Electric Circuits, 9th Edition P65 pptx

616 Fourier Series Example 16.2 Finding the Fourier Series of an Odd Function with Symmetry Find the Fourier series representation for the cur- rent waveform shown in Fig, 16.10. -T/2 Figure 16.10 • The periodic waveform for Example 16.2. In the interval Ost < 7/4, the expression for /(f) is m-f, Thus 8 f T/4 4I b k = — —risinkaitfdt 1 Jo * 321 m I sin ktoot t cos kco () t 2 2 T l \ kW 0 k(0() TfA Solution We begin by looking for degrees of symmetry in the waveform. We find that the function is odd and, in addition, has half-wave and quarter-wave symme- try. Because the function is odd, all the a coeffi- cients are zero; that is, a v = 0 and a k = 0 for all k. Because the function has half-wave symmetry, b k = 0 for even values of k. Because the function has quarter-wave symmetry, the expression for b k for odd values of k is 8j, /« . kir 2 7 sin — {k is odd). Kr £ TT The Fourier series representation of /(f) is i(t) =—r* 2J ~ sin —- sin «<W 0 / TT" « = 1,3.5, « 2 —r sin w Q t - - sin 3<o G t IT 1 V y r/4 b k =-= f /(f) sin ko) {] t dt. 1 -A) + — sm Statf - — sin 7^ + /ASSESSMENT PROBLEM Objective 1—Be able to calculate the trigonometric form of the Fourier coefficients for a periodic waveform 16.3 Derive the Fourier series for the periodic volt- age shown. \7V m « sin (mr/3) . Answer: v g (t) = —z— 2J z smnco 0 t. TT «=1,3,5, n MO v m -v m / \ 1 \ 0 T/6 T/3 r/s — 1 1 .27/3 57/6 / /T NOTE: Also try Chapter Problems 16.11 and 16.12. 16.4 An Alternative Trigonometric Form of the Fourier Series 617 16.4 An Alternative Trigonometric Form of the Fourier Series In circuit applications of the Fourier series, we combine the cosine and sine terms in the series into a single term for convenience. Doing so allows the representation of each harmonic of v(t) or i(t) as a single phasor quan- tity. The cosine and sine terms may be merged in either a cosine expres- sion or a sine expression. Because we chose the cosine format in the phasor method of analysis (see Chapter 9), we choose the cosine expres- sion here for the alternative form of the series. Thus we write the Fourier series in Eq. 16.2 as fit) = a v + ^A n cos(no) {) t - 6,X (16.38) «=i where A n and d„ are defined by the complex quantity a n - jb n = Vflg + bl/-6 n = A a /-9„. (16.39) We derive Eqs. 16.38 and 16.39 using the phasor method to add the cosine and sine terms in Eq. 16.2. We begin by expressing the sine functions as cosine functions; that is, we rewrite Eq. 16.2 as 00 /(/) = a v + 2X,cosncotf + b n cos(nco 0 t - 90°). (16.40) Adding the terms under the summation sign by using phasors gives SP{fl„ cos natf} = a n /0^ (16.41) and ®{b„ cos(nco () t - 90°)} = b n /-90° = -jb n . (16.42) Then 2P{«„ cos(«a>o* + b n cos(ti(OQt — 90°)} = a n - jb n = Val + %/-$„ = A„/-0„. (16.43) When we inverse-transform Eq. 16.43, we get a n cosno) () t + b n cos(nw 0 t - 90°) = ^jA,,/-6,,} = A n cos(no) 0 t - 6 n ). (16.44) Substituting Eq. 16.44 into Eq. 16.40 yields Eq. 16.38. Equation 16.43 corresponds to Eq. 16.39. If the periodic function is either even or odd, A n reduces to either a n (even) or b„ (odd), and 6 n is either 0° (even) or 90° (odd). The derivation of the alternative form of the Fourier series for a given periodic function is illustrated in Example 16.3. 618 Fourier Series Example 16.3 Calculating Forms of the Trigonometric Fourier Series for Periodic Voltage a) Derive the expressions for a k and b k for the peri- odic function shown in Fig. 16.11. b) Write the first four terms of the Fourier series representation of v(t) using the format of Eq. 16.38. J L T_ T_ yr T 5T 3T IT 2T 4 2 4 4 2 4 Figure 16.11 • The periodic function for Example 16.3. Solution a) The voltage v(t) is neither even nor odd, nor does it have half-wave symmetry. Therefore we use Eqs. 16.4 and 16.5 to find a k and b k . Choosing to as zero, we obtain a k = T 7*/4 PT V m cos ko) {) t dt + (0) cos ka> {) t dt JTjA 2V m smkcotf kcor 7/4 0 Vjn . kTJ^ kir 2 and 2 [ T/A b k = — I V m sin ktntf dt 1 Jo 2V m ( — cos koot k(o {) V m (. k-n• b) Tlie average value of v{t) is a„ = -— = -v The values of a k — jb k for k = 1,2, and 3 are «i - jb\ V, IT V, TT V2V, /-45°, V V (h — lb~> = 0 - / = / —90 , 3 " J IT TT L «3 _ jb 3 = -y,> 3TT Vzv, 3ir /-135' Thus the first four terms in the Fourier series representation of v(t) are V V2V V v(t) =^r + -—^-cos^t - 45°) + -f-cos(2co () t - 90°) 7T 77 V2V + ——^cos(3o>o* - 135°) + 3TT V U f /ASSESSMENT PROBLEM Objective 1—Be able to calculate the trigonometric form of the Fourier coeffiaents for a periodic waveform 16.4 a) Compute A\—A 5 and 0i~0 5 for the periodic function shown if V m — 9TT V. b) Using the format of Eq. 16.38, write the Fourier series for v(t) up to and including the fifth harmonic assuming T = 125.66 ms. Answer: (a) 10.4,5.2,0,2.6,2.1 V, and -120°, -60°, not defined, -120°, -60°; (b) v(t) = 21.99 + 10.4cos(50/ - 120°) + 5.2cos(100r - 60°) + 2.6 cos(200f - 120°) + 2.1 cos(250/ - 60°) V. 3 IT 3 4J 3 57' 3 IT NOTE: Also try Chapter Problem 16.22. 16.5 An Application Now we illustrate how to use a Fourier series representation of a periodic excitation function to find the steady-state response of a linear circuit. The RC circuit shown in Fig. 16.12(a) will provide our example. The circuit is energized with the periodic square-wave voltage shown in Fig. 16.12(b). The voltage across the capacitor is the desired response, or output, signal. The first step in finding the steady-state response is to represent the peri- odic excitation source with its Fourier series. After noting that the source has odd, half-wave, and quarter-wave symmetry, we know that the Fourier coefficients reduce to b k , with k restricted to odd integer values: b. T 4V rrk 774 V m sin kaj {] t dt (k is odd). Then the Fourier series representation of v„ is 4V IT CO -t - j? — sin nco {) t. (16.45) (16.46) Writing the series in expanded form, we have 4K„, . 4V m v., = sin cunt + —— sin 3oW 8 TT 3ir ^ 16.5 An Application (a) v„ -V, 27 37 (b) Figure 16.12 A An RC circuit excited by a periodic voltage, (a) The RC series circuit, (b) The square-wave voltage. 4V 4V in • c ^ v in • *7 ., —— sin 5(ti {) t + —— sin 7ct> 0 / + 5 7T 777 (16.47) Tlie voltage source expressed by Eq. 16.47 is the equivalent of infi- nitely many series-connected sinusoidal sources, each source having its own amplitude and frequency. To find the contribution of each source to the output voltage, we use the principle of superposition. For any one of the sinusoidal sources, the phasor-domain expression for the output voltage is V r v„ = 1 + jo)RC (16.48) All the voltage sources are expressed as sine functions, so we interpret a phasor in terms of the sine instead of the cosine. In other words, when we go from the phasor domain back to the time domain, we simply write the time-domain expressions as sin(atf + 6) instead of cos(wf + 6). The phasor output voltage owing to the fundamental frequency of the sinusoidal source is V,„ = {4V m /ir)/Qf Writing V„i in polar form gives cii where 1 + j(ti {) RC ' (4VJ/-/3, TrVl + (tilR 2 C r 0i = tan~ X (ti {) RC. (16.49) (16.50) (16.51) From Eq. 16.50, the time-domain expression for the fundamental fre- quency component of v ( , is 4V sin(ttiof - ft). (16.52) TTVI + (4RC 2 We derive the third-harmonic component of the output voltage in a simi- lar manner. The third-harmonic phasor voltage is (4V„ ; /377)/cy Y " 3 J3(OQRC 4V f=Zz&> (16.53) 3TTVI + 9<4R 2 C where j3 3 = tan~ l 3(o i} RC. (16.54) The time-domain expression for the third-harmonic output voltage is 4V V 0 3 = , '" = = =sin(3<o 0 f - j8 3 ). (16.55) 3TT VI + 9wg^ 2 C 2 Hence the expression for the &th-harmonic component of the output voltage is v ok = '" = sin(/cw,/ - j8jt) (& is odd), (16.56) where /3* = tan ~ ] k(o {) RC (k is odd). (16.57) We now write down the Fourier series representation of the output voltage: y «(0 = ^T 2; / =?