fourier_series_continued

fourier_series_continued - EXAMPLE 16.1 Find the Fourier...

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Unformatted text preview: EXAMPLE 16.1 Find the Fourier series :or the periodic voltage shown in Fig. 16.5. SOLUTION When using Eqs. (16.3)—(l6.5) to find av, ak, and bk, we may choose the value of to. For the periodic voltage of Fig. 16.5, the best choice for to IS zero. Any other choice makes the required integrations more Figure 16.5 The periodic voltage for Example 16.1. cumbersome. The expression for 110) between 0 and T is v0) = r. Vm The equation for an is T _ .1. V \T 1 T V l A re (A a 2 n m au=- —— tdt=—Vm. /}y”€féL I TNT) 2 - M...— 7// M5 3 av “ T This is clearly the average value of the waveform in Fig. 16.5. 1 , The equation for thMS-ef'dfi is ['k) L” , , ( ’ X) “.1: 2 T V ak=——f ‘(l)tcoskw0tdt T O T 2Vm 1 t . = —— COS + ‘66 Sln T O l = 2% [—(cos 277k — 1)] = 0 for all k. T2 mg 3/, bk. The equation for Wis 2 T V bk=? 0 (-Tfl)tsinkwotdt T 2Vm 1 s' k t t c k t =—— —— in a) —— 0s w T2 kzwg 0 kwo 0 0 2Vm T = 7 O———coserk T~ kwo —Vm r. / we)" st t 5 A “ka The Fourier series forut is J ’ ‘1 - v , . () K'! _'L EflsanvLD V Vm °° 1 110‘) = 3r: — 71- ; ; sin moot Vm Vm . Vm . Vm A =———--——sma)0t———Sin2th———Sln3wot—--~ 2 7t 27: 3.7 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 - ’- fCt) = f(—t). (16.13) Functions that satisfy Eq. (16.13) are said to be even because polyno- mial functions with only even exponents possess this characteristic. For even periodic functions, the equations for the Fourier coefficients reduce to 3 1 l (16.14) ‘ @ -T/2 ak = -/ f(t) cos kwot dt, (16.15) T 0 J (16.16) bk ; 0, for all k. 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. (l6.l4)—(16.16) follow directly from Eqs. (16.3)—(16.5). In each derivation, w select to = ——T 2 nd then break the interval of inte- gration into the range from ——T / 2 to 0 and 0 to T / 2, or 1 T/Z up? / ftt)dt —T/2 1 T/Z T 0 f(t)dt. (16.17) 0 Figure 16.5 An even peritzudic function, f(t) = ___ l f t d; f(-r)- T —T/2 f() + 1 Now we change the variable of integration in the first integral on the right—hand side of Eq. (16.17). Specifically, we let 1? = —x and 1 t . " ’ ‘ note that f(t) = f(—x) = f(x) because the function is even. We also \ ' " observe thatx = T/2 when t = ——T/2 and dt = —dx. Then T/2 0 0 1;: T ] f(r)dz= f(x)(—dx)=/ f(x)dx, (16.18) ( —T/2 r/2 0 from 0 to T/2; therefore Eq. (16.17) is the same as Eq. (16.14). The '- which shows that the integration from —T / 2 to O is identical to that \\ derivation of Eq. (16.15) proceeds along similar lines. Here, 2 O ak = -— / f(t) cos kwot dt T 4/2 2 T/2 + ‘T‘ f f0) COS kwot dt, (16.19) 0 (( but ' 0 0 i/fljm cos kwot dt =/ f(x) cos(—kwox)(—dx) . —T/2 T/ J / 7/2 = 1‘ f(x) cos kwox dx. (16.20) As before, the integration from —T / 2 to 0 is identical to that from 0 to T/2. Co:;1binii':-__;Eq. (16.20) with Eq. (16.19) yields Eq. (16.15). Aflthe b coefficientsgrgazgro when f (t) is an even eriodic func- ti_o_n, because the integration fromd—T/ 2 to 0 is the exact negative of the integration from O to T /2; that is. a} o o / f(t) sin kwot dt = f(x) sin(—kwox)(—dx) _T/2 T/2 g = f(x) sin kwox 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 f0) = —f(-t)- Functions that satisfy Eq. (16.22) are said to be odd because polyno- mial functions with only odd exponents have this characteristic. The expressions for the Fourier coefficients are (16.22) au 2 0; ak = 0, fOr all 4o {/2 T 0 (16.24) bk = f(t) Sin kwot dt. (16.25) Note that all the a coefficients are zero if the eriodic function is odd. Figure 16.7 shows an odd periodic function. We use the same process to derive Eqs. (16.23)—(16.25) that we used to derive Eqa‘. (l6.l4)—(16.l6). We leave the derivations to you J _ in Problem 16.4. The evenness, or oddness, of a periodic function can be destroyed by shifting the? function along the time axis. In other words, the judi- cious 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) — f {—r). fir): Figure 16.7 An odd periodic function f(t) = . _/ \‘v’ ,-',[ \v" (a) f(t)| x” \ I; \ _ _f_\. _.. K. _+ \V’: —A[|-— \V/ \ lb) (a) Figure 16.8 How the choice of where t = O 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. (cT'fhe triangular wave of (a) made odd by shifting the function along the z axis. ’— 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 ftr) = —f(r — T/2). (16.25) gas-m has li_r1:'—=.'