chapter 15 - Chapter 15: – Fourier Series Exercises Ex....

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Unformatted text preview: Chapter 15: – Fourier Series Exercises Ex. 15.3-1 Notice that f (t − T ) = f1 (t − T ) + f 2 (t − T ) = f1 (t ) + f 2 (t ) = f ( t ) Therefore, f(t) is a periodic function having the same period, T. Next f ( t ) = k1 f1 ( t ) + k2 f 2 ( t ) ∞ = k1 a10 + ∑ ( a1n cos ( n ω 0 t ) + b1n sin ( n ω 0 t ) ) n =1 ∞ + k2 a20 + ∑ ( a2 n cos ( n ω 0 t ) + b2 n sin ( n ω 0 t ) ) n =1 ∞ = ( k1 a10 + k2 a20 ) + ∑ ( ( k1 a1n + k2 a2 n ) cos ( n ω 0 t ) + ( k1 b1n + k2 b2 n ) sin ( n ω 0 t ) ) n =1 Ex. 15.3-1 f(t) = K is a Fourier Series. The coefficients are a0 = K; an = bn = 0 for n ≥ 1. Ex. 15.3-2 f(t) = Acosw0t is a Fourier Series. a1 = A and all other coefficients are zero. 15-1 Ex. 15.4-1 2π π π T = 4 = , ω0 = = 4 rad s T 8 2 Set origin at t = 0, so have an odd function; then an = 0 for n = 0,1, . . . Also, f(t) has half wave symmetry, so bn = 0 for n = even. For odd n, we have bn = 2 T2 20 2 T2 ∫−T 2 f (t ) sin( n ω0 t ) dt = − T ∫−T 2 sin( n ω0 t ) dt + T ∫0 sin ( n ω0 t ) dt T 4T = ∫0 2 sin ( n ω 0 t ) dt T 4 4 (1−cos( n ω 0 T )) = = n = 1, 3, 5, . . . 2π nf 0T nπ Finally, f (t ) = 4 N1 ∑ sin nω 0t; πnn n odd and ω 0 = 4 rad s Ex. 15.4-2 T = π , ω0 = 2π =2 T a0 = 0 , an = 0 for all n odd function with quarter wave symmety ⇒ n = even bn = 0 −2t 0<t <π π 6 8 π4 bn = ∫0 f (t ) sin nω 0t dt where f (t ) = 6 π π ≤t <π −2 6 4 −24 1 nπ Thus bn = 2 2 sin πn 3 −24 N 1 nπ so f (t ) = 2 ∑ 2 sin sin (2nt ) π n =1 n 3 odd n 15-2 Ex. 15.4-3 a) is neither even nor odd. f(t) will contain both sine and cosine terms 1 b) wave symmetry ⇒ no even harmonics 4 c) average value of f(t) = 0 ⇒ a0 = 0 Ex. 15.5-1 T = 2 s, ω 0 = Cn = 2π = π rad/s T − jnπ t 12 1 1 − jnπ t 12 f (t )e dt = ∫0 e dt − ∫1 e − jnπ t dt 2 ∫0 2 2 1 1 −e − jnπ +1+ e − j 2 nπ − e− jnπ = (1− e− jnπ ) = 2 jnπ jnπ 2 Cn = jnπ 0 n odd n even Finally, f (t ) = 2 jπ 1 1 jπ t 1 j 3π t 1 j 5π t e + e + e + ... − e − jπ t − e− j 3π t − e− j 5π t −... 3 5 3 5 Ex. 15.5-3 Cn = 1 T 4 − jω0nt 1 T − j 2π nt T ∫−T 4 e dt = T − j 2π n e T (n −1) (−1) 2 πn Cn = 0 12 T 4 −T = 4 1 e − jπ n / 2 − e jπ n / 2 − j 2π n n odd n even , n ≠ 0 n =0 15-3 Ex. 15.6-1 Use the “stem plot” in Matlab to plot the required Fourier spectra: % Fourier Spectrum of a Pulse Train A = 8; T = 4; d = T/8; pi = 3.14159; w0 = 2*pi/T; % pulse amplitude % period % pulse width %fundamental frequency N = 49; n = linspace(-N,N,2*N+1); x = n*w0*d/2; % Eqn.15.6-3. Division by zero when n=0 causes Cn(N+1) to be NaN. Cn = (A*d/T)*sin(x)./x; Cn(N+1)=A*d/T; % Fix Cn(N+1); sin(0)/0 = 1 % Plot the spectrum using a stem plot stem(n,Cn,'filled'); xlabel('n'); ylabel('|Cn|'); title('Fourier Spectrum of Pules Train with d = T/8'); 15-4 Ex. 15.8-1 ω 0 = 4 rad/s From Example 15.4-1: N vs (t ) = 3.24 ∑ n =1 odd n 1 nπ sin n2 2 1 1 1 sin nω 0t = 3.24 sin 4 t − sin 12 t + sin 20 t − sin 28 t 9 25 49 The network function of the circuit is Vo (ω ) 1 1 1 jω C = = Vs (ω ) R + 1 1 + jω C R 1 + jω 4 jω C Evaluating the network function at the frequencies of the input series 1 H ( n4 ) = n = 1,3,5... 1 + j 16 n n H(n4) 1 0.062∠-86° 3 0.021∠-89° 5 0.012∠-89° 7 0.0009∠-89° Using superposition 0.021 0.012 0.0009 vo (t ) = 3.24 ( 0.062 ) sin ( 4 t −86° )− sin (12 t −89° ) + sin ( 20 t −89° )− sin ( 28 t −89° ) 9 25 49 H (ω ) = = vo (t ) = ( 0.2009 ) sin ( 4 t − 86° ) − (.00756 ) sin (12 t − 89° ) + ( 0.00156 ) sin ( 20 t − 89° ) − ( 5.95 × 10−5 ) sin ( 28 t − 89° ) Discarding the terms that are smaller than 25 of the fundamental term leaves vo (t ) = ( 0.2009 ) sin ( 4 t − 86° ) − (.00756 ) sin (12 t − 89° ) 15-5 Ex. 15.9-1 f (t ) = e − at u (t ) F (ω ) = ∫ +∞ −∞ f (t ) e jω t ∞ dt = ∫ e e − at jω t 0 ∞ e( ) 1 = dt = − ( a + jω ) 0 a + jω − a + jω t Ex. 15.10-1 F { f ( at )} = ∫ ∞ −∞ f ( at ) e− jω t dt Let τ = at ⇒ t = F { f ( at )} = ∫ ∞ −∞ τ a f (τ ) e − jωτ a d τ a = 1∞ 1 ω − j (ω a )τ dτ = F ∫−∞ f (τ ) e a a a Ex. 15.10-2 f (t ) = 1 2π 1 ω ∫ ( 2πδ (ω ) A) e dt = 2π ∫ ( 2πδ (ω ) A) dt = A ∞ 0+ jt 0− −∞ Ex. 15.11-1 F -1 {δ (ω − ω 0 )} = 1 2π ∫ ∞ −∞ δ (ω − ω 0 ) e jω t dt = Take the Fourier Transform of both sides to get: e jω0t + e − jω0t F { A cos ω 0t} = F A 2 ( 1 jω0t e 2π ) F e jω0t = 2πδ (ω − ω 0 ) A A F e jω0t + F e − jω0t = ( 2πδ (ω − ω 0 ) + 2πδ (ω + ω 0 ) ) = 2 2 = Aπδ (ω − ω 0 ) + Aπδ (ω + ω 0 ) (( ) ( )) 15-6 Ex. 15.12-1 a) b) Vin (ω ) = Win = 1 Wout = ∴η = ∫ π ∞ 0 1 π ∫ 120 24 + jω ⇒ Vin (ω ) = 1202 14400 = 2 2 24 + ω 576 + ω 2 ∞ 14400 14000 1 −1 ω dω = 24 tan 24 = 300 J 2 576 + ω π 0 48 24 48 14400 14000 1 −1 ω dω = 24 tan 24 = 61.3 J 2 576 + ω π 24 Wout 61.3 ×100% = ×100% = 20.5% Win 300 Ex. 15.13-1 f + ( t ) = te − at f − ( t ) = te at ∴ F + (s) = Then F (ω ) = F + ( s ) s = jω + F − (s) f − ( −t ) = −te− at ⇒ 1 (s + a) s = − jω = = 2 and F − ( s ) = 1 (s + a) + 2 s = jω 1 ( a + jω ) 2 − −1 (s + a) 2 −1 (s + a) 2 1 ( a − jω ) 2 s = − jω = − j 4aω (a 2 +ω2 ) 2 15-7 Problems Section 15.3: The Fourier Series P15.3-1 T = 2 s ⇒ ω0 = 2π = π rad/s and f (t ) = t 2 for 0 ≤ t ≤ 2 . The coefficients of the Fourier 2 series are given by: 1 22 t dt = 4 3 2 ∫0 22 4 an = ∫0 t 2 cos nπ t dt = 2 2 ( nπ ) a0 = bn = ∴ f (t ) = −4 2 22 ∫0 t sin nπ t dt = nπ 2 4 4N1 4∞1 + 2 ∑ 2 cos nπ t − ∑ sin nπ t π n =1 n 3 π n =1 n 15-8 P15.3-2 an = T 2π 2π 2 T 4 2 ∫0 cos n t dt + ∫T 2 cos n t dt T T T 4 1 = nπ = 2π sin n T t T 4 0 T 2π 2 + 2 sin n t T T 4 1 nπ nπ (sin 2 − 0) + 2 (sin nπ ) −sin 2 nπ (−1) n +1 1 2 odd n nπ sin =− = nπ nπ 2 even n 0 () 2π 2 sin n t dt T T T 1 2π 4 2π 2 cos n t + 2 cos n t =− nπ T 0 T T 4 nπ 1 =− (2 cos (nπ ) −1) − cos 2 nπ 3 n is odd nπ 2 n = 2,6,10,… = − nπ n = 4,8,12,… 0 bn = 2 T 4 2π ∫ sin n T t dt + T0 T 2 T 4 ∫ 15-9 P15.3-3 a 0 = average value of f ( t ) = t f ( t ) = A 1 − T an = 2 T ∫ T 0 t 2π A 1 − cos n T T A 2 for 0 ≤ t ≤ T 2A T 2π t dt = ∫ 0 cos n T T 1T 2π t dt − ∫ t cos n T0 T t dt T 2π 2π 2π cos n t+n t s in n t 2A 1 TT T = 2 0 − T T 2π n T 0 −A = 2 2 cos ( 2nπ ) − cos ( 0 ) + 2nπ sin ( 2nπ ) − 0 2n π =0 bn = 2 T ∫ T 0 t 2π A 1 − sin n T T 2 A T 2π t dt = sin n T ∫0 T 1T 2π t dt − ∫ t sin n 0 T T t dt T 2π 2π 2π t−n t cos n t sin n 2A 1 TT T = 0− 2 T T 2π n T 0 −A = 2 2 ( sin ( 2nπ ) − sin ( 0 ) ) − ( 2nπ cos ( 2nπ ) − 0 ) 2n π A = nπ f (t ) = A∞A 2π sin n +∑ 2 n =1 n π T t 15-10 P15.3-4 T = 2 s, ω 0 = 2π = π rad/s , a 0 = average value of f ( t ) = 1 , 2 f (t ) = t 2 an = 2 ∫ 2 0 t cos ( n π t ) dt = = for 0 ≤ t ≤ 2 cos ( n π t ) + ( n π t ) sin ( n π t ) ( nπ ) 2 2 0 1 cos ( 2nπ ) − cos ( 0 ) + 2nπ sin ( 2nπ ) − 0 n π2 2 =0 2 bn = 2 ∫ 2 0 t sin ( n π t ) dt = sin ( n π t ) − ( n π t ) cos ( n π t ) (nπ ) 2 2 0 1 ( sin ( 2nπ ) − sin ( 0 ) ) − ( 2nπ cos ( 2nπ ) − 0 ) n π2 −2 = nπ = 2 ∞ f (t ) = 1 − ∑ n =1 2 2π si n n nπ T t Use Matlab to check this answer: % P15.3-4 pi=3.14159; A=2; T=2; % input waveform parameters % period w0=2*pi/T; tf=2*T; dt=tf/200; t=0:dt:tf; % % % % a0=A/2; v1=0*t+a0; % avarage value of input % initialize input as vector fundamental frequency, rad/s final time time increment time, s for n=1:1:51 % for each term in the an=0; % specify series bn=-A/pi/n; cn=sqrt(an*an + bn*bn); % convert thetan=-atan2(bn,an); v1=v1+cn*cos(n*w0*t+thetan); % add the Fourier series end Fourier series ... coefficients of the input to magnitude and angle form next term of the input 15-11 plot(t, v1,'black') % plot the Fourier series grid xlabel('t, s') ylabel('f(t)') title('P15.3-4') 15-12 Section 15-4: Symmetry of the Function f(t) 2π π = rad/s . 4 2 The coefficients of the Fourier series are: 15.4-1 T = 4 s ⇒ ωo = a0 = average value of vd ( t ) = 0 an = 0 because vd(t) is an odd function of t. bn = 14 π ∫0 ( 6 − 3 t ) sin n 2 t dt 2 4 34 π π = 3∫ sin n t dt − ∫ t sin n t dt 0 20 2 2 4 π − cos n 2 t 3 1 ππ π − = 3 2 2 sin n t − n t cos n t π 2 2 2 2n π n 4 0 2 0 6 = ( −1 + cos ( 2nπ ) ) − n26 2 ( sin ( 2nπ ) − 0 ) − ( 2 n π cos ( 2nπ ) ) nπ π 12 = nπ 4 ( ) The Fourier series is: ∞ vd ( t ) = ∑ n =1 12 π sin n t nπ 2 P15.4-2 ∞ vc ( t ) = vd ( t − 1) − 6 = −6 + ∑ n =1 ∞ 12 12 π π π sin n ( t − 1) = −6 + ∑ s in n t − n nπ 2 2 2 n =1 n π 15-13 P15.4-3 2π 1000 π π = rad/s = krad/s .006 3 3 The coefficients of the Fourier series are: 3× 2 1 V a0 = average value of va ( t ) = 2 = 6 2 bn = 0 because va(t) is an even function of t. T = 6 ms = 0.006 s ⇒ ω o = 2 0.001 1000π an = 2 t dt ∫0 ( 3 − 3000 t ) cos n 3 0.006 0.001 1 1000π 1000π t dt − ( 2 ×106 ) ∫ t cos n = 2000 ∫ cos n 0 0 3 3 1000π sin n 3 = 2000 n 1000π 3 t − 1000 n 2106 π 2 9 t dt 0.001 1000π 1000π 1000π t+n t sin n t cos n 3 3 3 0 3 9 π π π π = 2000 sin n 3 − 0 − n 2103 π 2 cos n 3 − 1 + n 3 sin n 3 − 0 n1000 π 6 π 18 π 6 π sin