626 Fourier Series /ASSESSMENT PROBLEM Objective 3—Be able to estimate the average power delivered to a resistor using a small number of Fourier coefficients 16.7 The trapezoidal voltage function in Assessment Problem 16.3 is applied to the circuit shown. If \2V m = 296.09 V and T = 2094.4ms, estimate the average power delivered to the 2 fi resistor. Answer: 60.75 W. NOTE: Also try Chapter Problems 16.34 and 16.35. 16.7 The rms Value of a Periodic Function The rms value of a periodic function can be expressed in terms of the Fourier coefficients; by definition, ^=^/ fitrdt. (16.79) Representing /(f) by its Fourier series yields t u +T a v + 2Ai COS (iwat - 0„) «=i dt. (16.80) The integral of the squared time function simplifies because the only terms to survive integration over a period are the product of the dc term and the harmonic products of the same frequency. All other products inte- grate to zero. There fore Eq. 16.80 reduces to F - = w r ,r + „?,!"" '4 + 2 v /r=l z 4+2(^¼). (16-81) Equation 16.81 states that the rms value of a periodic function is the square root of the sum obtained by adding the square of the rms value of each harmonic to the square of the dc value. For example, let's assume that a periodic voltage is represented by the finite series v = 10 + 30cos(w ( / - $ t ) + 20cos(2w 0 / - 0 2 ) + 5cos(3to 0 f - 0 3 ) + 2cos(5w ( / - 9 5 ). The rms value of this voltage is V = VlO 2 + (30/V2) 2 + (20/V2 - ) 2 + (5/V5) 2 + (2/V2) 2 = V7643 = 27.65 V. Usually, infinitely many terms are required to represent a periodic func- tion by a Fourier series, and therefore Eq. 16.81 yields an estimate of the true rms value. We illustrate this result in Example 16.5. 16.8 The Exponential Form of the Fourier Series 627 Example 16.5 Estimating the rms Value of a Periodic Function Use Eq. 16.81 to estimate the rms value of the voltage Therefore, in Example 16.4. Solution m / 2 fljmy /19T0V f9M\* fSAOY From Example 16.4, V V V5 J ^ V5 J ^ ^ V dc = 15 V, = 28.76 V. V, = 27.01/V2 V, the rms value of the fundamental, V 2 = 19.10/V2 V, the rms value of the second harmonic, From Example 16.4, the true rms value is 30 V. We nnn j m*r , , /-, ,.,, • approach this value by including more and more harmon- V% = 9.00/V2 V, the rms value of the third harmonic, . . n ,, 01 c . ., ^ . t . , • 1 ; ICS m Eq. 16.81. For example, if we include the harmonics V 5 = 5.40/V2 V, the rms value of the fifth harmonic. through k = 9, the equation yields a value of 29.32 V. NOTE: Assess your understanding of this material by trying Chapter Problems 16.36 and 16.39. 16.8 The Exponential Form of the Fourier Series The exponential form of the Fourier series is of interest because it allows us to express the series concisely. The exponential form of the series is oo /(0 = 2 CjP**, (16.82) ft =t-CX3 where C n = ^ / f(t)e~ ,,m 'dt. (16.83) To derive Eqs. 16.82 and 16.83, we return to Eq. 16.2 and replace the cosine and sine functions with their exponential equivalents: (16.84) 2/ • (W - 85) Substituting Eqs. 16.84 and 16.85 into Eq. 16.2 gives OO _ I /(0 = a v + ^—(e inco »' + e-'" bJ "') + —(e> n ^ - e ~ inmf ) «=i 2 2/' COSrtW()f = e //!w„r _|_ e -jnco {) i 2 e jnuytf _ g-jnwy/t = «„ + 2( " 9 " r'"^ + ( " 2 k* W - ( 16 - 86 ) Now we define C„ as C„ = rfa, - jb n ) = -f/-B,„ AZ = 1,2,3, (16.87) From the definition of C, c„-\ 2 /*'« +7 ' 7 r ' n+T f(t) cos na) () t dt — j — I /(f) sin nco () tdt / L i A, * «//« i r'u+T = — / f(t) (cos nw^t — jsm no) { )t) dt = ~/ f{t)e-> ,m * dt, (16.88) which completes the derivation of Eq. 16.83. To complete the derivation of Eq. 16.82, we first observe from Eq. 16.88 that C 0 = -/ ,J{t)dt = a Next we note that ./,,+7 (16.89) (16.90) c-« = ^ /" f{t)e^<dt = c;; = |(«„ + A). Substituting Eqs. 16.87,16.89, and 16.90 into Eq. 16.86 yields 00 /(0 = c 0 + ^(c,^' 1 + cy^) 00 DO = ^C n e jn ^ + 2)C>" /lwrf . (16.91) /7=0 //=1 Note that the second summation on the right-hand side of Eq. 16.91 is equivalent