + = Option (D) is correct
28. If A = xy âx+ x2 ây then ∮c A.dl over the path shown in the figure is
(A) 0 (B) 2/√3
(C) 1 (D) 2√3
GATE-2010
Soln. (28) Given
< = [ +[
Zℎ <.pZℎ^[Zℎ1ℎ
p = . [ + [
p = [ + [
<. p = +
From :- P – Q = , = 0
<N
. p = . . . = 2 ¡ `: `: `:
`:
= ¢: − :£ =
Q – R = : = 0
<. p = 26 = 066 − = 863
R – S = 6 = 0
GATE-2010
<P3
. p = 6 `:
`:= 62 = 62 6 − 06 = − 62
S – P = : = 0
<. p = 6:
= 6N
P − 6 = − 26
f <. p = <. p + <.3
N
p + <.P3
p + <NP
. p
= + ¥: − : − :
<. p =
Option C is correct
29. The residues of a complex function X(z) = (1-12z)/[z(z-1)(z-2)] at its poles are
(A) 1/2, -1/2 and 1 (B) 1/2, -1/2 and -1
(C) 1/2, -1 and -3/2 (D) 1/2, -1 and 3/2
Soln. (29) Given function
¦I = §§§ § To find residue at its poles
Residue at z = a of G(Z)= lim§⟶.I − [ vI
Poles are at 0, 1 & 2
For Pole at 0
Residue: u = lim§⟶- y yy y I =
For Pole at z = 1
GATE-2010
Residue: u = lim§⟶ y y yy y =
Pole at z =2
u: = limy⟶ y y yy y = −:
Option C is correct
30. Consider differential equation dy(x)/dx - y(x) = x with the initial condition y(0) = 0. Using Euler’s first order method with a step size of 0.1, the value of y (0.3) is
(A) 0.01 (B) 0.031
(C) 0.0631 (D) 0.1
31. Given f(t) = L-1[(3s+1)/(s3+4s2+(k-3)s)]. If Limx→∞ f(t) = 1, then the value of k is :
(A) 1 (B) 2 (C) 3 (D) 4
Soln. (31) limn⟶∞ lZ = limO⟶- 1A1
Given that
A1 = ¢ :OQO?QO¬: O£ Since limn⟶∞ lZ =
Thus, limO⟶- 1 ¢ :OQO?QOQ¬: O£ =
= limO⟶- ¢ :OQOQOQ¬: £ =
¬: =
− 6 =
®¯ ° = ±
Option (D) is correct
32. In the circuit shown, the switch S is open for a long time and is closed at t=0. The current i(t) for t≥0+ is
GATE-2010
(A) i(t)=0.5-0.125e-1000tA (B) i(t)=1.5-0.125e-1000tA
(C) i(t)=0.5-0.5e-1000tA (D) i(t)=0.375e-1000tA
Soln. (32) At t = 0-, the circuit is shown below
1.5A
10Ω
10Ω
10Ω
S
0.75A 0.75A
The current of 1.5A source will divide equally in both10Ω resistors
At t = 0+
1.5A 15mH
10Ω
10Ω
10Ω
S
Current in inductor will not change instantaneously, so the current of 0.75 mA which was flowing through first 10Ω resister will get divided to this branch and the switch branch
1 0Q = 0.7²2 = 0.67²<
∞ = 0.²<
The current will divide into three 10Ω branches (inductor is short circuited)
Z = ∞ − ∞ − 0Q 3n
Where R is equivalent resistor seen across L with current source opened.
GATE-2010
10Ω
10Ω10Ω R
R = 10 + (10 10)
R = 10 + 5 = 15 Ω
Z = 0.² − 0.² − 0.67² /n³/´-µ? Z = 0.² − 0.2²---nµ
Option A is the correct
33. The current I in the circuit shown is
(A) -j1A (B) J1A (C) 0A (D) 20A
Soln. (33) The given circuit is redrawn
0020
+
-
jωL
1
jωC
A
ω = 103rad/Sec.
1Ω
(L = 20mH , C = 50µH)
At node A
)¶-XY + )¶ + )¶ XY· = 0
GATE-2010
& ¢ XY + + \o£ = -XY
& ¢ X.-?.-.-µ? + + .0:.²0.0U£ = -X.-?.-.-µ?
