ece2031 tutorial 1 solutions

8/11/2019 ECE2031 Tutorial 1 Solutions

1/12

ECE2031 Tutorial 1 Solutions


2/12

0),( ttvO

FIND

C

1R

2R

KCLUSE0.tFORMODEL

0)()(0)( 2121

c

CCC vtdt

dvCRR

RR

vt

dt

dvC

sFCRR 6.0)10100)(106()(63

21

STEP 1

)(31)(

422)( tvtvtv CCO

STEP 20,)( 21

teKKtv

t

C 01K

INITIAL CONDITIONS. CIRCUIT IN STEADY STATE t


3/12

0),( ttvO

FIND

0,)( 21

teKKtv

t

o:1STEP

VOLTAGECAPACITORINITIAL:2STEP

)0(Cv

0t

1v

2v

2

12

12)0(

)12(3

1

vv

vv

C

V12

022

8

3

12: 1111

vvvv@KCL 6/*

][60488 11 Vvv

][10)0()0(][22 VvvVv CC

)0( OvDETERMINE:3STEP

0t

)0(Ov

V10

][5)10(22

2)0( VvO

Tute 1 Problem 2


4/12

)(OvDETERMINE:4STEP

)(Ov

][524)12(

52)( Vv

O

CONSTANTTIMEDETERMINE:5STEP

CRTH

:CircuitCapacitive

THR

5

44||1

THR

sFC5

82

21,KKDETERMINE:6STEP

][5

24)(1 VvK O

][5

1][5)0(

221

VKKKVvO

0];[5

1

5

24)( 5

8

tVetv

t

O:ANS

ORIGINAL CIRCUIT


5/12

Tute 1 Problem 3

vs(t)

0 1 2 3 4 5

-1

1

2

t

+

+

R = 2

vc(t)C = 1Fvs(t)

First Approach:

Consider the piecewisecontinuous input waveform shown above. We use the term

piecewise (continuousunderstood) to describe a waveform consisting of continuous

curves jointed together at a finite number of breakpoints. At these breakpoints, thewaveform may or may not be continuous. The waveform shown is a typical piecewise

waveform. A pulse, step, square, and triangular waves (for a finite time interval) can

also be regarded as special cases of piecewise waveforms.

Find the zero-state response vC(t)of the series RC circuit shown below to an

applied voltage vs(t)having the waveform illustrated in the same Figure. Sketch v

c(t)

and note its value at t=0, 4, 5,.


6/12

The peculiarity of the input waveform allows us to solve the problem piece-by-piece,

or from breakpoint to breakpoint. This approach is intuitive, and we shall describe it

by analyzing the above circuit as follows.

First time interval [0,4]

Within this time interval, the problem is obviously that of finding the step response

and the solution is -1 (step response). The circuit is in its zero-state at the beginning

of the interval. The step height is 1, and the natural frequency is1/2. The solution

can be written down by inspection:

At the end of this interval t=4, the capacitor voltage is vc(4)=0.8647.

Tute 1 Problem 3

, 1 2

, 1

, 1 2 2

2

2,

,

( )

( ) 1

(0) 0 1

( ) 1

1

( ) 1

( )

(1 )

t

c ste

t

c ste

p

c step

c step

c

t

p

c step

v t k k e

v k

v k k k

v t

t e

v t

e

v

By linearity property


7/12

Second time interval [4,5]

Within this time interval, the problem is to find the complete response with a step

input. It is expedient to do a time shiftt=(t-4) first. At the beginning of this interval,

the circuit is at a state of vc(t=4)=0.8647, which becomes the initial condition for

this time interval (helping to find the ZIR). The input is 2u(t). The solution can be

written down by inspection:

At the end of this interval t=5 (or t=1), the capacitor voltage is vc(5) = 0.2625.

Third time interval [5, ]

Within this time interval, the problem is to find the zero-input response. It is

expedient to do a time shift t=(t-5) first. At the beginning of this interval, the circuit

is at state vc(t=5) =

vc(t=0) = 0.2625. The solution can be written down byinspection:

The capacitor voltage decays to zero vc() = 0 as expected .

vc(t' ) 2(1 e

t' / 2) 0.8647 e

t' / 2; t' [0,1]

vc

(t") 0.2625e t"/ 2

; t"[0, ]

Tute 1 Problem 3

ZSR ZIR


8/12

0 1 2 3 4 5 6 7 8-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

time t (s)

voltagevc

(t)(V)


9/12

Second Approach:

First we decompose the input waveform in terms of unit step functions as follows.

Immediately we obtain from the LTIV properties of the circuit

Where the step response

and

This completes the derivation. The voltage vC(t) at any time t can be evaluated from

the equations above.

Remark

The step height at the interval [4,5] is 3, while in the First Approach, it was regarded

as 2. Why?

vs(t) u(t)3u(t 4) 2u(t 5)

vc(t) (t)3(t4)2 (t5)

(t)(1et/2)u(t)

-(t- ) /ts(t- ) = (1 - e )u(t- ), = 4,5.

Tute 1 Problem 3

2


10/12

0),( ttiFIND

Once the switch opens the circuit is RLC

series

0)()0()(2)(30

t

C dxxivt

dt

diti

0)(2

1)(

2

3)(

2

2

titdt

dit

dt

id

5.0,1

05.05.12

s

ss

:roots

:Ch.Eq.

0;)( 221

teKeKti

t

t

To find initial conditions use steady state analysis for t


11/12

0),(0 ttvFIND

For t>0 the circuit is RLC series

)(2)(0 titv

)(ti

KVL

0)(2)0()(3/2

1)(

2

1

0

tivdxxitdt

di

C

t

0)(3)(4)(2

2

titdt

dit

dt

id

3,1

0342

s

ss

:roots

:Eq.Ch.

0;)( 321

teKeKti tt

To find initial conditions we use steady

state analysis for t


12/12

0);(),( 00 ttvtiDETERMINE )(12)(18)( 00 Vtitv

KVL

012)(18)(2)0()(36/114

0

titdt

divdxxi

t

C

0)(18)(9)(2

2

titdt

dit

dt

id

6,3

01892

s

ss

:roots

:Eq.Ch.

0;)( 623

10

teKeKti tt

0)0( C

v AiL 5.0)0(

Analysis at t=0+

)(5.0)0()0(0 Aii L

)0()0()0( 0 dt

diL

dt

diLv

L

L )0(Lv

0)0(Cv

012)0(18)0(4 LL iv

210

210

5.0)0(

632/17)0(

KKi

KKdt

di

Steady state t 0

6 6 6

14;

6

1121 KK

Tute 1 Problem 6

ece2031 tutorial 1 solutions

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