Homework 6. April 11, H(s) = Y (s) X(s) = s 1. s 2 + 3s + 2

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1 Homework 6 April, 6 Problem Steady state of LTI systems The transfer function of an LTI system is H(s) = Y (s) X(s) = s s + 3s + If the input to this system is x(t) = + cos(t + π/4), < t <, what is the output y(t) in the steady state. Solution According to the eigenfunction property of LTI,the steady state response corresponding to the given x(t) = + cos(t + π/4) = cos(t) + cos(t + π/4) is Since y(t) = H(j) cos(t + H(j)) + H(j) cos(t + π/4 + H(j)) H(j) = H(s) s=j = ej H(j) = H(s) s=j = j + 3j = + j 5 = = Hence, y(t) = cos(t ) Problem Fourier series of sampling delta The periodic signal δ Ts (t) = m= δ(t mt s ) will be very useful in the sampling of continuous-time signal.

2 (a) Find the Fourier series of the signal-that is δ Ts (t) = k e jkωst k= find the Fourier coefficients k. (b) Plot the magnitude line spectrum of this signal. (c) Plot δ Ts (t) and its corresponding line spectrum k as functions of time and frequency. Are they both periodic? How are their periods related? Explain. Solution (a)a period of δ Ts (t) is δ(t ) for T s / t + T s /.The Fourier series coefficients of δ Ts (t) of period T s are given by Thus, k = T s L[δ(t )] s=jωsk = T s e jkωs δ Ts = k= T s e jkωs(t ) (b) The magnitude line spectrum of δ Ts (t) is a periodic signal of fundamental frequency Ω s = π/t s and magnitude /T s. (c) The magnitude of k and signal δ Ts (t) is periodic with T s and Ω s = π/t s. The angle of k is k = kω s,which is an odd function. Problem 3 Manipulation of periodic signals Let the following be the Fourier series of a periodic signal x(t) of period T (fundamental frequency Ω = π/t ): x(t) = X k e jω kt k= Consider the following functions of x(t), and determine if they are periodic and what are their periods if so: () y(t) = x(t) + 3 () z(t) = x(t ) + x(t) (3) w(t) = x(t ) Express the Fourier series coefficients Y k, Z k and W k in terms of X k. Solution 3 (a) If the periodic signal is x(t) = k X ke jω kt,then y(t) = x(t) + 3 = (X + 3) + k X k e jω kt

3 is also periodic of period T with Fourier coefficients { X + 3 k = Y k = X k k (b) The signal is z(t) = x(t ) + x(t) = k X k e jω k(t ) + k X k e jω kt = k [X k ( + e jω k )]e jω kt is periodic of period T and with Fourier series coefficients (c) The signal Z k = X k ( + e jω k ) w(t) = x(t ) = k = m even X k e jω k(t ) X m/ e jω mt e jω m/ is periodic of period T /,with Fourier series coefficients { Xk/ e W k = jω k/ k even otherwise problem 4:Windowing and music soundsmatlab In the computer generation of musical sounds,pure tones need to be windowed to make them more interesting.windowing mimics the way a musician would approach the generation of a certain sound.increasing the richness of the harmonic frequencies is the result of the windowing,as we will see in this problem.consider the generation of a musical note with frequencies around f A = 88Hz.Assume our musician while playing this note uses three strokes corresponding to a window w (t) = r(t) r(t T ) r(t T ) + r(t T ),so that the resulting sound would be the multiplication,or windowing,of a pure sinusoid cos(πf A t) by a periodic signal w(t),with w (t) a period that repeats every T = 5T where T is the period of the sinusoid.let T = T /4 and T = T /4. (a) Analytically determine the Fourier series of the window w(t) and plot its line spectrum using MATLAB.Indicate how you would choose the number of harmonics needed to obtain a good approximation to w(t) (b) Use the modulation or the convolution properties of the Fourier series to obtain the coefcients of the product s(t) = cos(πf A t)w(t).use Matlab to plot the line spectrum of this periodic signal and again determine how many harmonic frequencies you would need to obtain a good approximation to s(t). (c) The line spectrum of the pure tone p(t) = cos(πf A t) only displays one harmonic,the one

