Electric Circuits, 9th Edition P64 doc

606 Fourier Series 4) + 27/ Figure 16.1 A A periodic waveform. v{t) V m - v(t) V,„ - T (a) 27 7/2 (b) Figure 16.2 • Output waveforms of a nonfiltered sinu- soidal rectifier, (a) Full-wave rectification, (b) Half-wave rectification. Another practical problem that stimulates interest in periodic func- tions is that power generators, although designed to produce a sinusoidal waveform, cannot in practice be made to produce a pure sine wave. The distorted sinusoidal wave, however, is periodic. Engineers naturally are interested in ascertaining the consequences of exciting power systems with a slightly distorted sinusoidal voltage. Interest in periodic functions also stems from the general observation that any nonlinearity in an otherwise linear circuit creates a nonsinusoidal periodic function. The rectifier circuit alluded to earlier is one example of this phenomenon. Magnetic saturation, which occurs in both machines and transformers, is another example of a nonlinearity that generates a nonsinusoidal periodic function. An electronic clipping circuit, which uses transistor saturation, is yet another example. Moreover, nonsinusoidal periodic functions are important in the analysis of nonelectrical systems. Problems involving mechanical vibra- tion, fluid flow, and heat flow all make use of periodic functions. In fact, the study and analysis of heat flow in a metal rod led the French mathe- matician Jean Baptiste Joseph Fourier (1768-1830) to the trigonometric series representation of a periodic function. This series bears his name and is the starting point for finding the steady-state response to periodic exci- tations of electric circuits. V m - Figure 16.3 A The triangular waveform of a cathode-ray oscilloscope sweep generator. v(t) Vm 0 1/ v m n 7 27 (a) V( Vm 0 0 r 2 7 (c) Figure 16.4 A Waveforms produced by function generators used in laboratory testing, (a) Square wave, (b) Triangular wave, (c) Rectangular pulse. 16.1 Fourier Series Analysis: An Overview 607 16.1 Fourier Series Analysis: An Overview What Fourier discovered in investigating heat-flow problems is that a periodic function can be represented by an infinite sum of sine or cosine functions that are harmonically related. In other words, the period of any trigonometric term in the infinite series is an integral multiple, or har- monic, of the fundamental period T of the periodic function. Thus for peri- odic /(/), Fourier showed that /(/) can be expressed as oo f(t) = a v + 5X cos n(a d + b n sin/iwo^ (16.2) < Fourier series representation of a periodic «=i function where n is the integer sequence 1,2,3, In Eq. 16.2, a v , a fn and b n are known as the Fourier coefficients and are calculated from /(f). The term o> 0 (which equals 2TT/T) represents the fundamental frequency of the periodic function /(f). The integral multi- ples of OJ () —that is, 2&> 0 , 3a> 0 , 4w 0 , and so on—are known as the harmonic frequencies of/(/). Thus 2w 0 is the second harmonic, 3a> 0 is the third har- monic, and no) () is the «th harmonic of /(f). We discuss the determination of the Fourier coefficients in Section 16.2. Before pursuing the details of using a Fourier series in circuit analysis, we first need to look at the process in general terms. From an applications point of view, we can express all the periodic functions of interest in terms of a Fourier series. Mathematically, the conditions on a periodic function /(f) that ensure expressing /(f) as a convergent Fourier series (known as Dirichlet's conditions) are that 1. /(f) be single-valued, 2. /(/) have a finite number of discontinuities in the periodic interval, 3. /(/) have a finite number of maxima and minima in the periodic interval, 4. the integral / 1/(01* exists. Any periodic function generated by a physically realizable source satisfies Dirichlet's conditions. These are sufficient conditions, not necessary con- ditions. Thus if /(f) meets these requirements, we know that we can express it as a Fourier series. However, if/(f) docs not meet these require- ments, we still may be able to express it as a Fourier series. The necessary conditions on /(f) are not known. After we have determined /(f) and calculated the Fourier coefficients («,„ a,„ and b„), we resolve the periodic source into a dc source (a„) plus a sum of sinusoidal sources (a„ and /?