Question

(5) For the system described by the following difference equation y(n)= 0.9051y(n 1) 0.598y(n 2) -0.29y(n 3) 0.1958y(n - 4) +0.207r(n)0.413r(n 2)+0.207a(n - 4) (a) Plot the magnitude and phase responses of the above system. What is the type of this filter? (b) (b) Find and plot the response of the system to the input signal given by /6)sin(w2n +T /4) u(n), where w 0.25m and ws 0.45m a(n) 4cos(win -T = (c) Determine the steady-state output and hence find the transient response. (d) After how any samples does the output reach steady state? Write a program to find the beginning of the steady-state period (end of transient period). Plot the transient response and calculate the energy of the transient signal (e) Plot the magnitude spectra of the input and output signals. Hence, comment on the results

Answer 1

yn)- -0.905 1y (n-Τ-0.598y (n-2)-029y(n-3 )-01958y (n-4)+ 0.207α (n ) + 0.413a (n-2) +0.207π (n- 4)

Taking z-transform

Y(2) -2 0.207 0.413 0.207z4 10.90511 0.59822 0.2930.1958-4 H(z) X(z)

a)

w=-pi:0.001:pi;
z=exp(1i.*w);
H=(0.207+0.413.*z.^-2+0.207.*z.^-4)./(1+0.9051.*z.^-1+0.598.*z.^-2+0.29.*z.^-3+0.1958.*z.^-4);
subplot(211)
plot(w,abs(H));
xlabel('\omega');
ylabel('|H(e^{j\omega})|');
subplot(212)
plot(w,angle(H));
xlabel('\omega');
ylabel('\angle H(e^{j\omega})');

---------------

0.5 Malyl

b)

N=100;
n1=0:N;
w1=0.25.*pi;
w2=0.45.*pi;
x1=4.*cos(w1.*n1-pi./6).*sin(w2.*n1+0.25.*pi);
x=[0 0 0 0 x1];
y=zeros(size(x));
for k=5:length(y)
y(k)=-0.9051.*y(k-1)-0.598.*y(k-2)-0.29.*y(k-3)-0.1958.*y(k-4)+0.207.*x(k)+0.413.*x(k-2)+0.207.*x(k-4);
end
n=-4:N;
stem(n,y);
hold on
stem(n,x);
hold off
xlabel('n');
legend('y(n)','x(n)');
--------------

y(n) xin) 40 60 20 20 80 100

------------------

c)

(n4 cos(wn T/6) sin (w2n +/4) 7

r(n) 2(sin(wn T/4w1n T/6)sin(w2n T/4 -w1n T/6))

r(n) 2(sin(0.7nT/12) +sin(0.2n5T/12))

(n) 2(r1 (n) +x2(n))

Now

H(e 0. =0.524e2.64

and

H(e 0.2) 0.242e05

therefore the steady state output for x(n) is

s(n)2(0.524 sin(0.7mn+T/12+2.64)0.242 sin(0.2Tn 5T/12-0.53)

s(n) 1.05 sin(0.7Tn2.9) 0.484 sin(0.2n0.78)

--------------

N=100;
n1=0:N;
w1=0.25.*pi;
w2=0.45.*pi;
x1=4.*cos(w1.*n1-pi./6).*sin(w2.*n1+0.25.*pi);
x=[0 0 0 0 x1];
y=zeros(size(x));
for k=5:length(y)
y(k)=-0.9051.*y(k-1)-0.598.*y(k-2)-0.29.*y(k-3)-0.1958.*y(k-4)+0.207.*x(k)+0.413.*x(k-2)+0.207.*x(k-4);
end
n=-4:N;
y1=1.05.*sin(0.7.*pi.*n1+2.9)+0.484.*sin(0.2.*pi.*n1+0.78);
ys=[0 0 0 0 y1];
stem(n,(y-ys))
hold on
stem(n,ys);
hold off
xlabel('n');
ylabel('y(n)=y_{s}(n)');
legend('transient response','steady state response');
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transient response -steady state response 0.5 0 40 -20 20 60 80 100 (u)*K=(u)