• (16.58) ff /, = ul tt VI + (HW 0 i?C) 2 The derivation of Eq. 16.58 was not difficult. But, although we have an ana- lytic expression for the steady-state output, what v 0 (t) looks like is not imme- diately apparent from Eq. 16.58. As we mentioned earlier, this shortcoming is a problem with the Fourier series approach. Equation 16.58 is not useless, however, because it gives some feel for the steady-state waveform of v (> (t), if we focus on the frequency response of the circuit. For example, if C is large, 1/ncooC is small for the higher order harmonics. Thus the capacitor short cir- cuits the high-frequency components of the input waveform, and the higher order harmonics in Eq. 16.58 are negligible compared to the lower order har- monics. Equation 16.58 reflects this condition in that, for large C, 4V m °° 1 v <> ~ S^ 2 -^sin(Aio) 0 r - 90°) a ~ 2 -jcosnwof. (16.59) 7T(D()RC ,, = 1¾ It Equation 16.59 shows that the amplitude of the harmonic in the output is decreasing by 1/n 2 , compared with 1/n for the input harmonics. If C is so large that only the fundamental component is significant, then to a first approximation ~4V V»{t) « ;^cosw () f, (16.60) 16.5 An Application 621 and Fourier analysis tells us that the square-wave input is deformed into a sinusoidal output. Now let's see what happens as C —>0. The circuit shows that v () and v g are the same when C = 0, because the capacitive branch looks like an open circuit at all frequencies. Equation 16.58 predicts the same result because, as C —> 0, W m * 1 , v (> = >. — smno) {) t. (16.61) But Eq. 16.61 is identical to Eq. 16.46, and therefore v 0 —* v g as C —*• 0. Thus Eq. 16.58 has proven useful because it enabled us to predict that the output will be a highly distorted replica of the input waveform if C is large, and a reasonable replica if C is small. In Chapter 13, we looked at the distortion between the input and output in terms of how much mem- ory the system weighting function had. In the frequency domain, we look at the distortion between the steady-state input and output in terms of how the amplitude and phase of the harmonics are altered as they are transmitted through the circuit. When the network significantly alters the amplitude and phase relationships among the harmonics at the output rel- ative to that at the input, the output is a distorted version of the input. Thus, in the frequency domain, we speak of amplitude distortion and phase distortion. For the circuit here, amplitude distortion is present because the ampli- tudes of the input harmonics decrease as 1/rc, whereas the amplitudes of the output harmonics decrease as 1 1 n Vl + (na> Q RC) 2 ' This circuit also exhibits phase distortion because the phase angle of each input harmonic is zero, whereas that of the nth harmonic in the output sig- nal is - tan" 1 ri(o 0 RC. An Application of the Direct Approach to the Steady-State Response For the simple RC circuit shown in Fig. 16.12(a), we can derive the expres- sion for the steady-state response without resorting to the Fourier series representation of the excitation function. Doing this extra analysis here adds to our understanding of the Fourier series approach. To find the steady-state expression for v 0 by straightforward circuit analysis, we reason as follows. The square-wave excitation function alter- nates between charging the capacitor toward +V„, and —V m . After the circuit reaches steady-state operation, this alternate charging becomes periodic. We know from the analysis of the single time-constant RC circuit (Chapter 7) that the response to abrupt changes in the driving voltage is exponential. Thus the steady-state waveform of the voltage across the capacitor in the circuit shown in Fig. 16.12(a) is as shown in Fig. 16.13. The analytic expressions for v„{t) in the time intervals 0 < t < T/2 and T/2<t<T are Vo = V m + (V, - VJe^ RC , 0 < t < T/2; (16.62) Vo = ~V m + (V 2 + V m )e-^™ RC , T/2 < t < T. (16.63) We derive Eqs. 16.62 and 16.63 by using the methods of Chapter 7, as sum- marized by Eq. 7.60. We obtain the values of V\ and V 2 by noting from Eq. 16.62 that Vl = V m + (1/, - V m )e- TI2RC \ (16.64) Toward + V.„ Toward + V. \ \ Toward —V m Toward —V. Figure 16.13 • The steady-state waveform of v 0 for the circuit in Fig. 16.12(a). 622 Fourier Series Small C Figure 16.14 • The effect of capacitor size on the steady-state response. and from Eq. 16.63 that V 1 = -V ln + (V 2 + V m )e- T ? 