chap5_fourier_series_lr_circuit.doc 1/18 Consider the rectangular pulse train of example 3.2 of the text as the input to the series LR circuit.

Transcription

1 chap5_fourier_series_lr_circuit.doc /8 Consider the rectangular pulse train of eample 3.2 of the tet as the input to the series LR circuit. t () Rectangular pulse train (t) input signal. t (sec) VS ( t) = () t + - L + R VR ( t) = y() t - (a) + - R + - (b) (a) Series RL circuit and (b) phasor equivalent circuit. This pulse train (t), an even function, has a period of T = 2 s and fundamental 2π frequency of ω = = π (rad/s). The trigonometric Fourier series of (t) has T coefficients a =.5 for k = (DC term) and 2 a k = sin(.5 kπ ) for k =, 2, 3, (harmonics) kπ yielding 2 t ( ) = a + ak cos( kπt) =.5 + sin(.5 kπ)cos( kπt). kπ k= k=

2 chap5_fourier_series_lr_circuit.doc 2/8 The cosine-with-phase trigonometric form of the trigonometric Fourier series for (t) has coefficients a =.5 for k = (DC term), 2 Ak = ak = sin(.5 kπ ) for k =,2,3, kπ k even, = 2 k odd kπ and k = even &,5,9, k k = θk = ak = 2 =. sin(.5 kπ ) k =,2,3, π k = 3,5,, (where ak < ) kπ The cosine-with-phase trigonometric form of the trigonometric Fourier series for (t) is then t () = a + A cos( kπt+ θ ) k= k k. 2 2 =.5 + sin(.5 kπ ) cos kπt+ ( sin(.5 kπ) k= kπ kπ By voltage division, the frequency response H( ω ) of the series RL circuit was found to be H ( ω) =. + jω ( L R) The cosine-with-phase trigonometric form of the trigonometric Fourier series for the system output y(t) is then k= ( π θk ) y y y k cos k= ( k ) t ( ) = a H() + A H( kπ ) cos kπt+ θ + H( kπ) = A + A k t+ k

3 chap5_fourier_series_lr_circuit.doc 3/8 where the cosine-with-phase Fourier series coefficients for the output y(t) are and y A = a H() = a, y Ak = Ak H( kπ ) = Ak + jω L R ( ) y θk = θk + Hk ( π) = θk +. + jω ( L R) For eample, the table below shows the values of these various coefficients when the inductance to resistance ratio L/R = /π. k = ω k /ω ω = π H ( ωk ) = A + k jkπ L / R.5.5 (n/a) = j 2 π θ k, y A k y θ k = (n/a) + j2 = j3 2 3π = (n/a) + j4 = j5 2 5π Obviously, these calculations are much easier when a computer program (e.g., MATLAB) is used. Also, in the MATLAB file, we go to the N th harmonic instead of in the summation.

4 chap5_fourier_series_lr_circuit.doc 4/8 % chap5_rect_train_trig_fs.m % Chapter 5 frequency response eample- % Calculate frequency response of a series LR circuit % to a rectangular pulse train input (t) using trig. FS. clear;clc;close all; T = 2; w = pi; % Fundamental period and frequency t = -.5*T:T/:.5*T; % time range LR = /4/pi; % ratio L/R for series LR circuit N = 5; % Number of harmonics (ecluding DC term) % a =.5; H = ;% Input DC coeff. & DC freq. resp. N = a*ones(,length(t)); % Input dc component yn = a*h*ones(,length(t)); % Output dc component for k=:n, ak(k) = (2/k/w)*sin(.5*k*w); % Input FS coeff. Ak(k) = abs(ak(k)); % Input FS ampl. coeff. Thetak(k) = angle(ak(k)); % Input FS angle coeff. Hwk(k) =./(+j*k*w*lr); % System frequency response Aky(k) = Ak(k)*abs(Hwk(k)); % Output FS ampl. coeff. Thetaky(k) = Thetak(k) + angle(hwk(k)); % Output FS angle coeff. N = N + Ak(k)*cos(k*w*t+Thetak(k)); % Input FS yn = yn + Aky(k)*cos(k*w*t+Thetaky(k)); % Output FS % tack on DC terms for line spectra plots Ak =[a,ak]; Thetak =[,Thetak]; Hwk =[H,Hwk];Aky =[a*h,aky]; % Plot input amplitude spectra Ak k = ::N; wk = k*w; % indices and discrete freqs for FS stem(k,ak,'r.'),ais([- N+.75]), ylabel(['a_k^'],'fontsize',6,'fontname','times') title('fourier series amplitude coefficients for (t)',... 'fontsize',6,'fontname','times'); label('k = \omega/\omega_','fontsize',6,... 'fontname','times') for n=:n+, % Label stems if(ak(n)>.9), tet(k(n),ak(n)+.5,[' ' num2str(ak(n),3)],... 'HorizontalAlignment','center','VerticalAlignment','top') % Plot H(wk) magnitude of frequency response at wk figure,stem(k,abs(hwk),'r.'),ais([- N+.]), ylabel([' H(\omega_k ) '],'fontsize',6,'fontname','times') title({'frequency response at Fourier series harmonics for (t)';... ['L/R = ',num2str(lr),' and N = ',num2str(n)]},... 'fontsize',6,'fontname','times'); label('k = \omega/\omega_','fontsize',6,... 'fontname','times')

