Electric Circuits, 9th Edition P68 ppt

646 The Fourier Transform 17.1 The Derivation of the Fourier Transform We begin the derivation of the Fourier transform, viewed as a limiting case of a Fourier series, with the exponential form of the series: 00 f(0 = 2 C n e»'«»\ (17.1) n oa where 1 f r/2 C n = = / f(t)e-'"^ dt. (17.2) * J-T/2 In Eq. 17.2, we elected to start the integration at t 0 = -7/2. Allowing the fundamental period T to increase without limit accom- plishes the transition from a periodic to an aperiodic function. In other words, if T becomes infinite, the function never repeats itself and hence is aperiodic. As T increases, the separation between adjacent harmonic fre- quencies becomes smaller and smaller. In particular, Aw = {n + 1)0)() - HWQ = o>() = —, (17.3) and as 7 gets larger and larger, the incremental separation &co approaches a differential separation dxo. From Eq. 17.3, 1 doj „ T^ZTT" ^ 7- ^ 00, (17 ' 4 ) As the period increases, the frequency moves from being a discrete vari- able to becoming a continuous variable, or nco 0 ^u) as 7—»oo. (17.5) In terms of Eq. 17.2, as the period increases, the Fourier coefficients C„ get smaller. In the limit, C„—»0 as T—> oo. This result makes sense, because we expect the Fourier coefficients to vanish as the function loses its periodicity. Note, however, the limiting value of the product C„T; that is, C„T~* f(t)e~ jwt dt asr-*oo. (17.6) In writing Eq. 17.6 we took advantage of the relationship in Eq. 17.5. The integral in Eq. 17.6 is the Fourier transform of /(f) and is denoted / 00 f(t)e- ja " dt. (17.7) 00 17.1 The Derivation of the Fourier Transform 647 We obtain an explicit expression for the inverse Fourier transform by investigating the limiting form of Eq. 17.1 as T—» DO. We begin by multi- plying and dividing by T: /w= f^ c « r )^(f) (17.8) As 7*—»oo, the summation approaches integration, C n T —* F(co), n<x) {) —> OJ, and 1/7 —* dco/2TT. Thus in the limit, Eq. 17.8 becomes m 2ir F{co)e io)t dco. (17.9) -4 Inverse Fourier transform Equations 17.7 and 17.9 define the Fourier transform. Equation 17.7 trans- forms the time-domain expression f(t) into its corresponding frequency- domain expression F(co). Equation 17.9 defines the inverse operation of transforming F(co) into/(f). Let's now derive the Fourier transform of the pulse shown in Fig. 17.1. Note that this pulse corresponds to the periodic voltage in Example 16.6 if we let T —* oo.The Fourier transform of v(t) comes directly from Eq. 17.7: V(co) r/2 V „£->*" lit -r/2 (,->">' = v, (-jio) r/2 -r/2 o(t) -T/2 0 r/2 Figure 17.1 • A voltage pulse. V m ( n . • 0)T —— -2} sin — -jot \ 2 (17.10) which can be put in the form of (sin x)/x by multiplying the numerator and denominator by T.Then, V(co) = V m r sin COT/2 COT/2 (17.11) For the periodic train of voltage pulses in Example 16.6, the expression for the Fourier coefficients is C. V„,T sin nco () T/2 T nco{)T/2 (17.12) Comparing Eqs. 17.11 and 17.12 clearly shows that, as the time-domain function goes from periodic to aperiodic, the amplitude spectrum goes from a discrete line spectrum to a continuous spectrum. Furthermore, the envelope of the line spectrum has the same shape as the continuous spec- trum. Thus, as T increases, the spectrum of lines gets denser and the ampli- tudes get smaller, but the envelope doesn't change shape. The physical interpretation of the Fourier transform V{co) is therefore a measure of the frequency content of v(t). Figure 17.2 illustrates these observations. The amplitude spectrum plot is based on the assumption that r is constant and T is increasing. «w 0 C„ o.i v„ -4TT/T —2TT/T Ji u^_ 2TT/T Tvn^p (c) Figure 17.2 • Transition of the amplitude spectrum as/(/) goes