Question

For a sixth-order Butterworth high pass filter with cutoff frequency 3 rad/s, compute the following:

Answer 1

Solution :

Given :

A six order Butterworth high pass filter.

∴ n = 6, $w_c=1 \ rad/s$

a). The location at poles :

    $s^6-(w_c)^6=0$

   $s^6=(w_c)^6=1^6$

  ∴ $s^6 = 1$

Therefore, it has 6 repeated poles at s = 1.

b). The transfer function H(S) :

    Transfer function H(S) $=\frac{1}{1+j\left(\frac{w_c}{s}\right)^6}$

                                         $=\frac{1}{1-\left(\frac{w_c}{s}\right)^6}$

  ∴    H(S) $=\frac{s^6}{s^6-(w_c)^6}=\frac{s^6}{s^6-1}$

   H(S) $=\frac{Y(s)}{X(s)}=\frac{s^6}{s^6-1}$

c). The corresponding LCCDE description :

  $=\frac{Y(s)}{X(s)}=\frac{s^6}{s^6-1}$

   $Y(s)(s^6-1) = s^6 \times (s)$

   $Y(s)s^6-y(s).1 = s^6 \times (s)$

By taking inverse Laplace transformation on BS

   $L^{-1}[Y(s)s^6-Y(s)1]=L^{-1}[s^6 \times (s)]$

   $\frac{d^6y(t)}{dt^6}-y(t)=\frac{d^6 \times (t)}{dt^6}$

  Hence solved.