to summing C n e ina} "' from -1 to -co; that is, oo —oc «=i //=-1 Because the summation from -1 to -co is the same as the summation from -co to —1, we use Eq. 16.92 to rewrite Eq. 16.91: 00 —1 /(0 = ^cj*** + 2c>>"^ n=Q -co 00 = 2 C /^" W(,, > ( 16 - 93 ) —oo which completes the derivation of Eq. 16.82. We may also express the rms value of a periodic function in terms of the complex Fourier coefficients. From Eqs. 16.81,16.87, and 16.89, I °° a 2 + b 2 ^rms = \K + 2^V^' (16 - 94) , , Vfl?, + bl \C„\ = ~ 2 ", (16.95) Cl = a 2 , (16.96) Substituting Eqs. 16.95 and 16.96 into Eq. 16.94 yields the desired expression: F m ,= \Ci + 2^\C n \ 2 . (16.97) «=i Example 16.6 illustrates the process of finding the exponential Fourier series representation of a periodic function. 16.8 The Exponential Form of the Fourier Series 629 Example 16.6 Finding the Exponential Form of the Fourier Series Find the exponential Fourier series for the periodic voltage shown in Fig. 16.15. v(t) -T/2 0 T/2 T-T/2 T T+T/2 Figure 16.15 A The periodic voltage for Example 16.6. Solution Using -T/2 as the starting point for the integration, we have, from Eq. 16.83, C, T V, T/2 V m e-' ),m ^dt 72 -jnioat T \ -jnu>o T/2 -r/2 IV no}[)T 2V sin nco () T/2. na)()T Here, because v(t) has even symmetry, b„ = 0 for all n, and hence we expect C n to be real. Moreover, the amplitude of C n follows a (sin x)/x distribution, as indicated when we rewrite C„ = V m T sin (WQ> () T/2) T «W ( )T/2 We say more about this subject in Section 16.9. The exponential series representation of v(t) is v{t)= 2 T V, T V m r\ sin Qiaw/2) , na)QT/2 sin (H(JO 0 T/2) /ZO»()T/2 2 Jl'Ofil ,)nto (] t •ASSESSMENT PROBLEMS Objective 4—Be able to calculate the exponential form of the Fourier coefficients for a periodic waveform 16.8 Derive the expression for the Fourier coeffi- cients C n for the periodic function shown. Hint: Take advantage of symmetry by using the fact that C n = (a„ — jb n )/2. r(A) -2 -8-J T—r 1 r i—r 4 8 12 ITJ-20 24 2T32 36 40 44 f(ms) Answer: C„ = -/^(1 + 3 cos ,J f), n odd 16.9 a) Calculate the rms value of the periodic cur- rent in Assessment Problem 16.8. b) Using C\—C n , estimate the rms value. c) What is the percentage of error in the value obtained in (b), based on the true value found in (a)? d) For this periodic function, could fewer terms be used to estimate the rms value and still insure the error is less than 1 %? Answer: (a) V34 A; (b) 5.777 A; (c) -0.93 %; (d) yes; if C\-C<) are used, the error is -0.98 %. NOTE: Also try Chapter Problems 16.45 and 16.46. 630 Fourier Series 16.9 Amplitude and Phase Spectra A periodic time function is defined by its Fourier coefficients and its period. In other words, when we know a v , a„, /?„, and T, we can construct /(/), at least theoretically. When we know a„ and b n , we also know the amplitude (A n ) and phase angle (-0,,) of each harmonic. Again, we cannot, in general, visualize what the periodic function looks like in the time domain from a description of the coefficients and phase angles; nevertheless, we recognize that these quantities characterize the periodic function completely. Thus, with sufficient computing time, we can synthesize the time-domain wave- form from the amplitude and phase angle data. Also, when a periodic driv- ing function is exciting a circuit that is highly frequency selective, the Fourier series of the steady-state response is dominated by just a few terms. Thus the description of the response in terms of amplitude and phase may provide an understanding of the output waveform. We can present graphically the description of a periodic function in terms of the amplitude and phase angle of each term in the Fourier series of f(t). The plot of the amplitude of each term versus the frequency is called the amplitude spectrum of f(t), and the plot of the phase angle ver- sus the frequency is called the phase spectrum of /(/). Because the ampli- tude and phase angle data