~ .-.‘. I: M mine— :r. -. .r :a idfiillli.'-_1i to _——I—-—’l'-——_—-—'—‘—|—I-__._—-‘-_- .n..1l lung-nun Fur f.‘2.lllll‘i;’. ’:1-.' I'..ll'.'L";l\')I]$ shtgxsn nil-:31. in 7 and 10.6 have halt-wave symmetry, whereas those H1 Figs. io.5 and 16.6 do not. Pint-3 lli.:t 21.;Jllaaix'c ‘j. min-en". :0. ' If a periodic function has half-wave symmetry, both ak and bk are zero for even values of k. Moreover, av also is zero because the average value of a function with half—wave symmetry is zero. The expressions for the Fourier coefficients are 5.4 (in: :i (unt'litin -;'Ir' where. av = 0, (16.27) ak = O, for k even, (16.28) 772 _'__ ak = T f(t) cos kwot dt, for k odd, (16.29) 0 bk = O, for k even, (16.30) 4 7/2 X‘— bk = ? f(z) sin kwot dt, for k odd. (16.31) 0 We derive Eqs. (16.27)—(l6.31) by starting with Eqs. (16.3)—(16.5) and choosing the interval of integration as —T/2 to T /2. We then divide this range into the intervals —T/2 to O and O to T/2. For example, the derivation for ak is 2 F0+T ak=——/ COS/(mot dl‘ T to T/Z / f(t) cos kwot dt 2 T T/2 2 0 = — / f(t) cos kwot dt T 4/2 T ,"2 2 + — f(t) cos kwot dt. 16.32 T O ( ) Now we change a variable in the first integral on the right-hand side of Eq. (16.32). Specifically, we let tEx—Tfl. Then x=T/2, whent=0, x=0, whent=—T/2, dz? =dx. We rewrite the first integral as o / f(t) cos kwot dt —T/2 T/2 = f(x — T/2) cos kw0(x — T/Z) dx. (16.33) 0 Note that cos kwo(x — T/Z) = cos(kw0x — kn) = cos k7: cos kwox and that, by hypothesis, f(x — T/Z) = *f(x)- Therefore Eq. (16.33) becomes 0 / f0) cos kwot dt —T/2 T/2 . . =] [—f(x)] cos kn cos kwox dx. (16.34) 0 Incorporating Eq. (16.34) into Eq. (16.32) gives 2 T/2 ak = —f(1 — cos kn) 0 f(t) cos kwot dt. (16.35) But cos k7: 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. (1630) and (16.31) (sec Problem 16.5). We summarize our observations by noting that the Fourier series ) fmififimWfllb representation of a penOdlC funCtlon With half«wave symmetry has A zero average, or dc, value and contains only odd harmonics. <4 t' Cecivw ‘y‘y’ “Va 54 Ck fungi-gym lficrm ("E . I I I I I r I Brflneces‘gauly' p , (Sjl/“wAe/h7l /§ KM; I - ‘ _ 1 ’6‘ «av cm L « a? QWCGl wt\’)lt J v x V V / WV) The, 3 fvi-xmLu-M m 7332, an? if” W" Cb Vt?“ i” Lt} flu": Quarter-Wave Symmetry 5y Mum (1'7. arter-wave symmetry describes a periodic function that -- =- -:':-l m u Ina-"m- wmrnetg about the mid— :u The function illustrated . .g ‘ ._ _ J 1.2;. about the midpoint ofthe positive and negative half—cycles. The function in Fig. 16.9(b) does not have quarter-wave symmetry, although it does have half-wave symmetry. t A periodic function that lg; quarter-wave symmetry can always be ,4 )4 Ag; ( m__a_d_e either even or odd by the propercwe point where t = 0. “ a b fl L, (- For example, the function shown in Fig. 16.9(a) is odd and can be V made even by shifting the function T /4 units either right or left along the 1 axis. However, the function in Fig. 16.9(b) can never be made fit) either even or odd. To take advantage of quarter-wave symmetry in the calculation of the Fourier coefficients, you must choose the point D it here .-' : !.I tn make the function either even or odd. 3! '-?':° T:Ui!...!1<.‘i'| a ma e even, # ’— 1 The erm qu 1... .!. 1‘. '\ :_;-Ii--‘- . h. i- n- th- .-" _'.' :‘ltih. 1.1!. -. \‘;J".\ \"I'I‘. T T/Z (a) en T/2 :I. -: '1."_ because of half-wave symmetry, . «4 .x,;_ = I21 1' )r k even, because of half-wave symmetry, .‘5 = g / fa) cos kwoz dt, for k odd, i’ . -. (bl Figure 16.3 (a) A function that has quarter-wave symmetry. (b) A function that does not have quarter-wave symmetry. 91; z 41‘. fur all k‘, 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 superimposed on half—wave symmetry, so we can eliminate av and ak for k even. Comparing the expression for ak, k odd, in Eqs. (16.36) with Eq. (16.29) shows that combining quarter-wave symmetry with evenness allows the shortening of the range of integration from 0 to T/2 to 0 to T/4. We leave the derivation of Eqs. (16.36) to you in Problem 16.6. I the quarter wave symme ric :.:n-.=.I=.--:. a. _: ..: .-. -. av = 0, because the function 1- hid. ak = 0, for all k, because the i'n:».:::..