from periodic to aperiodic, (a) C„ versus /zw () , JT/T = 5; (b) C„ versus nw () , T/T = 10; (c) V(a>) versus to. 17.2 The Convergence of the Fourier Integral A function of time /(0 has a Fourier transform if the integral in Eq. 17.7 converges. If f(t) is a well-behaved function that differs from zero over a finite interval of time, convergence is no problem. Well-behaved implies that /(0 is single valued and encloses a finite area over the range of inte- gration. In practical terms, all the pulses of finite duration that interest us are well-behaved functions. The evaluation of the Fourier transform of the rectangular pulse discussed in Section 17.1 illustrates this point. If /(f) is different from zero over an infinite interval, the convergence of the Fourier integral depends on the behavior of /(0 as f—>oo. A single-valued function that is nonzero over an infinite interval has a Fourier transform if the integral [/(01 dt 17.2 The Convergence of the Fourier Integral 649 exists and if any discontinuities in /(0 are finite. An example is the decaying exponential function illustrated in Fig. 17.3. The Fourier trans- form of /(0 is F(co) = J f{t)e~ Jto ' dt = Ke- a 'e- ]a * dt fc e -(a+jco)t (a + jco) K o -(a + jw) K (0-1) a + ju) . a > 0. (17.13) Figure 17.3 • The decaying exponential function Ke- ( "u(t). A third important group of functions have great practical interest but do not in a strict sense have a Fourier transform. For example, the integral in Eq. 17.7 doesn't converge if/(0 is a constant. The same can be said if /(0 is a sinusoidal function, cos oj Q t, or a step function, Ku{t). These func- tions are of great interest in circuit analysis, but, to include them in Fourier analysis, we must resort to some mathematical subterfuge. First, we create a function in the time domain that has a Fourier transform and at the same time can be made arbitrarily close to the function of interest. Next, we find the Fourier transform of the approximating function and then evaluate the limiting value of F(w) as this function approaches f{t). Last, we define the limiting value of F(co) as the Fourier transform of f(t). Let's demonstrate this technique by finding the Fourier transform of a constant. We can approximate a constant with the exponential function /(0 = Ae- £{ ' 1 , e > 0. (17.14) As e —* 0, /(0 —* A. Figure 17.4 shows the approximation graphically. The Fourier transform of /(0 is F{(o) = / Ae a e~ joyl dt + / Ae^e'^ dt. (17.15) Carrying out the integration called for in Eq. 17.15 yields „, , A A 2eA F(to) = — + — = —. (17.16) e - j(o e + a) e 2 + a) 2 f(t) A£^^^ Ae^*^^- 0 e 2 < e l A -^^Ae^l ^^_A£2* Figure 17.4 A The approximation of a constant with an exponential function. The function given by Eq. 17.16 generates an impulse function at w = 0 as e —>0. You can verify this result by showing that (1) F(co) approaches infinity at co = 0 as € —» 0; (2) the width of F(eo) approaches zero as e —>• 0; and (3) the area under F(w) is independent of e. The area under F((o) is the strength of the impulse and is 2eA e L + (o l do) = 4eA dco o e 2 + a 2 = 277-/1. (17.17) 650 The Fourier Transform In the limit, f(t) approaches a constant A, and F(co) approaches an impulse function 2TTA8((O). Therefore, the Fourier transform of a constant A is defined as 2TTA8(co), or &{A) = 2TTA8((O). (17.18) In Section 17.4, we say more about Fourier transforms defined through a limit process. Before doing so, in Section 17.3 we show how to take advantage of the Laplace transform to find the Fourier transform of functions for which the Fourier integral converges. /ASSESSMENT PROBLEMS Objective 1—Be able to calculate the Fourier transform of a function 17.1 Use the defining integral to find the Fourier 17.2 The Fourier transform of f(t) is given by transform of the following functions: _,, . 