Question

Determine the transfer function for a 2nd order Chebyshev low pass filter with 3dB frequency of 100krad/sec, a maximum gain of OdB, and a passband ripple of 1dB. (40 points) (a) (b) A bandpass filter is made by cascading the filter described in part (a) with a 2nd order Chebyshev high pass filter with 3dB frequency of 1krad/sec, a maximum gain of OdB and passband ripple of 2dB. Determine the midband gain of the filter. (30 points) A Chebyshev bandpass filter (with different specifications from the one in (c)) has the following transfer function: (c) T(s)Gs (s'+2s+4) (s+50+2500) where frequency is in krad/sec and G is a gain parameter that affects the midband gain of the filter. Use Sallen-Key biquad circuits to design this filter. You do not have to design for a particular value of G or determine the value of G that your design produces. (30 points)

Answer 1

(a). Given : dB ripple = 1 dB

then the ripple parameter is given by

$\epsilon =\sqrt(10^{(dB/10)}-1)=0.5088$

Parameter h need to be found for finding the transfer function of the Chebyshev filter for the given order (n = 2) and the ripple parameter (0.5088)

then ,

$h=tanh\bigg(\frac{1}{n}sinh^{-1}\frac{1}{\epsilon}\bigg)=0.61317$

For an even order chebyshev filter, the transfer function is given by

$T(s)=\prod_{i=1}^{\frac{n}{2}}\frac{1}{(s/(a_i\omega_c))^2+(1/b_i)(s/a_i\omega_c)+1} \\ where, \\ a_i =\bigg(\frac{1}{1-h^2}-sin^2\theta_i\bigg)^\frac{1}{2}, for 1\leq i \leq \frac{n}{2}$

$b_i =\frac{1}{2}\bigg(1+\frac{1}{h^2tan^2\theta_i}\bigg)^\frac{1}{2}, for 1\leq i \leq \frac{n}{2}$

$\theta_i=\frac{2i-1}{n}* 90^\circ, for 1\leq i \leq\frac{n}{2}$

here, $n=2 \, and \, \omega_c=100*10^3 rad/sec$

therefore, $\theta_1=45$

$a_1=1.05$

$b_1=0.9565$

therefore, transfer function of chebyshev low pass filter is

$T(s)=\frac{1}{9.07*10^{-11}s^2+9.956*10^{-6}s+1}$