occur at discrete values of the frequency (that is, at <t> ( ), 2tt) ( ), 3w 0 , ), these plots also are referred to as line spectra. An Illustration of Amplitude and Phase Spectra Amplitude and phase spectra plots are based on either Eq. 16.38 (A„ and -0,,) or Eq. 16.82 (C„). We focus on Eq. 16.82 and leave the plots based on Eq. 16.38 to Problem 16.49. To illustrate the amplitude and phase spectra, which are based on the exponential form of the Fourier series, we use the periodic voltage of Example 16.6. To aid the discussion, we assume that V m - 5 V and r = 7/5. From Example 16.6, •r TTTS/ 1.0-H 0.6 0'.4 0!2 _L 10 -8 -6 -4 -2 -0.4 K.rm. 4 6 8 10 ft Figure 16.16 • The pLot of C„ versus n when r - T/5, for the periodic voltage for Example 16.6. Figure 16.17 A The plot of (sin x)/x versus x. C„ = V m T sin (rt« 0 r/2) T H(x) {) r/2 (16.98) which for the assumed values of V,„ and T reduces to C„= 1 sin (/277-/5) «77-/5 (16.99) Figure 16.16 illustrates the plot of the magnitude of C n from Eq. 16.99 for values of n ranging from -10 to +10. The figure clearly shows that the amplitude spectrum is bounded by the envelope of the |( sin -\:)/x| func- tion. We used the order of the harmonic as the frequency scale because the numerical value of 7 is not specified. When we know T, we also know o> 0 and the frequency corresponding to each harmonic. Figure 16.17 provides the plot of |(sinx)/x| versus x, where x is in radians. It shows that the function goes through zero whenever x is an integral multiple of TT. From Eq. 16.98, tuo () /777 T T /77T 77? (16.100) From Eq. 16.100, we deduce that the amplitude spectrum goes through zero whenever m/T is an integer. For example, in the plot, T/T is 1/5, and therefore the envelope goes through zero at n = 5, 10,15, 10, 15, and so 16.9 Amplitude and Phase Spectra 631 on. In other words, the fifth, tenth, fifteenth, harmonics are all zero. As the reciprocal of T/T becomes an increasingly larger integer, the number of harmonics between every 77 radians increases. If mr/T is not an integer, the amplitude spectrum still follows the |( sin x)/x\ envelope. However, the envelope is not zero at an integral multiple of w 0 . Because C„ is real for all n, the phase angle associated with C„ is either zero or 180°, depending on the algebraic sign of (sin /i7r/5)/(mr/5). For example, the phase angle is zero for n = 0, ±1, ±2, ±3, and ±4. It is not defined at n = ±5, because C±$ is zero. The phase angle is 180° at n - ±6, ±7, ±8, and ±9, and it is not defined at ±10. This pattern repeats itself as n takes on larger integer values. Figure 16.18 shows the phase angle of C„ given by Eq. 16.98. Now, what happens to the amplitude and phase spectra if f(t) is shifted along the time axis? To find out, we shift the periodic voltage in Example 16.6 t {) units to the right. By hypothesis, v(t) = 2 c^' jn&rf /1=-00 (16.101) Therefore V(t - '0) = 2 C n e J,H0 ^-^ = 2 C n e- }na> ^e j,lw «', (16.102) which indicates that shifting the origin has no effect on the amplitude spec- trum, because \C n \ = \C n e-''^% [16.103) However, reference to Eq. 16.87 reveals that the phase spectrum has changed to -(0,, + na) 0 t () ) rads. For example, let's shift the periodic voltage in Example 16.1 r/2 units to the right. As before, we assume that r = T/5: then the new phase angle d' n is B' n = -(0,, + nir/5). (16.104) We have plotted Eq. 16.104 in Fig. 16.19 for n ranging from -8 to +8. Note that no phase angle is associated with a zero amplitude coefficient. You may wonder why we have devoted so much attention to the amplitude spectrum of the periodic pulse in Example 16.6. The reason is that this particular periodic waveform provides an excellent way to illus- trate the transition from the Fourier series representation of a periodic function to the Fourier transform representation of a nonperiodic func- tion. We discuss the Fourier transform in Chapter 17. 