& ¢–X- + + X-£ = − & = −.&
# = )¶ =−< = −<
Option A is correct
34. In the circuit shown, the power supplied by the voltage source is
(A) 0W (B) 5W (C) 10W (D) 100W
Soln. (34) The given circuit is redrawn
ii+6
i+2
Since we have to find the power supplied by the voltage source.
So if we can find the current drawn from the source, power supplied can be evaluated. So the loop considered is as shown in the figure.
Let the current drawn be I Amp. Current in different branches is as per the figure.
Applying KVL to the given loop
GATE-2010
10 = (i+3)(1+1) + (i+2)(1+1)
⇒ 10 = 2i + 6 + 2i + 4
⇒ 4i = 0
⇒ i = 0
Hence power drawn from the supply
P = V . i = 10 . 0 = 0 Watts
Option (A) is correct
35. In a uniformly doped BJT, assume that NE, NB and NC are the emitter, base and collector dopings in atoms/cm3, respectively. If the emitter injection efficiency of the BJT is close unity, which one of the following conditions is TRUE?
(A) NE=NB=NC (B) NE >>NB and >NB>NC
(C) NE=NB and NB<NC (D) NE<NB<NC
Soln. (35) For proper transistor action the lightly doped base between the heavily doped emitter and moderately doped collector. Thus option B is the most nearest option
i.e. NE>>NB (Heavily doped emitter as compared to base)
NB >NC (it should be NB < NC)
Option B is correct
36. Compared to a p-n junction with NA=ND=1014/cm3, which one of the following statements is TRUE for a p-n junction with NA=ND=1020/cm3?
(A) Reverse breakdown voltage is lower and depletion capacitance is lower
(B) Reverse breakdown voltage is higher and depletion capacitance is lower
(C) Reverse breakdown voltage is lower and depletion capacitance is higher
(D) Reverse breakdown voltage is higher and depletion capacitance is higher
Soln. (36) In the given problem the doping concentration has been increased significantly. With increase in doping the breakdown voltage decreases. With more of doping depletion width (d) decreases. Depletion capacitance is given by C = εA/d, so it increases
Option C is correct
37. Assuming that flip-flops are in reset condition initially, the count sequence observed at QA in the circuit shown is
GATE-2010
(A) 0010111… (B) 0001011…
(C) 0101111… (D) 0110100…
Soln. (37) See the given diagram
Given all flip flops are in reset condition i.e. the outputs are 0 initially,
M¸ =M» =M4 = 0
= M» ¼M4 =
» = M¸ = 0 Since connected together
4 =M» = 0
After one clock pulse:
M¸ = (content of DA transfers to QA )
M» = 0 =M4
Now
= M» ¼M4 =
» = M¸ =
4 =M» = 0
After two clock pulses
M¸ = M» = M4 = 0
= M» ¼M4 = 0
» −M¸ = 4 =M» =
After three clock pulse
M¸ = 0 M» = M4 =
= M» ¼M4 =
» = M¸ = 04 =M» =
GATE-2010
So the sequence observed at
M¸ = 0000½ ½ ½ .. Option D is correct
38. The transfer characteristic for the precision rectifier circuit shown below is (assume ideal OP-AMP and practical diodes)
Soln. (38) The above circuit can be redrawn as follows:
GATE-2010
-+
R
4R
R
+20V
V1 V0
D1
D2
A
We choose the input voltage from one of the option say
V1 = -10 V
Point A is at the ground potential
),-3 = --3 +-- 3
&- =−² + 0 = ²&
Now at V1 = -5 V
),3 = --3 +-/ 3
or V0 = -5+5 = 0 V
For V1 > -5 V, both diodes are conducting
So, V0 = 0
Thus option (B) is correct
39. The Boolean function realized by the logic circuit shown is
(A) F=Σm(0,1,3,5,9,10,14) (B) F=Σm(2,3,5,7,8,12,13)
(C) F=Σm(1,2,4,5,11,14,15) (D) F=Σm(2,3,5,7,8,9,12)
Soln. (39) The given logic circuit having 4 X 1 M U X
GATE-2010
A = ff-#- +ff-# + ff-# +ff-#:
v f = < f- = B #- = # = # = #: =
Zℎq1 A< B = <B +<B + <B + <B
=<B¾ + ¿ +<B¾ + ¿ + <B¾ + ¿ + <B = <ÀBÁ + <ÀBÁÂ + <ÀB + <ÀBÀ + <BÁÀ + <BÁÀÂ + <BÀÂ
=e; 2 6 ² 7 8 5 2
Option (D) is correct
40. For the 8085 assembly language program given below, the content of the accumulator after the execution of the program is
(A) 00H (B) 45H (C) 67H (D) E7H
Soln. (40) Contents of various registers during the execution of the programs line by line are:
3000 MVI A, 45 H (Move 45 H to Acc)
(A) ⟶ 0100 0101
3002 MOV B,A (Move the content of Acc. To B)
(B) ⟶ 0100 0101
3003 STC (Set carry flag )
i.e. CS = 1
3005 CMC (complement the carry)
CS = 0
3005 RAR (Rotates the content of Acc. To right with carry)
0 0 1 0 0 0 1 0 1
i.e. 0010 0010
CS A7 A6 A1 A0 . . . .