n − 2 2 cos n − 1 − sin n = nπ 3 n π 3 nπ 3 18 π = − 2 2 cos n − 1 n π 3 The Fourier series is va ( t ) = 1 ∞ 18 nπ 1000 π + ∑ 2 2 1 − cos cos n 2 n =1 n π 3 3 t P15.4-4 1 ∞ 18 nπ π vb ( t ) = va ( t − 2 ) − 1 = −1 + + ∑ 2 2 1 − cos cos n ( t − 2 ) 2 n =1 n π 3 3 1 ∞ 18 nπ = − + ∑ 2 2 1 − cos 2 n =1 n π 3 2π 1000 π t−n cos n 3 3 15-14 P15.4-5 Choose t 0 = − π 2π =1 2π average value: a0 = 0 T = 2π , ω 0 = 2T f ( t ) sin nω 0 t dt T ∫0 an = 0 since have odd function bn = −π < t < π f (t ) = t bn = 2 2π π ∫ π t sin nt dt − π 1 sin nt t cos nt =2− n π n −π 1 π π b1 = + = 2 π 1 1 b2 = −1 b3 = 2 3 P15.4-6 T = 8 s, ω 0 = π 4 rad/s bn = 0 because f ( t ) is an even functon a0 = average = ( 2×2 ) − 2×1 = 1 4 8 4 T2 f ( t ) cos n ω 0 t dt T ∫0 π π 2 41 = ∫0 2 cos n t dt − ∫1 cos n t dt 4 4 8 2 nπ nπ = 3 sin 4 −sin 2 nπ a1 = .714, a2 = .955, a3 = .662 an = 15-15 P15.4-7 ω 0 = 2ω , T = π ω π ω 2ω 2A a0 = ∫−π A cos ω t dt = π π 2ω an = 2ω π π ∫ 2ω −π 2ω A cos ω t cos 2nω t dt π 2ω A sin ( 2n −1)ω t sin ( 2n +1)ω t 2ω = + 2( 2n +1)ω − π π 2( 2n −1)ω 2ω π π sin 2n −1) sin ( 2n +1) 2A ( 2+ 2 = 2n +1 π 2n −1 2A π π = 2 ( 2n +1) sin ( 2n −1) 2 −( 2n −1) sin ( 2n −1) 2 π ( 4n −1) =− 4A cos( nπ ) π ( 4n 2 −1) =− 4 A( −1) n π ( 4n 2 −1) bn = 0 due to symmetry P15.4-8 2π = 5 π rad s T 0 ≤ t ≤ .1 A cos ω 0t .1 ≤ t < .3 f (t ) = 0 A cos ω t .3 ≤ t ≤ .4 0 T = 0.4 s, ⇒ ω 0 = Choose period − .1 ≤ t ≤ .3 for integral 1 .1 A cos ω 0t = A π T ∫−.1 2 .1 an = ∫−.1 A cos ω 0t cos nω 0t dt T a0 = 15-16 .1 a1 = 5 A∫−.1 cos 2 ω 0t dt = A 2 .1 an = 5 A∫−.1 cos ω 0t cos nω 0t dt = 5 A∫−.1 1 [ cos 5π (1+ n)t + cos 5π (1− n)t ] dt 2 2 A cos (nπ / 2) n ≠1 = 1− n 2 π bn =0 because the function is even. .1 P15.4-9 a0 = 0 because the average value is zero an = 0 because the function is odd bn = 0 for even due to Next: nπ 8 sin T4 2 bn = ∫−T 4 t sin ( nω 0 t ) dt = 1 wave symmetry 4 nπ 8 − 4nπ cos 2 2 2 = n π n 2π 2 − 8 n2 π 2 for n = 1, 5, 9, ... for n = 3, 7,11, ... Section 15.5: Exponential Form of the Fourier Series P15.5-1 2π = 2π , the coefficients of the complex Fourier series are given by: 1 1 e jπ t − e − jπ t − j 2π n t 11 Cn = ∫ A sin (π t ) e − j 2π n t dt = ∫ A dt e 0 10 2j A 1 − jπ ( 2 n−1) t − jπ ( 2 n+1) t e dt = −e 2 j ∫0 T = 1 ⇒ ωo = ( ) 1 − jπ 2 n −1) t − jπ 2 n +1) t Ae ( e( −2 A = − = 2 2 j − jπ ( 2n − 1) − jπ ( 2n + 1) 0 π (4n − 1) where we have used e± j 2π n = 1 and e j π = e− j π . 15-17 P15.5-2 2π 1 1 A − j Cn = ∫ t e T T 0 T Recall the formula for integrating by parts: t2 ∫t 1 dv 2π − j nt = e T dt . nt dt = A 1 T2 ∫0 t e −j 2π nt T dt t2 t u dv = u v t2 − ∫ v du . Take u = t and t1 1 When n ≠ 0 , we get 2π t e− j T n t A Cn = 2 T − j 2π n T T + 0 T −j 1 e 2 π ∫0 j n T 2π − j nt − j 2π n AT e eT = + T − j 2 π n 2 π 2 n j T 2π nt T dt 0 T − j 2π n AT e e − j 2π n − 1 = + T − j 2 π n 2 π 2 n j T A =j 2π n Now for n = 0 we have C0 = 1 TA A ∫0 T t dt = 2 T Finally, f (t ) = A A +j 2 2π n =∞ 1 jn ∑ ne n =∞ 2π t T n≠0 15-18 P15.5-3 d /2 2π − jn t d /2 T dt e −d / 2 A Cn = ∫ T π − j n 2π t j nπ d − jn d T A e A e T eT = = − 2π T − j n 2π T j n 2π jn T −d / 2 T T π jnπ d − jn d T −e T A e = nπ 2j A nπ d = sin nπ T nπ d sin Ad T = T nπ d T P15.5-4 Cn = Let τ = t − td , then t = τ + td . 1 T 1 = T Cn = t0 +T −td ∫t −t 0 d t0 +T −td ∫t −t 0 d e − j n ω o td = T ( ( a f (τ ) + b ) e− j nω (τ +t )dτ o But ∫t −t 0 d be − j n ω oτ d ( a f (τ ) + b ) e− j nω τ e− j nω t dτ o t0 +T −td ∫t −t = a e − j n ω o td t0 +T −td 1 t0 +T ( a f ( t − td ) + b ) e− j nωot dt T ∫t0 0 d 1 )T ∫ od ( a f (τ ) + b ) e− j nω τ dτ o t0 +T −td t 0 −t d ( f (τ ) e − j n ωoτ dτ + e− j n ωotd t +T −td e − j nωoτ 0 dτ = b − j n ω o t −t 0 d 1 )T ∫ t0 +T −td t0 − t d b e− j n ωoτ dτ 0 n ≠ 0 so = b = 0 C0 = a C0 + b and C n = a e − j n ω o td C n n≠0 15-19 P15.5-5 T = 8 s, ω 0 = 2 × 2 − 2 (1× 1) 1 2π π = rad/s, C 0 = average value = = T 4 8 4 The coefficients of the exponential Fourier series are calculated as nπ nπ nπ t t t −j −j −j 1 2 1 −1 C n = ∫ − 1× e 4 dt + ∫ 2 × e 4 dt + ∫ − 1× e 4 dt −1 1 8 −2 −1 1 2 nπ nπ nπ −j −j −j t t t 4 4 4 1 e e e = −1 × + 2× + ( −1) × nπ nπ nπ 8 −j −j −j 4 −2 4 −1 4 1 −j = 2 nπ nπ π nπ π nπ j n4π −j j j − j n4 − j n2 2 4 −e +e −e 4 e − e − 2 e and C −n − nπ − nπ − nπ −j t −j t −j t 1 2 1 −1 4 4 = ∫ − 1× e dt + ∫ 2 × e dt + ∫ − 1× e 4 dt −2 −1 1 8 −1 1 2 nπ nπ nπ j t j t j t 4 4 4 1 e