180° 90° ++++ 15 -13-11 -9 -7 -5 -3 -1 -14-12-10 -8 -6 -4 -2 M M > M M 1 3 5 2 4 kkkk ] 9 11 13 15 8 10 12 14 16 0' fl 216° 144° 72° _L_ -8 -6 -4 -2 -72 c -144° -216° Figure 16.18 A The phase angle of C„. Figure 16.19 A The plot of B'„ versus n for Eq. 16.104. 632 Fourier Series • ASSESSMENT PROBLEM Objective 4—Be able to calculate the exponential form of the Fourier coefficients for a periodic waveform Answer: 16.10 The function in Assessment Problem 16.8 is shifted along the time axis 8 ms to the right. Write the exponential Fourier series for the periodic current. NOTE: Also try Chapter Problems 16.49 and 16.50. A 60 -I Ot) = " 2 -(1 + 3 cos ^) e -0V/2)(« + i) e ;^ A> 7T „_ w =-oo(odd) Practical Perspective Active High-Q Fitters Consider the narrowband op amp bandpass filter shown in Fig. 16.20(a). The square wave voltage shown in Fig. 16.20(b) is the input to the filter. We know that the square wave is comprised of an infinite sum of sinusoids, one sinusoid at the same frequency as the square wave and all of the remaining sinusoids at integer multiples of that frequency. What effect will the filter have on this input sum of sinusoids? 100 nF /? 3 |lOkn 100 nF «—• + (a) v g (V) 15.65TT •tips) -50TT -3'.5TT -25TT -1 .5TT 0 12 5-7T 25TT 37.5TT 50TT 15.6577 (b) Figure 16.20 • (a) narrowband bandpass filter; (b) square wave input. The Fourier series representation of the square wave in Fig. 16.20(b) is given by , s 4A £ 1 , mr vjt) = — V — sin —— cos ncont 77 «=fe„« 2 where A = 15.65TTV. The first three terms of this Fourier series are given by v g (t) = 62.6 cos oj 0 t - 20.87 cos 3«of + 12.52 cos 5co 0 t - The period of the square wave is 507r /AS SO the frequency of the square wave is 40,000 rad/s. The transfer function for the bandpass filter in Fig. 16.20(a) is H(S) = -= r s 2 + ps + oil where K = 400/313, p = 2000 rad/s, a> 0 = 40,000 rad/s. This filter has a quality factor of 40,000/2000 = 20. Note that the center frequency of the bandpass filter equals the frequency of the input square wave. Multiply each term of the Fourier series representation of the square wave, represented as a phasor, by the transfer function H(s) evaluated at the frequency of the term in the Fourier series to get the representation of the output voltage of the filter as a Fourier series: vM) = -80 cos ujtf - 0.5 cos 3u) 0 t + 0.17 cos 5<o Q t - Notice the selective nature of the bandpass filter, which effectively amplifies the fundamental frequency component of the input square wave and attenuates all of the harmonic components. Now make the following changes to the bandpass filter of Fig. 16.20(a) — let R x = 391.25 ft, R 2 = 74.4 O, R 3 = 1 kO, and C l = C 2 = 0.1 /xF . The transfer function for the filter, H(s), has the same form given above, but now K = 400/313, /3 = 20,000 rad/s, w 0 = 40,000 rad/s. The pass- band gain and center frequency are unchanged, but the bandwidth has increased by a factor of 10. This makes the quality factor 2, and the result- ing bandpass filter is less selective than the original filter. We can see this by looking at the output voltage of the filter as a Fourier series: v o(0 = — 80 cos a> 0 t — 5 cos 3w 0 r + 1.63 cos 5<o Q t — The fundamental frequency of the input has the same amplification fac- tor, but the higher harmonic components have not been attenuated as signif- icantly as they were when the filter with Q = 20 was used. Figure 16.21 plots the first three terms of the Fourier series representations of the input square wave and the resulting output waveforms for the two bandpass filters. Note the nearly perfect replication of a sinusoid in Fig. 16.21(b) and the distortion that results from the use of a less-selective filter in Fig. 16.21(c). 