Electric Circuits, 9th Edition P69 ppsx

Hence a rectangular pulse in the frequency domain must be gen-erated by a time-domain function of the form sin t/t.. Differentiation The Fourier transform of the first derivative of/f i

The following observations about F(co) are pertinent:

The real part of F(co) — that is, A(co)—is an even function of co; in

other words, A(co) = A{—co)

The imaginary part of F(co) —that is, B(co)—is an odd function of co;

in other words, 5(w) = — B(—(o)

The magnitude of F(co) —that is, y /42(w) + B 2 (co) —is an even

func-tion of a)

The phase angle of F(co) —that is, 0(<t>) = tan_15(w)/yl((w)—is an

odd function of co

Replacing co by -co generates the conjugate of F(co); in other words,

F(-co) = F\co)

Hence, if / ( / ) is an even function, F(co) is real, and if / ( / ) is an odd

function, F(to) is imaginary I f / ( 0 is even, from Eqs 17.37 and 17.38,

A(co) =2 f(t) cos cot dt (17.40)

and

B{to) = 0 (17.41)

If / ( / ) is an odd function,

A(co) = 0 (17.42)

and

B(oo) = - 2 / / ( / ) sin cot dt (17.43)

We leave the derivations of Eqs 17.40-17.43 for you as Problems 17.10

and 17.11

If / ( / ) is an even function, its Fourier transform is an even function,

and if / ( / ) is an odd function, its Fourier transform is an odd function

Moreover, if / ( / ) is an even function, from the inverse Fourier integral,

/ ( / ) = — / F(co)eJOJt dco = — / A(co)el<0' dco

l r

= — / A(co)(coscot + j sin cot) dco 2lT LOO

= - / '

2TT 7-C

A(co) cos cot dco + 0

2 r

= — / A(co) cos cot dco (17.44)

2?r A)

17.6 Operational Transforms 657

Now compare Eq 17.44 with Eq 17.40 Note that, except for a factor of

1/2TT, these two equations have the same form Thus, the waveforms of

A(a)) and / ( / ) become interchangeable if / ( / ) is an even function For

example, we have already observed that a rectangular pulse in the time

domain produces a frequency spectrum of the form (sin OJ)/CO. Specifically,

Eq 17.11 expresses the Fourier transform of the voltage pulse shown in

Fig 17.1 Hence a rectangular pulse in the frequency domain must be

gen-erated by a time-domain function of the form (sin t)/t We can illustrate

this requirement by finding the time-domain function / ( / ) corresponding

to the frequency spectrum shown in Fig 17.8 From Eq 17.44,

A{co)

M

-<u„/2 0 wo/2

Figure 17.8 A A rectangular frequency spectrum

<On/2

0

2TT M- sin <*Hf/2\

//2 J

— { M sin w^/2

We say more about the frequency spectrum of a rectangular pulse in

the time domain versus the rectangular frequency spectrum of (sin/)//

after we introduce Parseval's theorem

17.6 Operational Transforms

Fourier transforms, like Laplace transforms, can be classified as functional

and operational So far, we have concentrated on the functional

trans-forms We now discuss some of the important operational transtrans-forms With

regard to the Laplace transform, these operational transforms are similar

to those discussed in Chapter 12 Hence we leave their proofs to you as

Problems 17.12-17.19

Multiplication by a Constant

From the defining integral, if

94/(0} = F(«>),

then

&{Kf(t)} = KF(a>) (17.46)

Thus, multiplication o f / ( / ) by a constant corresponds to multiplying F(OJ)

by that same constant

Addition (Subtraction)

Addition (subtraction) in the time domain translates into addition

(sub-traction) in the frequency domain Thus if

94/2(0} = ^2(0)),

then

9{fi(t) - f 2 (t) + f 3 (t)} = F,(») - F 2 (a>) + F3(o,), (17.47)

which is derived by substituting the algebraic sum of time-domain

func-tions into the defining integral

Differentiation

The Fourier transform of the first derivative of/(f) is

®l^-\ = j(oF(u>) (17.48)

The nth derivative of /(f) is

d"f(t)

Equations 17.48 and 17.49 are valid if /(f) is zero at ±00

Integration

if

then

g(t)= / f(x)dx,

F((o) 9{g(t)\ = - M (17.50)

