Question

Question #1: a) An IIR lc w pass filter is designed with the Butterworth n CCLO 43 C14 method ej 1 Q2 1+0 provides the following pole plot with N-2 and Q = 0.56, S, = Oe'2 zero N Img Re 1 0-395903

kindly solve this question...
digital signal processing......

Answer 1

An IIR filter designed using Butterworth method is expressed as

1 )2N

Given that, ,
=0.56 c
, and
Sk2e2K+N+1)

The transfer function of the IIR filter is expressed as

H(s) N s-S k-1

Let's find all the poles by substituting the values for
K1,2, 3,4
in
Sp
expression.

When
K
,
s1 2e (5)- 0.56e(5)0.39598 j0.39598

When ,
2 e(0.56e(0.39598 j0.39598

When , $s_3 = \Omega_c e^{j\frac{\pi}{4}(9)} = 0.56e^{j\frac{\pi}{4}(9)} = 0.39598 +j0.39598$

When
K 4
, $s_4 = \Omega_c e^{j\frac{\pi}{4}(5)} = 0.56e^{j\frac{\pi}{4}(5)} = -0.39598 +j0.39598$

Note: We need to select the poles that lie on the left hand side of plane, which means, the poles with negative values in the real parts to calculate the transfer function.

The required poles are and .

=================================================================

Substituting the values for in the transfer function.

Sp

$H_c(s) = \frac{1}{(\frac{s+0.39598-j0.39598}{0.56})(\frac{s+0.39598+j0.39598}{0.56})}$

Taking the denominator term in the denominator section of H(s) into the numerator

$H_c(s) = \frac{0.56*0.56}{(s+0.39598-j0.39598)(s+0.39598+j0.39598)}$

We know that,

$H_c(s) = \frac{0.3136}{(s+0.39598)^2+0.39598^2}$

$H_c(s) = \frac{0.3136}{s^2+0.79196s+0.1568+0.1568}$

$H_c(s) = \frac{0.3136}{s^2+0.79196s+0.3136}$

================================================================

Alternatively, a second order normalized Butterworth filter is expressed as

$H_c(s) = \frac{1}{s^2+1.4142s+1}$

$H_c(s)|_{s = \frac{s}{\Omega_c}} = \frac{1}{(\frac{s}{\Omega_c})^2+1.4142\frac{s}{\Omega_c}+1}$

H(s) s21.4142cs 2

Placing in the above equation

=0.56 c

$H_c(s) = \frac{0.3136}{s^2+0.79196s+0.3136}$

================================================================