634 Fourier Series t(txs) Figure 16.21 A (a) The first three terms of the Fourier series of the square wave in Fig. 16.20(b); (b) the first three terms of the Fourier series of the output from the bandpass filter in Fig. 16.20(a), where Q = 20; (c) the first three terms of the Fourier series of the output from the bandpass filter in Fig. 16.20(a) with component values changed to give Q = 2. Summary • A periodic function is a function that repeats itself everv T seconds. Five types of symmetry are used to simplify the compu- tation of the Fourier coefficients: A period is the smallest time interval (T ) that a peri- odic function can be shifted to produce a function iden- tical to itself. (See page 604.) The Fourier series is an infinite series used to represent a periodic function. The series consists of a constant term and infinitely many harmonically related cosine and sine terms. (See page 607.) The fundamental frequency is the frequency determined by the fundamental period (/ 0 = 1/7 or o>o = 2TJ-/ 0 ). (See page 607.) The harmonic frequency is an integer multiple of the fundamental frequency. (See page 607.) The Fourier coefficients are the constant term and the coefficient of each cosine and sine term in the series. (See Eqs. 16.3-16.5.) (See page 608.) • even, in which all sine terms in the series are zero • odd, in which all cosine terms and the constant term are zero • half-wave, in which all even harmonics are zero • quarter-wave, half-wave, even, in which the series contains only odd harmonic cosine terms • quarter-wave, half-wave, odd, in which the series con- tains only odd harmonic sine terms (See page 611.) In the alternative form of the Fourier series, each har- monic represented by the sum of a cosine and sine term is combined into a single term of the form A n cos(nco {) t - 0,,). (See page 617.) Problems 635 For steady-state response, the Fourier series of the response signal is determined by first finding the response to each component of the input signal. The individual responses are added (super-imposed) to form the Fourier series of the response signal. The response to the individual terms in the input series is found by either frequency domain or s-domain analysis. (See page 619.) The waveform of the response signal is difficult to obtain without the aid of a computer. Sometimes the frequency response (or filtering) characteristics of the circuit can be used to ascertain how closely the output waveform matches the input waveform. (See page 620.) Only harmonics of the same frequency interact to pro- duce average power. The total average power is the sum of the average powers associated with each frequency. (See page 623.) The rms value of a periodic function can be estimated from the Fourier coefficients. (See Eqs. 16.81,16.94, and 16.97.) (See page 626.) The Fourier series may also be written in exponential form by using Euler's identity to replace the cosine and sine terms with their exponential equivalents. (See page 627.) The Fourier series is used to predict the steady-state response of a system when the system is excited by a periodic signal. The series assists in finding the steady- state response by transferring the analysis from the time domain to the frequency domain. Problems Sections 16.1-16.2 16.1 Find the Fourier series expressions for the periodic voltage functions shown in Fig. P16.1. Note that Fig. P16.1(a) illustrates the square wave; Fig. P16.1(b) illustrates the full-wave rectified sine wave, where v(t) = V m sm(<ir/T)t, 0 < t < T; and Fig. P16.1(c) illustrates the half-wave rectified sine wave, where v(t) = K m sin(27r/7)/,0 < t < T/2. Figure P16.1 v(t) -T V,n 0 -V It T IT IT (a) 16.2 Derive the Fourier series for the periodic voltage shown in Fig. P16.2, given that 2*7T v(t) = 100 sin — /V, 2TT v(t) = 60 sin — t V, 0 < t < t < T. T Figure P16.2 v(t) V T/4 T/2 37/4 T 16.3 For each of the periodic functions in Fig. P16.3, specify a) (o a in radians per second b) f 0 in hertz c) the value of a v d) the equations for a k and b k e) v(t) as a Fourier series . axis? To find out, we shift the periodic voltage in Example 16.6 t {) units to the right. By hypothesis, v(t) = 2 c^' jn&rf /1=-00 (16.101) Therefore V(t - '0) = 2 C n e J,H0 ^-^