\n w (bk : O, for k even, because (if T/4 bk = — f f (t) sin kwot dt, for k odd. '\ O A» can H N cram (Ude 7 ‘ Equations (16.37) are a direct consequence of quarter-wave symmetry Qufifi-{flt (AME [Mb TILYo and oddness. Again, quarter-wave symmetry allows the shortening of the interval of integration from 0 to T/2 to 0 to T/4. We leave the derivation of Eqs. (16.37) to you in Problem 16.7. Example 16.2 shows how to use symmetry to simplify the task of finding the Fourier coefficients. EXAMPLE 16.2 Find the Fourier series representation for the current waveform shown in Fig. 16.10. fit) I S 0 L U T l 0 N .X __ '. f _-‘_,-___ \ _ _,r. ‘.. _ —— —T :\ f ’ 'I" ‘ -' \L E" -? '- We begin by looking for degrees of symmetry in the waveform. We r’ find that the function is g‘édd and, in addition, h half-wave _a_n_d quarter-wave symmetry. Because the function is‘ all the a"'c'o- efficients are zero; that is, av = O and ak = 0 for all k. Because the Figure 16.10 The periodic waveform for Exam- function has half-wave symmetry, bk = O for even values of k. Be— pie 16'2- cause the function ha; quarter-wave symmetry, the expression for bk for odd values of k is "— 8 T/4 bk = —/ i(t) sin kwot dl‘. T 0 In the interval 0 5 t 5 T /4, the expression for i (t) is . m t = —t. t( ) T Thus W4 41 ~ bk = if -—mt sin kwot dt T 0 T . T/4 321m sm kwot ~ t cos kwot _ T2 k2wg kwo 0 81 , k7: . ' = 7:21:12 3m 7 (k 18 odd). , The Fourier series representation of i(t) is 81m °° 1 . mt . ' =— ——s1n—smna)t 1m n2 2 n2 2 o n=l,3,5. 81 1 l 1 m . - _ ‘ _ — . . . ' = —2— (Sm wot — 5 SID 3wot + 25 sm Swot 49 sm 76001 + > . 15.4 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 quantity. The cosine and sine terms may be merged in 00 either a cosine expression or a sine expression. Because we chose - v a at; 2: (an (OS Via/01' “1- the cosine format in the phasor method of analysis (see Chapter 9), 9 V ' we choose the cosine expression here for the alternative form of the series. Thus we write the Fourier series in Eq. (16.2) as f(t) = a, + Z A, cos(nw0t — H), (16.38) n=l where An and 6,, are defined by the complex quantity (16.39) We derive 1-H.) - '.'€ '1‘ .|!':-_i 5“, :.:>ilfg. !'J'I'.". phi-u -r tut-mod 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 f(r) = a, + Z a, cos nwot + b, cos(nw0t — 90°). (16.40) n=1 Adding the terms under the summation sign by using phasors gives flan cos moot} =anfl (16.41) and $91!)” cos(nwot — 900)} = b,[—90° = —jbn. {16-42) 4‘» “5"0‘0‘5 4' 6” 5’“ “"56 Then ; I a;"":[ [L n (as "wat- W 4" 1 '5 *6» fP{a,l cos nwot + b,l cosmth — 90°)} = an — an A” 4' 9’», 5,5 meat] remit—.61 . m , I - v ' ,5 :AnL—fl. (15-43) Kg“), 9": 4‘" an an Wh we inverse—transform Eq. (16.43), we get '- en '7 a, Cosnw-l’ 1' 5n 5M "Wof' an cos moot + b” cos(nw0t — 90°): CP"1{A,, [—Bn} : Afiicosgn as nwof = An cos(na)ot — 9"). (16.44) + 9:“ 9” S‘ ” nw‘ij Substituting Eq. (16.44) into Eq. (16.40) yields (16.38). Equa- : An COS ( “moi _ 9" ‘) tion (16.43) corresponds to Eq. (16.39). If the periodic function is w either even or odd, A” reduces to either an (even) or 19,, (odd), and 2 foe) : Q + 2 A“ £05 (n “gt_ 1% 9,, is either 0° (even) or 90° (odd). I I I V The derivation of the alternative form of the Fourier series for a given periodic function is illustrated in Example 16.3. E X A M P L E 1 6 .3 a) Derive the expressions for ak and bk for the periodic 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. (16.3%). SOLUTION a) The voltage v(t) is neither even nor odd, nor does it have half- wave symmetry. Therefore we use Eqs. (1614) and (16.5) to find ak and bk. Choosing to as zero, we obtain 2 r/4 T ak = — / Vm cos kwot dt + (0) cos kwot dt T 0 T 4 / _2Vm sinkwoflT/4_ Vm . kn _ T kwo $0 —kfl $111— and 2 nT/4 Vm sin kwot dt U4 0 > 21',” I -- cos kwot kwo b) The average value of 110) is Vm(T/4) Vm “"2 i T T' The values of ak — jb;c for k = 1, 2, and 3 are ‘V V 2v A : ai“}b1=—m-- 'l=f m —45°, 1 7t 71 IT .Vm Vm 0 A21 az—Jb2=0“1?-7L—‘90’ —Vm Vm Jivm , ._ =———— '-—-= —1350. A3”? M 371' 371' 3:1 L— Thus the first four terms in the Fourier series representation of v(t) are L’Vlfizcar V1,“th V ‘ 2V ‘ V v(t) = l +@ cos(a)0t — 45°) +@cos(2wot — 90°) 4 7: 71V fivm 03 t—135°+~~. + —-——737rfl c s( we ) \I 511x“ 5T/4 3T/2 7T/4 2T 0 TM 772 3T/4 T Figure 16.11 The periodic function for Exam- ple 16.3. 15.5 AN APPLICATION Now we illustrate how to use a Fourier series re Lesentation of a pgfigfligfimmmifu. action to find the steadv-state response of a WHe 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 {he dear-Ed response, or output, signal. The first step in finding the steady—state response is to represent the periodic excitation source with its Fourier series. After noting that the source has o_dd, half-wave, and .; thil'T'JZ " ' ‘ the Fourier coefficients reduce I_. ..... values: 9/ \ L‘. 