ECET 307

11/15/200611/15/2006 Ch 7 System Consideration- Paul LinCh 7 System Consideration- Paul Lin 11

ECET 307ECET 307

Analog Networks Signal ProcessingAnalog Networks Signal Processing

Ch 7 System ConsiderationsCh 7 System Considerations2 of 32 of 3

Fall 2006Fall 2006

http://www.etcs.ipfw.edu/~linhttp://www.etcs.ipfw.edu/~lin


Ch 7: System ConsiderationsCh 7: System Considerations

Transfer Function ExamplesTransfer Function Examples Differential Equation and Differential Equation and

Transfer FunctionTransfer Function Step and Impulse ResponsesStep and Impulse Responses


Example 7-1 Example 7-1

(a) Determine the transfer function and (a) Determine the transfer function and the impulse response of the circuit as the impulse response of the circuit as shown below. The input excitation is v1(t) shown below. The input excitation is v1(t) and the output response is v2(t).and the output response is v2(t).

0.25 F

2 +

-

v2(t)v1(t)

+

-


Example 7-1 SolutionExample 7-1 Solution

The transfer function is defined as The transfer function is defined as output/input, or output/input, or )(/)()( 12 sVsVsG

2

2

)(

)()(

2

2)(

42

4)(4

2

4)(

)()()(

42

)(

/1

)()()(

/42/1)()(

1

2

111

2

11

ssV

sVsG

s

sV

s

sV

s

ssV

sZsIsV

s

sV

sCR

sVsIsI

ssCRsZRsZ

cc

cR

c

2 +

-

V2(s)V1(s)

+

-

4/s


Example 7-1 Solution Example 7-1 Solution (cont.)(cont.)

The impulse response of the circuit The impulse response of the circuit is g(t) = Lis g(t) = L-1-1G(s) G(s)

tetg 21- 2]2s

2[L)(

2e-2t

v1(t) i2(t)



The response v2(t)The response v2(t)

From the definition of From the definition of G(s) = V2(s)/V1(s)G(s) = V2(s)/V1(s)

4

10

2

25]2sin5[)(

][sin

2221

22

sstLaplacesV

stLaplace

t

s

etv

ssV

sss

sA

s

B

s

A

sssssVsGsV

2'2

'2

2222

2

2212

2

5)(

,2

1

2

5)(

2

5

8

20

4)2(

20|4

20

)4)(2(

)2(2042

)4)(2(

20

4

10

2

2)()()(


Example 7-1 Solution Example 7-1 Solution (cont.)(cont.) Use the trick formulaUse the trick formula

The time function V(s)The time function V(s)

Mj

jsQ

jss

sQ

s

sQsV

45

2

10450

22

20

458

020

458

20

22

20)2(

2,2

20)(

4

)()(

2,1

2''2

)452sin(2

5)452sin(

22

10

)sin()(''2 ttt

Mtv

)452sin(2

5

2

5)()()( 2''

2'22

tetvtvtv t



Determine the transfer function of Determine the transfer function of the circuit. The input is v1(t) and the the circuit. The input is v1(t) and the desired output is i2(t).desired output is i2(t).

2 H 2 H

0.25 F 2

+

-

i2(t)

v1(t)i2(t)i1(t)



1. Draw the transformed circuit1. Draw the transformed circuit

2.Write a pair of simultaneous mesh equations2.Write a pair of simultaneous mesh equations

2 H 2 H

2

+

-

I2(s)

V1(s)I2(s)I1(s)

4/s

0)()234()(

4

)()(4

)()4

2(

21

121

sIss

sIs

sVsIs

sIs

s



Rewrite the equations in the matrix form circuitRewrite the equations in the matrix form circuit

3.Find I2(s) using determinants3.Find I2(s) using determinants

0

)(

)(

)(

)23/4(/4

/4)/42( 1

2

1 sV

sI

sI

sss

sss

255.1

)(

82046

4)(8

2046

4)(

16]8

1216

468[

4)(

)]/4)(/4[()]23/4)(/42[(

)]()/4[(

)23/4(/4

/4)/42(

0/4

)()/42(

)(

231

231

2

1

222

1

1

1

2

sss

sV

sss

sV

sss

ssV

sssss

ssV

ssssss

sVs

sss

sss

s

sVss

sI



4. Find the transfer function4. Find the transfer function

255.1

1

)(

)()(

231

2

ssssV

sI

input

outputsG

1.5s3+s2+5s+2

1V1(s) I2(s)


Example 7-3 Butterworth Low-Example 7-3 Butterworth Low-Pass Filter Pass Filter

Determine the transfer function for the Determine the transfer function for the following active filter:following active filter:

• A second-order Butterworth low-pass filter A second-order Butterworth low-pass filter using an Op-amp, normalized to a cutoff using an Op-amp, normalized to a cutoff frequency of 1 radian/sec with 1-Ω.frequency of 1 radian/sec with 1-Ω.

