Electric Circuits, 9th Edition P70 pot

17.8 Parseval's Theorem 667 Example 17.4 Applying Parseval's Theorem to an Ideal Bandpass Filter The input voltage to an ideal bandpass filter is The total 1 Cl energy available at the

\F(o>)\ 2

•<y 2 ~ u> \ 0 CO ] <x> 2

Figure 17.11 • The graphic interpretation of Eq 17.68

l r2

W m = - \F(co)\ 2 da>

Note that expressing the integration in the frequency domain as

277 \F{to)\ 2 da)

instead of

- I \F(co)\ 2 dco

^ Jo

allows Eq 17.67 to be written in the form

W m = ~ \F(a>)\ 2 dco + — / \F{a>)\ 2 do>

(17.67)

(17.68)

Figure 17.11 shows the graphic interpretation of Eq 17.68

Examples 17.3-17.5 illustrate calculations involving Parseval's theorem

Example 17.3 Applying Parseval's Theorem

The current in a 40 Cl resistor is

i = 20e- 2l u(t) A

What percentage of the total energy dissipated in

the resistor can be associated with the frequency

bandO < w < 2 V 3 r a d / s ?

Solution

The total energy dissipated in the 40 il resistor is

W

mi 40 / 400e~./0 4' dt

16.0()0

- 4 = 4000 J

We can check this total energy calculation with

Parseval's theorem:

Therefore

F(co) =

\F(»)\ =

20

2 + jo)

20

VT +

oj-and

Warn = —

400

TV 4 + co 2

doo

16,000/1

77 tan ]—

8 0 0 0 Y ? 1 = 4000J

The energy associated with the frequency band

0 < o) < 2 V3 rad/s is

Wwu =

Hence the percentage of the total energy associated with this range of frequencies is

v

8000/3

4000 x 100 = 66.67%

17.8 Parseval's Theorem 667

Example 17.4 Applying Parseval's Theorem to an Ideal Bandpass Filter

The input voltage to an ideal bandpass filter is The total 1 Cl energy available at the output of

the filter is

v(t) = 120<T24'«(f) V

The filter passes all frequencies that lie between

24 and 48 rad/s, without attenuation, and completely

rejects all frequencies outside this passband

a) Sketch |V(o»)|2 for the filter input voltage

b) Sketch |K„(«)|2 f o r t h e f i l t e r output voltage

c) What percentage of the total 1 fi energy content

of the signal at the input of the filter is available

at the output?

1 r4 8 14,400 , 600 _, co

W 0 = — / ~ dco = tan l

48

24

600 / _< _ , , 600/ 7T TT\

= (tan l 2 tan H) = —

-7T 77 \2.84 4 /

61.45 J

The percentage of the input energy available at the output is

Solution

a) The Fourier transform of the filter input voltage is V = ^§- X 100 = 20.48%

V(a>) = 120

24 + j(o

Therefore

|K(a,)|2 = 14,400

576 + co 2 '

Fig 17.12 shows the sketch of |V(a>)|2 versus co

b) The ideal bandpass filter rejects all frequencies

outside the passband, so the plot of 1^(6))12

ver-sus co appears as shown in Fig 17.13

- 6 0

1

- 4 0

in

25-/10 /15

X 10

5

1

- 2 0 0

W )| 2

1

20

1

40

~~ 1

60 co (rad/s)

Figure 17.12 • |V(a/)|2 versus co for Example 17.4

c) The total 1 ft energy available at the input to the

filter is

W/ =

-IT Jo 576 + w

14,400 14.400 ( 1 , co

«w = - — t a n —

24 24

6 0 0 7T

7T 2 = 300 J

1

- 6 0

\V 0 (co)\ 2

25

20

15

- /

4 0

-10

-20 0

1

20

1

40

1

60 co (rad/s)

Figure 17.13 A | Vo(w)|2 versus co for Example 17.4

Example 17.5 Applying Parseval's Theorem to a Low-Pass Filter

Parseval's theorem makes it possible to calculate

the energy available at the output of the filter even

if we don't know the time-domain expression for

v a (t) Suppose the input voltage to the low-pass RC

filter circuit shown in Fig 17.14 is

v,(t) = 15<T5'«(f)V

a) What percentage of the 1 Cl energy available in

the input signal is available in the output signal?

b) What percentage of the output energy is

associ-ated with the frequency range 0 < co < 10 rad/s?