531-)". [ilci 8 T/4 bk 2 —— / Vm sin kwot dt T 0 4V = —’1 (k is odd). (16.45) Ifk Then the Fourier series representation of 118 is 4V 1 m ~ sin moot. (16.46) (b) Figure 16.12 An RC circuit excited by a periodic voltage. (a) The RC series circuit. (b) The square- wave voltage. C V ~.——(/ . . . . (/gK) V ~ _——S——’ C Writing the series in expanded form, we have 17-“) _; [ S J ' g {u .CLg fl E 4V 4Vm . g t/ .( 2.1) =—msinwt+——sm3wt ' g J vg 7r 0 37: 0 ,5 a (d w) / —&—'gzc 4V 4V . 0 H'J + ——’“ sin 5th + J sin 76001? + . . -. (16.47) 571 77r The voltage source expressed by Eq. (16.47) is the eguivalent of infinitel many series-connected sinusoidal smitten. each source hav- ing its own amplitude and frequency. ":a_. ; ’ "2-3117: i=1 -:' suth source to the output voltage, we use tl';. - For any one of the sinusoidal sources, the phasor-domain expres- sion for the output voltage is -‘s.-..i-'. Vs = ——.—. (16.48) l-l— JwRC V0 All the voltage sources are expressed as $12? 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(a)t + 9) instead of cos(a)r + 9). a The phasor output voltage owing to the fundamental frequency of the smusordal source 15 (4Vm/7flfl V = ————. . "1 1+jcuoRC (1649) Writing V01 in polar form gives (7:; (4V ) —fl1 V01: _.___’"_L;, (15.50) 7r,/1+ ng2C2 * where fi1=tan_1w0RC. (15.51) From Eq. (16.50), the time-domain expression for the fundamental frequency component of 1),, is 4 Vm —————— sin(wot — ,31). 7TV/1 + cugRZC2 We derive the third—harmonic component of the output voltage in a Similar manner. The third-harmonic phasor voltage is (Wm/mg): " 1+ j3woRC 4Vm ‘ fl—fl, 3m/l + 9w5R2C2 v01 = (16.52) V03 (16.53) where 53 =tan_13woRC. (15.54) . The time-domain expression for the third-harmonic output voltage is 4Vm v03 = ———-—— 3n,/1+ 9w5R2C2 Hence the expression for the kth—harmonic component of the output sin(3w0t —— 133). (16.55) voltage is 2 4Vm , _ vok = ————~——— sin(ka)ot — M) (k 13 odd), (15.56) kin/1+ kzngzC2 where at =tan‘1kw0RC (k is odd). (15.57) We now write down the Fourier series representation of the output voltage: — fl 4Vm I _ n W = Z 7f n: y ny/l + (nwoRC)‘ The derivation of Eq. (16.58) was not difficult. But, although we have an analytic expression for the steady-state output, what 1100) looks like is not immediately apparent from Eq. (16.58). As we men— tioned 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 1200), if we focus on the frequency response of the circuit. For example, if C is large, 1 /anC is small for the higher order harmonics. Thus the capacitor short cir- “19 til-zit-r':‘::.:1:_:::i-.:. ' :Faa‘ :reforrn, and the " ..-._.;i:_-11I“.';:. compared to u'ics. Equation (lode, rcllects this condition oo 13 -_:-"|‘_1.r“-:-n_‘r;'..: -_-: .':‘-‘ .,\ .5... . '. 1-; uidt, luf 144.0» 00 4Vm i . m — Sin 1 — 90° v0 eroRC "22:5,". n2 (MOO ) —4v °° 1 w m Z —2 cos moot. (16.59) IrwoRC “:1” n Equation (1659) shows that the amplitude of the harmonic in the out- " ‘ ' 5 " -rupared with 1/ n for the input harmon- ica ff I'.‘ :5 ~ - -..::-;-_ ‘. '. 1511;. 1: then to a first approximation I) a) (15 60) % CO I . 00) 7rco0RC S 0 ’ and Fourier analvsis tells us that the S uare-wave ' . e ormed iant. Now let’s see what happens as C —> O. The circuit shows that 110 and pg 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, 00 4V 1 , 120:?“ E Esmnwot. (16.61) n=l,3,5,... But Eq. (16.61) is identical to Eq. (16.46), and therefore v0 ——> vg as C —> O. Thus Eq. (16.58) has proven useful because it enabled us to predic: ‘ 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 dgggion between the input and output in terms of how much memory the system weighting functi_on_had. In the freguency . ‘—_—‘._—l—"'——__-‘ . -—-" . domain, we look at the distortion between the steady-state input anc- output in terms of '1 ' altered as they are 2 7.51.3 '1I‘-_"E‘."1 . I ‘--~ 19.! phase of the harmonics are c :ircuit. When the network significantly alters iiic' grim: relationships among the harmonics at the output relative to that at the input, the output is 2': ,\ '- -: " inf)“: Thu: :'r:xl._l::rit:'-I domain, Wt: " ‘WJ P .t__._.l_ _ ac- dtstwiwvt / For Lire circa: here, arrip-iimde amplitudes of the input harmonics _... ..t. . . plitudes of the output harmonics decrease as ' aecause the reas the am- 1 . 