7W Equation 17.50 is valid if

f f(x)dx = 0

J-00

Scale Change

Dimensionally, time and frequency are reciprocals Therefore, when time

is stretched out, frequency is compressed (and vice versa), as reflected in

the functional transform

^{/(«0} = ~F(~\ a > °- (17-51)

Note that when 0 < a < 1.0, time is stretched out, whereas when a > 1.0,

time is compressed

Translation in the Time Domain

The effect of translating a function in the time domain is to alter the phase

spectrum and leave the amplitude spectrum untouched Thus

®{f(t - «)} = e-> w "F(aj) (17.52)

If a is positive in Eq 17.52, the time function is delayed, and if a is

nega-tive, the time function is advanced

Translation in the Frequency Domain

Translation in the frequency domain corresponds to multiplication by the

complex exponential in the time domain:

9{e**f(t)} =H<0 -<*>)• (17.53)

Modulation

Amplitude modulation is the process of varying the amplitude of a

sinu-soidal carrier If the modulating signal is denoted f(t), the modulated

car-rier becomes f(t) cos w(/.The amplitude spectrum of this carrier is one-half

the amplitude spectrum of /(f) centered at ±a) lh that is,

??{f(t) cos co()t} = - F(w - w0) + - F(co + w()) (17.54)

Convolution in the Time Domain

Convolution in the time domain corresponds to multiplication in the

fre-quency domain In other words

y(t) = J x(X)h(t - A) dk

becomes

®{y(t)\ = Y((o) = X(<o)H(a>) (17.55)

Equation 17.55 is important in applications of the Fourier transform,

because it states that the transform of the response function Y(o)) is the

product of the input transform X((o) and the system function H(co) We

say more about this relationship in Section 17.7

Convolution in the Frequency Domain

Convolution in the frequency domain corresponds to finding the Fourier

transform of the product of two time functions Thus if

then

F((o) = — / FI(M)F-,(6J - u) du, (17.56)

27T /_c

Table 17.2 summarizes these ten operational transforms and another

operational transform that we introduce in Problem 17.18

TABLE 17.2 Operational Transforms

/(')

Kf{t)

/ i ( 0 ~ flit)

d"f(t)/dt"

[ f(x)dx

«/-00

/(*)

fit - a)

e*"f(t)

f(t) COS O) 0 t

/

x{K)h{t-/1(0/2(0

/7(0

+ /3(0

A)rfA

F(a>) KF((o) Fi((o) - F 2 (a>) + F 3 ((o) (JoTFUo)

F((o)/jco

e-> m F{a)) Fio) - o>0 )

-F(u) - (o () ) + -F(o) + too)

X((o)H{to)

1 r

— / F[(u)F->(o) — u)du

2lT J-00 d n F((o) (i)"

17.7 Circuit Applications 661

/ " A S S E S S M E N T P R O B L E M S

Objective 1—Be able to calculate the Fourier transform of a function

17.4 Suppose f(t) is defined as follows:

/(0 2A t + A,

2A

- - < t < 0,

2

0 < t =s

2 ' /(r) = 0, elsewhere

a) Find the second derivative of /(f)

b) Find the Fourier transform of the second

derivative

c) Use the result obtained in (b) to find the

Fourier transform of the function in (a)

(Hint: Use the operational transform of

differentiation.)

Answer: (a) 2 A 7 T

—8 \ t +

+ — 8 { t

-r V 2

—8(t)

17.5 The rectangular pulse shown can be expressed

as the difference between two step voltages;

that is,

Vmii[t + T-\ -Vmii(t- T-)W

Use the operational transform for translation in the time domain to find the Fourier transform

of v{t)

v(t)

- T / 2 0 r/2

., , 4A ( (OT

( b )_ fc o !