GATE-2010
3006 XRA B (Content of B is to be Ex-ored with Acc.)
(B) 0100 0101
(A) 0010 0010
A ¼ B
⟶ 011 00111 (67H)
Option (C) is correct
41. A continuous time LTI system is described by
d2y(t)/dt2 + 4dy(t)/dt + 3y(t) = 2dx(t)/dt + 4x(t)
Assuming zero initial conditions, the response y(t) of the above system for the input x(t)=e-
2tu(t) is given by
(A) (et-e3t)u(t) (B) (e-t-e-3t)u(t)
(C) (e-t+e-3t)u(t) (D) (et+e3t)u(t)
Soln. (41) A CTLTI system is described by
n + 0 n + 6Z + 2 n n + 0Z
Since the system is continuous time LTI system we take Laplace transform on both sides, assuming zero initial condition.
1 + %1 + 01%1 + 6%1 = 21¦1 + 0. ¦1
RO VO = OQOQOQ: = OQ OQ OQ: Given that input is Z = nqZ
¦1 = OQ
%1 = OQ OQ OQ: OQ = OQ OQ: = OQ − OQ:
Taking inverse Laplace transform on both sides
Z = n −:n qZ
Option B is correct
GATE-2010
42. The transfer function of a discrete time LTI system is given by
ÃÄ = ÅƱǵƱǵQÈǵÅ
Consider the following statements:
S1: The system is stable and causal for ROC:|z|>½
S2: The system is stable but not causal for ROC:|z|<¼
S3: The system is neither stable nor causal for ROC: ¼<|z|<½
Which one of the following statements is valid?
(A) Both S1 and S2 are true (B) Both S2 and S3 are true
(C) Both S1 and S3 are true (D) S1, S2 and S3 are all true
Soln. (42) LÉ = ?ʧµ(?ʧµ(Q(˧µ
= (ʧµ(Q(§µ((ʧµ((§µ(
LÉ = ³ − ³2 I^− + ³ − ³0I^−
ROC due to first term.
Í IÍ > *I* >
2nd term given rise to
Í IÍ > *I* >
Intersection of these ROCs
< *I* <
For ROC *I* > the system is stable and causal
for ROC (composite ROC)
i.e. < *I* <
In this case ROC is not exterior of *Ç* > ³Å so not causal
ROC does not include the unit circle so not stable.