1 n x/l + (nwoRC)2' This circuit also exhibits phase distortion because the phase angle 0:? each input harmonic is zero, whereas that of the nth harmonic in the output signal is —— tan“1 nwoRC. M An Application of the Direct approach to the Steady-State Response f“ M For the simple RC circuit shown in Fig. 16.l2(a). we can derive the expression 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 no by straightforward circuit analysis, we reason as follows. The square-wave excitation function alternates between charging- =he LT:3;‘.-:ictmr +Vm and —Vm. Af- ter the circuit reaches arr: 'i‘.'-<I:1re ran-Inuit]. Cl'H'i alternate chargi-ri—g Worries periodic. We know from the analysrs 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. l6.12(a) is as shown in Fig. 16.13. The analytic expressions for 1100‘) in the time intervals 0 5 t g T / 2 and T/2 51‘s T are Toward — V," Toward — Vm v0 = Vm + (V: ~ nae—“RC, 0 :2 s T/2, (18.62) Figure 16.13 The steady-state waveform of 12,, for ' _[1_(T mn/RC the circui in Fig. 16.12(a). v0 = —m + (V2 + lee /‘ , T/2 S t s T. (15.83) t ~(t~'{)/rzc ~zvmlp e i We derive Eqs. (16.62) and (16.63) by using the methods of Chapter 7, as summarized by Eq. (7.60). We obtain the values of V1 and V2 by noting from Eq. (16.62) that V2 = Vm + (V1 — tam-Wm, - (16.64) and from Eq. (16.63) that V1: -Vm + (V; + nae-“MC. (16.65) Solving Eqs. (16.64) and (16.65) for V1 and V2 yields Vm(1_ e—T/ZRC) V = —V = ——————-— 16.£‘6 Substituting Eq. (16.66) into Eqs. (16.62) and (16.63) gives 2V", _ . U0=Vm—- W8 t/RC, f and v — —V + —-—2Y’”———e‘[’—(T/2)1/RC r/2 < t < T (16 468) 0" m 1+e—T/2RC ’ — — ' ' Equations (16.67) and (16.68) indicate that'vo(t) has half-wave sym- metry and that therefore the average value of no is .2339. This result agrees with the Fourier series solution for the steady-state response— namely, that because the excitation function has no zero frequency cmt, the response can have no such component. Equatian (16.67) and (16.68) also show the effect of changing the size of the capacitor. If C is small, the exponential functions quickly vam'sh, v0 = Vm between 0 and T / 2, and 110 = —Vm between T /2 and T. In other words, no ——> pg as C —> 0. If C is large, the output waveformbe- comes triangular in shape, as Fig. 16.14 shows. Note tha@$ we may approximate the exponential terms e"/RC and e‘["(T/2)l/RC by the linear terms 1 — (t/RC) and 1 — -[- -- - ' "‘ tively. Equation (16.59) gives the waveform. M 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 (x. Figure 16.14 The effect of capacitor size on the steady-state response. periodic function described by Eqs. (16.67) and (16.68). We have already noted that the periodic voltage response has half-wave sym- metry. Therefore the Fourier series contains only odd harmonics. For k odd, 4 T/2 2Vm8—t/RC Gk = fI/O (Vm — COS [C6001 dt _ —-8RCV,,, ‘ T[1+(kwoRC)2] 4 772 2Vme"/RC .. bk 2 -]—‘V/Ov (Vm "‘ Sin kaI (it 4v,,, _ Skwova3C2 _ krr T[1+(kaRC)2] (k is odd), (16.69) (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 ag+b2—— —’"— (16.71) 'k’ kn \/1+(kw0RC)2’ and that a_k = —ka)oRC. (16.72) bk We leave you to verify Eqs. (16.69)—(16.72) in Problems 16.19 and 16.20. 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 16.21. With this illustrative circuit, we showed how to use the Fourier se- ries in conjunction with the p’ri‘ngiple of superposition to obtain the steady-state response to _a periodic driving function. Again, the prin- cipal shortcoming of the Fourier series approach is the difficulty of ascertaining the waveform of the response. However, by thinking in terms of a Circuit’s frequency 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 Prob— lems 16.25 and 16.27.) 3 "5 11‘7me E—FK'E‘GU‘ETW‘rv—Tm-T — —— — PERIODIC FUNCTIONS If we have the Fourier series representation of the voltage and current at a pair of terminals in a linear 1urnped~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. (1638), we write the periodic voltage and current at the terminals of a network as 00 v = VdC + Z V” COS(nw0t — elm) n=l (16.73) 00 i = 1dc + Z 1,, cos(nw0t — 9,"). (16.74) n=1 The notation used in Eqs. (16.73) and (16.74) is defined as follows: Vdc = the amplitude of the _dc voltage component, V" = the amplitude of the nth-harmonic voltage, 9U” = the phase angle of the nth~harmonic voltage. [dc = the amplitude of the dc current component, 1,, z: the amplitude of the nth-harmonic current, em = the phase angle of the nth-harmonic current. We assume that the current reference is in the direction of the reference voltage drop across the terminals (using the passive sign convention), so that the instantaneous power at the terminals is vz. The average ower is 1 t0+T 1 t0+T =?