5 chap5_fourier_series_lr_circuit.doc 5/8 for n=:n+, % Label stems if(abs(hwk(n))>.9), tet(k(n),abs(hwk(n))+.7,[' ' num2str(abs(hwk(n)),3)],... 'HorizontalAlignment','center','VerticalAlignment','top') % Plot output amplitude spectra Aky figure,stem(k,aky,'r.'),ais([- N+.75]), ylabel(['a_k^y'],'fontsize',6,'fontname','times') title({'fourier series amplitude coefficients for y(t)';... ['L/R = ',num2str(lr),' and N = ',num2str(n)]},... 'fontsize',6,'fontname','times'); label('k = \omega/\omega_','fontsize',6,... 'fontname','times') for n=:n+, % Label stems if(aky(n)>.9), tet(k(n),aky(n)+.5,[' ' num2str(aky(n),3)],... 'HorizontalAlignment','center','VerticalAlignment','top') % Plot (t) and cosine/w phase trig. Fourier series for (t) eact = [ ]; teact = [ ]; figure,plot(t,n,'b-',teact,eact,'k:',[-3 3],[,],'k-',... [ ],[-.,.],'k-'), ais([ ]), title(['fourier series for (t), N = ',num2str(n)],... 'fontsize',8,'fontname','times'); label('time (sec)','fontsize',6,'fontname','times'); ylabel(['_{',num2str(n),'}(t) (Volts)'],'fontsize',6,... 'fontname','times'); % Plot (t) & cosine/w phase trig. Fourier series for y(t) figure, plot(t,yn,'r-',teact,eact,'k:',[-3 3],[,],'k-',... [ ],[-.,.],'k-'), ais([ ]), title({'fourier series for y(t)';... ['L/R = ',num2str(lr),' and N = ',num2str(n)]},... 'fontsize',6,'fontname','times'); label('time (sec)','fontsize',6,'fontname','times'); ylabel(['y_{',num2str(n),'}(t) (Volts)'],'fontsize',6,... 'fontname','times') set(findobj('type','line'),'linewidth',.5) set(findobj('type','tet'),'fontname','times') set(findobj('type','aes'),'linewidth',2) set(findobj('type','line'),'markersize',4)

6 chap5_fourier_series_lr_circuit.doc 6/8 L/R = /(4π)..7 Fourier series amplitude coefficients for (t) A k k = ω/ω Frequency response at Fourier series harmonics for (t) L/R = and N = H(ω k ) k = ω/ω

7 chap5_fourier_series_lr_circuit.doc 7/8 Fourier series amplitude coefficients for y(t) L/R = and N = A k y k = ω/ω

8 chap5_fourier_series_lr_circuit.doc 8/8 Fourier series for (t), N = (t) (Volts) Time (sec) Fourier series for y(t) L/R = and N = 5.8 y 5 (t) (Volts) Time (sec)

9 chap5_fourier_series_lr_circuit.doc 9/8 Net L/R = /π. Fourier series amplitude coefficients for (t) A k k = ω/ω Frequency response at Fourier series harmonics for (t) L/R =.383 and N = H(ω k ) k = ω/ω

10 chap5_fourier_series_lr_circuit.doc /8.7 Fourier series amplitude coefficients for y(t) L/R =.383 and N = A k y k = ω/ω Fourier series for y(t) L/R =.383 and N = 5.8 y 5 (t) (Volts) Time (sec)

11 chap5_fourier_series_lr_circuit.doc /8 Net L/R = /π. Fourier series amplitude coefficients for (t) A k k = ω/ω Frequency response at Fourier series harmonics for (t) L/R = 3.83 and N = 5.8 H(ω k ) k = ω/ω

12 chap5_fourier_series_lr_circuit.doc 2/8.7 Fourier series amplitude coefficients for y(t) L/R = 3.83 and N = A k y k = ω/ω Fourier series for y(t) L/R = 3.83 and N = 5.8 y 5 (t) (Volts) Time (sec)

13 chap5_fourier_series_lr_circuit.doc 3/8 Alternately, we could do this using the comple eponential Fourier series (actually easier to implement in the MATLAB code). Here, since (t) is even, the comple eponential Fourier series coefficients are and c k c = a =.5 for k = (DC term) ak = = sin(.5 kπ ) for k = ±, ±2, ±3, (harmonics). 2 kπ The comple eponential Fourier series for (t) is then jk t t () = c e π. k= The comple eponential Fourier series coefficients for the output y(t) are simply obtained by multiplying the (t) coefficients by the frequency response H( ω ) of the series RL circuit at the frequencies corresponding to the harmonics, i.e., H( ω = ωk = kπ) =. + jkπ ( L R) This yields y c = c H() = a() =.5 for k = (DC term) and y ck = ck H( kπ) = sin(.5 kπ) kπ jkπ ( L + R) for k = ±, ±2, ±3, (harmonics). The comple eponential Fourier series for y(t) is then y jk t yt () = c e π. k= Once again, in the MATLAB file, we go to the N th harmonic instead of in the summation. k k

