the solution :-
the bandedge of the coresponding elliptics
The band edge and filter order of LPF are same as taht of
high pass filter:
For LPF:
%Specifications
Rp=0.25;Rs=40;
Fs=44000;%Assumption
fp=6000;fs=4000;
wp=2*fp/Fs;ws=2*fs/Fs;
[N1,wc]=buttord(wp,ws,Rp,Rs)
[b1,a1]=butter(N1,wc);
[H1,f]=freqz(b1,a1,1024,Fs);
tf1=tf(b1,a1)
%Frequency response
figure(1)
subplot(211)
plot(f,20*log(abs(H1)),'r')
xlabel('f')
ylabel('|H(f)| in db')
title('Log magnitude response-butterworth')
subplot(212)
plot(f,angle(H1),'g')
xlabel('f in Hz')
ylabel('<H(f)>')
title('Phase response-butterworth')

COmmand window:
Butterworth Low pass filter
N1 = 14
wc = 0.24925
the bandedge of elliptics lpf
______________________________________________________________________-
Band edge of elliptic LPF:
disp('Elliptic Low pass Filter with the lowest edge of
BPF')
%Specifications
Rp=0.25;Rs=40;
fp=[30000];fs=[25000];
Fs=2*(10^5)
wp=2*fp/Fs;ws=2*fs/Fs;
[N2,wp] = ellipord(wp,ws,Rp,Rs)
[b2,a2] = ellip(N2,Rp,Rs,wp);
[H2,f]=freqz(b2,a2,1024,Fs);
tf2=tf(b2,a2)
%Frequency response
figure(1)
subplot(211)
plot(f,20*log10(abs(H2)),'r')
xlabel('f in Hz')
ylabel('|H(f)| in db')
title('Log magnitude response-elliptic')
subplot(212)
plot(f,angle(H2),'g')
xlabel('f in Hz')
ylabel('<H(f)>')
title('Phase response-elliptic ')

Command window:
Elliptic Low pass Filter with the lowest edge of BPF
Fs = 200000
N2 = 6
wp = 0.30000
therefore ...............the answers was given and sloved
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