e e = −1× + 2× + ( −1) × nπ nπ nπ 8 j j j 4 −2 4 −1 4 1 = j 2 nπ nπ − j n4π −j −e 2 e nπ −j j nπ − 2 e 4 −e 4 π nπ j j n2 +e −e 4 = −C n The function is represented as ∞ −∞ n =1 n = −1 f ( t ) = C0 + ∑ Cn e j n ω0 t + ∑ C− n e − j n ω0 t 15-20 This result can be checked using MATLAB: pi = 3.14159; N=100; T = 8; t = linspace(0,2*T,200); c0 = 1/4; w0 = 2*pi/T; % % % % period time average value fundamental frequency for n = 1: N C(n) = -j*((exp(+j*n*pi/4)-exp(+j*n*pi/2))-2*(exp(-j*n*pi/4)exp(+j*n*pi/4))+(exp(-j*n*pi/2)-exp(-j*n*pi/4)))/(2*pi*n); end for i=1:length(t) f(i)=c0; for n=1:length(C) f(i)=f(i)+C(n)*exp(j*n*w0*t(i))+C(n)*exp(-j*n*w0*t(i)); end end plot(t,f,'black'); xlabel('t, sec'); ylabel('f(t)'); 15-21 Alternately, this result can be checked using Mathcad: N := 15 n := 1 , 2 .. N T := 8 ω := 2 d := T C := π T i := 1 , 2 .. 400 200 ⌠ ⌡ m := 1 , 2 .. N −1 t := d ⋅ i i ⌠ −1⋅ exp( −j⋅ n ⋅ ω⋅ t ) dt + ⌡ 1 2 −1 −2 ⌠ 2⋅ exp( −j ⋅ n ⋅ ω⋅ t) dt + −1 exp( −j ⋅ n ⋅ ω⋅ t ) dt ⌡ 1 n T ⌠ ⌡ C := −1 ⌠ −1⋅ exp( j ⋅ m⋅ ω⋅ t ) dt + 1 ⌡ −2 ⌠ 2⋅ exp( j ⋅ m⋅ ω⋅ t) dt + ⌡ −1 −1 exp( j ⋅ m⋅ ω⋅ t ) dt 1 m T N f ( i) := 2 ∑ N ( )∑ C ⋅ exp j ⋅ n ⋅ ω⋅ t + n i n =1 ( C ⋅ exp −1⋅ j ⋅ m⋅ ω⋅ t m ) i m=1 C 0.357 0.357 1.685 0.477 0.477 1.745 0.331 0.331 1.807 0 0 1.856 0.199 0.199 1.88 0.159 0.159 1.872 0.051 0.051 1.831 0 0 1.767 0.04 0.04 1.693 0.095 0.095 1.628 0.09 0.09 1.589 0 0 1.589 0.076 0.076 1.633 0.068 0.068 1.717 0.024 0.024 1.825 n 2 f ( i) 0 2 100 200 i 300 400 m = f ( i) = C= 1.643 15-22 P15.5-6 The function shown at right is related to the given function by v ( t ) = −v1 ( t + 1) − 6 (Multiply by –1 to flip v1 upside-down; subtract 6 to fix the average value; replace t by t+1 to shift to the left by 1 s.) From Table 15.5-1 v1 ( t ) = ∞ ∑ n =−∞ ∞ j A ( −1) j n ω 0 t j 6 ( −1) j n π t e =∑ e2 nπ nπ n = −∞ n n Therefore v ( t ) = −6 − ∞ ∑ n =−∞ n n ∞ j 6 ( −1) j n π (t +1) j 6 ( −1) j n π j n π t 2 e e 2 e 2 = −6 − ∑ nπ nπ n = −∞ The coefficients of this series are: j 6 ( −1) j n π C0 = −6 and Cn = − e2 nπ This result can be checked using Matlab: n pi = 3.14159; N=100; A = 6; T = 4; t = linspace(0,2*T,200); c0 = -6; w0 = 2*pi/T; % amplitude % period % time % average value % fundamental frequency for n = 1: N C(n) = (-j*A*(-1)^n/n/pi)*exp(+j*n*pi/2); D(n) = (+j*A*(-1)^n/n/pi)*exp(-j*n*pi/2); end for i=1:length(t) f(i)=c0; for n=1:length(C) f(i)=f(i)+C(n)*exp(j*n*w0*t(i))+D(n)*exp(-j*n*w0*t(i)); end end plot(t,f,'black'); xlabel('t, sec'); ylabel('f(t)'); title('p15.5-6') 15-23 15-24 P15.5-7 Represent the function as 1 − e −5 t 0 ≤ t ≤1 f ( t ) = −5 ( t −1) −5 1≤ t ≤ 2 −e e (Check: f ( 0 ) = 0, f (1) = 1 − e −5 1, f ( 2 ) = e−5 − e −5 = 0 ) T = 2 s, ω 0 = 2π 1 = π , also C0 = average value = 2 2 The coefficients of the exponential Fourier series are calculated as ( ) 2 1 1 −5 ( t −1) −5 t − j nπ t −5 − j nπ t ∫ 0 (1 − e ) e dt + ∫1 e − e e dt 2 1 1 2 − 5+ j n π t 2 1 ) dt − e −5 ∫ e− j nπ t dt = ∫ e − j nπ t dt − ∫ e −5 t e − j nπ t dt + e5 ∫ e ( 0 1 1 2 0 1 2 1 2 − 5+ j nπ ) t −( 5+ j nπ ) t 1 e− j nπ t e( e − j nπ t + e5 e = − − e −5 2 − j n π 0 − ( 5 + j n π ) 0 − ( 5 + j n π ) 1 − j n π 1 Cn = ( = )( ) − 5+ j n π ) 2 − 5+ j n π ) 1 e − j nπ − 1 e −5 e − j nπ − 1 5 e ( −e ( e− j nπ 2 − e− j nπ − − e −5 +e 2 − j n π − (5 + j nπ ) − (5 + j nπ ) − j nπ − j nπ 2 1 e − j nπ − 1 e −5 e − j nπ − 1 e−5 e− j 2 nπ − e− j nπ − e − j nπ −5 e = − −e + 2 − j n π − (5 + j nπ ) − (5 + j nπ ) − j nπ n n n n −5 −5 1 ( −1) − 1 e ( −1) − 1 e − ( −1) −5 1 − ( −1) = − −e + 2 − j n π − (5 + j nπ ) − (5 + j nπ ) − j n π The terms that include the factor e −5 = 0.00674 are small and can be ignored. n − ( −1)n 1 ( −1) − 1 −1 C n = − + 2 − j n π − (5 + j n π ) − (5 + j n π ) 1 1 odd n − = j nπ 5 + j nπ 0 even n 5 j nπ 5 + j nπ )( ) = ( 0 odd n even n 15-25 This result can be checked using Matlab: pi = 3.14159; N=101; T = 2; t = linspace(0,2*T,200); c0 = 0.5; w0 = 2*pi/T; % period % time % average value % fundamental frequency for n = 1:2:N if n == 2*(n/2) C(n) = 5/((+j*pi*n)*(5+j*pi*n)); D(n) = 5/((-j*pi*n)*(5-j*pi*n)); else C(n)=0; D(n)=0 end end for i=1:length(t) f(i)=c0; for n=1:length(C) f(i)=f(i)+C(n)*exp(j*n*w0*t(i))+D(n)*exp(-j*n*w0*t(i)); end end plot(t,f,'black'); xlabel('t, sec'); ylabel('f(t)'); title('p15.5-7') 15-26 Section 15-6: The Fourier Spectrum P15.6-1 Average value = 0 ⇒ a0 = 0 half − wave symmetry 4 T 2 4A 4A 2π an = T ∫0 − T t cos n T t dt = − 2 2 (cos (nπ ) −1) nπ ⇒ f (t ) T bn = 4 ∫0 2 − 4 A t sin n 2π t dt = − 2 A (1− cos ( n π ) ) T nπ T T 2 n Cn = a n + bn 2 bn an θ n = tan −1 1 1.509 ⋅ A −57.5° 20 0 3 0.434 ⋅ A −78.0° 40 0 5 0.257 ⋅ A −82.7° 60 0 7 0.183 ⋅ A −84.8° 15-27 P15.6-2 Mathcad spreadsheet (p15_6_2.mcd): N := 100 n := 1 , 2 .. N T := 32 ω0 := 2 π T Calculate the coefficients of the exponential Fourier series: T T ⌠4 4 C1 := ⋅ n T ⌡3⋅ T ⌠2 4 2⋅ π ⋅ t exp( −j⋅ n⋅ ω0⋅ t) dt C2 := ⋅ sin n T T T ⌡ 16⋅ t − 3 exp( −j⋅ n ⋅ ω0⋅ t) dt T 16 4 3⋅ T ⌠4 4 C3 := ⋅ n T ⌡11⋅ T T 4⌠ 2⋅ π ⋅ t exp( −j⋅ n ⋅ ω0⋅ t) dt C4 := ⋅ sin n T T T ⌡3⋅ 11 − 16⋅ t exp( −j⋅ n ⋅ ω0⋅ t) dt T 4 16 C := C1 + C2 + C3 + C4 n n n n n Check: Plot the function using it's exponential Fourier series: d := T 200 N i := 1 , 2 .. 400 t := d ⋅ i i f ( i) := ∑ N ( ∑ i) C ⋅ exp −j⋅ n ⋅ ω0⋅ t C ⋅ exp j⋅ n ⋅ ω0⋅ t + n n =1 n ( ) i n =1 5 f ( i) 0 5 10 20 30 40 50 60 ti 15-28 Plot the magnitude spectrum: 1.5 1 Cn 0.5 0 1 2 3 4 5 6 7 8 9 10 n That’s not a very nice plot. Here are the values of the coefficients: C n ( n) = arg C ⋅ 1.385 180 π = -115.853 0 -90 0.589 22.197 0 -24.775 0.195 -113.34 0 106.837 0.139 66.392 0 -78.232 0.082 -69.062 0 -48.814 0.039 109.584 0 90.415 0.027 -25.598 0 1.226·10 -3 63.432 0 163.724 78.14 15-29 P15.6-3 Use Euler’s formula to convert the trigonometric series of the input to an exponential series: vi ( t ) = 10 cos t + 10 cos 10 t + 10 cos 100 t V e−t + e−t e−10 t + e−10 t e−100 t + e−100 t = 10 + 10 + 10 2 2 2 −100 t −10 t −t 10 t t = 5e + 5 e + 5 e + 5 e + 5 e + 5 e100 t The corresponding Fourier spectrum is: Evaluating the network function at the frequencies of the input: ω, rad/s 1 10 100 |H(ω)| 1.923 0.400 0.005 ∠ H(ω), ° -23 -127 -174 Using superposition: vo ( t ) = 19.23 cos ( t − 23° ) + 4.0 cos (10 t − 127° ) + 0.05 cos (100 t − 174° ) V Use Euler’s formula to convert the trigonometric series of the output to an exponential series: ) ) ) ) +e ( ) e( +e ( e( +e ( vo ( t ) = 19.23 + 4.0 + 0.05 V 2 2 2 =19.23e j 174°e − j t + 4.0 e j 127° e − j 10 t + 19.23e j 23° e − j t + 19.23e− j 23° e j t + 4.0 e − j 127° e j 10 t 19.23e− j 174° e j t e j ( t − 23° ) − j t − 23° j 10 t −127° − j 10 t −127° j 100 t −174° − j 100 t −174° 15-30 P15.6-4 2π 1 = 2 π rad/s, C 0 = T 2 f (t ) = 1− t when 0 ≤ t < 1 s T = 1 s, ω 0 = The coefficients of the exponential Fourier series are given by Cn = 1 1 11 − j 2π n t dt = ∫ e − j 2 π n t dt − ∫ t e − j 2 π n t dt ∫0 (1 − t ) e 0 0 1 Evaluate the first integral as 1 − j 2π n t e 0 ∫ e − j 2π n t dt = − j 2π n 1 = 0 e − j 2π n − 1 =0 − j 2π n To evaluate the second integral, recall the formula for integrating by parts: t2 ∫t 1 t2 u dv = u v t2 − ∫ v du . Take u = t and dv = e − j 2π nt dt . Then t 1 1 ∫0 t e t1 − j 2π n t t e − j 2π n t dt = − j 2π n 1 + 0 1 − j 2π n t 1 dt ∫0 e j 2π n e − j 2π n e − j 2π n t = + − j 2 π n ( j 2 π n )2 1 = 0 e − j 2π n e − j 2π n − 1 1 + =j 2 2π n − j 2π n ( j 2π n ) Therefore 1 2 Cn = −j 2π n n=0 n≠0 To check these coefficients, represent the function by it’s Fourier series: 1 n=∞ − j j 2π n t j − j 2π n t f (t ) = + ∑ e e + 2 n=1 2 π n 2π n Next, use Matlab to plot the function from its Fourier seris (p15_6_4check.m): pi = 3.14159; N=20; T = 1; t = linspace(0,2*T,200); % period % time 15-31 c0 = 1/2; w0 = 2*pi/T; % average value % fundamental frequency for n = 1: N C(n) = -j/(2*pi*n); end for i=1:length(t) f(i)=c0; for n=1:length(C) f(i)=f(i)+C(n)*exp(j*n*w0*t(i))-C(n)*exp(-j*n*w0*t(i)); end end plot(t,f,'black'); xlabel('t, sec'); ylabel('f(t)'); This plot agrees with the given function, so we are confident that the coefficients are correct. The magnitudes of the coefficients of the exponential Fourier series are: 1 2 Cn = 1 2π n n=0 n≠0 Finally, use the “stemplot” in Matlab to plot the Fourier spectrum (p15_6_4spectrum.m): pi = 3.14159; N=20; n = linspace(-N,N,2*N+1); Cn = abs(1/(2*pi)./n); % Division by 0 when n=0 causes Cn(N+1)= NaN. 15-32 Cn(N+1)=1/2; % Fix Cn(N+1); C0=1/2 % Plot the spectrum using a stem plot stem(n,Cn,'-*k'); xlabel('n'); ylabel('|Cn|'); 15-33 Section 15.8: Circuits and Fourier Series P15.8-1 The network function of the circuit is: 100 Vo (ω ) 1 jω H (ω ) = = = 100 ω Vi (ω ) 10 + 1+ j jω 10 Evaluating the network function at the harmonic