Señales y sistemas michael j roberts 1ed solutions

http://www.elsolucionario.blogspot.com LIBROS UNIVERISTARIOS Y SOLUCIONARIOS DE MUCHOS DE ESTOS LIBROS LOS SOLUCIONARIOS CONTIENEN TODOS LOS EJERCICIOS DEL LIBRO RESUELTOS Y EXPLICADOS DE FORMA CLARA VISITANOS PARA DESARGALOS GRATIS M J Roberts - 7/12/03 Chapter - Mathematical Description of Signals Solutions If g( t) = 7e −2 t − write out and simplify (a) g( 3) = 7e −9 (b) g(2 − t) = 7e −2( − t ) − = 7e −7 + t (c) t − −11  t  g + 4 = 7e  10  (d) g( jt) = 7e − j t − (e) (f) g( jt) + g(− jt) e − j 2t + e j 2t = 7e −3 = 7e −3 cos(2 t) 2  − jt − 3  jt − 3 g   + g     e − jt + e jt =7 = cos( t) 2 If g( x ) = x − x + write out and simplify (a) g( z) = z − z + (b) g( u + v ) = ( u + v ) − ( u + v ) + = u + v + uv − u − v + (c) g(e jt ) = (e jt ) − e jt + = e j t − e jt + = (e jt − 2) (d) g(g( t)) = g( t − t + ) = ( t − t + ) − ( t − t + ) + 2 2 g(g( t)) = t − t + 20 t − 16 t + (e) g(2) = − + = What would be the numerical value of “g” in each of the following MATLAB instructions? (a) t = ; g = sin(t) ; (b) x = 1:5 ; g = cos(pi*x) ; (c) f = -1:0.5:1 ; w = 2*pi*f ; g = 1./(1+j*w) ; 0.1411 [-1,1,-1,1,-1] Solutions 2-1 M J Roberts - 7/12/03 0.0247 + 0.0920 +    0.0920 − 0.0247 − j 0.155  j 0.289     j 0.289  j 0.155  Let two functions be defined by 1 , sin(20πt) ≥ x1 ( t) =  −1 , sin(20πt) < and t , sin(2πt) ≥ x ( t) =  − t , sin(2πt) < Graph the product of these two functions versus time over the time range, −2 < t < x(t) -2 t -2 For each function, g( t) , sketch g(− t) , − g( t) , g( t − 1) , and g(2t) (a) (b) g(t) g(t) t -1 t -3 g(-t) g(-t) -g(t) -g(t) t -2 -1 t -3 g(t-1) g(t-1) g(2t) 4 t t t -1 t -3 g(2t) -3 A function, G( f ) , is defined by Solutions 2-2 t -1 2 -3 t M J Roberts - 7/12/03  f G( f ) = e − j 2πf rect    2 Graph the magnitude and phase of G( f − 10) + G( f + 10) over the range, −20 < f < 20  f + 10   f − 10  − j 2π f +10 G( f − 10) + G( f + 10) = e − j 2π ( f −10) rect    + e ( ) rect      |G( f )| -20 f 20 Phase of G( f ) π -20 f 20 -π Sketch the derivatives of these functions (All sketches at end.) (a) g( t) = sinc( t) (b) g( t) = (1 − e −t g′ ( t) = π t cos(πt) − π sin(πt) πt cos(πt) − sin(πt) = πt (πt) e − t , t ≥ 0 − t g′ ( t) =   = e u( t) 0 , t < 0 ) u(t) (a) (b) x(t) x(t) 1 -4 t -1 -1 dx/dt 1 -1 t -1 dx/dt -4 t -1 t -1 Sketch the integral from negative infinity to time, t, of these functions which are zero for all time before time, t = Solutions 2-3 M J Roberts - 7/12/03 g(t) g(t) 1 t ∫ g(t) dt 3 t ∫ g(t) dt 1 2 t t Find the even and odd parts of these functions (a) g( t) = t − 3t + t − 3t + + 2(− t) − 3(− t) + t + 12 g e ( t) = = = 2t + 2 2 t − 3t + − 2(− t) + 3(− t) − −6 t g o ( t) = = = −3t 2 (b) π  g( t) = 20 cos 40πt −   4 π π   20 cos 40πt −  + 20 cos −40πt −     4 g e ( t) = Using