Signal Processing and Linear Systems Solution Ch.7

Chapter 7 7.1-1 H(j*omega) = (j*omega + 2)/((j*omega) ^ 2 + 5j*omega + 4) = (j*omega + 2)/((4 - omega ^ 2) + j5omega) H(j omega)|= sqrt((omega ^ 2 + 4)/((4 - omega ^ 3) ^ 2 + (5omega) ^ 2)) = sqrt((omega ^ 2 + 4)/(omega ^ 4 + 17omega ^ 2 + 16)) angle H(j*omega) = arctan(omega/2) - arctan((5omega)/(4 - omega ^ 2)) (a) f(t) = 5cos(2t + 30 deg) Here omega = 2and |H(j2)|= sqrt(2/25) = (sqrt(2))/5 ( H(j*omega) = arctan(- arctan(∞)) = 45 deg - 90 deg = - 45 deg y(t) = 5 * (sqrt(2))/5 * cos(2t + 30 deg - 45 deg) = sqrt(2) * cos(2t - 15 deg) (b) f(t) = 10sin(2t + 45 deg) y(t) = 10((sqrt(2))/5] * sin(2t + 45 deg - 45 deg) = 2sqrt(2) * sin 2t (e) f(t) = 10cos(3t + 40 ^ pi) Here omega = 3 |H(j*omega)| = sqrt(13/250) = 0.228 atıd angle H(j3)= 56.31 deg - 168.43 deg = - 52.12 deg y(t) = 10(0.228) * cos(3t + 49 deg - 52.12 deg) = 2.28cos(3t - 12.12 deg) Therefore 7.1-2 H(j*omega) = (j*omega + 3)/((j*omega + 2) ^ 2) |H(j*omega)| = (sqrt(omega ^ 2 + 9))/(omega ^ 2 + 4) and angle H(j*omega) = arctan(omega/3) - arctan(omega/2) (a) f(t) = 10n(t) =10e^ j otimes t u(t), Here omega = 0 and H (j0)=1 Therefore y(t) = 1 * 10e ^ (j0t) * u(t) = 10u(t) (b) f(t) = cos(2t + 60 deg) * u(t) Here omega = 2 |H(j2)|= (sqrt(- 33))/8 and Therefore angle H(j2)= 33.69 deg - 90 deg = - 56.31 deg y(t) = (sqrt(pi))/8 * cos(2t + 60 deg - 56.31 deg) * u(t) = (sqrt(13))/8 * cos(2t + 3.69 deg) * u(t) (e) f(t)=* in(3t - 45 deg) * u(t) Here omega = 3 and |H(j3)|= (sqrt(28))/15 and angle H(j3)= 45 deg - 112.62 ^ pi = - 67.62 deg Therefore (d) f(t) = e ^ (j3t) * v(t) y(t) = (sqrt(18))/13 * sin(3t - 45 deg - 67.62 deg) * u(t) = (sqrt(15))/13 * sin(3t - 112.62 deg) * u(t) y(t)=H(j3)e^ j:3 =[H(j3)[e^ j[H+/H(j3)] u(t) = (sqrt(18))/13 * e ^ (j[H - 67.62 ^ 5]) * u(i) 99 7.1-3 H(j*omega) = (- (j*omega - 10))/(j*omega + 10) - (10 - j*omega)/(10 + j*omega) |H(j*omega)| = sqrt((omega ^ 2 + 100)/(omega ^ 2 + 100)) = 1 6H(j*omega) = arctan(- omega/10) - arctan\langleomega/10\rangle =-2 tan^ -1 langle omega 10 ) (n) f(t) = e ^ (i*omega*t) y(t) = H(j*omega) * e ^ (j*omega*t) =[H(j omega)]e^ j[ omega i+ tilde omega H(j omega)] =e^ j[ omega t-2 i hbar n^ -1 ( omega/10j] (b) f(t) = cos(omega*t + y) w(t) = cos[omega*t + theta - 2arctan(omega/10)] (c) f(t) = cos t Here omega = 1 | H (j1)|=1| angle H(j*omega) =-2 tan^ -1 ( 1 10 rangle=-11.42^ pi y(t) = cos(t - 11.42 deg) (d) Here omega = 2 f(t) = sin 2t | H (j2)|=1 angle H(j2)= - 2arctan(2/16) = - 22.62 deg y(t) = sin(2t - 22.62 deg) (e) f(t) = cos 10t Here omega = 10 | H (j10)|=1 angle H(j10)= - 2arctan(10/10) = - 90 deg y(t) = cos(10t - 90 deg) = sin 10t (f) f(t) = cos 100t Hero omega = 100 | H (j100)|=1 angle H(j100)= - 2arctan(160/10) = - 163.58 ^ u y(t) = cos(100t - 168.58 deg) 7.2-1 (a) The transfer function can be expressed as H(s) = 100/(2 * 20) * (a(4/100 + 1))/((4/2 + 1)(4/20 + 1)) = 2.5 * (a(4/100 + 1))/((4/2 + 1)(s/20 + 1)) The amplitude response: The horizontal axis where the asymptotes begin is 2.5, which is 7.96 dh. We draw the asymptotes at omega = 1 20 dB/dec.), 2 (-20 dB/dec.), 20 (-20 dB/dec.), and 100 (20 dB/dec.) as shown in Fig. 57.2-la. The corrections are applied at various points as discussed in Examples 7.3 and 7.4. to obtain the Bode plot for amplitude response. We follow the similar procedure for phase response. (b) The transfer function can be expressed as H(s) = (10 * 20)/100 * ((s/10 + 1)(s/20 + 1))/(s ^ 2 * (s/100 + 1)) = 2 * ((s/10 + 1)(s/100 + 1))/(s ^ 2 * (s/100 + 1)) The amplitude response: The horizontal axis where the asymptotes begin is 2, which is 6 db. Asymptotes start omega=1( - 40 * d/de (B) * c .). 10 (20 * d/de (B) * c) 20 (20 dB/dec.), and 100 (-20 dB/dec.). The corrections are applied at discussed 7.3 and 7.4. to obtain the Bode plot. (c) The transfer function can be expressed as 100