Question

A digital low pass IIR filter is to be designed with Butterworth approximation using the Bilinear transformation technique having the following specifications:

(i)             Passband magnitude is constant within 1 dB for frequencies below 0.2 π.

(ii)           Stopband attenuation is greater than 15 dB for frequencies between 0.3 π to π.

Determine the order of the filter, cutoff frequency, poles location and transfer function of digital filter in order to meet the above specifications.     

Answer 1

$$ \begin{aligned} &\text { Ap }=1 \mathrm{~dB}, \text { As }=15 \mathrm{~dB} \\ &\text { wp }=0.2 \pi, \text { ws }=0.3 \pi \\ &\text { Finding characteristics of analog filter, } \\ &\text { For the bilinear transformation } \\ &\Omega=2 / \text { Ts * tan (w/2) } \\ &\text { The value of Ts is not given so }=1 \text { second } \\ &\text { So, } \Omega=2 \text { * } \tan (w / 2) \\ &\Omega p=2 \text { * } \tan (w p / 2)=2 \text { * } \tan (0.2 \pi / 2)=0.65 \\ &\text { rad/sec } \\ &\Omega s=2 \text { * } \tan (w s / 2)=2 \text { * } \tan (0.3 \pi / 2)=1.02 \\ &\text { rad/sec } \\ &\text { Finding order using below equation, } \\ &N=(1 / 2) *\left[\operatorname { l o g } \left(\left(10^{0.1 * A s}-1\right) /\left(10^{0.1 * A p}-1\right)\right.\right. \\ &)] /[\log (\Omega s / \Omega p)] \\ &N=(1 / 2) *\left[\log \left(\left(10^{0.1 *} 15_{-} 1\right) /\left(10^{0.1 * 1}-1\right)\right)\right. \\ &] /[\log (1.02 / 0.65)] \\ &N=1 / 2 *(2.073 / 0.65) \\ &N=5.297 \end{aligned} $$

Answer 2

Ap = 1dB, As = 15 dB

wp = 0.2$\pi$, ws = 0.3$\pi$

Finding characteristics of analog filter,

For the bilinear transformation

$\Omega$ = 2/Ts * tan (w/2)

The value of Ts is not given so = 1 second

So,  $\Omega$ = 2 * tan (w/2)

$\Omega$p = 2 * tan (wp/2) =  2 * tan (0.2$\pi$/2) = 0.65 rad/sec

$\Omega$s = 2 * tan (ws/2) =  2 * tan (0.3$\pi$/2) = 1.02 rad/sec

Finding order using below equation,

N = (1/2) * [ log((10^0.1*As - 1) / (10^0.1*Ap - 1) ) ] / [log($\Omega$s/$\Omega$p)]

N = (1/2) * [ log((10^0.1*15 - 1) / (10^0.1*1 - 1) ) ] / [log(1.02/0.65)]

N = 1/2 * (2.073/0.65)

N = 5.297

So the order of the filter = N = 6..ANS