„). Because the periodic source is driv- ing a linear circuit, we may use the principle of superposition to find the steady-state response. In particular, we first calculate the response to each source generated by the Fourier series representation of/(f) and then add the individual responses to obtain the total response. The steady-state response owing to a specific sinusoidal source is most easily found with the phasor method of analysis. The procedure is straightforward and involves no new techniques of circuit analysis. It produces the Fourier series representation of the steady-state response; consequently, the actual shape of the response is 608 Fourier Series unknown. Furthermore, the response waveform can be estimated only by adding a sufficient number of terms together. Even though the Fourier series approach to finding the steady-state response does have some draw- backs, it introduces a way of thinking about a problem that is as important as getting quantitative results. In fact, the conceptual picture is even more important in some respects than the quantitative one. 16.2 The Fourier Coefficients After defining a periodic function over its fundamental period, we deter- mine the Fourier coefficients from the relationships a v = f. /( ? ) dt > ( 16 - 3 ) 2 / Fourier coefficients • a k = —I f(t) cos kco 0 t dt, (16.4) Tj t(3 2 f'o+T b k = r fit) sin ka) Q t dt. (16.5) 1 J t 0 In Eqs. 16.4 and 16.5, the subscript k indicates the &th coefficient in the integer sequence 1,2,3, Note that a v is the average value of /(/), a k is twice the average value of f{t) cos kcotf, and b k is twice the average value of f(t) sin kcotf. We easily derive Eqs. 16.3-16.5 from Eq. 16.2 by recalling the follow- ing integral relationships, which hold when m and n are integers: sin mw 0 f dt = 0, for all m, (16.6) 'o t 0 +T cos mcootdt = 0, for all m, (16-7) I Jtr, t n +T cos moitf sin nco^t dt = 0, for all m and n, (16.8) t 0 +r sin mo)()t sin no) {) t dt = 0, for all m # n, = —, for m - n, (16.9) 2 to+T cos mo) 0 t cos nco G t dt = 0, for all m # «, T = —, for m = n. (16.10) We leave you to verify Eqs. 16.6-16.10 in Problem 16.5. 16.2 The Fourier Coefficients 609 To derive Eq. 16.3, we simply integrate both sides of Eq. 16.2 over one period: rt {) +T rh+T / oo \ / f(t)dt = / f a v 4- ^ a n cos n(x) () t + b lt sinna) {) t jdt Jio Ji,) \ «=l / /.?„+'/' oo rh+T = / a v dt + 2 / ( fl « cos nw o* + ^« sm '""of) & J/ tl 11= 1 Jfy a v T + 0. (16.11) Equation 16.3 follows directly from Eq. 16.11. To derive the expression for the A:th value of A„, we first multiply Eq. 16.2 by cos kio {) t and then integrate both sides over one period of /(f): f„+r ,-tu+T f{t) cos kco {) t dt = / a v cos k(o {) t dt + 2)/ («/, cos «o> ( / cos &w () f + /}„ sin «6> ( / cos /ca> () f) d/ «=1 A, '« = 0 + a A |y) (16.12) Solving Eq. 16.12 for a k yields the expression in Eq. 16.4. We obtain the expression for the kXh. value of b n by first multiplying both sides of Eq. 16.2 by sin kcotf and then integrating each side over one period of/(f). Example 16.1 shows how to use Eqs. 16.3-16.5 to find the Fourier coefficients for a specific periodic function. Example 16.1 Finding the Fourier Series of a Triangular Waveform with No Symmetry Find the Fourier series for the periodic voltage shown in Fig. 16.5. -7 0 T IT Figure 16.5 • The