/ pdt=—T—/ uidt. to to 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, tit}; appear: to be a formidable task. because the product vi requires (16.75) -uI_'-.'-.' E“ u - -_.I_;:"- t tan:- tr..'*.:'.:=-_- "l‘.!'.'i_‘ . (Fir-1;" 1:1.- Uhnn _1J'r." 2.5:; C-I'i'1_)'::L'i'- ‘-\!i.-_::.' .mt': -_'. -=-_. =. '-' uen A revrew or Eqs. (lam—(10.10) should Nuance ,cu oi the validity of this observation. Therefore Eq. (16.75) reduces to [0+T / VnIn cos(nwor — 9m) P = ——Vdcldct[ +2: ? r 0 n=1 0 X cos(nwot — 6)") dt. (16.76) Now, using the trigonometric identity 1 1 cos oz cos )3 = E cos(a — 6) + E cos(oz + )3), we simplify Eq. (16.76) to (GMT/m1 1 00 ann [gel-T M P = Vac-Ida + T 2 [costevn — an) + cos(2nw0t —- 6W, — 61-0] dt. (16.77) 5":1223 tr.- ZCT’T. so |ii‘.Ll=.‘f i:'._' The second ‘-_- (16.78) 4, FZVL Equation (16.78) is particularly important because it states that in the case of an interaction between a periodic voltage and the corre— sponding periodic current, the total average power is the sum of the ‘ average powers obtained from the interactholtages oWncy Currents and voltages of different frequencies do not interact to produce average power. Therefore, in average—power calculations involving periodic functions, the total average power is the superposition of the average powers associated with each har— monic voltage and current. Example 16.4 illustrates the computation ~ of average power involving a periodic voltage. \(nIn_ \Q.I__r\:fl/n 2, r 01 v2 WL‘RN :Ph 00 P: P¢¢+éfpn (’65 ( 87% “(96/13 Vdcldc: rPdo ) EXAMPLE16.4 ,2“ L {[H‘ - / \ \ Assume that the periodic square-wave voltage in Example 16.3 is A Jr 1 t . applied across the terminals of a 15 S2 resistor. The value of Vm is A 3/ 5'32 3 I R L 60 v, and that O@ a) Write the first fi ve nonzero terms of the Fourier series representa— ] . tion of 00‘). Use the trigonometric 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 Q 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><T/4>r av = 15 V. T :i m 29-? .' From Example i' 6.3 we have V ’ —- _ — I - —_ A1=~/§60/7r=27.01V, 4, 3 1 1r J7r 71' L. 61:45°, A2 = 60/7: = 19.10 v, 62 = 90°, ' A3 = 20fi/n = 9.00 v, 63 =1350, A4 =0, 94 = 0°, A5 = 5.40 V, 95 =45°, 2 27f 1000 we = 37.1 = —-(5——) = 400:: rad/s. Thus, using the first five nonzero terms of the Fourier series, 0(1) = 15 + 27.01 cos(400m — 45°) + 19.10 cos(800nt — 90°) + 9.00 cos(12007rt — 135°) + 5.40 005(2000m — 45°) + . . . V, b) '1 he voltage is applied to the terrriimle- of a resistor, so we can find the power associated with each term as follows: ,7 1 '2 Pdc = 1—35 = 15 w, < a l’\ k L 2 f > 1 1 27.01 4' ‘ P1 — — == 24.32 W, n i V (1/) . p 2 15 ’ K ‘ 1 19.102 ‘ P2=- _ ==12.16W [73 V313" 2 13 r" 7’ I 92 . L _ __ = '7 P3—215 2..0W, pm : Vn 1 5.42 *' P5 = —— = 0.97 W. 41K 2 l C) To obtain the total average power delivered to the 15 8'2 resistor, we first calculate the rms value of v(t): 8‘41 fox Ulve fnm) 60 2 T 4 V“: I guizrmm + The total average power delivered to the 15 Q resistor is V (M - '302 A ' — (l )9: T=-——=:60W. 15 V 1 -" d) The total power delivered by the first five nonzero terms is {L MS 3 - ’ T =Pdc+P1+P2+I3+P5=53.15W. This is (55.15/60)(100), or 91.92% of the total. 5" J) )9” {7110+ Zion (09(Qvn “(57514) 5)“, /P Vex/way” {Zests’t’l‘b’g N 0:721) ow: 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, 1 10—1-1" Fm: \/— I my d:. (16.79) T .0 Representing f (t) by its Fourier series yields 2 1 t0+T 0° Frms = ? / av + E An cos(nwot — 9,.) dl‘. (16.80) to n=1 The integral of the squared time function simplifi 95 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 integrate to zero. Therefore Eq. (16.80) reduces to 1 2 I 00 T 2 Fm: ? avT—t-ZEAn \ n=l .00 ’ An =\a%+Z—2- n=1 1‘ z 7. I 6?” 