written 2.4 years ago by | • modified 2.4 years ago |
Design an analog Butterworth filter that has −2 passband attenuation at frequency of 20rad/sec& at least −10dB stop band attenuation at 30rad/sec.
written 2.4 years ago by | • modified 2.4 years ago |
Design an analog Butterworth filter that has −2 passband attenuation at frequency of 20rad/sec& at least −10dB stop band attenuation at 30rad/sec.
written 2.4 years ago by | • modified 2.4 years ago |
Solution:
Given:
Ap=−2dBAs=−10dBfΩp=20rad/sec.Ωs=30rad/sec.
Ap=20logδq−2=20logδ7∴δ7=0.7943
As=20logδ2−10=20logδ2
Ωc=Ωp(1/δ21−1)1/2N=
((301/0.79432−1)1/8)
Ωc=32.079rad/sec
sik=Ωcejπ/2ej(2k+1)π/2N=32.079ejπ/2ej(2k+1)π/8
S0=32.079ejπ/2ejπ/8=32.079ej5π/8=32.079[cos(5π/8)+jsin(5π/8)]=−12.276+j29.637
S1=32.079ejπ/2ej3π/8=32.079ej7π/8=32.079[cos(7π/8)+jsin(7π/8)]=−29.637+j12.276
S3=32.079ejπ/2ej>π/8=32.079ejππ/8=32.079[cos(11π/8)+jsin(11π/8)]=−12.276−j29.637
H(S)=S0S1S2S3(S−S0)(S−S1)(S−S2)(S−S3)=(−12.276+j29.637)(−12.276−j29.637)[S−(−12.276+j29.637)][S−(−12.276−j29.637)][S−(−29.637−j12.276)][S−(−29.637+j12.276)]
=[(2.276)2+(29.637)2][(29.637)2+(12.276)2][(s+12.276)2+(29.637)2][(s+29.637)2+(12.276)2]=1.0589×106[(5+12.276)2+(29.637)2][(s+29.637)2+(2.276)2]