4 corresponding to the f A = 88Hz frequency.how many more harmonics does s(t) have?to listen to the richness in harmonics use the MATLAB function sound to play the sinusoid p(t) and s(t)(use F s = 88Hz to play both). (d) Consider a combination of notes in a certain scale; for instance, let p(t) = sin(π 44t) + sin(π 55t) + sin(π 66t) Use the same windowing w(t),and let s(t) = p(t)w(t).use MATLAB to plot p(t) and s(t) and to compute and plot their corresponding line spectra.use sound to play p(nt s ) and s(nt s ) using F s =. Solution 4:if T = T /4 (a) Since window w(t) is a periodic function with period T = 5T,where T = /f A = /88 sec., and T = T /4,T = T /, w == r(t) r(t T ) r(t T ) + r(t T ),then the Fourier series of the window w(t) is W k = T W (s) s=jkω = s ( e st /4 e st / + e st ) s=jkω = ( cos(kπ/) cos(kπ) + j sin(kπ/)) Ω k Ω = π = 354πrad/sec T (b) The signal s(t) = cos(πf A t)w(t),by the convolution properties of the Fourier series,the s(t) is obtained by shifting the harmonics of w(t) to a central frequency πf A c l e a r a l l ; c l o s e a l l ; c l c ; c l e a r a l l ; c l c ; 3 Fs =88; T=/Fs ; 5 T=5 T; T=T/4;T=T/; Ts=/76; % o v e r a l l sample p eriod 7 t =:Ts : T Ts ; Nn=l e n g t h ( t ) ; x=[ t ( : c e i l (Nn/4) ) t ( c e i l (Nn/4) +) ones (, c e i l (Nn/4) ) ] ; 9 x=x ( c e i l (Nn/4) +) ones (, c e i l (Nn/4) ) t ( : c e i l (Nn/4) ) ; wind=[x x ] ; f i g u r e ( ) ; 3 p l o t ( t, wind ) ; g r i d on ; t i t l e ( wind ( t ) ) ; x l a b e l ( t ) ; y l a b e l ( magnitude ) ; 5 %a syms t 7 N=; x=t h e a v i s i d e ( t ) (t T/4) h e a v i s i d e ( t T/4)... 9 (t T/) h e a v i s i d e ( t T/)+... ( t T) h e a v i s i d e ( t T) ; [ X, w]= f o u r i e r s e r i e s ( x, T,N) ; f i g u r e ( ) 3 subplot (,, ) ;

5 w= f l i p l r (w) ; 5 w=[w w ] ; stem (w, [ f l i p l r ( abs (X) ) abs (X) ] ) ;%t i t l e ( Magnitude l i n e spectrum ) ; 7 t i t l e ( The l i n e spectrum o f w( t ) a ) ; subplot (,, ) ; 9 stem (w,[ f l i p l r ( angle (X) ) angle (X) ] ) ;%t i t l e ( Phase l i n e spectrum ) ; 3 %b s=cos ( pi Fs t ) x ; 33 [ X, w]= f o u r i e r s e r i e s ( s, T,N) ; f i g u r e ( 3 ) 35 subplot (,, ) ; w= f l i p l r (w) ; 37 w=[w w ] ; stem (w, [ f l i p l r ( abs (X) ) abs (X) ] ) ;%t i t l e ( Magnitude l i n e spectrum ) ; 39 t i t l e ( The l i n e spectrum o f s ( t ) b ) ; subplot (,, ) ; 4 stem (w,[ f l i p l r ( angle (X) ) angle (X) ] ) ;%t i t l e ( Phase l i n e spectrum ) ; 43 %c p=cos ( pi Fs t ) ; 45 [ X, w]= f o u r i e r s e r i e s (p, T,N) ; f i g u r e ( 4 ) 47 subplot (,, ) ; w= f l i p l r (w) ; 49 w=[w w ] ; stem (w, [ f l i p l r ( abs (X) ) abs (X) ] ) ;%t i t l e ( Magnitude l i n e spectrum ) ; 5 t i t l e ( The l i n e spectrum o f p ( t ) c ) ; subplot (,, ) ; 53 stem (w,[ f l i p l r ( angle (X) ) angle (X) ] ) ;%t i t l e ( Phase l i n e spectrum ) ; 55 t =:Ts : T Ts ; p s=cos ( pi Fs. t ) ; 57 sound ( p s ) ; pause ( ) ; s s=cos ( pi Fs. t. wind ) ; 59 sound ( s s ) ; 6 %d Fs =; 63 p=s i n ( pi 44 t )+s i n ( pi 55 t )+s i n ( pi 66 t ) ; s=p x ; 65 T p=/; f i g u r e ( 5 ) ; 67 subplot (,, ) ; e z p l o t (p, [ T p ] ) ; t i t l e ( p ( t ) ) ; 69 subplot (,, ) ; e z p l o t ( s, [ T ] ) ; t i t l e ( s ( t ) ) ; 7 t i t l e ( the f i g u r e o f p ( t ) and s ( t ) ) ; [ X, w]= f o u r i e r s e r i e s (p, T p,n) ; 73 f i g u r e ( 6 ) ; 75 subplot (,, ) ; w= f l i p l r (w) ; 77 w=[w w ] ; stem (w, [ f l i p l r ( abs (X) ) abs (X) ] ) ;%t i t l e ( Magnitude l i n e spectrum ) ; 79 t i t l e ( The l i n e spectrum o f p ( t ) d ) ;