• An actual circuit is derived from the normalized An actual circuit is derived from the normalized circuit by scaling the frequency and resistance circuit by scaling the frequency and resistance levels.levels.

• v1(t) – inputv1(t) – input• v2(t) - outputv2(t) - output

C1

1.4124 F

C20.7071 F

R1

1

R2

1 +V2V3

V4

V1

+



1. Circuit Identification1. Circuit Identification The op-amp is configured as a “voltage-follower”, Av = The op-amp is configured as a “voltage-follower”, Av =

1, or vout = vin.1, or vout = vin. No-current is assumed to flow into the input + or non-No-current is assumed to flow into the input + or non-

inverting terminal: Zin = ∞.inverting terminal: Zin = ∞. Four nodes: V1(s), V2(s), V3(s), V4(s)Four nodes: V1(s), V2(s), V3(s), V4(s)

• V2(s) = V4(s)V2(s) = V4(s)

R1

1

R2

1

V2(s)

V3(s)

V4(s)

V1(s)

0.7071/s

1.4142/s

+

-


Example 7-3 Solution Example 7-3 Solution (cont.)(cont.)3. Write the simultaneous equation and solve for 3. Write the simultaneous equation and solve for

G(s)G(s)

)227()()(

2170

7071.01)(

1

)()(

20701

)()(

4142.11

)()(

1

)()(

42

434

432313

sVsVs

sVsVsV

sVsV

s

sVsVsVsV

R1

1

R2

1

V2(s)

V3(s)

V4(s)

V1(s)

0.7071/s

1.4142/s

+

-


Example 7-3 Solution Example 7-3 Solution (cont.)(cont.)Rearrange the equations, by substituting (7-22) Rearrange the equations, by substituting (7-22)

into (7-21) and eliminate V4(s):into (7-21) and eliminate V4(s):

Then substitute this equation and equation (7-22) Then substitute this equation and equation (7-22) into (7-20):into (7-20):

0)(4142.1

)()( 232 sVs

sVsV )()4142.1

1()( 23 sVs

sV

0)()()4142.1

1()]()([7071.0

)()()4142.1

1(

01

)()(

1

)()(

7071.01

)()(

01

)()(

4142.1

1)()(

1

)()(

222312

232313

432313

sVsVs

sVsVs

sVsVs

sVsVsVsVssVsV

sVsV

s

sVsVsVsV


Example 7-3 Solution Example 7-3 Solution (cont.)(cont.)Simplify the above equation to obtain:Simplify the above equation to obtain:

And finally the transfer function G(s):And finally the transfer function G(s):

14142.1

)()(

21

2

ss

sVsV

14142.1

1

)(

)()(

21

2

sssV

sVsG

s2+1.4142s+1

1V1(s) V2(s)



Determine the response resulting from Determine the response resulting from an excitation x(t) = 5 sin t, assume thatan excitation x(t) = 5 sin t, assume that

The input to a certain system is x(t)The input to a certain system is x(t) The output is y(t)The output is y(t) The impulse response of the system g(t) = 10eThe impulse response of the system g(t) = 10e--

tt sin 2t sin 2t

g(t) = 10e-tsin2tx(t) = 5sin t

y(t)


Example 7-4 Example 7-4 (cont.)(cont.)

1. Find the transfer function G(s)1. Find the transfer function G(s)

)1)(52(

100

)1)(412(

100

)1](4)1[(

100

1

5

4)1(

20)()()(

1

15)(

4)1(

20

4)1(

210]2sin10[)(

,2

2]2[sin

)()]([

2sin10)(

2222

2222

2

22

22

ssssss

sssssXsGsY

ssX

ssteLaplacesG

ands

tLaplace

asFtfeLaplace

tetg

t

at

t



2. Use MATLAB to find poles and zeros??2. Use MATLAB to find poles and zeros??>> den1 = [1 2 5];>> den1 = [1 2 5];

>> den2 = [1 0 1];>> den2 = [1 0 1];

