Question

ENGINEERING SIGNALS AND SYSTEMS:

Design a notch filter that meets the specifications below. Plot the impulse response and the frequency response. You may ignore plotting any delta functions. (a) It eliminates 100Hz (b) lh (t) | .001 for t 0.5 s

Answer 1

The expression for the angular frequency is, 2T f Substitute 100 for f _ a27(100) 200T rad/s The impulse response of the ideal notch filter is, h(t)(t)-Ae cos(at+0)u(t) Here (1) 4 A = 4c and e=tan 2a Consider the following condition: h(t0.001 for t> 0.5 s It means that the impulse response of the filter should decay to less than 0.001 within 0.5 seconds. Hence, the cosine component is no larger than 0.001 at t 0.5 s Determine the value of the selectivity a that satisfies the condition Therefore . Ae 0.00 4. -0.5a0,001 e 2 4c2 + Substitute 2007 for a. -0.5a<0,001 4a2 (2007) (2)

Solve equation (2) using MATLAB Execute the following code in MATLAB: solve (' (sqrt ( (4*x^2) (-0.5*x))0.001') (x^4) / (200 pi)^2)) *exp ans = -0.00049987504685413729488244730018934 Therefore, the value of the selectivity is, a> 0.0004998 Select the value of 100 for a. The transfer function of the notch filter is, (2as a (s+ a)a Substitute 100 for a H(s) 1 - 2 and 200 for az. +(100)) (2(100)s 1 - Hs (s+100)2(2007) (200s 104) 1- s200s10* 394784.176 + (200s 10) 1- 2 200s 404784.176 S- 200(s 50) s2200s404784.176 Therefore, the transfer function of notch filter that eliminates 100 Hz is 200(s50) H(s) 1-- s2200s 404784 .176

Recall equation (1) for the impulse response of the filter h(t))Aecos (at +0)u(t) 5(t) cos ttan 2. 2ca and 200T for a. Substitute 100 for 200nt IJa(t00. (00 (2007) 4 |u(t)- -100t e h(t)() 100 cos -1 tan 2(2007 cos 200ttan(0.0796)]u(t) cos(2007t 0.0794)u(t) -100t 6(t)-200.632e -100t =6(t)-200.632e Execute the following MATLAB code to plot the impulse response of the filter > t-0:0.001:0.5; >> h-dirac (t)-200.632.*exp ( 100 t).cos (200 pi*t+0.0794).*heaviside (t); plot (t,h) %plot the impulse response >> >> xlabel ('t (s)') ylabel ('h (t) ') >>

Substitute ja for s and obtain the frequency response (jo) (ja)200(j)+ 404784.176 394784.176 H(a) - 394784.176 (-a404784.176)+ j (200a) The magnitude response of the notch filter is, -a394784.176 -o)H (-404784.176) +(200a) 2 Execute the following MATLAB code to plot the frequency response of the filter >>w0 200*pi; >>w-3*w0:0.001: 3*w0; > H= (-w.^2+394784.176)./ (sqrt ( (- w. 2+404784.176).^2+(200.w).^2)); > plot (w, H) > xlabel 'w (rad/s)') ylabel 'H (w) ') >>

Obtain the plot for the frequency response of the notch filter as shown in Figure 2 0.5 0 -0.5 -2000 -1000 0 1000 2000 w(rad/s) Figure 2 H(w)

MATLAB CODE FOR SELECTIVITY:

>> solve('(sqrt((4*x^2)+((x^4)/(200*pi)^2))*exp(-0.5*x))==0.001')

MATLAB CODE FOR IMPULSE RESPONSE OF THE FILTER:

>> t=0:0.001:0.5;
>> h=dirac(t)-200.632.*exp(-100*t).*cos(200*pi*t+0.0794).*heaviside(t);
>> plot(t,h) %plot the impulse response
>> xlabel('t(s)')
>> ylabel('h(t)')

MATLAB CODE FOR FREQUENCY RESPONSE OF THE FILTER:

>> w0=200*pi;
>> w=-3*w0:0.001:3*w0;
>> H=(-w.^2+394784.176)./(sqrt((-w.^2+404784.176).^2+(200.*w).^2));
>> plot(w,H)
>> xlabel('w(rad/s)')
>> ylabel('H(w)')