6 subplot (,, ) ; 8 stem (w,[ f l i p l r ( angle (X) ) angle (X) ] ) ;%t i t l e ( Phase l i n e spectrum ) ; [ X, w]= f o u r i e r s e r i e s ( s, T,N) ; 87 f i g u r e ( 7 ) ; subplot (,, ) ; 89 w= f l i p l r (w) ; w=[w w ] ; 9 stem (w, [ f l i p l r ( abs (X) ) abs (X) ] ) ;%t i t l e ( Magnitude l i n e spectrum ) ; t i t l e ( The l i n e spectrum o f s ( t ) d ) ; 93 subplot (,, ) ; stem (w,[ f l i p l r ( angle (X) ) angle (X) ] ) ;%t i t l e ( Phase l i n e spectrum ) ; Prob 44.m

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9 Solution 4:if T =3 T /4 (a) The fundamental period of the sinusoid is T = /88 so that T = 5T = 5/88 sec., T = T /4 and T = 3T /4. The Fourier series coefficients of ω(t) are W k = T W (s) s=jkω = s ( e st/4 e s3t/4 + e st ) s=jkω = e st/ s (e st/ e st/4 e st/4 + e st/ ) s=jkω = e jkπ Ω (cos(πk) cos(πk/)) Ω k = π T = 35π rad/sec (b) The signal s(t) = w(t) cos(πf A t) shifts the harmonics of w(t) to a central frequency πf A. The number of harmonic frequencies is increased by windowing the sinusoid or sum of sinusoids. %Problem 4.4 clear all;clc; % parameters Fs = 88; T = /Fs; %frequency %sample period T = 5*T; T = T/4; T = T*3/4; % a) % define one period of w signal -> w Ts=/76; t=:ts:t-ts; Nn=length(t); % overall sample period x=[t(:ceil(nn/4)) t(ceil(nn/4)+)*ones(,ceil(nn/4))]; x=fliplr(x); w=[x x]; N=length(w); % plot line spectrum figure() FSeries(w,Ts,T,N); %number of samples in w % harmonics H to approximate w? w=*pi/t; DC=.65; H=3; for k=:h, a(k)=*(cos(k*w*t/4)-cos(k*w*t/))/t/kˆ/wˆ; a=[dc *a]; figure() for N=:H+, x=a()*ones(,length(t)); for k=:n, x=x+a(k)*cos((k-)*w.*(t+t/)); plot(t,x); axis([ max(t).*min(x).*max(x)]);

10 hold on; plot(t,w, r ) ylabel( Amplitude );xlabel( t (sec) ); leg( aprox w(t), w(t) );grid; hold off pause(.); clear x; clear H a x % b) s = cos(*pi*fs*t).*w; %modulated signal s figure(3); FSeries(s,Ts,T,N); % line spectrum of s % how many harmonics H are necessary to approximate s? w=*pi/t; DC=.65;H=3; for k=:h, a(k)=*(cos(k*w*t/4)-cos(k*w*t/))/t/kˆ/wˆ; a=[dc *a]; figure(4) for N=:H+, x=a()*ones(,length(t)); for k=:n, x=x+a(k)*cos((k-)*w.*(t+t/)); x=x.*cos(5*w.*t); plot(t,x); axis([ max(t).*min(s).*max(s)]); hold on; plot(t,s, r ) ylabel( Amplitude ); xlabel( t (sec) ); leg( aprox s(t), s(t) ); grid; hold off pause(.); clear x; % c) p = cos(*pi*fs*t); %signal p figure(5); FSeries(p,Ts,T,N); %line spectrum of p %sound(p); pause(); sound(w*s);pause(); % d) p = sin(*pi*44*t)+sin(*pi*55*t)+sin(*pi*66*t); %signal p s = p.*w; %modulated signal s

11 figure(6); FSeries(s,Ts,T,N); %line spectrum of s % sound(*p); pause(); sound(*w*s); x 3 Periodic signal x 4 x 3 Periodic signal.5 4 aprox w(t) w(t) x(t).5 x(t) t (sec) x 3 Magnitude Line Spectrum x 3 Phase Line Spectrum Amplitude t (sec) x 4 Magnitude Line Spectrum 5 x 4 Phase Line Spectrum X k <X k X k <X k w (rad/sec) x 4 w (rad/sec) x 4 t (sec) x 3 w (rad/sec) x 4 w (rad/sec) x 4 Figure : Right to left: window and line spectra, approximation of window, windowed sinusoid and its line spectra..5 x 3 aprox s(t) Periodic signal x 3 Periodic signal s(t) x(t) x(t).5 4 Amplitude t (sec).5.4 Magnitude Line Spectrum 5 x 6 Phase Line Spectrum t (sec) x 4 Magnitude Line Spectrum.5 x 3 Phase Line Spectrum X k.3. <X k X k 3 <X k t (sec) x 3 w (rad/sec) x 4 w (rad/sec) x 4 w (rad/sec) x 4 w (rad/sec) x 4 Figure : Right to left: approximation of windowed signal, line spectra of sinusoid, windowed sum of sinusoids and line spectra.