2RC . Solving Eqs. 16.64 and 16.65 for V\ and V 2 yields y, = -y. = —™i L Substituting Eq. 16.66 into Eqs. 16.62 and 16.63 gives 2V, V ° "j" j |_ e -f/2RC ->/ RC , 0 < t < T/2. (16.65) (16.66) (16.67) and •V m + 2V, 1 + e -r/ac ff-(y/2)]/RC F/2 S / =S 7. (16.68) Equations 16.67 and 16.68 indicate that v w (0 has half-wave symmetry and that therefore the average value of v 0 is zero. This result agrees with the Fourier series solution for the steady-state response —namely, that because the excitation function has no zero frequency component, the response can have no such component. Equations 16.67 and 16.68 also show the effect of changing the size of the capacitor. If C is small, the exponential functions quickly vanish, v a = V m between 0 and T/2, and v a = —V m between T/2 and T. In other words, v a —* v% as C —> 0. If C is large, the output waveform becomes triangular in shape, as Fig. 16.14 shows. Note that for large C, we may approximate the exponential terms e~' /RC and C ,-['-(772)1/KC by the Hnear terms j _ ( t /RC) and 1 - {[t - (T/2)]/RC}i respectively. Equation 16.59 gives the Fourier series of this triangular waveform. Figure 16.14 summarizes the results. The dashed line in Fig. 16.14 is the input voltage, the solid colored line depicts the output voltage when C is small, and the solid black line depicts the output voltage when C is large. Finally, we verify that the steady-state response of Eqs. 16.67 and 16.68 is equivalent to the Fourier series solution in Eq. 16.58. To do so we simply derive the Fourier series representation of the periodic function described by Eqs. 16.67 and 16.68. We have already noted that the periodic voltage response has half-wave symmetry. Therefore the Fourier series contains only odd harmonics. For k odd, «* = T!1 ( 2V e' t/IiC (y _ ZK "' C ; — -8RCV,,, cos kco {) t dt T[\ + (kco {) RC) : (k is odd), (16.69) 4 ^/ 2V m e-l« c b k = - J \V m - - + e _ T/2RC ) 4V, $kco ( y m R 2 C 2 kir T[\ + (kiOuRC) 2 ] (k is odd). (16.70) To show that the results obtained from Eqs. 16.69 and 16.70 are consistent with Eq. 16.58, we must prove that 4V„, 1 Vol + b 2 k = k7T Vl + (ka> {) RC) 2 ' and that — = ~ko)[)RC. (16.71) (16.72) 16.6 Average-Power Calculations with Periodic Functions 623 We leave you to verify Eqs. 16.69-16.72 in Problems 16.23 and 16.24. Equations 16.71 and 16.72 are used with Eqs. 16.38 and 16.39 to derive the Fourier series expression in Eq. 16.58; we leave the details to you in Problem 16.25. With this illustrative circuit, we showed how to use the Fourier series in conjunction with the principle of superposition to obtain the steady- state response to a periodic driving function. Again, the principal short- coming of the Fourier series approach is the difficulty of ascertaining the waveform of the response. However, by thinking in terms of a circuit's fre- quency response, we can deduce a reasonable approximation of the steady-state response by using a finite number of appropriate terms in the Fourier series representation. (See Problems 16.27 and 16.29.) SSESSMENT PROBLEM Objective 2—Know how to analyze a circuit's response to a periodic waveform 16.5 The periodic triangular-wave voltage seen on the left is applied to the circuit shown on the right. Derive the first three nonzero terms in the Fourier series that represents the steady- state voltage v 0 if V m = 281.25ir 2 mV and the period of the input voltage is 200-7T ms. Answer: 2238.83 cos(10; - 5.71°) + 239.46 cos(30/ - 16.70°) + 80.50 cos(50f - 26.57°) + mV 16.6 The periodic square-wave shown on the left is applied to the circuit shown on the right. a) Derive the first four nonzero terms in the Fourier series that represents the steady- state voltage v 0 if V„, = 210-77 V and the period of the input voltage is 0.277 ms. b) Which harmonic dominates the output voltage? Explain why. 1 i y,n 0 ~v m 1 T/2 1 T 100 kft -^vw— + 100 nF o a Answer: (a) 17.5 cos(10,000r + 88.81°) + 26.14cos(30,000^ - 95.36°) + 168cos(50,0000 + 17.32 cos(70,000/ + 98.30°) + V; (b) The fifth harmonic, at 10,000 rad/s, because the circuit is a bandpass filter with a center frequency of 50,000 rad/s and a quality factor of 10. 10 kH !20nF + 20 mH v„ NOTE: Also try Chapter Problems 16.27 and 16.28. 