(x) T

cor

1 - cos —

2 Answer: V(co) = V mr

sin(wr/2)

K/2)

NOTE: Also try Chapter Problem 17.19

17.7 Circuit Applications

The Laplace transform is used more widely to find the response of a

cir-cuit than is the Fourier transform, for two reasons First, the Laplace

trans-form integral converges for a wider range of driving functions, and second,

it accommodates initial conditions Despite the advantages of the Laplace

transform, we can use the Fourier transform to find the response The

fun-damental relationship underlying the use of the Fourier transform in

tran-sient analysis is Eq 17.55, which relates the transform of the response

Y((o) to the transform of the input X(o)) and the transfer function H(a)) of

the circuit Note that H (o>) is the familiar H(s) with s replaced by jo)

Example 17.1 illustrates how to use the Fourier transform to find the

response of a circuit

662 The Fourier Transform

Use the Fourier transform to find /,,(0 in the circuit

shown in Fig 17.9 The current source ig(t) is the

signum function 20 sgn(0 A

Q

1 H

Figure 17.9 • The circuit for Example 17.1

Evaluating K\ and K2 gives

Therefore

*-?-»

K, = ^ = - 1 0

- 4

7 ^ 1 0 1 0

I0(co) = -

yo> 4 + ya) The response is

Solution

The Fourier transform of the driving source is

/ » = ^ { 2 0 s g n ( 0 }

( —

40

jto'

The transfer function of the circuit is the ratio of I,

to L; so

//(w) = ^ =

k 4 + /w

The Fourier transform of /,,(0 is

/,(0)) = Ig(co)H(w)

40 /o>(4 + ;'&>) Expanding /„(ft>) into a sum of partial fractions yields

/ » = — +

/,(0 = ®-[[i<M)

= 5sgn(r) - 10e_4 '//(0-Figure 17.10 shows the response Does the solu-tion make sense in terms of known circuit behav-ior? The answer is yes, for the following reasons The current source delivers —20 A to the circuit between - o o and 0 The resistance in each branch governs how the - 2 0 A divides between the two branches In particular, one fourth of the - 2 0 A

appears in the i a branch; therefore i () is - 5 for t < 0

When the current source jumps from - 2 0 A to

+20 A at ( = 0, ia approaches its final value of +5 A exponentially with a time constant of \ s

An important characteristic of the Fourier transform is that it directly yields the steady-state response to a sinusoidal driving function The

rea-son is that the Fourier transform of cos OJ{)( is based

on the assumption that the function exists over all time Example 17.2 illustrates this feature

5 sgn(r)

i 0

Ut)

5

0

- 1 0

( A ) 5 sgn(r)

^•^c

'y^r*

/ft) 4 + jo) Figure 17.10 • The plot of /,,(0 versus t

17.7 Circuit Applications 663

Example 17.2 Using the Fourier Transform to Find the Sinusoidal Steady-State Response

The current source in the circuit in Example 17.1

(Fig 17.9) is changed to a sinusoidal source The

expression for the current is

iH(t) = 50 cos 3f A

Use the Fourier transform method to find ia(t)

Solution

The transform of the driving function is

I g ((o) = 50TT[8(O) - 3 ) + 8(io + 3)]

As before, the transfer function of the circuit is

H(co) = - - ^ -

4 + )o>

The transform of the current response then is

8(a) - 3 ) + 8(a> + 3) 1„(OJ) =

50ir-4 + j(D

Because of the sifting property of the impulse func-tion, the easiest way to find the inverse transform of

I0(o)) is by the inversion integral:

i 0 (t)=^{I 0 (co)}

50TT

2TT

8(aj - 3 ) + 8(a) + 3)

4 + j(o eiwl dco

= 25

= 25

,pt -j si

+

4 + /3 4

,,/3/-y36.87

-I

/3,

5 5

= 5[2cos(3/ - 36.87°)]

= 10cos(3/ - 36.87°)

We leave you to verify that the solution for /,,(7) is identical to that obtained by phasor analysis

/ A C C C C C M C M T r> o n D I C M C

Objective 2—Know how to use the Fourier transform to find the response of a circuit

17.6 The current source in the circuit shown delivers

a current of 10 sgn (r) A The response is the

voltage across the 1 H inductor Compute

(a) 1,(0)); (b) ff(/»); (c) V » ; (d) v a (t);

(e) *,(0T); (f) *i(0+); (g) /2(0"); (h) *2(0+);

( i ) ^ , ( 0 - ) ; a n d ( j ) ^ ( 0+)

Answer: (a) 20/jw;

(b) 4 ; V ( 5 + / » ) ;

(c) 80/(5 + / « ) ;

(d) $Qe~5tu(t) V;

(e) - 2 A;

(f) 18 A;

( g ) 8 A ;

( h ) 8 A ; (i) OV;