Option (C) is correct
GATE-2010
43. The Nyquist sampling rate for the signal s(t) = sin(500πt)/πt x sin(700πt)/πt is given by
(A) 400 Hz (B) 600 Hz
(C) 1200Hz (D) 1400 Hz
Soln. (43) Given
On iÏÐÑ/--Tn Tn ´ ÏÐÑ8--Tn Tn
Using the intently
2sinA sinB = cos(A-B) – cos(A+B)
so, 1Z = Tn Fcos200ÕZ − cos200ÕZ G here \Ö = 200Õ Ö = --TT = _00Ly
So, lO =2lÖ = 2 ´ _00 = 200Ly
Option C is correct
44. A unity negative feedback closed loop system has a plant with the transfer function G(s) = 1/(s2+2s+2) and a controller Gc(s) in the feed forward path. For a unit set input, the transfer function of the controller that gives minimum steady state error is
(A) Gc(s) = (s+1)/(s+2) (B) Gc(s) = (s+2)/(s+1)
(C) Gc(s) = [(s+1)(s+4)]/[(s+2)(s+3)] (D) Gc(s) = 1 + 2/s + 3s
Soln. (44) Block diagram of the system is as follows
zO = u1 − 1
zO = u1 − z1 . v1 . v·1
3bQrO rcO
GATE-2010
Since input is unity
Then u1 = O Given v1 = OQOQ
×imOØ- v1 = -Q-Q = = 0.²
Steady state error OO = limOØ- 1z1
= limOØ- O.3O QrO rcO = limOØ- O.3O QrO rcO = QÙÐÚbØ, rO ÙÐÚbØ, rcO = Q-./.ÙÐÚbØ, rcO
For ess to be minimum Gc(s) should be maximum. In option (D) ÛÜÝ = + ÅÝ + ÆÝ (since ÞßàÝØáÛÜÝ = â)
Option (D) is correct
45. X(t) is a stationary process with the power spectral density Sx(f)>0 for all f. The process is passed through a system shown below.
Let Sy(f) be the power spectral density of Y(t). Which one of the following statements is correct?
(A) Sy(f)>0 for all f (B) Sy(f)=0 for |f|>1kHz
(C) Sy(f)=0 for f=nfo, fo=2kHz, n any integer
(D) Sy(f)=0 for f=(2n+1)fo=1kHz, n any integer
Soln. (45) From the given system diagram
ãZ = n FZ + Z − ä G Taking Fourier on either side
ãl = ål +XTæç.l è\
GATE-2010
Lãl = æ æ = \å +XTæçè L^′l = \F + o12Õlä − 12ÕläG *L^′l *^2 = \^2F2 + 2o12ÕläG = 〖2\〗^2F + o12ÕläG f- = *Ll *. fg f- = 2\F + o12ÕläG. fg
+ o12ÕläG = 0ll = 2 + ³2ä
fl = 0ll = 2 + l-where f0 = 1 KHz and n an integer
Option D is correct
46. A plane wave having the electric field component Ei = 24cos(3x108t-βy)âz V/m and traveling in free space is incident normally on a lossless medium with µ=µo and ε=9ε0 which occupies the region y≥0. The reflected magnetic field component is given by
(A) [cos(3x108t+y)âx] /10π A/m (B) [cos(3x108t+y)âx] /20π A/m
(C) -[cos(3x108t+y)âx] /20π A/m (D) -[cos(3x108t+y)âx] /10π A/m
Soln. (46) Given
zg = 20 cos6 ´ 0¥Z − = [y
Above given Electric field component is in Z direction and the wave is travelling in y-direction.
We can write the magnetic field component which is normal to y
i.e. Lg = | ¾[ ´ zêg¿
= 20o1 ³20Õ6 ´〖0〗^8Z − = [_
Lg = /T . cos 6 ´ 0¥Z − = [
íîí> = |(||(Q| = sï,ð(sï,ðsï,ð(Qsï,ð
= `ñ–`ñ(`ñQ`ñ( = `9–``9Q`
= 313Q1 = 2
4= 1
2
Thus, Lk = -T cos6 ´ 0¥Z + = [
This indicates that the reflected wave is travelling in –y direction
Option A is correct
GATE-2010
47. In the circuit shown, all the transmission line sections are lossless. The Voltage Standing Wave Ration (VSWR) on the 60W line is
(A) 1.00 (B) 1.64 (C) 2.50 (D) 3.00
Soln. (47) The circuit is redrawn
z0=60Ω z0=30³`2 zL=30Ω
z2
z1
z0=30Ω
λ/4
λ/8
short
The expression for the input impedance of the line is
Ig =I- §òQX§, óôÑõ§,QX§ò óôÑõ
Input impedance due to shorted λ/8 stub is
I = 60 -QX óôÑö ÷Ë:-Q- ¡ I = 60
Input impedance due to
Load I= 60ø
I = 60`2 :-QX:-` óôÑ÷ ÷Ê:-`QX:- óôÑ÷ ÷Ê¡
= 60`2ù ?,úûü QX:-`?,`úûü QX:- ý
GATE-2010
É = _0ø
Now load impedance is
I =I +I = 60 + _0
Magnitude of reflection coefficient
*þ* = ͧò–§,§òQy,Í = ÍU-QX:-U-U-QX:-QU-Í = Í X:--QX:-Í = Í XQXÍ = 1`16Q1 = 1`17 &fu = Q**** = Q (`( (`( = ._0
Option B is correct
COMMON DATA QUESTIONS: 48 & 49
Consider the common emitter amplifier shown below with the following circuit parameters:
β=100, gm=0.3861 A/V, r0=∞ , rp=259 W, RS=1kW, RB=93KW, RC=250 W, RL=1kW, C1=∞ and C2=4.7mF.
