hw13sol - Rsfinal = Km*Rs = 708.5 (ohm) Cfinal = 0.5 uF...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Rsfinal = Km*Rs = 708.5 (ohm) Cfinal = 0.5 uF Lfinal = Km*L/Kf = 0.1255H %%ECE homework Problem 50 % Part a Amax = 0.8; wp = 2*pi*600; ws = 2*pi*2200; Amin1 = 20; Amin2 = 25; while (Amin2-Amin1>0.001) n1 = buttord(wp, ws, Amax, Amin1, 's'); n2 = buttord(wp, ws, Amax, Amin2, 's'); if (n1==3) && (n2==3) Amin3 = Amin2 + 0.5*(Amin2 - Amin1); n3 = buttord(wp, ws, Amax, Amin3, 's'); end if (n3 == 3) Amin1 = Amin2; Amin2 = Amin3; elseif (n3 > 3) Amin4 = 0.5 * (Amin3 + Amin2); n4 = buttord(wp, ws, Amax, Amin4, 's'); Amin3 = Amin4; while (n4 ~= 3) Amin4 = 0.5 * (Amin3 + Amin2); Amin3 = Amin4; n4 = buttord(wp, ws, Amax, Amin4, 's'); end if (n4 == 3) Amin1 = Amin2; Amin2 = Amin4; end end end %% Result Amin1 = Amin2 = 26.92; %% So the largest integer value of Amin is 26, so that the filter order is 3. % Part b Amin = 26 wcmin = wp/(10^(0.1*Amax)-1)^(1/6); wcmax = ws/(10^(0.1*Amin)-1)^(1/6); fcmin = wcmin/2/pi; fcmax = wcmax/2/pi; %% Result: Allowable Wc is from 4920.5 to 5098.6 rad/s; %% Result: Allowable fc is from 783.1254 to 811.4680 Hz; % Part c syms s; H(s) = 1/(s^3+2*s^2+2*s+1); n = 3; [z,p,k]=buttap(n); %% Result %%! p = %%! ! %%! ! %%! %% Result Ploes: -0.5000 + 0.8660i -0.5000 - 0.8660i -1.0000 No zeros; % To check all the poles are on the unit circle abs(p) %% Results: %%! ! ans = %%! ! 1.0000 %%! ! 1.0000 %%! ! 1.0000 %%So all the three results above is 1, which means all the poles %%are on the unit circle w = 0:0.01:7; num = [1]; den = [1 2 2 1]; h = freqs(num, den, w); plot(w,abs(h)); grid on; % Part d H(s) = 1/((s+1)*(s^2+s+1)) % Part e Alpha = [wp/wcmin, 1, ws/wcmin]; h1 = 1/((i*Alpha(1)+1)*((i*Alpha(1))^2+i*Alpha(1)+1)); h2 = 1/((i*Alpha(2)+1)*((i*Alpha(2))^2+i*Alpha(2)+1)); h3 = 1/((i*Alpha(3)+1)*((i*Alpha(3))^2+i*Alpha(3)+1)); A1 = -20*log10(abs(h1)); A2 = -20*log10(abs(h2)); A3 = -20*log10(abs(h3)); %% Result: A1=0.8; A2=3; A3=26.92; %% so the filter response meet the given brickwall specs at fp and fs. % Part f fcmin = wcmin/2/pi; f = 0:4400; num = [1]; den = [1 2 2 1]; hf = freqs(num, den, f/fcmin); Aw = -20*log10(abs(hf)); plot(f, Aw); grid on; % Part g; f = 0:4400; num = [1]; den = [1 2 2 1]; hf = freqs(num, den, f/fcmin); plot(f, abs(hf)); grid on; H1(s) = 1/(s+1) H2(s) = 1/(s^2 + s +1) R1final = R2final = 10,000 (ohm) C1final = C1/(Km*Kf) = 20 nF RLfinal = 200 (ohm) C2final = C2/(Km*Kf) = 100 nF Lfinal = Km*L/Kf = 0.04H We could use separate magnitude scaling, because the op amp circuit is decoupled from the passive part which does not load down the output. Rfinal = 1355 ohm C1final = C2final = 150 nF Lfinal = Km*L/Kf = 0.55 H ** Profile: "SCHEMATIC1-ece202" Date/Time run: 03/26/10 14:29:19 500mV [ N:\Desktop\ffff-PSpiceFiles\SCHEMATIC1\ece202.sim ] Temperature: 27.0 (A) ece202 (active) 400mV (783.333,354.229m) 300mV 200mV 100mV 0V 0Hz V(R2:2) 0.5KHz 1.0KHz 1.5KHz 2.0KHz Frequency Page 1 2.5KHz 3.0KHz 3.5KHz 4.0KHz Date: March 26, 2010 Time: 14:33:00 ...
View Full Document

This document was uploaded on 04/02/2010.

Ask a homework question - tutors are online