periodic voltage for Example 16.1. Solution When using Eqs. 16.3-16.5 to find a v , %, and b k , we may choose the value of f 0 - FOT the periodic voltage of Fig. 16.5, the best choice for f () is zero. Any other choice makes the required integrations more cum- bersome. The expression for v(t) between 0 and 7 is ly 1 The equation for a v is 1 f T fV a - = H KTr = i v ' This is clearly the average value of the waveform in Fig. 16.5. The equation for the kth value of a n is (lk = T I 1*7^ ' 2V 1 * •Z—z COS k(Ont + ~ SmkiOnt T 2 Kk 2 ^ kc*a 2V, T 2 ——z ( cos 2-rrk - 1) 0 for all k. 610 Fourier Series The equation for the A:th value of b„ is bk = T I \^r L ) tsinko)[)tdt 2V "'f l i t i \ T o 2V m ( T \ = 0 0 , coslirk T 2 \ kcou J -v m irk y m = — - y y m 2 The Fourier series for v(t) is V °9 1 v in ^r\ l . X~ smtt&w K V V sm (o {) t - -— sin 2oj () t - —— sin 3a> 0 ? - • • •. 1" 27T 37T I/ASSESSMENT PROBLEMS Objective 1—Be able to calculate the trigonometric form of the Fourier coefficients for a periodic waveform 16.1 Derive the expressions for «„, a k , and b k for the periodic voltage function shown if V m = 9ir V. v„ 3 27/ 3 4T 3 57/ 3 IT f sin^ V, Answer: a v = 21.99 V, NOTE: Also try Chapter Problems 16.1-16.3. cos^) V. 16.2 Refer to Assessment Problem 16.1. a) What is the average value of the periodic voltage? b) Compute the numerical values of ay — a 5 and b\ - b 5 . c) If T = 125.66 ms, what is the fundamental frequency in radians per second? d) What is the frequency of the third harmonic in hertz? e) Write the Fourier series up to and including the fifth harmonic. Answer: (a) 21.99 V; (b) -5.2 V,2.6V,0V,-1.3, and 1.04 V; 9 V, 4.5 V, 0 V, 2.25 V, and 1.8 V; (c) 50 rad/s; (d) 23.87 Hz; (e) v(t) = 21.99 - 5.2 cos 50? + 9 sin 50? + 2.6 cos 100? + 4.5 sin 100* - 1.3 cos 200? + 2.25 sin 200? + 1.04 cos 250? + 1.8 sin 250? V. Finding the Fourier coefficients, in general, is tedious. Therefore any- thing that simplifies the task is beneficial. Fortunately, a periodic function that possesses certain types of symmetry greatly reduces the amount of work involved in finding the coefficients. In Section 16.3, we discuss how symmetry affects the coefficients in a Fourier series. 16.3 The Effect of Symmetry on the Fourier Coefficients 611 16.3 The Effect of Symmetry on the Fourier Coefficients Four types of symmetry may be used to simplify the task of evaluating the Fourier coefficients: • even-function symmetry, • odd-function symmetry, • half-wave symmetry, • quarter-wave symmetry. The effect of each type of symmetry on the Fourier coefficients is discussed in the following sections. Even-Function Symmetry A function is defined as even if fit) = /(-0- (16.13) -< Even function Functions that satisfy Eq. 16.13 are said to be even because polynomial functions with only even exponents possess this characteristic. For even periodic functions, the equations for the Fourier coefficients reduce to 2 f T/2 (16.14) T/2 f{t) cos ktotfdt, 1 Jo (16.15) b k = 0, for all k. (16.16) Note that all the b coefficients are zero if the periodic function is even. Figure 16.6 illustrates an even periodic function. The derivations of Eqs. 16.14-16.16 follow directly from Eqs. 16.3-16.5. In each derivation, we select t {) = -T/2 and then break the interval of integration into the range from -T/2 to 0 and 0 to T/2, or a r -T/2 -T/2 0 /(0 dt T/2 T/2 "fl f(t)dL (16.17) -T m 0 Figure 16.6 • An even periodic function, /(0=/(-0- Now we change the variable of integration in the first integral on the right-hand side of Eq. 16.17. Specifically, we let t = -x and note that /(0 = f(~ x ) = /(•*) because the function is even. We also observe that x — T/2 when t = -T/2 and dt = —dx. Then /(0 dt - T/2 J'T/2 0 /-7/2 f(x)(-dx) = / f(x) dx, (16.18) 612 Fourier Series which shows that the integration from -T/2 to 0 is identical to that from 0 to 7/2; therefore Eq. 16.17 is the same as Eq. 16.14. The derivation of Eq. 16.15 proceeds along similar lines. Here, a k = 2 f° /(/) cos ka) {) t dt T/2 V — / /(/) cos ko) {) t dt, ' h) (16.19) but •4) ,.