48. The resistance seen by the source Vs is
(A) 258 Ω (B) 1258 Ω
(C) 93 KΩ (D) ∞
Soln. (48) Equivalent model for the given circuit is
GATE-2010
ℎg = õ
= ---.:¥U = 2²5ø
Resistance seen by the source Vs
ugã =uO +u»* *ℎg
= 1000+ 93000´25993000Q259
ugã = 000 + 2²8 = 2²8ø
Option B is correct.
49. The lower cut-off frequency due to C2 is
(A) 33.9 Hz (B) 27.1 Hz
(C) 13.6 Hz (D) 16.9 Hz
Soln. (49) Lower cutoff frequency due to C2
l = T3cQ3ò 4
= T/-Q--- ´.8´-µ l = 27.L§
Option B is correct
GATE-2010
COMMON DATA QUESTIONS: 50 & 51
The signal flow graph of a system is shown below.
50. The state variable representation of the system can be
Soln. (50) Soln. (50) From signal flow graph
= 2q1 −
= −
= + ´ 0.² = 0.² +
⇒ FG = F0.²0.²G ¢£ = − +
= − + 2q1
"$ = ¢− − 0£ ¢£ + ¢02£ q
Thus, Option (D) is correct
GATE-2010
51. The transfer function of the system is
(A) (s+1)/(s2+1) (B) (s-1)/(s2+1)
(C) (s+1)/(s2+s+1) (D) (s-1)/(s2+s+1)
Soln.(51)
ä1 = RO O =e N¬
¬
Where
PK is the gain of the forward path between U(s) and Y(s)
∆ is the determinant of the graph
= − e ^[1 + eZqoℎp^[1Z[Z[Z[Z; − eZqoℎp^[1Z[Zℎ[Z[Z; +. . . . . . . . . = − e + e − e : + e ½ ½ ½ ½ ½ ½. ¬= − e p^[Z;11Zℎ[ZZqoℎZℎnl[^[Zℎ.
= 2.1 .1 . . 0.² = 1
= 2. O . . 0.² = O
= − O =− O
= −¢− O − O£ = + O + O
1= 12= 2
RO O = (b..Q(b.Q(bQ (b
= OQOQOQ
Option (C) is correct
GATE-2010
Linked Answer Question: Q. 52 to Q. 53 Carry Two Marks Each
Statement of Linked Answer Questions: 52 & 53
The silicon sample with unit cross-sectional area shown below is in thermal equilibrium. The following information is given: T=300K, electronic charge=1.6x10-19C, thermal voltage=26mV and electron mobility = 1350cm2/V-s
52. The magnitude of the electric field at x=0.5 µm is
(A) 1kV/cm (B) 5kV/cm (C) 10 kV/cm (D) 26kV/cm
Soln. (52) For the given silicon sample the electric field will be given
z = )Ö
= 1
10µ6 =106 V/m
z = 0 V/cm
This electric field will remain the same throughout the given sample.
Option C is correct.
53. The magnitude of the electron drift current density at x=0.5 µm is
(A) 2.16x104 A/cm2 (B) 1.08x104 A/cm2
(C) 4.32x103 A/cm2 (D) 6.48x102 A/cm2
Soln. (53) Electronic drift current in given by
= ´ ´ ~ ´ z
=1016 ´ 1.6 ´ 1019 ´ 135 ´ 10 ´ 103 = 2.16 ´ 104 A/cm2
Option A is correct
GATE-2010
Statement of Linked Answer Questions: 54 & 55
Consider a baseband binary PAM receiver shown below. The additive channel noise n(t) is whit with power spectral density SN(f)=N0/2=10-20 W/Hz. The low-pass filter is ideal with unity gain and cutoff frequency 1MHz. Let Yk represent the random variable y(tk).