frequencies: 1 20 20 π nπ = = ∠ − tan −1 Hn = 2 1 + j n π 20 + j n π 20 400 + n 2 π 2 20 From problem 15.4-2, the Fourier series of the input voltage is ∞ vc ( t ) = −6 + ∑ n =1 12 π π si n n t − n nπ 2 2 Using superposition, the Fourier series of the output voltage is ∞ vo ( t ) = −6 + ∑ n =1 π π nπ sin n t − n + tan −1 20 n π 400 + n 2 π 2 2 2 240 P15.8-2 The network function of the circuit is: R2 H (ω ) = j ω C1 R 2 1 + j ω C2 R 2 Vo (ω ) =− =− 1 Vi (ω ) (1 + j ω C1R1 ) (1 + j ω C2 R 2 ) R1 + j ω C1 =− =− j ω (10−6 ) ( 2000 ) (1 + j ω (10 ) (1000) ) (1 + j ω (10 ) ( 2000 )) −6 j −6 ω 500 ω ω 1 + j 1 + j 1000 500 Evaluating the network function at the harmonic frequencies: 15-34 2π 1000 π 3 Hn =− π 2π 3 1 + j n 1 + j n 3 3 From problem 15.4-4, the Fourier series of the input voltage is jn 1 ∞ 18 2π nπ 1000 π vb ( t ) = − + ∑ 2 2 1 − cos t−n cos n 2 n =1 n π 3 3 3 Using superposition, the Fourier series of the output voltage is 1000 π 18 × H n 3 1 2π 1000 π nπ 1000 π vb ( t ) = − + ∑ 1 − cos 3 cos n 3 t − n 3 + ∠H n 3 2 2 2 n =1 nπ ∞ P15.8-3 H (ω ) = Vo (ω ) Ho = Vi (ω ) 1 + j ω p When ω = 0 (dc) −5 = − When ω = 100 rad/s R ( 2 ) ⇒ R = 25 kΩ 104 135° = ∠H (ω ) = 180° − tan −1 (ω C R ) ⇒ tan ( 45° ) = (100 ) C ( 25000 ) ⇒ C = 0.4 µF 25000 104 = 3.032 c4 = ( 5 ) H ( 400 ) = ( 5 ) 1 + j ( 400 ) ( 0.4 ×10−6 ) ( 25000 ) θ 4 = 45° + ∠H ( 400 ) = 45° + 180 − tan −1 ( 400 × 0.4 × 10−6 × 25000 ) = 149° 15-35 P15.8-4 When ω = 0 (dc) 5 = H o ( 2 ) ⇒ H o = 2.5 V/V When ω = 25 rad/s ω 25 −45° = ∠H (ω ) = − tan −1 ⇒ tan ( 45° ) = ⇒ p p p = 25 rad/s 2.5 = 3.03 100 1+ j 25 100 θ 4 = 45° + ∠H (100 ) = 45° − tan −1 = −31° 25 c4 = ( 5 ) H (100 ) = ( 5 ) P15.8-5 H (ω ) = When ω = 0 (dc) R2 R1 + R 2 = R2 Vo (ω ) = Vi (ω ) R1 + R 2 + jω C R1 R 2 3.75 2.25 2.25 ⇒ R1 = R2 = (500) = 300 Ω 6 3.75 3.75 When ω = 1000 rad/s R1 R 2 ( 300 ) ( 500 ) −20.5° = ∠H (ω ) = − tan −1 ω C ⇒ tan ( 20.5° ) = (1000 ) C R1 + R 2 800 ⇒ C = 2 µF 500 ( 5∠45° ) = 2.076∠ − 3.4 c3∠θ 3 = 800 + j ( 3000 ) ( 2 ×10−6 ) ( 500 ) ( 300 ) 15-36 P15.8-6 ˆ Rather than find the Fourier Series of v(t ) directly, consider the signal v(t ) shown above. These two signals are related by ˆ v(t ) = v (t − 1) − 6 since v(t ) is delayed by 1 ms and shifted down by 6 V. ˆ The Fourier series of v (t ) is obtained as follows: 2π radians π = rad/ms 4 ms 2 ˆ ˆ a n = 0 because the average value of v (t ) = 0 14 π ˆ ˆ bn = ∫0 ( 6−3t ) sin n t dt because v (t ) is an odd function. 2 2 34 4 π π = 3∫0 sin n t dt − ∫0 t sin n t dt 2 2 2 T = 4 ms ⇒ ω 0 = 4 π − cos n t 2 =3 4 1 π nπ π 3 − sin n t − t cos n t π 2 n 2π 2 2 2 2 0 n 4 2 0 6 12 = ( −1+ cos( 2nπ ) ) − n26 2 ( sin ( 2nπ ) − 0 ) − ( 2nπ −cos( 2π ) − 0 ) = nπ π nπ ( ) 12 π si n n t n =1 nπ 2 ∞ ˆ v (t ) = ∑ Finally, ˆ The Fourier series of v (t ) is obtained from the Fourier series of v (t ) as follows: ∞ 12 12 π π π sin n ( t −1) = − 6 + ∑ si n n t − n n =1 nπ 2 2 2 n =1 nπ ∞ v(t ) = − 6 + ∑ where t is in ms. Equivalently, 15-37 v(t ) = − 6 + ∞1 π π 3 ∑ sin n 10 t − n π n =1 n 2 2 12 where t is in s. R s L Next, the transfer function of the circuit is H ( s ) = = . 1 1 2R + Ls + R s + s+ Cs L LC R jω 104 jω L The network function of the circuit is H (ω ) = = . R 1 2 (108 − ω 2 ) + 104 jω − ω + jω L LC We see that H(0) = 0 and R j 20nπ π H ( n ω 0 ) = H n 10 3 = = 22 2 ( 400 − n π ) + j 20nπ Finally, v0 ( t ) = 12 π ∞ ∑ n =1 1 ( 400−n π ) 2 22 + 400n 2π 2 e 20 nπ j 90 − tan −1 400 − n 2π 2 π π 20nπ sin n 103 t − n + 90° − tan −1 2 2 2 400 − n π 2 n ( 400−n π ) 2 22 + 400n 2π 2 15-38 P15.8-7 ˆ Rather than find the Fourier Series of v(t ) directly, consider the signal v(t ) shown below. ˆ These two signals are related by v(t ) = v ( t − 2 ) − 1 ˆ Let's calculate the Fourier Series of v (t ), taking advantage of its symmetry. 