() /(/) cos ku>()tdt = / f(x) cos (-k00^)(-dx) T/2 JT/2 T/2 f(x) cos k(x) {) x dx. (16.20) As before, the integration from -T/2 to 0 is identical to that from 0 to T/2. Combining Eq. 16.20 with Eq. 16.19 yields Eq. 16.15. All the b coefficients are zero when /(/) is an even periodic function, because the integration from —T/2 to 0 is the exact negative of the inte- gration from 0 to T/2; that is, 0 ni) /(/) sin ka) {) t dt = / f(x)sm(-ko) {) x)(—dx) -T/2 JT/2 T/2 f(x) sin koi()X dx. (16.21) When we use Eqs. 16.14 and 16.15 to find the Fourier coefficients, the interval of integration must be between 0 and T/2. Odd-Function Symmetry A function is defined as odd if Odd function • Figure 16.7 • An odd periodic function /(0 = -/(-0. m = -/(-0. (16.22) Functions that satisfy Eq. 16.22 are said to be odd because polynomial functions with only odd exponents have this characteristic. The expres- sions for the Fourier coefficients are a v = 0; a k = 0, 6 * = f for all k\ rT/2 / /(/) sin ka) {) t dt. /o (16.23) (16.24) (16.25) Note that all the a coefficients are zero if the periodic function is odd. Figure 16.7 shows an odd periodic function. 16.3 The Effect of Symmetry on the Fourier Coefficients 613 We use the same process to derive Eqs. 16.23-16.25 that we used to derive Eqs. 16.14-16.16. We leave the derivations to you in Problem 16.7. The evenness, or oddness, of a periodic function can be destroyed by shifting the function along the time axis. In other words, the judicious choice of where t - 0 may give a periodic function even or odd symmetry. For example, the triangular function shown in Fig. 16.8(a) is neither even nor odd. However, we can make the function even, as shown in Fig. 16.8(b), or odd, as shown in Fig. 16.8(c). Half-Wave Symmetry A periodic function possesses half-wave symmetry if it satisfies the constraint /(f) = -f(t - T/2). (16.26) Equation 16.26 states that a periodic function has half-wave symmetry if, after it is shifted one-half period and inverted, it is identical to the original function. For example, the functions shown in Figs. 16.7 and 16.8 have half- wave symmetry, whereas those in Figs. 16.5 and 16.6 do not. Note that half- wave symmetry is not a function of where t = 0. If a periodic function has half-wave symmetry, both a k and b k are zero for even values of k. Moreover, a v also is zero because the average value of a function with half-wave symmetry is zero. The expressions for the Fourier coefficients are a v = 0, a k = 0, (16.27) 7/2 1 -/() h = o, h = y7 T/2 fit) cos ko) 0 t dt, i fit) sin k(i) {) t dt, for k even; for k odd; for k even; for k odd. (16.28) (16.29) (16.30) (16.31) (c) Figure 16.8 A How the choice of where t = 0 can make a periodic function even, odd, or neither, (a) A periodic triangular wave that is neither even nor odd. (b) The triangular wave of (a) made even by shifting the function along the t axis, (c) The triangular wave of (a) made odd by shifting the function along the t axis. We derive Eqs. 16.27-16.31 by starting with Eqs. 16.3-16.5 and choos- ing the interval of integration as —T/2 to T/2. We then divide this range into the intervals —T/2 to 0 and 0 to T/2. For example, the derivation for a k is a k = Jt, t fit) cos kcjQt dt T.L T/2 fit) cos kco {) t dt T L T/2 .0 = — / fit) cos kcoot dt T/2 T/2 + ;=r / /(f) cos ka) {] t dt. ' -A) (16.32) 614 Fourier Series Now we change a variable in the first integral on the right-hand side of Eq. 16.32. Specifically, we let Then t = x - T/2. x = T/2, when t = 0; x = 0, when t = —T/2; dt = dx. We rewrite the first integral as /(0 cos kto () tdt =/ f(x - T/2)coskto {) (x - T/2)dx. (16.33) -T/2 