16.6 Average-Power Calculations with Periodic Functions If we have the Fourier series representation of the voltage and current at a pair of terminals in a linear lumped-parameter circuit, we can easily express the average power at the terminals as a function of the harmonic voltages and currents. Using the trigonometric form of the Fourier series expressed in Eq. 16.38, we write the periodic voltage and current at the terminals of a network as 00 v = V dc + 2Xcos(/io>of - Q m )< (16.73) DO ' = 'dc + ^,I a COS(na) {r t - B tn ). (16.74) /(=1 The notation used in Eqs. 16.73 and 16.74 is defined as follows: V dc = the amplitude of the dc voltage component, V n = the amplitude of the nth-harmonic voltage, Q vn - the phase angle of the nth-harmonic voltage, / dc = the amplitude of the dc current component, / n = the amplitude of the nth-harmonic current, d in = the phase angle of the nth-harmonic current. We assume that the current reference is in the direction of the refer- ence voltage drop across the terminals (using the passive sign conven- tion), so that the instantaneous power at the terminals is w'.The average power is j rh+T j ft tt +T P = ~ / P dt = T Vt dL (16 - 75) 1 Jk 1 At To find the expression for the average power, we substitute Eqs. 16.73 and 16.74 into Eq. 16.75 and integrate. At first glance, this appears to be a for- midable task, because the product vi requires multiplying two infinite series. However, the only terms to survive integration are the products of voltage and current at the same frequency. A review of Eqs. 16.8-16.10 should convince you of the validity of this observation. Therefore Eq. 16.75 reduces to y^dc^dcf t 0 +T oo_ i fh+T v„i„co$(na) {) t - e m ) DO -i ri n=\ l Jt a X cos(nw ( / - B in )dt. (16.76) Now, using the trigonometric identity 1 1 cos a cos (3 = -cos(a -/3)+ — cos(a + /3), we simplify Eq. 16.76 to 1 °° V I f' 0+T p = v dc / dc + 7S-^r 1 / I cos (0- - *bd 1 ;i=i z A, + cos(2nft) 0 f - 6 m - d in )]dt. (16.77) The second term under the integral sign integrates to zero, so P = ^ dc / dc + 2-^cos(0,„ - 0 in ). (16.78) Equation 16.78 is particularly important because it states that in the case of an interaction between a periodic voltage and the corresponding periodic current, the total average power is the sum of the average powers obtained from the interaction of currents and voltages of the same frequency. Currents and voltages of different frequencies do not interact to produce average 16.6 Average-Power Calculations with Periodic Functions 625 power. Therefore, in average-power calculations involving periodic func- tions, the total average power is the superposition of the average powers associated with each harmonic voltage and current. Example 16.4 illustrates the computation of average power involving a periodic voltage. Example 16.4 Calculating Average Power for a Circuit with a Periodic Voltage Source Assume that the periodic square-wave voltage in Example 16.3 is applied across the terminals of a 15 H resistor. The value of V m is 60 V, and that of T is 5 ms. a) Write the first five nonzero terms of the Fourier series representation of v(t). Use the trigono- metric form given in Eq. 16.38. b) Calculate the average power associated with each term in (a). c) Calculate the total average power delivered to the 15 O resistor. d) What percentage of the total power is delivered by the first five terms of the Fourier series? Solution a) The dc component of v(t) is (60)(7/4) T = 15 V. From Example 16.3 we have A ] = V2 6O/77 = 27.01 V, 0i = 45°, A 2 = 60/TT = 19.10 V, e 2 = 90°, A 3 = 20 V2/TT = 9.00 V, 0 3 = 135°, A 4 = 0, 04 = 0°, A 5 = 5.40 V, 05 = 45°, 2TT 277(1000) w () = 40077 rad/s. Thus, using the first five nonzero terms of the Fourier series, 17(f) = 15 + 27.01 cos(40077/ - 45°) + 19.10COS(800T7/ - 90°) + 9.00COS(1200T7/ - 135°) + 5.40COS(2000T7/ - 45°) + V. b) The voltage is applied to the terminals of a resis- tor, so we can find the power associated with each term as follows: 15 2 Pdc= 15" = 15W ' 1 9 2 P 3 = __ = 2 . 70W , 1 5.4 2 ,, = -__ = 0.97 W. c) To obtain the total average power delivered to the 15 0 resistor, we first calculate the rms value of v(t): V = r rms /(60) 2 (774) T = V900 = 30 V. The total average power delivered to the 15 (1 resistor is 30 2 , P T = — = 60 W. d) The total power delivered by the first five nonzero terms is P = P dc + P { + P 2 + P 3 + P 5 = 55.15 W. This is (55.15/60)(100), or 91.92% of the total.