2π rad π = rad ms 6 ms 3 3.2 1 ˆ ao = average value of v(t ) = 2 = V 6 2 ˆ bn = 0 because v(t ) is an even function T = 6 ms ⇒ ω 0 = π 2 1 an = 2 ∫0 ( 3−3t ) cos n t dt 3 6 1 π 1 π an = 2∫0 cos n t dt − 2∫0 t cos n t dt 3 3 π sin n 3− 1 = 2 π n 2π 2 n 3 9 = 1 π π π cos n t + n t sin n t 3 3 3 0 π 18 π6 π π 6 18 sin n − 2 2 cos n −1 + sin n = − 2 2 cos n −1 3 n π 3 nπ 3 3 nπ nπ so ˆ v(t ) = 1∞ 18 π π 1− cos n cos n t +∑ 2 2 2 n =1 n π 3 3 ˆ v ( t ) = v ( t − 2 ) −1 = − nπ 1∞ 18 1− cos +∑ 2 2 2 n =1 n π 3 2π π cos n t − n 3 3 15-39 where t is in ms. Equivalently, 1 ∞ 18 nπ v(t ) = − + ∑ 2 2 1− cos 2 n =1 n π 3 2π π 3 cos n 10 t − n 3 3 where t is in s. The network function of the circuit is: R2 − jω C1 R2 1+ jω C2 R2 = H (ω ) = 1 (1+ jω R1C1 )(1+ jω R2C2 ) R1 + jω C1 Evaluate the network function at the harmonic frequencies of the input to get. − jn π π 3 H ( nω 0 ) = H n 103 = 3 1+ jn π 1+ jn 2π 3 3 The gain and phase shift are H( n ω0 ) ∠H( n ω 0 ) n π nπ 3 = 22 22 n π 4 n π (9+ n π ) ( 9+ 4n2π 2 ) 1+ 1+ 9 9 π 2π = −90° − tan −1 n + tan −1 n 3 3 = 2 2 The output voltage is ∞ v0 ( t ) = ∑ nπ 181−cos 3 n =1 2π π 2π π 3 − 90° − tan −1 n − tan −1 n cos n 10 t − n 3 3 3 3 22 22 22 n π ( 9 + n π ) ( 9 + 4n π ) At t = 4 ms =0.004 s v0 (.004 ) nπ 2π 4π ° −1 π −1 2π 181− cos cos n − n − 90 − tan n − tan n 3 3 3 3 3 =∑ 22 22 22 n =1 n π ( 9 + n π ) ( 9 + 4n π ) ∞ 15-40 Section 15.9 The Fourier Transform P15.9-1 Let g ( t ) = e − at u ( t ) − e at u ( −t ) . Notice that f ( t ) = lim g ( t ) . Next a →0 ∞ ∞ G (ω ) = ∫ e e − at − jω t 0 dt − ∫ e e at − jω t −∞ Finally F (ω ) = lim G (ω ) = lim a →0 0 a →0 0 e− ( a + jω )t e( a − jω ) t dt = − − ( a + jω ) 0 ( a − jω ) −∞ −2 jω 2 = 2 2 a +ω jω −2 jω 1 1 = 0− − 0 = 2 − a +ω2 − ( a + jω ) ( a − jω ) P15.9-2 ∞ F (ω ) = ∫ Ae u ( t ) e − at −∞ − jω t ∞ ∞ dt = ∫ Ae e 0 − at − jω t Ae − ( a + jω ) t A A = 0− = dt = − ( a + jω ) 0 − ( a + jω ) a + jω P15.9-3 First notice that 2 AT T 2 ωT − AT 2 ωT Sa Then, from line 6 of Table 15.10-2: F { f1( t )} = − = Sa 2 2 4 4 4 AT 2 2 ωT d From line 7 of Table 15.10-2: F { f ( t )} = F f1( t ) = jω F { f1( t )} = − jω Sa 4 dt 4 2 ωT 2 sin AT 4 = 4 A sin 2 ωT This can be written as: F { f ( t )} = − jω 2 4 jω 4 ωT 4 15-41 P15.9-4 {( First notice that: F −1 δ ω − ω { } 1 1 − jω t = δ (ω − ω ) e dω = e 0 )} 2π ∫ 0 2π ∞ − jω t 0 −∞ Therefore F e − jω0t = 2πδ (ω − ω 0 ) . Next, 10 cos 50t = 5 e j 50t + 5 e − j 50t . Therefore F {10 cos 50t} = F {5 e j 50t } + F {5 e − j 50t } = 10πδ (ω − 50) + 10πδ (ω + 50) . P15.9-5 2 F (ω ) = −2 ∫ e 1 − jω t −2e − jω t dt = − jω 2 = 1 2 − j 2ω − jω ( e − e ) = j2 ( ( cos 2ω − j sin 2ω ) − ( cos ω − j sin ω ) ) jω ω = 2j ω 2 ( cos ω − cos 2ω ) + ( sin ω − sin 2ω ) ω P15.9-6 B F (ω ) = ∫ B 0 A − jω t A e − jω t A e − jω B 1 t e dt = jω B − 1) − 2 − jω t − 1) = 2( 2( B B ( − jω ) ω 0 B −ω = A − Be − jω B e− jω B 1 + 2 − 2 B jω ω ω = 1 ( e j 2ω − e− j 2ω ) − j1 ( e jω − e− jω ) jω ω P15.9-7 2 F (ω ) = ∫ e −2 − jω t 1 dt − ∫ e − jω t −1 e − jω t dt = − jω 2 e− jω t − − jω −2 1 −1 = P15.12-1 2 ω ( sin 2ω − sin ω ) is ( t ) = 40 signum ( t ) 2 80 I s (ω ) = 40 = jω jω I (ω ) 1 = H (ω ) = I s (ω ) 4 + jω I (ω ) = H (ω ) I s (ω ) = 1 80 20 20 × = − 4 + jω jω jω 4 + jω ∴ i ( t ) = 10 signum ( t ) − 20 e −4t u ( t ) 15-42 P15.12-2 is ( t ) = 100 cos 3t A I s (ω ) = 100π δ (ω − 3) + δ (ω + 3) I (ω ) 1 H (ω ) = = I s (ω ) 4 + jω δ (ω − 3 ) + δ ( ω + 3 ) I (ω ) = 100π 4 + jω e − j 3t 100π ∞ δ (ω − 3) + δ (ω + 3) jω t e j 3t + i (t ) = e dω = 50 2π ∫−∞ 4 + jω 4 − j3 4 + j3 = 10 e − j ( 3t −36.9 ) +e j ( 3t −36.9 ) = 10 cos ( 3t − 36.9 ) P15.12-3 v ( t ) = 10 cos 2t V (ω ) = 10π δ (ω + 2 ) + δ (ω − 2 ) 1 Y (ω ) = 2 + jω I (ω ) = Y (ω ) V (ω ) = i (t ) = 10π δ (ω + 2 ) + δ (ω − 2 ) 2 + jω 10π 2π δ (ω + 2 ) + δ (ω − 2 ) jω t e − j 2t e j 2t + e dω = 5 ∫−∞ 2 + jω 2 − j2 2 + j2 ∞ = 5 e − j ( 2 t − 45 ) +e j ( 2 t − 45) = 5 cos ( 2t − 45 ) A 15-43 P15.12-4 v ( t ) = e t u ( −t ) + u ( t ) F {e u ( −t )} = ∫ e u ( −t ) e ∞ t t − jω t −∞ dt = ∫ e e t − jω t −∞ e( ) dt = 1 − jω 1− jω t 0 = −∞ 1 1 − jω 1 jω F {u ( t )} = πδ (ω ) + ∴V (ω ) = 0 1 1 + πδ (ω ) + jω 1 − jω 1 1 1 1 1 2 + jω 2 jω , H (ω ) = = = 11 1 2 + jω 3 + jω 1+ + 2 jω 2 + jω 1 1 1 − 11 1 3 πδ ( ω ) 12 4 Vo (ω ) = 1 − jω + πδ (ω ) + jω = 3 + jω + 1 − jω + jω + 3 + jω 3 + jω πδ (ω ) 1 ∞ πδ (ω ) − jω t 1 F −1 = ∫−∞ 3 + jω e dω = 6 3 + jω 2π ∴ vo ( t ) = − 1 −3t 1 1 1 e u ( t ) + et u ( −t ) + signum ( t ) + 12 4 3 6 P15.12-5 vs ( t ) = 15e−5t u ( t ) V ⇒ V (ω ) = Ws = ∫ ∞ (15e u ( t ) ) dt = ∫ (15e ) dt = 22.5 −5t −∞ 15 5 + jω 2 ∞ −5t 2 0 J 1 1 jω C H (ω ) = = RC 1 1 R+ + jω jω C RC C =10 µ F. Try R =10 kΩ. Then 10 15 Vo (ω ) = × 10 + jω 5 + jω Wo = 1 π ∫ ∞ 0 2 10 15 1 × dω = π 10 + jω 5 + jω ∫ ∞ 0 2 300 300 dω = 15 J − 2 2 25 + ω 100 + ω 15-44 P15.12-6 H (ω ) = 4 4 + jω 8 8 − jω Vs (ω ) = F {8u ( t ) − 8u ( t − 1)} = 8πδ (ω ) + − 8πδ (ω ) + e jω jω 8 Vs (ω ) = (1 − e− jω ) since