cos( z1 + z2 ) = cos( z1 ) cos( z2 ) − sin( z1 ) sin( z2 ) g e ( t) =     π  π  20 cos( 40πt) cos −  − sin( 40πt) sin −   4       +20 cos(−40πt) cos − π  − sin(−40πt) sin − π           4       π  π    20 cos( 40πt) cos  + sin( 40πt) sin    4       +20 cos( 40πt) cos π  − sin( 40πt) sin π           4     g e ( t) = Solutions 2-4 M J Roberts - 7/12/03 20 π  g e ( t) = 20 cos  cos( 40πt) = cos( 40πt)  4 π π   20 cos 40πt −  − 20 cos −40πt −    4 4 g o ( t) = Using cos( z1 + z2 ) = cos( z1 ) cos( z2 ) − sin( z1 ) sin( z2 )     π  π  20 cos( 40πt) cos −  − sin( 40πt) sin −   4       −20 cos(−40πt) cos − π  − sin(−40πt) sin − π           4     g o ( t) = g o ( t) =   π  π    20 cos( 40πt) cos  + sin( 40πt) sin    4       −20 cos( 40πt) cos π  − sin( 40πt) sin π           4     20 π  g o ( t) = 20 sin  sin( 40πt) = sin( 40πt)  4 (c) t − 3t + g( t) = 1+ t t − 3t + t + 3t + + 1− t + t g e ( t) = (2t g e ( t) = g e ( t) = − 3t + 6)(1 − t) + (2 t + 3t + 6)(1 + t) (1 + t)(1 − t) t + 12 + t + t = − t2 2(1 − t ) t − 3t + t + 3t + − 1− t + t g o ( t) = Solutions 2-5 M J Roberts - 7/12/03 (2t g o ( t) = − 3t + 6)(1 − t) − (2 t + 3t + 6)(1 + t) (1 + t)(1 − t) −6 t − t − 12 t 2t + g o ( t) = = −t − t2 2(1 − t ) sin(πt) sin(−πt) + t π −πt = sin(πt) g e ( t) = πt (d) g( t) = sinc( t) (e) g( t) = t(2 − t )(1 + t ) g o ( t) = g( t) = {t (2 − t )(1 + t ) 4 odd 12312 even even Therefore g( t) is odd, g e ( t) = and g o ( t) = t(2 − t )(1 + t ) (f) g( t) = t(2 − t)(1 + t) g e ( t) = t(2 − t)(1 + t) + (− t)(2 + t)(1 − t) g e ( t) = t g o ( t) = t(2 − t)(1 + t) − (− t)(2 + t)(1 − t) g o ( t) = t(2 − t ) 10 Sketch the even and odd parts of these functions Solutions 2-6 M J Roberts - 7/12/03 g(t) g(t) 1 t 1 t -1 g e(t) g e(t) 1 t 1 t -1 g o(t) g o(t) 1 t 1 t -1 (a) (b) 11 Sketch the indicated product or quotient, g( t) , of these functions (a) (b) 1 -1 -1 t -1 t -1 g(t) g(t) Multiplication -1 t Multiplication -1 t -1 g(t) g(t) 1 -1 -1 t -1 -1 Solutions 2-7 t M J Roberts - 7/12/03 (c) (d) 1 t -1 t g(t) g(t) Multiplication 1 Multiplication t t g(t) g(t) -1 t -1 -1 t (e) (f) 1 -1 t t -1 g(t) -1 g(t) Multiplication Multiplication -1 t t -1 g(t) g(t) -1 1 t t -1 -1 (g) (h) 1 t -1 -1 -1 t g(t) Division 1 g(t) π Division t g(t) t g(t) t -1 -1 t 12 Use the properties of integrals of even and odd functions to evaluate these integrals in the quickest way Solutions 2-8 M J Roberts - 7/12/03 1 −1 −1 even ∫ (2 + t)dt = ∫ (a) 1 −1 odd 2{ dt + ∫ {t dt = ∫ dt = (b) 20 20 ∫ [4 cos(10πt) + sin(5πt)]dt = ∫ − 20 − 20 ∫ (c) − 20 20 cos(10πt) dt + 14243 even 20 20 − 20 odd t{ cos(10πt) dt = 1424 odd even 142 43 odd 1   10 10 cos(10πt)   cos(10πt) (d) ∫ t{ sin(10πt) dt = ∫ t sin(10πt) dt = − t + dt  424 10π ∫0 10π odd odd   − 142 43   10 even 10 10   10 sin(10πt)   = (10π3t)dt == 2 + ∫1 odd{t sin  