Solution

a) The 1 H energy in the input signal to the filter is

= 22.5 J

- 1 ( ¾

W(= / (15e-502rfr = 225——

Jo - 1 0

The Fourier transform of the output voltage is

V () (co) = VMH(wl

where

15

H(co) =

5 + jo) l/RC l/RC + jta ~ 10 + j(o

10

Hence

150

(5 + ;o>)(10 + jco)

22,500

2 \ '

(25 + 6/)(100 + OJ 1 )

The 1 H energy available in the output signal of

the filter is

W n = 1 22,500

irJo (25 + ^/)(100 + co z ) do)

10 kO + +

e

Figure 17.14 • The low-pass RC filter for Example 17.5

We can easily evaluate the integral by expand-ing the kernel into a sum of partial fractions:

(25 + w2)(100 + o) 2 ) 25 + w2 100 +

Then

K I./() 25 + o?

do)

300

5 V 2 / 10 V 2

0 100 + o) 1

= 15 J

The energy available in the output signal there-fore is 66.67% of the energy available in the input signal; that is,

V = 15 22.5 (100) = 66.67%

b) The output energy associated with the frequency

range 0 < w < 10 rad/s is

W' 300 r •in do)

n [ J0 25 + o) 2 J {) 100 + o) 1

10

3 0 0 / 1

\.5 13.64 J

- t a n " 1 5 J_ -] 10 A _ 30 /5 10 t a n 1 0 / " TT U 8 4 "" 4 2TT TT

The total 1 fi energy in the output signal is 15 J,

so the percentage associated with the frequency range 0 to 10 rad/s is 90.97%

The Energy Contained in a Rectangular Voltage Pulse

We conclude our discussion of Parseval's theorem by calculating the energy associated with a rectangular voltage pulse In Section 17.1 we found the Fourier transform of the voltage pulse to be

V{0)) = V„,T sin COT/2

17.8 Parseval's Theorem 669

To aid our discussion, we have redrawn the voltage pulse and its Fourier

transform in Fig 17.15(a) and (b), respectively These figures show that, as

the width of the voltage pulse (T) becomes smaller, the dominant portion

of the amplitude spectrum (that is, the spectrum from -2TT/T to 2TT/T)

spreads out over a wider range of frequencies This result agrees with our

earlier comments about the operational transform involving a scale

change, in other words, when time is compressed, frequency is stretched

out and vice versa To transmit a single rectangular pulse with reasonable

fidelity, the bandwidth of the system must be at least wide enough to

accommodate the dominant portion of the amplitude spectrum Thus the

cutoff frequency should be at least 2TT/T rad/s, or 1/T HZ

We can use Parseval's theorem to calculate the fraction of the total

energy associated with v(t) that lies in the frequency range 0 ^ co :£ 2TT/T

From Eq 17.69,

W = 1 f

2jr/T , sin2a>r/2

Vlr 2 - dco

(cor/2) 2

To carry out the integration called for in Eq 17.70, we let

COT

(17.70)

(17.71)

v(t)

- T / 2 0

(a)

T / 2

Figure 17.15 • The rectangular voltage pulse and its Fourier transform, (a) The rectangular voltage pulse,

(b) The Fourier transform of v (t)

noting that

dx = - dco

and that

If we make the substitutions given by Eqs 17.71-17.73, Eq 17.70 becomes

n Jo r

We can integrate the integral in Eq 17.74 by parts If we let

(17.74)

u = SHTJC

dx CIV = — r ,

X

(17.75) (17.76)

then

and

du = 2sin.v:cosx dx = sin2x dx, (17.77)

(17.78)

Hence

"• • 2 - 2

sin x , sin x

— T — d x =

0 X '

— sin 2x dx

rsin2.v ,

0 + I dx

o x

(17.79)

W-—l—dx (17.80)

To evaluate the integral in Eq 17.80, we must first put it in the form of

sin y/y We do so by letting y = 2x and noting that dy = 2 dx, and y = 2ir

when x = 77- Thus Eq 17.80 becomes

W = — — / — - dy (17.81)

