666 The Fourier Transform \F(o>)\ 2 •<y 2 ~ u> \ 0 CO ] <x> 2 Figure 17.11 • The graphic interpretation of Eq. 17.68. l r 2 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 < 2V3rad/s? Solution The total energy dissipated in the 40 il resistor is W mi 40 / 400e~ 4 ' dt ./0 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 w 4() Warn = — 400 TV 4 + co 2 doo 16,000/1 77 tan ] — 8000 Y? 1 = 4000J. IT \2 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<T 24 '«(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 for the filter 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 r 48 14,400 , 600 _, co W 0 = — / ~ dco = tan l —- TrJ 2 A 576 + o) 2 v 24 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))1 2 ver- sus co appears as shown in Fig. 17.13. -60 1 -40 in 25- /10 /15 X 10 5 1 -20 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 = - —tan — 24 24 600 7T 7T 2 = 300 J. 1 -60 \V 0 (co)\ 2 25 20 15 -/ -40 - 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. 668 The Fourier Transform 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<T 5 '«(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 -5 0 2 rfr = 225—— Jo -10 The Fourier transform of the output voltage is V () (co) = VMH(wl where 15 VM = H(co) = 5 + jo) l/RC l/RC + jta ~ 10 + j(o 10 Hence W,M\ 2 = 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: 22,500 300 500 (25 + w 2 )(100 + o) 2 ) 25 + w 2 100 + Then W = do) 300 K I./() 25 + o? do) 300 5V2/ 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 [ J 0 25 + o) 2 J {) 100 + o) 1 10 300/1 \.5 13.64 J. -tan" 15 J_ -] 10 A _ 30 / 2TT TT 5 10 tan 10/ " TT U.84 "" 4 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 O)T/2 (17.69) 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 , sin 2 a>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 2 (17.72) and that x = 7T, when co = 2TT/T. (17.73) If we make the substitutions given by Eqs. 17.71-17.73, Eq. 17.70 becomes W = ^f^ rf , 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—dx = 0 X' — sin 2x dx 0 .70 x r sin2.v , 0 + I dx. o x (17.79) Substituting Eq. 17.79 into Eq. 17.74 yields 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 W, = V 2 m r. (17.83) 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 ^ASSESSMENT PROBLEMS 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) ^ V3 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. 672 The Fourier Transform Summary The Fourier transform gives a frequency-domain descrip- tion of an aperiodic time-domain function. 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 function, 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-time functions and therefore is suited to problems described in terms of events that start at f = -co. 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 -10 < 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 if/(0 is an even function. a) /(/) = /4 sin |/, /(0 = 0, b) /(0-^r* + A 2A -2 < / < 2; elsewhere. < / < 0; 2 /(0 = f + A 0 £*<=-; jW T 2 /(/) = 0, elsewhere. Sections 17.3-17.5 17.4 Derived {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) /(/) = /Vl' 1 ; c) /(f) = e"" 1 ' 1 cos »„/; d) /(/) = e^'sino),/; 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 ./-00 J-00 <?-;'"" ///. 674 The Fourier Transform Next, reverse the order of integration and then make a change in the variable of integration; that is, let u = t — A. 17.17 Given /(f) = fi(t)f 2 (t), show that F(<o) = (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 F((o) do' 1 nm)Y b) Use the result of (a) to find each of the following Fourier transforms (assuming a > 0): ^{|/|e~ a| "K 17.19 Suppose that /(f) = /i(f)/ 2 (f)* where /,(0 = cosa)()f, / 2 (0 = 1, -r/2<t<r/2; / 2 (0 = 0, elsewhere. a) Use convolution in the frequency domain to find F((D). 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 p s piCE 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 17.21 Repeat Problem 17.20 except replace /,,(0 with v a (t). P5PICE MULTISIM 17.22 a) Use the Fourier transform method to find v a (t) M y S L "" 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 17.23 Repeat Problem 17.22 if the input voltage (t? g ) is PSPICE changed to 100 sgn(f). MULTISIM 17.24 a) Use the Fourier transform to find i a in the circuit 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 17.25 Repeat Problem 17.24 except replace i 0 with v ( 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 17.27 Repeat Problem 17.26 except replace v t) {t) with /„(/), 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<T S|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 17.31 a) Use the Fourier transform method to find i a in the "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 5mH 120 H 17.32 a) Use the Fourier transform method to find v 0 in the circuit shown in Fig. P17.32. The voltage source generates the voltage v g = 45<r 5l) W V. b) Calculate V o (0~), v o {0 + ), and v 0 (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 - 4H- v„ ^800 0 17.29 a) Use the Fourier transform to find i a in the circuit PSPICE i n 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 OmH loo n 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,{Ml00mF T 17.34 a) Use the Fourier transform method to find v 0 in PSPICE MULnSIM the circuit in Fig. PI7.34 when v g = 36e 4t u(-t) - 36e~ 41 u(t) V. b) Find v„(0"). c) Find v o (0 + ).