Yk=Nk if transmitted bit bk=0
Yk=a+Nk if transmitted bit bk=1
Where Nk represents the noise sample value. The noise sample has a probability density function, PNk(n)=0.5αe-α|n| (This has mean zero and variance 2/α2). Assume transmitted bits to be equiprobable and threshold z is set to a/2=10-6V.
54. The value of the parameter α (in V-1) is
(A) 1010 (B) 107
(C) 1.414x10-10 (D) 2x10-20
Soln. (54) Given:
Power spectral density
fhl = h, = 0- W/Hz
In time domain
u = h, ´ E
Power = 2 ´ 0
2´ 1 ´ 106
= 2 ´ 1014 Variance
Which is equal to power since mean is zero
t = 08
Option (B) is correct
GATE-2010
55. The probability of bit error is
(A) 0.5xe-3.5 (B) 0.5xe-5
(C) 0.5xe-7 (D) 0.5xe-10
Soln. (55) Probability of
Bit error = 0.² ta**
= 0.²-´-µ
= 0.²-
Option (D) is correct
Question No. 56-60 One Mark Each
56. Which of the following options is the closest in meaning to the world below: Circuitous
(A) Cyclic (B) Indirect
(C) Confusing (D) Crooked
Soln. (56) Circuitous means deviating from a straight course ⇒ Indirect
(A) Cyclic : Recurring in cycle (B) Indirect : Not leading by a straight line (C) Confusing : Lacking Clarity (D) Crooked : Irregular shape
Option (B) is correct
57. The question below consists of a pair of related of related words followed by four pairs of words. Select the pair that best expresses the relation in the original pair.
Unemployed: Worker
(A) fallow : land (B) unaware : sleeper
(C) wit : jester (D) renovated : house
Soln. (57) Unemployed worker ⇒ Here one is opposite to other
(A) Fallow : land ⇒ Undeveloped Land⇒ Appropriate choice.
(B) Unaware : steeper , both mean the same
GATE-2010
(C) Wit : Jester to make jokes and jester is a joker
(D) Renovated : Homes ⇒ House can be renovated.
Option (A) is correct
58. Choose the most appropriate word from the options given below to complete the following sentence:
If we manage to ________ our natural resources, we would leave a better planet for our children.
(A) uphold (B) restrain
(C) Cherish (D) conserve
Soln. (58) (A) Uphold : cause to remain ⇒ not appropriate
(B) Restrain : Keep under control ⇒ not appropriate
(C) Cherish : Be fond of ⇒ not related
(D) Conserve : Keep in safety and project from harm decay,
Loss or destruction : Most appropriate
Option (D) is correct
59. Choose the most appropriate word from the options given below to complete the following sentence:
His rather casual remarks on politics _________ his lack of seriousness about the subject.
(A) masked (B) belied
(C) betrayed (D) suppressed
Soln.(59) (A) Masked : Hide under a false appearance ⇒ opposite.
(B) Belied : Be in contradiction with ⇒ Not appropriation
(C) Betrayed : Reveal unintentionally ⇒ most appropriate
(D) Suppressed: To put down by force or authority ⇒ irrelevant
Option (C) is correct
GATE-2010
60. 25 persons are in a room. 15 of them play hockey, 17 of them play football and 10 of them play both hockey and football. Then the number of persons playing neither hockey nor football is:
(A) 2 (B) 17 (C) 13 (D) 3
Soln. (60) Using set theory.
n(A) : Number of people who play hockey =15
n(B) : Number of people who play football =17
n(A∩B) : persons who play both hockey and football = 10
n(A∪B) = persons who play either hockey or football or both.
Formula
< ∪ B = < + B − < ∩ B
< ∪ B = ² + 7 − 0 = 22
Thus the people who play neither hockey nor football = 25 – 22 = 3
Other method
Hence, neither hockey nor football = 3 players.
Option (D) is correct
Question No. 61-65 Carry Two Mark Each
61. Modern warfare has changed from large scale clashes of armies to suppression of civilian populations. Chemical agents that do their work silently appear to be suited to such warfare; and regretfully, there exist people in military establishments who think that chemical agents are useful tools for their cause.
which of the following statements best sums up the meaning of the above passage:
(A) Modern warfare has resulted in civil strife.