77 Jo y

The value of the integral in Eq 17.81 can be found in a table of sine

integrals.1 Its value is 1.41815, so

2V 2 T

W = ——(1.41815) (17.82)

The total 1 O energy associated with v(t) can be calculated either from

the time-domain integration or the evaluation of Eq 17.81 with the upper

limit equal to infinity In either case, the total energy is

The fraction of the total energy associated with the band of frequencies

between 0 and 2TT/T is

W

_ 2I/?„7(1.41815)

= 0.9028 (17.84)

Therefore, approximately 90% of the energy associated with v(t) is

con-tained in the dominant portion of the amplitude spectrum

1 M Abramowitz and I Stegun, Handbook of Mathematical Functions (New York: Dover,

l%5),p.244

Practical Perspective 671

^ A S S E S S M E N T P R O B L E M S

Objective 3—Understand Parseval's theorem and be able to use it

17.8 The voltage across a 50 Cl resistor is

v = 4te~'u(t) V

What percentage of the total energy dissipated

in the resistor can be associated with the

fre-quency band 0 < a) ^ V 3 rad/s?

17.9 Assume that the magnitude of the Fourier

transform of v(t) is as shown This voltage is

applied to a 6 kfi resistor Calculate the total energy delivered to the resistor

\V(jco)\

-20007T 0 2000 it (o (rad/s)

Answer: 94.23%

NOTE: Also try Chapter Problem 17.4(1

Answer: 4 J

Practical Perspective

Filtering Digital Signals

To understand the effect of transmitting a digital signal on a telephone line,

consider a simple pulse that represents a digital value of 1, using 5 V, as shown

in Fig 17.15(a), with V m = 5 V and T = 1 fxs The Fourier transform of this

pulse is shown in Fig 17.15(b), where the amplitude V m r = 5 /JLM and the first

positive zero-crossing on the frequency axis is 2TT7T = 6.28 Mrad/s = 1 MHz

Note that the digital pulse representing the value 1 is ideally a sum of

an infinite number of frequency components But the telephone line cannot

transmit all of these frequency components Typically, the telephone has a

bandwidth of 10 MHz, meaning that it is capable of transmitting only those

frequency components below 10 MHz This causes the original pulse to be

distorted once it is received by the computer on the other end of the

tele-phone line, as seen in Fig 17.16

Figure 17.16 • The effect of sending a square voltage

pulse through a bandwidth-limited filter, causing

distor-tion of the resulting output signal in the time domain

Summary

The Fourier transform gives a frequency-domain

descrip-tion of an aperiodic time-domain funcdescrip-tion Depending on

the nature of the time-domain signal, one of three

approaches to finding its Fourier transform may be used:

• If the time-domain signal is a well-behaved pulse of

finite duration, the integral that defines the Fourier

transform is used

• If the one-sided Laplace transform of /(f) exists and

all the poles of F(s) lie in the left half of the s plane,

F(s) may be used to find F(a>)

• If /(f) is a constant, a signum function, a step

func-tion, or a sinusoidal funcfunc-tion, the Fourier transform is

found by using a limit process

(See page 646.)

Functional and operational Fourier transforms that are

useful in circuit analysis are tabulated in Tables 17.1 and

17.2 (See pages 655 and 660.)

• The Fourier transform of a response signal y(t) is

Y(a>) = X{(o)H(co),

where X(co) is the Fourier transform of the input signal

x(t), and H{co) is the transfer function H(s) evaluated at

s = jco (See page 660.)

• The Fourier transform accommodates both negative-time and positive-negative-time functions and therefore is suited

to problems described in terms of events that start at

f = - c o In contrast, the unilateral Laplace transform is suited to problems described in terms of initial

condi-tions and events that occur for t > 0

• The magnitude of the Fourier transform squared is a measure of the energy density (joules per hertz) in the frequency domain (Parseval's theorem).Thus the Fourier transform permits us to associate a fraction of the total energy contained in /(f) with a specified band of fre-quencies (See page 664.)