4:5,] .00 A 2 A : ag+z<7£> (16.81) \ n=1 2 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 rrns 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 1) = 10 + 30 cos(a)0t — 91) + 20 c030.th —- 92) + 5 CUJ(3U)0t — 93) + 2 COS(5a)0t —~ 95). The rms value of this voltage is v = V/102 + <30N§>2 + (20%)2 + (s/Jifl + <2N§>2 = M = 27.65 v. Usually, infinitely many terms are required to represent a periodic . function by a Fourier series, and therefore Eq. (16.81) yields an esti- mate of the true rms value. We illustrate this result in Example 16.5. EXAMPLE153 Use Eq. (16.81) to estimate the rms value of the voltage in Exam- ple 16.4. SOLUTlON From Example 16.4, VdC = 15 V, V1 = 27.01 /«/2 V, the rms value of the fundamental, V2 = 19.10/«/2 the rms value of the second harmonic, V3 = 9.00M? v, the rms value of the third harmonic, V5 = 5.40 / J2 V, the _rms value of the fifth harmonic. Therefore, 1s2+<a5>2+<e>2+<a>2+<129>2 = 28.76 V. From Example 164, fie ‘trg rms value is 30 V. We approach this value by including more and more harmonics in Eq. (16.81). For example, if we include the harmonics through k = 9, the equation yields a value of 29.32 V. 15.3 TH E! EXPO MERIT] FORM OF TH E FOU R|Efi which completes the derivation of Eq. (1683). To complete the deri- . vation of Eq. (16.82), we first observe from Eq. (16.88) that S ERIES 1 1 fo-i-T The exponential form of the Fourier series is of interest because it ‘3 C0 = — f f (t) dt 2 av. (16.89) allows us to express the series concise-1y. The exponential form of the T t0 genes 15 1 Next we note that 00 1 = w jnmol < 1 _ t0+T , 1 f“) 2 We ’ “632) C-) = — / f(r)e1”“’0‘ dt = c; = —(a,. + jbn). (16.90) n=—oo _ T )0 2 where S . . . . ubstituting Eqs. (16.87), (16.89), and (16.90) into Eq. (16.86) yields 1 10+T . t 00 C = —- 1‘: ‘1’“) dt. 16.83 » . . " T f.) f )8 ( ) ftr) = C0 + Z (CneJWO’ + Cze‘m‘) n21 To derive Eqs. (16.82) and. (16.83), we return to Eq. (16.2) and replace Do 00 the cosine and sine functions with their exponential equivalents: : C” 61'”th + Z cze-jnwot. (16.91) ejnwcit + e-jnwot :0 ":1 003 "(00’ = (15-84) Note that the second summation on the right—hand side of Eq. (16.91) ejwot # [flaw is equivalent to summing Cnejna’o‘ from —1 to —00; that is, sin moot = (16.85) 00 ‘ _OO 1 - Z cge‘lmo‘ = Z CneJWO‘. (16.92) Substituting Eqs. (16.84) and (1685) into Eq. (16.2) gives 71:1 n=—l 00 a I I b . V Because the summation from —1 to —00 is the same as the summation f(t) = av + Z él-(elm’o' + e’lnw‘”) + 2—;(ejnw0t - eflwor) from —00 to —l, we use Eq. (16.92) to rewrite Eq. (16.91): :1 noo . . 0° . '1 . = av + 2CD; _ Jig) ejnwot + e—jnwoz- (16.86) = Z Cnelnwot + z Cnejnwot n=1 2 2 ":0 —Oo 00 1 Now we define C" as = Z Cnejnmot, (16.93) —OO 1 - A" e 16 87 Cu = Etan — an) : ‘2' ‘ m ” =1’2v3W ' ‘ ~ ( ' ) which completes the derivation of Eq. (16.82). _——’-—~ We may also express the rrns value of a periodic function in terms From the definition of C”, of the complex Fourier coefficients. From Eqs. (16.81), (16.87), and (16.89), 1 2 t0+T 2 to+T Cn = — —/ f(t) cos moot dt — j—/ f(t) sin moot dt 2 T ’0 T 2‘0 2 a2 + by: Frms = a); + Z 2 , 1 to” . . =1 = . ./ f(t)(cos moot — J sm moot) dz N n T to [12 + b2 (Cm: ———” ", . (16.95) 1 [0” 1' 0’d (1688) _’ — I 8— mo ti ' T [.0 fl ) C2=a2 (16.96) Substituting Eqs. (16.95) and (16.96) into Eq. (16.94) yields the de- sired expression: N Fm; = C5 + 2 2 10,12. (16.97) \J . n=l Example 16.6 illustrates the process of finding the exponential Fourier series representation of a periodic function. E X A M P L E 1 6 .6 Find the exponential Fourier series for the periodic voltage shown in Fig. 16.15 V“) Vm S 0 L U T I 0 N — I _ . —1/2 0 7/2 7—7/2 T T+1l2 t Using ——r/2 as the starting pomt for the integration, we have, from Eq. (16.83), Figure 16.15 The periodic voltage for Exam- ple 16.6. {/2 r C” = 4:] Vme—jnwot df 1 —r/2 Vm (e‘jWO’) “2 T ——]na)0 _r/2 _j Vm (e-jnwor/Z _ ejnwor/Z) ‘ ‘ non *5 [VM T: VMT/ Sm “Wot/2 2V , .__. E ‘ ' r r : ——-——-— E =-—”—L—s1nnwor/2. - .C 5’“ ’V‘u’o /Z __‘- [AU/Z} .z/L moor _ mm 2 / H t! m ' 5 ’ “ 1 Here, because v(t) has even symmetry, [9,, = 0 for an n, and hence a 5 Ha . V we expect C" to be real. Moreover, the amplitude of Cn follows a ‘ X (sin x) /x distribution, as indicated when we rewrite C _ er sin(nwor/2) "-— T moor/2 We say more about this subject in Section 16.9. The exponential series representation of u(t) is _d n:=—oo : er i sin(nwot/2) ejnwoz- T moor/2 7L=——OO (>0 . V—\ V r sm(nwor/2) -n v<r>= .