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 /4 2 (w) + B 2 (co) —is an even func- tion of a). The phase angle of F(co) —that is, 0(<t>) = tan _1 5(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. If/(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)e JOJt dco = — / A(co)e l<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. /(0 = W„ Ua ~**-2pc-r) <On/2 0 2TT M- sin <*Hf/2\ //2 J — { M sin w ^/ 2 2TT I 0)(//2 (17.45) 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 transforms. 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 of/(/) 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>) + F 3 (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) 7 W 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 - w 0 ) + - 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 /(0 = /t(0/ 2 (0, 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 l M" >o 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 /"ASSESSMENT PROBLEMS 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 /(0 = -—t + A, < 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) 2A7 T — 8 \ t + - T \ 2 2A J 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, V m ii[t + T -\ -V m ii(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) _f co ! (x) T cor 1 - cos — 2 Answer: V(co) = V m r 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 Example 17.1 Using the Fourier Transform to Find the Transient Response Use the Fourier transform to find /,,(0 in the circuit shown in Fig. 17.9. The current source i g (t) is the signum function 20 sgn(0 A. Q 1 H Figure 17.9 • The circuit for Example 17.1. Evaluating K\ and K 2 gives Therefore *-?-». K, = ^ = -10. -4 7^ 10 10 I 0 (co) =- . yo> 4 + ya) The response is Solution The Fourier transform of the driving source is /» = ^{20sgn(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)) = I g (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 -oo and 0. The resistance in each branch governs how the -20 A divides between the two branches. In particular, one fourth of the -20 A appears in the i a branch; therefore i () is -5 for t < 0. When the current source jumps from -20 A to +20 A at ( = 0, i a 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 -10 (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 i H (t) = 50 cos 3f A. Use the Fourier transform method to find i a (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 I 0 (o)) is by the inversion integral: i 0 (t)=^{I 0 (co)} 50TT 2TT 8(aj -3)+ 8(a) + 3) 4 + j(o e iwl 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. /ACCCCCMCMT r> o n D I CMC 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-);and(j)^(0 + ). Answer: (a) 20/jw; (b) 4;V(5 + /»); (c) 80/(5 + /«); (d) $Qe~ 5t u(t) V; (e) -2 A; (f) 18 A; (g)8A; (h)8A; (i) OV; (j) 80 V. 17.7 The voltage source in the circuit shown is gen- erating the voltage v 8 = e l u(-t) + u(t) V. a) Use the Fourier transform method to find v„. b) Compute v„(0~), v a (0 + ), and v^ 00 )- Answer: (a) v a -e l u(-t)-—e-\i(t) + - 1 + -sgn(r)V; 1 1 1 (b) - V. - V. - V. w 4 4 3 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 function. 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)\ 2 dco. ^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 7-00 /(0 — / F((o)e f °*dw 277 ./-CO 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 )l0t dt The interior integral is F(-w), so Eq. 17.61 reduces to dm. (17.61) />oc 1 p.oo J_J 2 (t)dt = —jj(co)F(-co)dco. (17.62) 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 „00 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 = / e m dt + 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 la (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)\ 2 df 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