Problems

Sections 17.1-17.2

17.1 a) Find the Fourier transform of the function shown

inFig.P17.1

b) Find F{co) when w = 0

c) Sketch \F(O))\ versus co when A = 1 and T = 1

Hint: Evaluate \F(co)\ at co = 0,2, 4, 6,8, 9,10,

12,14, and 15.5 Then use the fact that |.F(ft>)| is

an even function of co

Figure PI7.1

w

17.2 The Fourier transform of /(f) is shown in Fig P17.2

a) Find/(f)

b) Evaluate/(0)

c) Sketch /(f) for - 1 0 < f < 10 s when A = 20IT

and COQ = 2 rad/s Hint: Evaluate /(f) at f = 0,

1, 2 , 3 , , 10 s and then use the fact that /(f)

is even

Figure P17.2

F(co)

Problems 673

17.3 Use the defining integral to find the Fourier

trans-form of the following functions:

17.11 Show that i f / ( 0 is an even function

a) / ( / ) = /4 sin | / ,

/ ( 0 = 0,

b) / ( 0 - ^ r * + A

2A

-2 < / < 2;

elsewhere

- - < / < 0;

2 / ( 0 = f + A 0 £ * < = - ;

/ ( / ) = 0, elsewhere

Sections 17.3-17.5

17.4 D e r i v e d {sin wo/}

17.5 Find the Fourier transform of each of the following

functions In all of the functions, a is a positive real

constant and — oo < / < oo

a) / ( 0 = I**4*1:

b) / ( / ) = / V l '1;

c) /(f) = e""1'1 cos »„/;

d) / ( / ) = e ^ ' s i n o ) , / ;

e) / ( / ) = § ( / - / „ )

17.6 If / ( / ) is a real function of /, show that the inversion

integral reduces to

l r

/(/) = — / [y4(a)) cos to/ — B(co) sin cot] da)

27TJ-00

17.7 If / ( / ) is a real, odd function of t, show that the

inversion integral reduces to

l r

/*(/) = - — / 5(a)) sin a)/ doo

2TT /_«,

17.8 Use the inversion integral (Eq 17.9) to show that

&~ ] {2/j<o} = sgn(f) Hint: Use Problem 17.7

17.9 Find S> { cos o)()/} by using the approximating function

/(0 = - - « w COS 0)()/, where e is a positive real constant

17.10 Show that if / ( / ) is an odd function,

A(co) = 0,

/l(w) = 2 / / ( / ) cos o)/rf/,

Jo

5(o>) = 0

Section 17.6

17.12 a) Show that &{df(t)/dt} = jcoF(co), where

F(o)) = ^ { / ( / ) } Hint: Use the defining integral

and integrate by parts

b) What is the restriction on / ( / ) if the result given

in (a) is valid?

c) Show that ?${d"f(t)/dt"} = (ja>) n F(a>), where F(a>) = 9{f(f)}

17.13 a) Show that

<;9 f(x)dx > =

where F(a>) = SF {/(#)}• Hint: Use the defining

integral and integrate by parts

b) What is the restriction on f(x) if the result given

in (a) is valid?

c) If f(x) = e~ ax u(x), can the operational

trans-form in (a) be used? Explain

17.14 a) Show that

* { / < * ) } - ^ ( f ) *>0

b) Given that f(at) = e"^ for a > 0, sketch

F(OJ) = ®{f(at)} for a = 0.5,1.0, and 2.0 Do

your sketches reflect the observation that

compression in the time domain corresponds to stretching in the frequency domain?

17.15 Derive each of the following operational transforms:

a) &{f(t - a)} = e- iioa F{OJ)\

b) 9{e**f(t)} = F(o) - o)());

C) &{f(t)cOS00 0 t} = \F((I) - O)0) + \F{00 + 0)(,)

17.16 Given

y(0 J-a

x(\)h(t - A) //A,

show that Y(o)) = &{y(()} = X(oo)H(w), where

X(OJ) = &{x(t)} and H(<*>) = &{h(t)} Hint: Use

the defining integral to write

B(m) = - 2 / / ( / ) sin cotdt

Jo

/.00 P /.00

9{y(t)} = / / x(X)h(t -A)//A

/ - 0 0 J - 0 0

<?-;'"" / / /

Next, reverse the order of integration and then

make a change in the variable of integration; that is,

let u = t — A

(I/277) / Fi{u)F 2 {oi - u) du Hint: First, use the

J-co

defining integral to express F(a>) as

Figure P17.20

F(o>) f fi(t)f 2 (t)e-l**dt

Second, use the inversion integral to write

27T ,/_oo

Third, substitute the expression for fi(t) into the

defining integral and then interchange the order

of integration

17.18 a) Show that

07 d n do' F((o) 1 nm)Y

b) Use the result of (a) to find each of the following

Fourier transforms (assuming a > 0):

