Electric Circuits, 9th Edition P52 pptx

13.6 The Transfer Function and the Convolution Integral The convolution integral relates the output yt of a linear time-invariant circuit to the input x{t of the circuit and the circu

486 The Laplace Transform in Circuit Analysis

ASSESSMENT PROBLEM

Objective 3—Understand the definition and significance of the transfer function; be able to derive a transfer function 13.9 a) Derive the numerical expression for the

transfer function V 0flg for the circuit shown

b) Give the numerical value of each pole and

zero of H(s)

NOTE: Also try Chapter Problem 13.50

Answer: (a) H(s) = 10(5 + 2)/(s 2 + 2s + 10);

(b) - / ; , = - 1 + / 3 , -p 2 = - 1 - / 3 ,

-z = -2

2a

1 H

0.1 F

The Location of Poles and Zeros of H(s)

For linear lumped-parameter circuits, H(s) is always a rational function

of s Complex poles and zeros always appear in conjugate pairs The poles of H(s) must lie in the left half of the s plane if the response to a

bounded source (one whose values lie within some finite bounds) is to

be bounded The zeros of H(s) may lie in either the right half or the left half of the s plane

With these general characteristics in mind, we next discuss the role

that H(s) plays in determining the response function We begin with the partial fraction expansion technique for finding y(t)

13.5 The Transfer Function

in Partial Fraction Expansions

From Eq 13.93 we can write the circuit output as the product of the trans-fer function and the driving function:

We have already noted that H(s) is a rational function of s Reference to Table 13.1 shows that X(s) also is a rational function of s for the excitation

functions of most interest in circuit analysis

Expanding the right-hand side of Eq 13.96 into a sum of partial

frac-tions produces a term for each pole of H(s) and X(s) Remember from

Chapter 12 that poles are the roots of the denominator polynomial; zeros are the roots of the numerator polynomial The terms generated by the

poles of H(s) give rise to the transient component of the total response, whereas the terms generated by the poles of X(s) give rise to the

steady-state component of the response By steady-steady-state response, we mean the response that exists after the transient components have become negligi-ble Example 13.2 illustrates these general observations

Example 13.2 Analyzing the Transfer Function of a Circuit

The circuit in Example 13.1 (Fig 13.31) is driven by

a voltage source whose voltage increases linearly

with lime, namely, v g = 50tu(t)

a) Use the transfer function to find v a

b) Identify the transient component of the

response

c) Identify the steady-slate component of the

response

d) Sketch v a versus / for 0 < t =£ 1.5 ms

Solution

a) From Example 13.1,

10()0(5 + 5000)

H(S) = -:

' s 2 + 6000^ + 25 X 106

The transform of the driving voltage is 50/.v~;

therefore, the v-domain expression for the

out-put voltage is

K, = 1000(5 + 5000) 50

(s 2 + 6000s + 25 x 106) v2 *

The partial fraction expansion of V a is

V ° ~ s + 3000 - /4000

K i K 2 K 3

+ s + 3000 + /4000 I 2 V '

We evaluate the coefficients K^ K 2 , and / t3 by

using the techniques described in Section 12.7:

/<, = 5V5 x l ( r4/ 7 9 7 0 ° ;

K\ = 5V5 x I Q -4 / - 7 9 7 0 ° ,

K 2 = 10,

The time-domain expression for v a is

v a = [10V5 X 10""4e~3,)()()/cos (4000/ + 79.70°)

+ 10? - 4 X 10~ 4 }u(t) V

b) The transient component of v a is

10V5 X 10-V3 0 0 0'cos (4000/ + 79.70°)

Note that this term is generated by the poles (-3000 4- /4000) and (-3000 - /4000) of the transfer function

c) The steady-state component of the response is

(10/ - 4 X 10"4)z<(/)

These two terms are generated by the

second-order pole (K/s 2 ) of the driving voltage

d) Figure 13.33 shows a sketch of v <7 versus / Note that the deviation from the steady-state solution 10,000/ - 0.4 mV is imperceptible after approxi-mately 1 ms

M m V )

16

1 4

-12

8 h

6

4

2

-(10.()00/- 0.4) mV,

V

0.2 0.4 0.6 0.8 1.0 1.2 1.4 t (ms)

Figure 13.33 A The graph of i\, versus t for Example 13.2

488 The Laplace Transform in Circuit Analysis

/ A S S E S S M E N T PROBLEMS

Objective 4—Know how to use a circuit's transfer function to calculate the circuit's impulse response, unit step response, and steady-state response to sinusoidal input