Ml Note that cos ktO()(x — T/2) — cos (ko) {) x — kir) = cos kir cos kco () x and that, by hypothesis, f(x - T/2) = -f(x). Therefore Eq. 16.33 becomes A 7/4 -A 7/2 37/4 7 (a) /(0 /I 0 - i 7/4 -A I 7/2 37/4 7 (b) Figure 16.9 • (a) A function that has quarter-wave symmetry, (b) A function that does not have quarter- wave symmetry. ./-7/2 f(t) cos kco () t dt T/2 [-f(x)] cos k 77 cos ko) {) x dx. (16.34) t Incorporating Eq. 16.34 into Eq. 16.32 gives 7/2 a k = -(1 — cos kit) / f(t) cos &G> 0 f rf/. (16.35) But cos kTT is 1 when k is even and -1 when k is odd. Therefore Eq. 16.35 generates Eqs. 16.28 and 16.29. We leave it to you to verify that this same process can be used to derive Eqs. 16.30 and 16.31 (see Problem 16.8). We summarize our observations by noting that the Fourier series rep- resentation of a periodic function with half-wave symmetry has zero aver- age, or dc, value and contains only odd harmonics. Quarter-Wave Symmetry The term quarter-wave symmetry describes a periodic function that has half-wave symmetry and, in addition, symmetry about the midpoint of the positive and negative half-cycles. The function illustrated in Fig. 16.9(a) 16.3 The Effect of Symmetry on the Fourier Coefficients 615 has quarter-wave symmetry about the midpoint of the positive and nega- tive half-cycles. The function in Fig. 16.9(b) does not have quarter-wave symmetry, although it does have half-wave symmetry. A periodic function that has quarter-wave symmetry can always be made either even or odd by the proper choice of the point where t = 0. For example, the function shown in Fig. 16.9(a) is odd and can be made even by shifting the function 7/4 units either right or left along the t axis. However, the function in Fig. 16.9(b) can never be made either even or odd. To take advantage of quarter-wave symmetry in the calculation of the Fourier coefficients, you must choose the point where t = 0 to make the function either even or odd. If the function is made even, then a v = 0, because of half-wave symmetry; a k = 0, for k even, because of half-wave symmetry; 8 f T/4 a k = — I f(t) cos kaitf dt, for k odd; T Ju b k = 0, for all /c, because the function is even. (16.36) Equations 16.36 result from the function's quarter-wave symmetry in addition to its being even. Recall that quarter-wave symmetry is super- imposed on half-wave symmetry, so we can eliminate a r and a k for k even. Comparing the expression for a k , k odd, in Eqs. 16.36 with Eq. 16.29 shows that combining quarter-wave symmetry with evenness allows the shorten- ing of the range of integration from 0 to 7/2 to 0 to 7/4. We leave the der- ivation of Eqs. 16.36 to you in Problem 16.9. If the quarter-wave symmetric function is made odd, a v = 0, because the function is odd; a k = 0, for all k, because the function is odd; b k = 0, for k even, because of half-wave symmetry; 8 f T/4 b k = ^ / f(t) sin kco {) tdt, for k odd. (16.37) Equations 16.37 are a direct consequence of quarter-wave symmetry and oddness. Again, quarter-wave symmetry allows the shortening of the inter- val of integration from 0 to 7/2 to 0 to 7/4. We leave the derivation of Eqs. 16.37 to you in Problem 16.10. Example 16.2 shows how to use symmetry to simplify the task of find- ing the Fourier coefficients. . another example. Moreover, nonsinusoidal periodic functions are important in the analysis of nonelectrical systems. Problems involving mechanical vibra- tion, fluid flow, and heat flow all make. name and is the starting point for finding the steady-state response to periodic exci- tations of electric circuits. V m - Figure 16.3 A The triangular waveform of a cathode-ray oscilloscope. meets these requirements, we know that we can express it as a Fourier series. However, if/(f) docs not meet these require- ments, we still may be able to express it as a Fourier series. The