Question

BC:9.4 A LTI discrete time system has an impulse response h[n] = (−0.6)nu[n] + (0.95)nu[n − 1] Find the transfer function, Hˆ (e jωˆ ), in the normalized frequency domain. Use Matlab to plot the magnitude and phase (in degrees) of Hˆ (e jωˆ ) in the range of −π ≤ ωˆ ≤ π. Attach your Matlab source code with the plots.

BC:9.4 A LTI discrete time system has an impulse response h[n] = (-0.6)"u[n] + (0.95)"u[n-1] Find the transfer function, H(e^), in the normalized frequency domain. Use Matlab to plot the magni- tude and phase (in degrees) of H(e^) in the range of-r ώ < π. Attach your Matlab source code with the plots.

Answer 1

H[n] = (- 0.6)^n u(n) + (0.95)^n u (n - 1) h(n) = (_.36)^n u(n) + .95 (0.95)^n-1 u(n - 1) apply D T F T on both sides of above equation implies H(e^j w) = z/z + 0.6 + 0.95/z - 0.95 where z = e^j w. therefore H(e^j w) = e^j w/w^j w + 0.6 + 0.95/e^j w - 0.95 = e^j w (e^j w - 0.95) + 0.95 (e^j w + 0.6)/(e^j w + 0.6) (e^j w - 0.95) = 0.57 + e^j w/- 0.57 - 0.35 e^j w + e^j 2 w multiply num & dem with e^-j 2 w, we get H(e^j w) = 1 + 0.57 e^-j 2 w/1 - 0.35 e^-j w - 0.57 e^j 2 w matlab is used to plot magnitude & phase of H(e^j w) MATLAB code: clc: close all: clear all: % define w from - pi to pi w = -pi: 0.01: pi % define the transfer function H = (1 + 0.57

MATLAB code:

clc;
close all;
clear all;

% define w from -pi to pi
w = -pi:0.01:pi;

% define the transfer function
H = (1+0.57*exp(-2j*w))./(1-0.35*exp(-1j*w)-0.57*exp(-2j*w));

% now plot the magnitude and phase spectra as below
figure;
subplot(211);
plot(w/pi,20*log10(abs(H)));grid on; xlabel('normalized w = wn = w/pi');
ylabel('Magnitude in dB');
subplot(212);
plot(w/pi,angle(H)*57.3);grid on; xlabel('normalized w = wn = w/pi');
ylabel('Degrees');