(j) 80 V

17.7 The voltage source in the circuit shown is

gen-erating the voltage

v8 = elu(-t) + u(t) V

a) Use the Fourier transform method to find v„ b) Compute v„(0~), v a (0 + ), and v ^0 0)

-Answer: (a) v a elu(t)—e\i(t) +

-1 + - s g n ( r ) V ;

1 1 1 (b) - V - V - V

NOTE: Also try Chapter Problems 17.20,17.28, and 17.30

17,8 Parseval's Theorem

ParsevaFs theorem relates the energy associated with a time-domain

func-tion of finite energy to the Fourier transform of the funcfunc-tion Imagine that

the time-domain function /(f) is either the voltage across or the current in

a 1 £1 resistor The energy associated with this function then is

W m = f{t)dt (17.57)

ParsevaFs theorem holds that this same energy can be calculated by an

integration in the frequency domain, or specifically,

[ f(t)dt = ^-[ f{t)dt = — \F(w)\2dco

^TT J-C

(17.58)

Therefore the 1 il energy associated with /(f) can be calculated either by

integrating the square of f(t) over all time or by integrating 1/2TT times

the square of the magnitude of the Fourier transform of/(f) over all

fre-quencies Parseval's theorem is valid if both integrals exist

The average power associated with time-domain signals of finite

energy is zero when averaged over all time Therefore, when comparing

signals of this type, we resort to the energy content of the signals Using a

1 £1 resistor as the base for the energy calculation is convenient for

com-paring the energy content of voltage and current signals

We begin the derivation of Eq 17.58 by rewriting the kernel of the

integral on the left-hand side as /(f) times itself and then expressing one

/(f) in terms of the inversion integral:

/

OO /.00

f\t)dt= / f(t)f(t)dt

OO 7 - 0 0

/(0 — / F((o)ef°*dw

277 / - C O

dt (17.59)

We move /(f) inside the interior integral, because the integration is with

respect to w, and then factor the constant 1/277- outside both integrations

Thus Eq 17.59 becomes

.oo r /-oo

/2(f)</f = - - / / F(m)f(t)e&da

oo ^7T J—oo L J-oo

dt (17.60)

We reverse the order of integration and in so doing recognize that F(ot))

can be factored out of the integration with respect to f Thus

1

f\t)dt = -~-f F(co)

3 277 J-c

f(t)e)l0tdt

The interior integral is F ( - w ) , so Eq 17.61 reduces to

dm (17.61)

/>oc 1 p.oo

In Section 17.6, we noted that F(—<o) = F"(<y) Thus the product F(co)F(-io) is simply the magnitude of F(a>) squared, and Eq 17.62 is

equivalent to Eq 17.58 We also noted that |F(w)| is an even function of w Therefore, we can also write Eq 17.58 as

„ 0 0

f\t) dt = - I \F(oj)\ 2 do> (17.63)

A Demonstration of Parseval's Theorem

We can best demonstrate the validity of Eq 17.63 with a specific example If

f(t) = e""1'1, the left-hand side of Eq 17.63 becomes

-2aM,/, =

dt = / emdt + 2at 2a! dt J—oc

e 2at

la

1

la

0

+

—oc

1 + —

la

./o

e ~2at

—la

1

a (17.64)

la The Fourier transform of f(t) is

F<«) =

(T + (O

and therefore the right-hand side of Eq 17.63 becomes

71 Jo (a + a)-)

4a 2 _ 4a 2 1 fc) 1 _,ft)

» r H— tan —

- - ( o + f •

77- V l a

(17.65)

Note that the result given by Eq 17.65 is the same as that given by Eq 17.64

The Interpretation of Parseval's Theorem

Parseval's theorem gives a physical interpretation that the magnitude of the Fourier transform squared, |F(a>)|2, is an energy density (in joules per hertz).To see it, we write the right-hand side of Eq 17.63 as

77

- / \F(l7rf)\ 2 l7rdf = l \F(2irf)\ 2 df, (17.66) where \F(lirf)\2df is the energy in an infinitesimal band of frequencies (df), and the total 1 0 energy associated with f(t) is the summation

(inte-gration) of |F(27r/)|2<:/f over all frequencies We can associate a portion

of the total energy with a specified band of frequencies In other words,

the I fi energy in the frequency band from OD\ to a)2 is