13.10 Find (a) the unit step and (b) the unit impulse

response of the circuit shown in Assessment

Problem 13.9

Answer: (a) [2 + (10/3)e"f cos (3/ - 126.87°)]w(0 V;

(b) 10.54*?-'cos (3/ - 1 8 4 3 > ( 0 V

NOTE: Also try Chapter Problems 13.79(a) and (b)

13.11 The unit impulse response of a circuit is

v()(t) = 10,000e_70 'cos(240r + 6) V, where tan 6 = ^

a) Find the transfer function of the circuit

b) Find the unit step response of the circuit

Answer: (a) 9600s/(s 2 + 140s + 62,500);

(b)40<T70'sin240r V

Observations on the Use of H(s) in Circuit Analysis

Example 13.2 clearly shows how the transfer function H(s) relates to the

response of a circuit through a partial fraction expansion However, the example raises questions about the practicality of driving a circuit with an increasing ramp voltage that generates an increasing ramp response Eventually the circuit components will fail under the stress of excessive voltage, and when that happens our linear model is no longer valid The ramp response is of interest in practical applications where the ramp function increases to a maximum value over a finite time interval If the time taken to reach this maximum value is long compared with the time constants of the circuit, the solution assuming an unbounded ramp is valid for this finite time interval

We make two additional observations regarding Eq 13.96 First, let's look at the response of the circuit due to a delayed input If the input is

delayed by a seconds,

£{x(t - a)u(t - a)} = e- ai X(s),

and, from Eq 13.96, the response becomes

Y(s) = H(s)X(s)e-as

Uy(t) = %- l {H(s)X(s)}, then, from Eq 13.97,

y(t - a)u{t - a) = ^T 1 {B{s)X{s)e- as }

(13.97)

(13.98)

Therefore, delaying the input by a seconds simply delays the response function by a seconds A circuit that exhibits this characteristic is said to

be time invariant

Second, if a unit impulse source drives the circuit, the response of the circuit equals the inverse transform of the transfer function Thus if

x(t) = 5(0, then X{s) = 1

and

Hence, from Eq 13.99,

y(t) = //(/), (13.100)

where the inverse transform of the transfer function equals the unit

impulse response of the circuit Note that this is also the natural response

of the circuit because the application of an impulsive source is equivalent

to instantaneously storing energy in the circuit (see Section 13.8) The

subsequent release of this stored energy gives rise to the natural response

(see Problem 13.90)

Actually, the unit impulse response of a circuit, h(t), contains enough

information to compute the response to any source that drives the circuit

The convolution integral is used to extract the response of a circuit to an

arbitrary source as demonstrated in the next section

13.6 The Transfer Function and the

Convolution Integral

The convolution integral relates the output y(t) of a linear time-invariant

circuit to the input x{t) of the circuit and the circuit's impulse response

h(t) The integral relationship can be expressed in two ways:

y(t) = / h(X)x(t - X)dX = / h(t - X)x(X)dX (13.101)

We are interested in the convolution integral for several reasons

First, it allows us to work entirely in the time domain Doing so may be

beneficial in situations where x(t) and h{t) are known only through

experimental data In such cases, the transform method may be awkward

or even impossible, as it would require us to compute the Laplace

trans-form of experimental data Second, the convolution integral introduces

the concepts of memory and the weighting function into analysis We will

show how the concept of memory enables us to look at the impulse

response (or the weighting function) h{t) and predict, to some degree,

how closely the output waveform replicates the input waveform Finally,

the convolution integral provides a formal procedure for finding the *(0 *j K f) ")'(*)

inverse transform of products of Laplace transforms

We based the derivation of Eq 13.101 on the assumption that the

cir-cuit is linear and time invariant Because the circir-cuit is linear, the principle F i9u r e 13*34 A A block d i a g r a m of a g e n e r a l d K u f L

of superposition is valid, and because it is time invariant, the amount of

the response delay is exactly the same as that of the input delay Now

con-sider Fig 13.34, in which the block containing h{t) represents any linear

time-invariant circuit whose impulse response is known, x(t) represents

the excitation signal and y(t) represents the desired output signal

We assume that x(t) is the general excitation signal shown in

Fig 13.35(a) For convenience we also assume that x(t) = 0 for t < (T

Once you see the derivation of the convolution integral assuming

x{t) = 0 for t < 0~, the extension of the integral to include excitation

functions that exist over all time becomes apparent Note also that we

permit a discontinuity in x(t) at the origin, that is, a jump between 0~

a n d 0

490 The Laplace Transform in Circuit Analysis

x(t)

x(t)