12 problem 5:Compution of π MATLAB As you know,π is an irrational number that can only be approximated by a number with a finite number of decimals.how to compute this value recursively is a problem of theoretical interest.in this problem we show that the Fourier series can provide that formulation. (a) Consider a train of rectangular pulses x(t),with a period x (t) = [u(t +.5) u(t.5)].5 t.5 and period T =.Plot the periodic signal and nd its trigonometric Fourier series. (b) Use the above Fourier series to nd an innite sum for π. (c) If π N is an approximation of the innite sum with N coefficients,and π is the value given by MATLAB,find the value of N so that π N is 95% of the value of π given by MATLAB.

13 Solution 5 π) so that (a) The signal x(t) is zero-mean. The Fourier series coefficients are obtained from (Ω = π/t = ˆX (s) = s (e.5s e.5s ) s=jkπ = sin(πk/) πk/ x(t) = = k=, k= sin(πk/) e jkπt πk/ sin(πk/) πk/ cos(πkt) Since x() =, letting t = in both sides and then multiplying by π we get that If we let we use MATLAB to find the value of N. % Problem 4.5 clear all; clc N=; x = ; for k=:n %DC value x = x + 4*sin(k*pi/)/k; xx(k) =x; format long π = 4 π N = 4 k= N k= display( Error of approximation for ) N e = pi-x ppi=pi*ones(,length(xx)); n=:n; figure() sin(πk/) k sin(πk/) k =.95π subplot(); plot(n,xx); title( Approximation of PI ); axis([ N 4.5]) hold on; plot(n,ppi, r ); grid; hold off; leg( approximate, pi ) subplot(); plot(n,xx-pi); title( Error of approximation ); grid % N for 95 percent of pi sum = ; for i = :N, %initial approximation sum = sum + 4*cos(i*pi/-pi/)/i; if (sum >=(.95*pi))&&(sum<=pi) break; display([numstr(i) coeficients needed of Fourier series for 95% approx. of PI ])

14 Error of approximation for N = e = coeficients needed of Fourier series for 95\% approx. of PI 4 3 Approximation of PI n Error of approximation approximate pi n Figure 4: Approximation of π by series, error of approximation

15 problem 5:Compution of π MATLAB As you know,π is an irrational number that can only be approximated by a number with a finite number of decimals.how to compute this value recursively is a problem of theoretical interest.in this problem we show that the Fourier series can provide that formulation. (a) Consider a train of rectangular pulses x(t),with a period x (t) = [u(t +.5) u(t.5)].5 t.5 and period T =.Plot the periodic signal and nd its trigonometric Fourier series. (b) Use the above Fourier series to nd an innite sum for π. (c) If π N is an approximation of the innite sum with N coefficients,and π is the value given by MATLAB,find the value of N so that π N is 95% of the value of π given by MATLAB.

16 Solution 5 π) so that (a) The signal x(t) is zero-mean. The Fourier series coefficients are obtained from (Ω = π/t = ˆX (s) = s (e.5s e.5s ) s=jkπ = sin(πk/) πk/ x(t) = = k=, k= sin(πk/) e jkπt πk/ sin(πk/) πk/ cos(πkt) Since x() =, letting t = in both sides and then multiplying by π we get that If we let we use MATLAB to find the value of N. % Problem 4.5 clear all; clc N=; x = ; for k=:n %DC value x = x + 4*sin(k*pi/)/k; xx(k) =x; format long π = 4 π N = 4 k= N k= display( Error of approximation for ) N e = pi-x ppi=pi*ones(,length(xx)); n=:n; figure() sin(πk/) k sin(πk/) k =.95π subplot(); plot(n,xx); title( Approximation of PI ); axis([ N 4.5]) hold on; plot(n,ppi, r ); grid; hold off; leg( approximate, pi ) subplot(); plot(n,xx-pi); title( Error of approximation ); grid % N for 95 percent of pi sum = ; for i = :N, %initial approximation sum = sum + 4*cos(i*pi/-pi/)/i; if (sum >=(.95*pi))&&(sum<=pi) break; display([numstr(i) coeficients needed of Fourier series for 95% approx. of PI ])

17 Error of approximation for N = e = coeficients needed of Fourier series for 95\% approx. of PI 4 3 Approximation of PI n Error of approximation approximate pi n Figure 4: Approximation of π by series, error of approximation