14 chap5_fourier_series_lr_circuit.doc 4/8 % chap5_rect_train_comple_ep_fs.m % Chapter 5 frequency response eample- % Calculate frequency response of a series LR circuit % to a rectangular pulse train input (t) using comple % eponential Fourier series. clear;clc;close all; T = 2; w = pi; % Fundamental period and frequency t = -.5*T:T/:.5*T; % time range LR = /4/pi; % ratio L/R for series LR circuit N = 5; % Number of harmonics (ecluding DC term) % c =.5; % Input DC coeff N = zeros(,length(t)); % initialize input vector yn = zeros(,length(t)); % initialize output vector for k = ::2*N+, ktmp = k-(n+); % Count from -N to +N % Calculate comple ep. coeff. for input and output if(ktmp == ), ck(k) = c; % DC coeff. Hwk(k) =.; % System DC freq. response cky(k) = c*hwk(k); else ck(k) = (/ktmp/w).*sin(.5*ktmp*w); Hwk(k) =./(+j*ktmp*w*lr); % System freq. response cky(k) = ck(k)*hwk(k); N = N + ck(k)*ep(j*ktmp*w*t); yn = yn + cky(k)*ep(j*ktmp*w*t); % Plot input amplitude spectra ck k = -N::N; wk = k*w; % indices and discrete freqs for FS stem(k,abs(ck),'r.'),ais([-n- N+.55]), ylabel([' c_k^ '],'fontsize',6,'fontname','times') title('comple ep. Fourier series for (t)',... 'fontsize',6,'fontname','times'); label('k = \omega/\omega_','fontsize',6,... 'fontname','times') for n=:2*n+, % Label stems if(abs(ck(n))>.4), tet(k(n),abs(ck(n))+.35,[' ' num2str(abs(ck(n)),2)],... 'HorizontalAlignment','center','VerticalAlignment','top') % Plot H(wk) magnitude of frequency response at wk figure,stem(k,abs(hwk),'r.'),ais([-n- N+.5]), ylabel([' H(\omega_k ) '],'fontsize',6,'fontname','times')

15 chap5_fourier_series_lr_circuit.doc 5/8 title({'freq. resp. at comple ep. Fourier series harmonics for (t)';... ['L/R = ',num2str(lr),' and N = ',num2str(n)]},... 'fontsize',6,'fontname','times'); label('k = \omega/\omega_','fontsize',6,... 'fontname','times') % Plot output amplitude spectra cky figure,stem(k,abs(cky),'r.'),ais([-n- N+.55]), ylabel([' c_k^y '],'fontsize',6,'fontname','times') title({'comple ep. Fourier series for y(t)',... ['L/R = ',num2str(lr),' and N = ',num2str(n)]},... 'fontsize',6,'fontname','times'); label('k = \omega/\omega_','fontsize',6,... 'fontname','times') for n=:2*n+, % Label stems if(abs(cky(n))>.2), tet(k(n),abs(cky(n))+.35,[' ' num2str(abs(cky(n)),2)],... 'HorizontalAlignment','center','VerticalAlignment','top') % Plot (t) and comple ep. Fourier series for (t) eact = [ ]; teact = [ ]; figure,plot(t,n,'b-',teact,eact,'k:',[-3 3],[,],'k-',... [ ],[-.,.],'k-'), ais([ ]), title(['comple ep. Fourier series for (t), N = ',num2str(n)],... 'fontsize',8,'fontname','times'); label('time (sec)','fontsize',6,'fontname','times'); ylabel(['_{',num2str(n),'}(t) (Volts)'],'fontsize',6,... 'fontname','times'); % Plot (t) & comple ep. Fourier series for y(t) figure, plot(t,yn,'r-',teact,eact,'k:',[-3 3],[,],'k-',... [ ],[-.,.],'k-'), ais([ ]), title({'comple eponential Fourier series for y(t)';... ['L/R = ',num2str(lr),' and N = ',num2str(n)]},... 'fontsize',6,'fontname','times'); label('time (sec)','fontsize',6,'fontname','times'); ylabel(['y_{',num2str(n),'}(t) (Volts)'],'fontsize',6,... 'fontname','times') set(findobj('type','line'),'linewidth',.5) set(findobj('type','tet'),'fontname','times') set(findobj('type','aes'),'linewidth',2) set(findobj('type','line'),'markersize',6)

16 chap5_fourier_series_lr_circuit.doc 6/8 Comple ep. Fourier series for (t) c k k = ω/ω Freq. resp. at comple ep. Fourier series harmonics for (t) L/R = and N = H(ω k ) k = ω/ω

17 chap5_fourier_series_lr_circuit.doc 7/8 Comple ep. Fourier series for y(t) L/R = and N = c k y k = ω/ω

18 chap5_fourier_series_lr_circuit.doc 8/8 Comple ep. Fourier series for (t), N = (t) (Volts) Time (sec) Comple eponential Fourier series for y(t) L/R = and N = 5.8 y 5 (t) (Volts) Time (sec)