*(X () )

(a) liYX V V ( X ; ? )

x 0 \ j x 2 x? • • K •

(b)

x(X () ) AX

Xo X ! ^

r ~

-*>

2x

7 *

3

9>

7

s

X •i

(c)

Figure 13.35 • The excitation signal of x(t) (a) A

general excitation signal, (b) Approximating x(t) with a

series of pulses, (c) Approximating x(t) with a series

of impulses

h(t)

0

y(')

(a)

^ ,N\^ Approximation of v(f)

(b)

Figure 13.36 A The approximation of y(f) (a) The

impulse response of the box shown in Fig 13.34

(b) Summing the impulse responses

Now we approximate x(t) by a series of rectangular pulses of

uni-form width AA, as shown in Fig 13.35(b) Thus

x(t) = x0(t) + Xl(t) + ••• + xt{t) + (13.102)

where x^t) is a rectangular pulse that equals x(A,) between A/ and A(+1 and is zero elsewhere Note that the /th pulse can be expressed in terms of step functions; that is,

Xi(t) = x(Xi){u(t - A/) - u[t - (A, + AA)]}

The next step in the approximation of x(t) is to make AA small

enough that the /th component can be approximated by an impulse func-tion of strength x(Aj)AA Figure 13.35(c) shows the impulse representa-tion, with the strength of each impulse shown in brackets beside each

arrow The impulse representation of x(t) is

x(t) = x(A0 )AA5(f - An) + x(X^)AXS(t - A,) +

+ jc(Af)AA5(/ - A/) + • • • (13.103)

Now when x{t) is represented by a series of impulse functions

(which occur at equally spaced intervals of time, that is, at An, Aj, A2, ) ,

the response function y{t) consists of the sum of a series of uniformly

delayed impulse responses The strength of each response depends on the strength of the impulse driving the circuit For example, let's assume that the unit impulse response of the circuit contained in the box in Fig 13.34 is the exponential decay function shown in Fig 13.36(a) Then

the approximation of y{t) is the sum of the impulse responses shown in

Fig 13.36(b)

Analytically, the expression for y(t) is

y(t) = x(X{))AXh(t - A() ) + x(M)AXk(t - X{) + x(X 2 )AXh(t - A2) +

+ x(A,:)AA/j(/ - A/) + • (13.104)

As A A ^ - 0 , the summation in Eq 13.104 approaches a continuous integration, or

oo r°°

^x(Ai)h(t- A;) A A - > I x{X)h{t - X)dX (13 i=0 J0

105)

Therefore,

If x{t) exists over all time, then the lower limit on Eq 13.106 becomes

- o o ; thus, in general,

which is the second form of the convolution integral given in Eq 13.101

We derive the first form of the integral from Eq 13.107 by making a

change in the variable of integration We let u = t — A, and then we note

that du = — dX, u — - c o when A = oo, and u = + oo when A = - c o

Now we can write Eq 13.107 as

or

But because ti is just a symbol of integration, Eq 13.108 is equivalent to

the first form of the convolution integral, Eq 13.101

The integral relationship between y(t),h(t), and x(t), expressed in

Eq 13.101, often is written in a shorthand notation:

y(t) = h(t)*x{t)^x(t)*h(t), (13.109)

where the asterisk, signifies the integral relationship between h(t)

and x(t) Thus h(t) * x(t) is read as "//(f) is convolved with * ( / ) " and

implies that

h(t)*x(t) = J h(X)x(t - X)dX, whereas x(t) * //(/) is read as "x(t) is convolved with //(0" and implies that

x(\)h(t - X)dX

The integrals in Eq 13.101 give the most general relationship for the

convolution of two functions However, in our applications of the

convolu-tion integral, we can change the lower limit to zero and the upper limit to

/•.Then we can write Eq 13.101 as

y(t) = / h(X)x(t - X)dX = / x(X)h(t - X)dX (13.110)

Jo Jo

We change the limits for two reasons First, for physically realizable

circuits, h(t) is zero for t < 0 In other words, there can be no impulse

response before an impulse is applied Second, we start measuring

time at the instant the excitation x{t) is turned on; therefore x{t) = 0

for t < 0"