> w w: AL’IPLlTUDE AND PHASE SPECTRA A periodic time function is defined by its Fourier coefficients and its period. In other words, when we know av, an, bn, and T, we can con- .. struct f (t), at least theoretically. When we know an and b", We also ’- know the amplitude (An) and phase angle (—6”) 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 waveform from the amplitude and phase angle data. Also, when a periodic driving 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 _, g ‘C r / phase may provide an understanding of the output waveform. m " ‘7': ’ 5/ l in terms of the am litude and hasc an 1e of each term in the Fourier - % J” . P P g :)C_551hD—e5(/) h We can present graphically the description of a periodic function series of f (t). The plot of the amplitude of each term versus the . ~ ‘5’— T frequency is called the amplitude spectrum of f (t), and the plot of the phase angle versus the frequency is called the phase spectrum of f (of) f (t). Because the amplitude and phase angle data occur at discrete values of the frequency (that is, at we, 2w0, 3600, . . .), these plots also are referred to as line spectra. An Illustration of Amplitude and Phase Spectra . nth. (‘flzo mace ?,ir§i2/T,r3r5 Amplitude and phase spectra plots are based on either Eq. (16.38) (An and —9,,) or Eq. (16.82) (Cn). We focus on Eq. (16.82) and leave the plots based on Eq. (16.38) to Problem 16.45. To illustrate the W amplitude and phase spectra, which are based on the “Wm to, of the Fourier series, we use the periodic voltage of xample 16.6. ’ L To aid the discussion, we assume that Vm = 5V and t : T/S. From Example 16.6, V 1: sin(na)0t 2 _ _ _ _ a _ 4 _ _ - anm /)’ 109876543276121234567891: T moor/2 -0.4 which for the assumed values of Vm and r reduces to Figure 1615 The plot of _Cssn_ versus H . < /5) . when r = T/5, for the periodic voltage for Sin rm 1 16.6. C,, =1——-——. (16.99) Exam" 8 INT/5 Figure 16.16 illustrates the plot of the magnitude of C, 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 1(sin x) /xl function. We used the order of the harmonic as the frequency scale because the numerical value of T is not specified. When we know T, we also know too and the frequency corresponding to each harmonic. Figure 16.17 provides the plot of l(sin x)/x[ versus x, Where x is ‘2” “1'5” '7 *0-5” 0 1 0-5" " 1°" 27' in radians. It shows that the function goes through zero whenever x is an integral multiple of mFrom Eq. (16.98), Figure 16.17 The plot of (sin x)/x versus x. 1 111 :1 H11 _ the. I..." UflL.a._Y_+.+ _+_+_+_§,1_.J_LL1_+_+_+_+_1_L n —15 —13 —11 —9 —7 —5 -3 —T 1 2 3 4 5 6 7 8 910111213141516 -14 -12 -10 -8 —6 —4 -2 (16.100) From Eq. (16.100), we deduce that the amplitude spectrum goes through zero whenever nr/ 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, and so on. In other words, the fifth, tenth, fifteenth, . . . harmonics are all zero. As the reciprocal of r/T becomes an in- creasingly larger integer, the number of harmonics between every 71' 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 too. Because C" is real for all n, the phase angle associated with C n is either zero or 180°, depending on the algebraic sign of (sin nrt/S)/ (mt/5). For example, the phase angle is zero for n = 0, i1, i2, 21:3, and i4. It is not defined at n = :1:5, because Cis is zero. The phase angle is 180° at n 2 11:6, i7, i8, and i9, and it is not defined at 21:10. This pattern repeats itself as n takes on larger integer values. Figure 16.18 shows the phase angle of Cn 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 to units to the right. By hypothesis, 0C) , . v(t) = :4 Cnel’m’o’. n=—-oo (16.101) Therefore 00 00 tie—:0): Z Cnej"w°(t_t°):= Z cne-J’wotoeiwot, (16.102) n=-OO Viz—00 which indicates that shifting the origin has no effect on the amplitude spectrum, because 1Cn1=lCne""“’0‘°l. (16.103) However, reference to Eq. (1687) reveals that the phase spectrum has changed to —(9n + ntwoto) rads. For example, let’s shift the periodic voltage in Example 1.6.6 r/ 2 units to the right. As before, we assume that r = T / 5; then the new phase angle 6;, is 9;, = ~(6n—1—mt/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 illustrate the transition from the Fourier series representation of a peri- odic function to the Fourier transform representation of a nonperiodic function. We discuss the Fourier transform in Chapter 17. —216' m Figure 16.19 The plot of egsn versus n for Eq. (16.104). ...
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fourier_series_continued - EXAMPLE 16.1 Find the Fourier...

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