424 100π (10π )  50π odd − 142  43  10 10 even (e) 1 [ −t −t −t −t ∫ e{ dt = 2∫ e dt = 2∫ e dt = −e −1 even ] = 2(1 − e −1 ) ≈ 1.264 (f) ∫ −t t{ e{ dt = odd2 even −1 odd 13 Find the fundamental period and fundamental frequency of each of these functions (a) g( t) = 10 cos(50πt) f = 25 Hz , T0 = s 25 (b) π  g( t) = 10 cos 50πt +   4 f = 25 Hz , T0 = s 25 (c) g( t) = cos(50πt) + sin(15πt)  15  = 0.4 s f = GCD 25,  = 2.5 Hz , T0 =   2.5 (d) (5π3t)dt = ∫ cos(10πt)dt = ∫ 81sin 424 10π 3π   g( t) = cos(2πt) + sin( 3πt) + cos 5πt −   4 Solutions 2-9 M J Roberts - 7/12/03 z − 1.3942 z + 0.5392 z z − 1.3942 z + 0.5392 z = 9762 z − 2.404 z + 1.864 z − 0.4602 z − z − 1.404 z + 0.4602 H −1 ( z) = 0.9762 2.5 z 1.999 z 0.4756 z − + z − z − 0.8818 z − 0.5219 H −1 ( z) = [ ] h −1[ n ] = 2.5 − 1.999(0.8818) + 0.4756(0.5219) u[ n ] n h- 1(t) n Unit Step Response 2.451 0.1885 t Bilinear-z Unit Sequence Response h [n] -1 2.4541 30 (c) n s2 + H( s) = s + 12 s + 32 [ ] h( t) = δ ( t) − 17e −8 t + 5e −4 t u( t) s2 + 0.125 0.04688 0.3215 0.2656 = − + − s s( s + 12 s + 32) s s s+4 s+8 H −1 ( s) = [ ] h −1 ( t) = 0.125 t − 0.04688 + 0.3215e −4 t − 0.2656e −8 t u( t) H( z) = 0.6617 z − 1.9752 z + z − 1.25 z + 0.3802 z − 1.9752 z + z z − 1.9752 z + z = 6617 z − 2.25 z + 1.631z − 0.3802 z − z − 1.25 z + 0.3802 H −1 ( z) = 0.6617 H −1 ( z) = [ 0.125 z 1.08 z 1.617 z − + z −1 z − 0.7285 z − 0.5219 ] h −1[ n ] = 0.125 − 1.08(0.7285) + 1.617(0.5219) u[ n ] n 12-52 n M J Roberts - 7/12/03 h- 1(t) Unit Step Response 2.3562 t Bilinear-z Unit Sequence Response h- 1[n] 0.662 30 n 32 Design a digital-filter approximation to each of these ideal analog filters by sampling a truncated version of the impulse response and using the specified window In each case choose a sampling frequency which is 10 times the highest frequency passed by the analog filter Choose the delays and truncation times such that no more than 1% of the signal energy of the impulse response is truncated Graphically compare the magnitude frequency responses of the digital and ideal analog filters using a dB magnitude scale versus linear frequency % Program to generate the FIR impulse responses of some approximations to % ideal filters and plot a comparison of the magnitude frequency responses of % the approximate and ideal filters close all ; % % Bandpass, Rectangular Window % fc = [10 20] ; type = 'BP' ; fs = 10*fc(2) ; h = FIRDF(type,fc,fs,'RE',0.01) ; N = length(h) ; Ts = 1/fs ; n = [0:N-1]' ; F = [0:0.001:1/2]' ; [H,F] = DTFT(n,h,F) ; subplot(2,2,1) ; p = xyplot(F*fs,20*log10(abs(H)),[0,fs/2,-140,0],'\itf ', '|H(\ite^{{\itj}2{\pi}{\itf}{\itT}_s} )|', 'Times',18,'Times',14, 'Bandpass - Rectangular Window','Times',24,'n','c') ; subplot(2,2,2) ; p = xyplot(F*fs,20*log10(abs(H)),[0,fs/10,-5,0],'\itf ', '|H(\ite^{{\itj}2{\pi}{\itf}{\itT}_s} )|', 'Times',18,'Times',14, 'Bandpass - Rectangular Window','Times',24,'n','c') ; % % Bandpass, Blackman Window % h = FIRDF(type,fc,fs,'BL',0.01) ; N = length(h) ; Ts = 1/fs ; n = [0:N-1]' ; F = [0:0.001:1/2]' ; [H,F] = DTFT(n,h,F) ; subplot(2,2,3) ; p = xyplot(F*fs,20*log10(abs(H)),[0,fs/2,-140,0],'\itf ', '|H(\ite^{{\itj}2{\pi}{\itf}{\itT}_s} )|', 'Times',18,'Times',14, 'Bandpass - Blackman Window','Times',24,'n','c') ; 12-53 M J Roberts - 7/12/03 subplot(2,2,4) ; p = xyplot(F*fs,20*log10(abs(H)),[0,fs/10,-5,0],'\itf ', '|H(\ite^{{\itj}2{\pi}{\itf}{\itT}_s} )|', 'Times',18,'Times',14, 'Bandpass - Blackman Window','Times',24,'n','c') ; % Function to design a digital filter using truncation of the % impulse response to approximate the ideal impulse response The % user specifies the type of ideal filter, % % LP lowpass % BP bandpass % % the cutoff frequency(s) fcs (a scalar for LP and HP a 2-vector % for BP and BS), the sampling rate, fs, the type of window, % % RE rectangular % VH von Hann % BA Bartlett % HA Hamming % BL Blackman % % and the allowable truncation error, % err, as a fraction of the total impulse response signal energy % % The function returns the filter coefficients as a vector % % function h = FIRDF(type,fcs,fs,window,err) % function h = FIRDF(type,fcs,fs,window,err) if fs == | fs == inf | fs == -inf, disp('Sampling rate is unusable') ; else fs Ts = 1/fs ; type = upper(type) ; window = upper(window) ; switch type case 'LP', fc = fcs(1) ; zc1 = 1/(2*fc) ; N = 200*zc1/Ts ; n = [0:N]' ; Etotal = sum(sinc(2*fc*n*Ts).^2) E = ; N = ; while abs(E-Etotal) > Etotal*err, N = 2*N ; n = [0:N]' ; E = sum(sinc(2*fc*n*Ts).^2) ; end delN = floor(N/10) ; while abs(E-Etotal)1, N = N-delN ; n = [0:N]' ; E = sum(sinc(2*fc*n*Ts).^2) ; delN = floor(delN/2) ; end N = ceil(N/2)*2 ; % Make number of pts even nmid = (N/2-1) ; 12-54 M J Roberts - 7/12/03 w = makeWindow(window,n,N) ; h = w.*sinc(2*fc*(n-nmid)*Ts) ; h = h/sum(h) ; case 'BP' fl = fcs(1) ; fh = fcs(2) ; fmid = (fh + fl)/2 ; df = abs(fh-fl) ; zc1 = 1/df ; N = 200*zc1/Ts ; n = [0:N]' ; Etotal = sum((2*df*sinc(df*n*Ts).* cos(2*pi*fmid*n*Ts)).^2) ; E = ; N = ; while abs(E-Etotal) > Etotal*err, N = 2*N ; n = [0:N]' ; E = sum((2*df*sinc(df*n*Ts).* cos(2*pi*fmid*n*Ts)).^2) ; end delN = floor(N/10) ; while abs(E-Etotal)1, N = N-delN ; n = [0:N]' ; E = sum((2*df*sinc(df*n*Ts).* cos(2*pi*fmid*n*Ts)).^2) ; delN = floor(delN/2) ; end N = ceil(N/2)*2 ; % Make number of pts even nmid = (N/2-1) ; n = [0:N-1]' ; w = makeWindow(window,n,N) ; h = w.*2.*df.*sinc(df*(n-nmid)*Ts).* cos(2*pi*fmid*(n-nmid)*Ts) ; h = h/sum(h.*cos(2*pi*fmid*(n-nmid)*Ts)) ; end end function w = makeWindow(window,n,N) switch window case 'RE' w = ones(N,1) ; case 'VH' w = (1-cos(2*pi*n/(N-1)))/2 ; case 'BA' w = 2*n/(N-1).*(0