6 F(o)) = 0, -oo < a) < -3; a) f(t) = -A, -r/2 < t < 0; F((o) = ^ _ 3 < <, < _ 2; f{t) « A 0<f<r/2; F(C) = 1, -2<O,<2; fit) = 0 elsewhere. F(ft)) m 4> 2 < co < 3; b) fit) = 0, t < 0; F{a)) = 0i 3 < w < oo. / . /2A\f. O>T\ Answer: (a) -/I — II 1 - cos— I; 1 1 (b) —«. Answer: f(?) = —(4 sin 3t - 3 sin 2f). (rt + JO))" TTt NOTE: Also try Chapter Problems 17.1 and 17.2. 17.3 Using Laplace Transforms to Find Fourier Transforms We can use a table of unilateral, or one-sided, Laplace transform pairs to find the Fourier transform of functions for which the Fourier integral con- verges. The Fourier integral converges when all the poles of F(s) lie in the left half of the s plane. Note that if F(s) has poles in the right half of the s plane or along the imaginary axis,/(/) does not satisfy the constraint that Xoo 1/(01* exists. The following rules apply to the use of Laplace transforms to find the Fourier transforms of such functions. 1. If /(/) is zero for t ^ 0~, we obtain the Fourier transform of f(t) from the Laplace transform of /(f) simply by replacing 5 by jco. Thus *{/(0) «2{/(0W ( 17 - 19 ) 17.3 Using Laplace Transforms to Find Fourier Transforms 651 For example, say that f(0 = 0, t ^ 0~ fit) = e " f cosw ( /, 0 + . Then *{/(/)} s + a jco + a (s + a) 2 + (nl x =j lo (jco + a) 2 + (x)f) 2. Because the range of integration on the Fourier integral goes from — oo to +oo, the Fourier transform of a negative-time function exists. A negative-time function is nonzero for negative values of time and zero for positive values of time. To find the Fourier trans- form of such a function, we proceed as follows. First, we reflect the negative-time function over to the positive-time domain and then find its one-sided Laplace transform. We obtain the Fourier trans- form of the original time function by replacing s with -jco. Therefore, when f(t) = 0 for t > 0 + , Hf(*)} = -^1/(-01.=- ;w (17.20) For example, if fit) = 0, (for t > 0 + ); then /(/) = Aosfttf, (for / < 0"). /(-0 - o. (for t < 0"); /(-0 = <T'"cosw ( /, (for t > 0 + ). Both /(/') and its mirror image are plotted in Fig. 17.5. The Fourier transform of /(/) is 9{f{t)} = W(~0h= /a -jco + a s + a (s + a) + COQ s=- /w (-jo + a) 2 + &»§ Figure 17.5 • The reflection of a negative-time function over to the positive-time domain. 3. Functions that are nonzero over all time can be resolved into positive- and negative-time functions. We use Eqs. 17.19 and 17.20 to find the Fourier transform of the positive- and negative-time functions, respec- tively. The Fourier transform of the original function is the sum of the two transforms. Thus if we let then and A/) =/(0 (for r>0), /-(/) = /(0 (for/<0), /(o = no + /"(o = ^{/ + (0},=ya> + ie{/-(-0}.v=- ;w . (17.21) An example of using Eq. 17.21 involves finding the Fourier trans- form of e _a ''L For the original function, the positive- and negative- time functions are / + (0 = e~ ul and /"(/) = e'". Then 2{/ + (0} i S + a 1 s + a Therefore, from Eq. 17.21, &{ e -<to\} 1 s + a 1 + S= I (it + s + a 1 s=—ju) jo) + a —jo) + a 2a 2 '< 2" o) + a If/(0 i s even, Eq. 17.21 reduces to 9{f(*)) = %{f(t)}.s= jt o + ^{/(01,=-/0,- If/(/) is odd, then Eq. 17.21 becomes 9{f{t)} = %{f(t)} s=ia - 2{/(/)},~/ (17.22) (17.23) 17.4 Fourier Transforms in the Limit 653 /ASSESSMENT PROBLEM Objective 1—Be able to calculate the Fourier transform of a function 17.3 Find the Fourier transform of each function. In each case, a is a positive real constant. a) f{t) = 0, fit) = e "'smcoyt, b) f(t) = 0, /(0 = -te at , c) f(t) = te- (,t , /(0 = to", t < 0, t > 0. t > 0, t < 0. t > 0, t s 0. Answer: (a) (b) OiQ (c) (a + jco) 2 + col' 1 (a - jco) 2, -j4aa) NOTE: Also try Chapter Problem 17.5. 