>> den = conv(den1, den2)>> den = conv(den1, den2)

den =den =

1 2 6 2 51 2 6 2 5

Factor out the Y(s) by using the reside functionFactor out the Y(s) by using the reside function

>> num =[0 0 0 100];>> num =[0 0 0 100];

>> [r, p, k] = residue(num, den)>> [r, p, k] = residue(num, den)

r =r =

5.0000 + 2.5000i 5.0000 - 2.5000i5.0000 + 2.5000i 5.0000 - 2.5000i

-5.0000 -10.0000i -5.0000 +10.0000i-5.0000 -10.0000i -5.0000 +10.0000i

5262

100)(

234

sssssY



p =p =

-1.0000 + 2.0000i-1.0000 + 2.0000i -1.0000 - 2.0000i -1.0000 - 2.0000i

-0.0000 + 1.0000i-0.0000 + 1.0000i -0.0000 - 1.0000i -0.0000 - 1.0000i

The Y(s)??The Y(s)??

Well! Try the Trick formula”Well! Try the Trick formula”

)(

105

)(

105

)21(

5.25

)21(

5.25)(

is

i

is

i

is

i

is

isY



Well! Try the Trick formula”Well! Try the Trick formula”

)57.1162sin(18.11)57.26sin(36.22)()()(

)57.1162sin(2

36.22)(

2,1);sin()(

57.11636.2257.1164721.4

0100

1)21(

100

1

100)21(

)57.26sin(36.22)57.26sin(1

36.22)sin()(

1;57.2636.22

57.264721.4

0100

24

100

521

100

52)(

100

52

100)(

)1](4)1[(

100)(

21

2

2

222

1

221

22

tettytyty

tety

teM

ty

jsjsQ

tttM

ty

M

jjjjssjsQ

sssY

t

t

t



Determine the transfer function G(x) for the Determine the transfer function G(x) for the following system description:following system description:

x(t) = 5x(t) = 5 y(t) = 10ey(t) = 10e-2-2t + 5et + 5e-t-t sin 2t sin 2t

SolutionSolution

)52)(2(

)73(2

)52)(2(

)73(10

5)(

)()(

5)(

)52)(2(

)73(10

)52)(2(

)2(10)52(10

2)1(

25

2

110)(

2

2

2

2

2

2

2

2

22

sss

sss

sss

sss

sX

sYsG

ssX

sss

ss

sss

sss

sssY


7-2 7-2 Differential Equation and Transfer Differential Equation and Transfer

FunctionFunction General differential equation description of the General differential equation description of the

systemsystem The highest derivative of the y determine the The highest derivative of the y determine the

order of the systemorder of the system m ≥ nm ≥ n

The transform of a highest-order derivativeThe transform of a highest-order derivative

xadt

yda

dt

ydayb

dt

ydb

dt

ydb

n

n

nn

n

nm

m

mm

m

m 0101

1

1 ......

)0(...)0()(][ )1(1 kkkk

k

yyssYsdt

yd


Example 7-6Example 7-6

(a). Determine a differential equation expressing (a). Determine a differential equation expressing the input-output relationship for the circuit the input-output relationship for the circuit below.below.

(b). determine the transfer function for the (b). determine the transfer function for the circuitcircuit

8 F

4 F

1/2 H1/6 v2(t)v1(t)

+

-

+

-


Example 7-6 SolutionExample 7-6 Solution

This circuit will be solved using nodal equations This circuit will be solved using nodal equations Summing the current leaving the node, ∑ I = 0Summing the current leaving the node, ∑ I = 0

Differentiate all terms to remove the integral Differentiate all terms to remove the integral sign, and move v1 terms to the right-hand side sign, and move v1 terms to the right-hand side of the equationof the equation

8 F

4 F

1/2 H1/6 v2(t)v1(t)

+

-

+

-

t

dtvvvvdt

dv

dt

dv

0

121222 0)(2)(864

121

2

22

22

2

1221

2

22

22

22

2

282612

0228864

vdt

vdv

dt

dv

dt

vd

vvdt

vd

dt

vd

dt

dv

dt

vd



Assume all I.C.s are zero, and take Laplace Assume all I.C.s are zero, and take Laplace transformtransform

The transfer function isThe transfer function is

)()28()()2612( 1222 sVssVss

136

14

26212

28

)(

)()(

2

2

2

2

1

2

ss

s

s

s

sV

sVsG

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ECET 307