A graphic interpretation of the convolution integrals contained in

Eq 13.110 is important in the use of the integral as a computational

tool We begin with the first integral For purposes of discussion, we

assume that the impulse response of our circuit is the exponential decay

function shown in Fig 13.37(a) and that the excitation function has the

waveform shown in Fig 13.37(b) In each of these plots, we replaced t

with A, the symbol of integration Replacing A with -A simply folds the

excitation function over the vertical axis, and replacing -A with t — X

slides the folded function to the right See Figures 13.37(c) and (d).This

h{\)

A

0 (a)

x{\)

M

0 (b) -v(-X)

M

T i T i 0

(c)

.v(/ - \)

M

t—Tl t

(d)

h(k)x(

MA

0

t - \ ) y(t) = Area

(e)

Figure 13.37 A A graphic interpretation of the

convolution integral f' ] h(X)x{t - X)d\ (a) The

impulse response, (b) The excitation function, (c) The folded excitation function, (d) The folded excitation function displaced t units, (e) The product

h(X)x(t - A)

folding operation gives rise to the term convolution At any specified value of f, the response function y(t) is the area under the product

func-tion /j(A).v(f - A), as shown in Fig 13.37(e) It should be apparent from this plot why the lower limit on the convolution integral is zero and the

upper limit is t For A < 0, the product h{k)x(t - A) is zero because h{k)

is zero For A > r, the product h(X)x{t — A) is zero because x(t - A)

is zero

Figure 13.38 shows the second form of the convolution integral Note that the product function in Fig 13.38(e) confirms the use of zero for the

lower limit and t for the upper limit

Example 13.3 illustrates how to use the convolution integral, in conjunction with the unit impulse response, to find the response of

a circuit

(b)

/2(-0

-x)

A

(c)

h(t - X)

A

0

— ^>

t

(d)

h(t - X)x(X)

MA

y — >

^ 1

X

i >'(/) = Area

X

u (e) Figure 13.38 • A graphic interpretation of the

convolu-tion integral J {) h(t - \)x{X)d\ (a) The impulse

response, (b) The excitation function, (c) The folded

impulse response, (d) The folded impulse response

displaced / units, (e) The product h(t - A)x(A)

Example 13.3 Using t h e Convolution Integral t o Find an Output Signal

T h e excitation voltage v- t for the circuit shown in

Fig 13.39(a) is shown in Fig 13.39(b)

a) U s e the convolution integral to find v 0

b) Plot v a over the r a n g e of 0 ^ t ^ 15 s

20 V

(a)

5 10 (b)

f(s)

Figure 13.39 A The circuit and excitation voltage for

Example 13.3 (a) The circuit, (b) The excitation voltage

Impulse response

Figure 13.40 • The impulse response and the folded excitation function for Example 13.3

Solution

a) T h e first step in using the convolution integral is

to find the unit impulse response of the circuit

We obtain the expression for V () from the

s-domain equivalent of the circuit in Fig 13.39(a):

K = V,

s + 1

(1)-W h e n v t is a unit impulse function 5(f),

v„ = h(t) = e~'ii(t),

from which

//(A) = e' A u(A)

Using the first form of the convolution integral

in E q 13.110, we construct the impulse response

and folded excitation function shown in

Fig 13.40, which are helpful in selecting the

lim-its o n the convolution integral Sliding the

folded excitation function to the right requires

breaking the integration into three intervals:

0 < t < 5; 5 < t < 10; a n d 10 < t < oo T h e

breaks in the excitation function at 0 , 5 , and 10 s

dictate these break points Figure 13.41 shows

the positioning of the folded excitation for each

of these intervals T h e analytical expression for

Vi in the time interval 0 < / < 5 is

Vt = 4/, 0 < t < 5 s

/ i ( \ )

1.0

0

(t - 10) [t - 5) 0 t 5

^ (t - \)

20

5 s s r « 1 0

[t - 10) 0 (/ - 5) 5 t 10

v t (t - X)

20

10 =£ /=¾ x

0 ( / - 1 0 ) 5 ( / - 5 ) 1 0 / Figure 13.41 A The displacement of ?;,(/ - A) for three different time intervals

494 The Laplace Transform in Circuit Analysis

Hence, the analytical expression for the folded

excitation function in the interval t — 5 =£ A

< f is

Vi(t - A) = 4(r - A), t - 5 < A < t

We can now set up the three integral

expressions for v a For 0 ^ t < 5 s:

MV)

V n = 4(t - \)e~xdX

For 5 < t

= 4(e~' + f - 1)V

10 s,

v„ = 20e~xd\ + / 4(r - A)e_AdA

J f - 5

= 4(5 + e~ l - e'{'~5)) V

And for 10 < t < oo s

= 4(eT' - e"('-5) + 5e~ {t~m) V

,-A 20t>_ArfA + / 4 ( 7 - \)e AdX

10 i f - 5

Figure 13.42 A The voltage response versus time for

Example 13.3

b) We have computed v 0 for 1 s intervals of time,

using the appropriate equation The results arc

tabulated in Table 13.2 and shown graphically in

Fig 13.42

NOTE: Assess your understanding of convolution by trying Chapter Problems 13.62 and 13.63

TABLE 13.2

t

1

2

3

4

5

6

7

8

Numerical Values of v 0(t)

*0

1.47 4.54 8.20 12.07 16.03 18.54 19.56 19.80

t

9

10

11

12

13

14

15

v0

19.93 19.97 7.35 2.70 0.99 0.37 0.13

v,(t-k)

Future (will happen)

Past (has happened)

(t - 10) (/

Figure 13.43 • The past, present, and future values of

the excitation function

The Concepts of Memory and the Weighting Function

We mentioned at the beginning of this section that the convolution inte-gral introduces the concepts of memory and the weighting function into circuit analysis The graphic interpretation of the convolution integral is the easiest way to begin to grasp these concepts We can view the folding and sliding of the excitation function on a timescale characterized as past, present, and future The vertical axis, over which the excitation function

x(t) is folded, represents the present value; past values of x(t) lie to the

right of the vertical axis, and future values lie to the left Figure 13.43

shows this description of x(t) For illustrative purposes, we used the

exci-tation function from Example 13.3

When we combine the past, present, and future views of x(t — r) with

the impulse response of the circuit, we see that the impulse response

weights x(t) according to present and past values For example, Fig 13.41

shows that the impulse response in Example 13.3 gives less weight to past

values of x(t) than to the present value of x(t) In other words, the circuit retains less and less about past input values Therefore, in Fig 13.42, v a

quickly approaches zero when the present value of the input is zero (that

is, when t > 10 s) In other words, because the present value of the input

receives more weight than the past values, the output quickly approaches the present value of the input

The multiplication of x{t — A) by /z(A) gives rise to the practice of

referring to the impulse response as the circuit weighting function The

weighting function, in turn, determines how much memory the circuit has

Memory is the extent to which the circuit's response matches its input For

example, if the impulse response, or weighting function, is flat, as shown in

Fig 13.44(a), it gives equal weight to all values of x(t), past and present

Such a circuit has a perfect memory However, if the impulse response is

an impulse function, as shown in Fig 13.44(b), it gives no weight to past

values of x(t) Such a circuit has no memory Thus the more memory a

cir-cuit has, the more distortion there is between the waveform of the

excita-tion funcexcita-tion and the waveform of the response funcexcita-tion We can show this

relationship by assuming that the circuit has no memory, that is,

h(t) — A8(t), and then noting from the convolution integral that

/i(0

A8(X)x(t - \)dX

Equation 13.111 shows that, if the circuit has no memory, the output is a

scaled replica of the input

The circuit shown in Example 13.3 illustrates the distortion between input

and output for a circuit that has some memory This distortion is clear when we

plot the input and output waveforms on the same graph, as in Fig 13.45

13.7 The Transfer Function and the

Steady-State Sinusoidal Response

Once we have computed a circuit's transfer function, we no longer need to

perforin a separate phasor analysis of the circuit to determine its

state response Instead, we use the transfer function to relate the

steady-state response to the excitation source First we assume that

1.0

(a)

h(t)

1.0

(b)

Figure 13.44 • Weighting functions, (a) Perfect

mem-ory, (b) No memory

10 12 14

Figure 13.45 • The input and output waveforms for

Example 13.3

x(t) = A cos (<ot + (f>), (13.112)

and then we use Eq 13.96 to find the steady-state solution of y(t) To find

the Laplace transform of x(t), we first write x(t) as

x(t) = A cos lot cos 4> - A sin wt sin (/>, (13.113) from which

X{s) (A cos 4>)s (A sin (f>)(o

$ + co s + or Ais cos 4> - OJ sin 4>)