17A Fourier Transforms in the Limit As we pointed out in Section 17.2, the Fourier transforms of several prac- tical functions must be defined by a limit process. We now return to these types of functions and develop their transforms. {a 1 + <«/) 2\2' The Fourier Transform of a Signum Function We showed that the Fourier transform of a constant A is 2TTA8((O) in Eq. 17.18. The next function of interest is the signum function, defined as + 1 for / > 0 and -1 for t < O.The signum function is denoted sgn(0 and can be expressed in terms of unit-step functions, or sgn(0 - u(t) - u(-t). (17.24) Figure 17.6 shows the function graphically. To find the Fourier transform of the signum function, we first create a function that approaches the signum function in the limit: sgn(r) = lim[e~ €t u(t) - e et u(-t)], e > 0. £->() (17.25) The function inside the brackets, plotted in Fig. 17.7, has a Fourier trans- form because the Fourier integral converges. Because /(0 is an odd func- tion, we use Eq. 17.23 to find its Fourier transform: 9^(/(0} 1 S + € 1 1 s=)to S + € 1 s=—]u> joo + e —jco + e _ -2jco a) 2 + e 2 ' As e -> 0, /(0 -» sgn(0, and ^{/(0} ~* 2/>. Therefore, ^{sgn(0} = —• (17.26) sgn(/) 1.0 -1.0 Figure 17.6 • The signum function. eM-t) -1.0 (17 27) Rw* 17.7 A A function that approaches sgn(r) as e approaches zero. The Fourier Transform of a Unit Step Function To find the Fourier transform of a unit step function, we use Eqs. 17.18 and 17.27. We do so by recognizing that the unit-step function can be expressed as "(') = - + -sgn(f). (17.28) Thus, ^2 ( + H^ s § n ( r ) = TT8(O)) + —. (17.29) 7*> The Fourier Transform of a Cosine Function To find the Fourier transform of cos o) () t, we return to the inverse-transform integral of Eq. 17.9 and observe that if F(<o) = 2TT8(O) - a>o), (17.30) then l r /(?) = — / \2TTS{O) - o) {) )]e"°' do). (17.31) Using the sifting property of the impulse function, we reduce Eq. (17.31) to f(t) = e M>r . (17.32) Then, from Eqs. 17.30 and 17.32, <${e<^} = 2ir8(a)-a) {) ). (17.33) We now use Eq. 17.33 to find the Fourier transform of cos w () r, because e hvf + e ~m\t cos ay = . (17.34) Thus, ^{cosa> 0 /} =-(3=(^} + ^{e' j0 ^}) — [2TT8((O — wo) + 2TT8((O + £%)] = TT8((O — ojo) + TT8(O) + G>o). (17.35) 17.5 Some Mathematical Properties 655 The Fourier transform of sin co 0 t involves similar manipulation, which we leave for Problem 17.4. Table 17.1 presents a summary of the transform pairs of the important elementary functions. We now turn to the properties of the Fourier transform that enhance our ability to describe aperiodic time-domain behavior in terms of frequency- domain behavior. TABLE 17.1 Fourier Transforms of Elementary Functions Type impulse constant signum step positive-time exponential negative-time exponential positive- and negative-time exponential complex exponential cosine sine /(') 8(t) A sgn(/) u(t) e- in u{t) e m u(-t) e -«\'\ e/w COS 0)()1 sin corf F(co) I 2TTA8(CO) 2//oi 7T8((X)) + 1//(1) l/(n + /ftj), a > 0 l/(« - /«), a > 0 2a/(a 2 + or), a > 0 2TT8{(0 — too) 7r[5(o> + (0()) + (5((0 - /V[5(o) + (o () ) — 5(a; • mi)] ~ «,,)] 17.5 Some Mathematical Properties The first mathematical property we call to your attention is that F(co) is a complex quantity and can be expressed in either rectangular or polar form. Thus from the defining integral, / 00 f{t)e-' Mt dt CXJ •J /(/)(cos cot - j sin cot) dt L f(t) cos cot cit - j f (t) sin cot dt. (17.36) Now we let M&) = / f(0 COS cot cit (17.37) -X fi(o)) = -/ /(/) sin mtdt. (17.38) J—no Thus, using the definitions given by Eqs. 17.37 and 17.38 in Eq. 17.36, we get F(co) = A(a>) + /£(<«>) = |F(ft>)|e^ M . (17.39)