(B) Chemical agents are useful in modern warfare.
(C) Use of chemical agents in warfare would be undesirable.
(D) People in military establishments like to use chemical agents in war.
Soln. (61) (A) Modern warfare has resulted in civil strife : Not appropriate
(B) chemical agents are useful in modern war fare : Not appropriate
GATE-2010
(C) Use of chemical agents as war fare would be undesirable : Not appropriate or wrong.
(D) people in military establishments like to use chemical agents in war : correct choice.
Option (D) is correct
62. If 137+276=435 how much is 731+672?
(A) 534 (B) 1403
(C) 1623 (D) 1513
Soln. (62) Given 137+276=435
From the problem it is obvious that the addition is not in base 10, since the numbers add up to different total in base 10 (which we use in daily life)
Intution: Consider units place and consider the total of 7+6 (base 10), which is 13 (base 10). Since we have 5 in units place, the intuition would be that this would be base 8. Lets check the sum for base 8 and the full sum matches., i.e
(137)8 +(276)8 = (435)8
Now solve the addition in base 8
(731)8 +(672)8 = (1623)8
fB Ø 2 + = 6
Z Ø 6 + 7 = 2Zℎo[
Z Ø 7 + _ = ²Zℎo[
So option (C) is correct
63. 5 skilled workers can build a wall in 20 days; 8 semi-skilled worker can build a wall in 25days; 10 unskilled workers can build a wall in 30 days. If a team has 2 skilled, 6 semi-skilled and 5 unskilled workers, how long will it take to build the wall?
(A) 20 days (B) 18 days
(C) 16 days (D) 15 days
Soln. (63) Per day work (Rate of 5 skilled workers)
= 120
So, Per day work of one skilled worker
GATE-2010
= 1
5´20 Per day work of rate of 8 semi-skilled workers
= 125
Per day work of rate of one semi-skilled worker
= 1
8´25 = 1200
Per-day work of 10 skilled workers
= 130
Per-day work of one semi-skilled worker
= 1
10´30 = 1300
Thus total per day work of 2 skilled, 6 semi-skilled and 5 unskilled workers
= 2100
+ 6200
+ 5300
= 12Q18Q10600
= 40600
= 115
Thus time to complete the work is 15 days
Option (D) is correct
64. Given digits 2,2,3,3,3,4,4,4,4 how many distinct 4 digit numbers greater than 3000 can be formed?
(A) 50 (B) 51
(C) 52 (D) 54
Soln. (64) Given digits are
2, 2, 3, 3, 4, 4, 4, 4
Numbers greater than 3000, will have either 3 or 4 in the thousands place. We can break this into two parts, i.e. either thousands digit is 3 or it is 4.
GATE-2010
Ist case First digit = 3
Rest 3 digits can be combination of
234, or 3! = 6
223, or 3!/2! = 3
224, or 3!/2! = 3
332, or 3!/2! = 3
334 or 3!/2! = 3
442 or 3!/2! = 3
443 or 3!/2! = 3
444 or 1 = 1
So total combination = 25
II Case First digit = 4
Rest 3 digit may be combination of
234 or 6
223 or 3
224 or 3
332 or 3
334 or 3
442 or 3
443 or 3
444 or 1
333 or 1
Total combination = 26
Required answer = 25+26=51
Option (B) is correct
GATE-2010
65. Hari (H), Gita (G), Irfan (I) and Saira (S) are sibiligs (i.e. brothers and sisters). All were born on 1st January. The age difference between any two successive siblings (that is born one after another) is less than3 years. Given the following facts:
i. Hair’s age + Gita’s age > Irfan’s age + Saira’s age.
ii. The age difference between Gita and Saira is 1 year. However, Gita is not the oldest and Saira is not the youngest.
iii. There are not twins.
In what order were they born (Oldest first)?
(A) HSIG (B) SGHI
(C) IGSH (D) IHSG
Soln. (65) As given
H+G> I +S
⇒ I – H< G – S ___________________ (1)
And G – S = 1 ___________________ (2)
From equations. (1) and (2), we get
I – H < 1
⇒ H > I
So order is SGHI
Option B is correct