^{|/|e~ a| "K

/ , ( 0 = cosa)()f,

/2( 0 = 1, -r/2<t<r/2;

/2( 0 = 0, elsewhere

a) Use convolution in the frequency domain to

b) What happens to F(a>) as the width of /2(f)

increases so that /(f) includes more and more

cycles on/j(f)?

Section 17.7

17.20 a) Use the Fourier transform method to find

psp i C E i 0 (t) in the circuit shown in Fig PI7.20 if

MULTISIM -, y- / s , r

v g = 36 sgn(f) V

b) Does your solution make sense in terms of

known circuit behavior? Explain

2/xF

P5PICE MULTISIM

MySL""M in the circuit shown in Fig P17.22 The initial

value of v () {t) is zero, and the source voltage is

100M(r) V

b) Sketch v () (t) versus t

Figure P17.22

5H

PSPICE changed to 100 sgn(f)

MULTISIM

in Fig P17.24 if i g = 200 sgn(f) /xA

b) Does your solution make sense in terms of known circuit behavior? Explain

PSPICE MULTISIM

Figure P17.24

-/,,(0

0.5 /AF

PSPICE

MULTISIM

PSPICE MULTISIM

17.26 The voltage source in the circuit in Fig P17.26 is

given by the expression

Vg = 3 sgn(f) V

a) Find vjf)

b) What is the value of v o (0~)?

c) What is the value of v o (0 + )?

d) Use the Laplace transform method to find v 0 (t)

for f > 0+ e) Does the solution obtained in (d) agree with

vJt) for f > 0+ from (a)?

Problems 675

Figure P17.26

0.5 n

250 mF

PSPICE

MULTISIM

17.28 a) Use the Fourier transform to find v (> in the

cir-PSPICE cuit in Fig PI 7.28 if L equals 3<TS|fl A

MULTISIM lS

b) Find 1^(0-)

c) Find?;o(0+)

d) Use the Laplace transform method to find v 0

for t > 0

e) Does the solution obtained in (d) agree with v 0

for t > ()+ from (a)?

Figure PI 7.28

0.1 F

"pice circuit in Fig P17.31 if v g = 125 cos 40,000f V

MULTISIM

b) Check the answer obtained in (a) by finding the

steady-state expression for i 0 using phasor domain analysis

Figure P17.31

5 m H

120 H

the circuit shown in Fig P17.32 The voltage source generates the voltage

v g = 45<r5l)W V

b) Calculate V o (0~), v o {0 + ), and v0(oo)

c) Find /L((r); /L(0+); i;c(0"); and v c (0 + )

d) Do the results in part (b) make sense in terms of known circuit behavior? Explain

PSPICE

HULTISIM

Figure P17.32

1 /xF

+ IV

PSPICE in Fig P17.28 if L equals 3e~5|f| A

MULTISIM

b) Find /„(()")

c) Find/o(0+)

d) Use the Laplace transform method to find i ()

for t > 0

e) Does the solution obtained in (d) agree with /„

for t > 0 + from (a)?

17.30 Use the Fourier transform method to find /'„ in the

PSPICE circuit in Fig P17.30 if v„ = 300 cos 5000r V

MULTTSIM

Figure P17.30

800 nF

17.33 The voltage source in the circuit in Fig P17.33 is

PSPICE generating the signal

MULTISIM

v g = 5 sgn(f) - 5 + 30e"5l«(0 V

a) Find ^,(0-) and v o (0 + )

b) Find ^ ( 0-) and i o (0 + )

c) Find v 0

Figure P17.33

5 a

-vw-I VVV—;—

y , { M l 0 0 m F

T

PSPICE

v g = 36e 4t u(-t) - 36e~ 41 u(t) V

b) Find v„(0")

c) Find v o (0 + )