δ (ω ) e− jω = δ (ω ) jω Vo (ω ) = Next use 4 8 × (1 − e− jω ) = j8 − 4 +8jω − j8 − 4 +8jω e− jω 4 + jω jω ω ω 1 1 = + πδ (ω ) − πδ (ω ) to write jω jω 1 8 1 8 − jω Vo (ω ) = 8 + πδ (ω ) − πδ (ω ) − + πδ (ω ) − πδ (ω ) − − 8 e 4 + jω jω 4 + jω jω 1 8 1 8 − jω = 8 + πδ (ω ) − + πδ (ω ) − − 8 e 4 + jω jω 4 + jω jω ( vo ( t ) = 8u ( t ) − 8e −4t u ( t ) − 8u ( t − 1) − 8e = 8(1 − e −4t )u ( t ) − 8(1 − e −4( t −1) −4( t −1) u ( t − 1) ) )u ( t − 1) V 15-45 PSpice Problems SP 15-1 Vin R1 1 1 .tran .four .probe .end 0 0 pulse 1 (25 5 0 0 0 4 5) 0.01 5 0.2 v(1) FOURIER COMPONENTS OF TRANSIENT RESPONSE V(1) DC COMPONENT = HARMONIC NO 1 2 3 4 5 6 7 8 9 8.960000E+00 FREQUENCY (HZ) NORMALIZED COMPONENT 7.419E+00 6.030E+00 4.061E+00 1.935E+00 8.000E-02 1.182E+00 1.704E+00 1.537E+00 8.954E−01 2.000E-01 4.000E-01 6.000E-01 8.000E-01 1.000E+00 1.200E+00 1.400E+00 1.600E+00 1.800E+00 FOURIER COMPONENT 1.000E+00 8.127E-01 5.473E-01 2.609E-01 1.078E-02 1.593E-01 2.297E-01 2.072E-01 1.207E-01 PHASE (DEG) 1.253E+02 1.606E+02 -1.642E+02 -1.289E+02 -9.360E+01 1.217E+02 1.570E+02 -1.678E+02 -1.325E+02 NORMALIZED PHASE(DEG) 0.000E+00 3.528E+01 -2.894E+02 -2.542E+02 -2.189E+02 -3.600E+00 3.168E+01 -2.930E+02 -2.578E+02 SP 15-2 Vin R1 1 1 .tran .four .probe .end 0 0 pulse (1 -1 -0.5 1 0 0 1) 1 0.1 1 1 v(1) FOURIER COMPONENTS OF TRANSIENT RESPONSE V(1) DC COMPONENT = HARMONIC NO 1 2 3 4 5 6 7 8 9 1.299437E-02 FREQUENCY (HZ) 1.000E+00 2.000E+00 3.000E+00 4.000E+00 5.000E+00 6.000E+00 7.000E+00 8.000E+00 9.000E+00 FOURIER COMPONENT NORMALIZED COMPONENT 6.364E-01 3.180E-01 2.117E-01 1.585E-01 1.264E-01 1.051E-01 8.972E-02 7.817E-02 6.916E-02 1.000E+00 4.996E-01 3.326E-01 2.490E-01 1.987E-01 1.651E-01 1.410E-01 1.228E-01 1.087E-01 PHASE (DEG) -1.777E+02 4.679E+00 -1.730E+02 9.366E+00 -1.683E+02 1.407E+01 -1.636E+02 1.880E+01 -1.588E+02 NORMALIZED PHASE (DEG) 0.000E+00 1.823E+02 4.682E+00 1.870E+02 9.376E+00 1.917E+02 1.409E+01 1.965E+02 1.883E+01 Verification Problems 15-46 VP 15-1 t f (t ) = 2 + cos ⇒ a0 = 2 , a1 = 1 and all other coefficients are zero. 2 The computer printout is correct. VP 15-2 Table 15.4-2 shows that the average value of a full wave rectified sinewave is 2A 2(400) where A is the amplitude of the sinewave. In this case a0 = = 255. π π Unfortunately the report says, "half-wave rectified." The report is not correct. Design Problems DP 15-1 For sinusoidal analysis, shift horizontal axis to average, which is 6 V. Now we have an odd function so an = 0 T = π s , ω 0 = 2π / π = 2 rad/s 2×2 T / 2 f (t ) sin nω 0t dt T ∫0 Need third harmonic : 4 4 π /2 π /2 b3 = ∫0 sin 6t dt = − cos 6t 0 = 0.424 6π T v1 ( t ) = 0.424 sin ( 6t ) =0.424 cos(6t −90°) V ⇒ V1 (ω ) = 0.424∠−90° bn = Zc = −j −j = for third harmonic ω C 6C ∴ transfer function is H (3ω 0 ) = V2 (ω ) = H ( 3ω 0 ) V1 (ω ) = 16 16 − ( H(3ω 0 j 6C ) ∠H (3ω 0 ) ) (0.424∠−90° ) Choose V2 (ω ) = 1.36 ⇒ so H ( 3ω 0 ) = 3.2 This requires C = 1 16 = 3.2∠64.9° F. Then H ( 3ω 0 ) = 205 16− j 34 ∴ third harmonic of v2 ( t ) = 1.36 sin(6t + 64.9° ) V 15-47 DP 15-2 Refer to Table 15.4-2. 2A N 4A 1 −∑ v (t ) = cos(2nω0t ) s π n = 1 π 4n 2 − 1 In our case: 360 N 640 1 v (t ) = −∑ cos(2n377t ) s π n = 1 π 4n 2 − 1 N N n =1 n =1 Let vs (t ) = vs0 + ∑ vsn (t ) and v0 (t ) = vo0 + ∑ von (t ) We require ripple ≤ 0.04 ⋅ dc output ( N max ∑ von (t ) n =1 ) ≤ 0.04 ⋅ v o0 ⇒ vo1 (t ) ≤ 0.04 vo0 but vo0 = vs0 because the inductor acts like a short at dc. R Next, using the network function of the circuit gives Von = Vsn . R + jω 0 n L For n=1: Vo1 = R 1 640 1 640 so V01 = Vs1 = Vs1 , but Vs1 = R + jω 0 L 1+ j 377 L 1+ j 377 L 3π π (3) We require Vo1 ≤ 0.04 vo0 and vo0 = vs0 = 360 π . Then 1 1+ (377) 2 L 2 ⋅ 640 360 ≤ 0.04 3π π Solving for L yields L > 1.54 mH 15-48 DP 15-3 From Table 15.5-1, the Fourier series can represent the input to the circuit as: vs ( t ) = 1 π + ∞ j jω 0 t j jω 0 t 1 +e +∑ e e j nω 0 t 2 4 4 even n = 2 π (1 − n ) The transfer function of the circuit is calculated as Vo1 = Zp R Vs1 where Z p = Z L +Z p 1+ jω RC So 1 LC Vo == 1 1 Vs + ( jω ) 2 + ( jω ) RC LC The gain at dc, ω = 0 , is 1 so 1 v =v = o0 s0 π For n = 1 Vo1 = 1 1 1 vo0 = vs0 = 20 20 20 π 1 / LC ⇒ 2 ω 1 + RC LC = 2 1 20 π ω 4 + We are given ω =800π and R =75 kΩ. Choosing L =0.1 mH yields C =0.1 F 15-49 ...
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