pass band ripple: < dB

pass band edge: 4 kHz

stop band attenuation: - 40dB

stop band edge: 6kHz

Sample rate : 24 kHz

Use bilinear transformation.

Use Bilnear transformation :-

a) Calculation of $w_p$ and $w_s$.

we have, $f_{PB}=\dfrac {F_{PB}}{F_s}=\dfrac {4khz}{24khz}=0.166=\dfrac 16 $

$$w_p=2\pi F_{PB}=2\pi\times 0.166=1.043 $$

$=f_{SB}= \dfrac {F_{SB}}{F_s} =\dfrac {6khz}{24khz}= 0.25=\dfrac 14 \\ w_s=2\pi f_{SB}=2\pi\times 0.25=1.570 $

b) Calculation of $Ω_P, Ω_s$

$$\Omega_p =\dfrac 2T \tan (\dfrac {w_p}2)=\dfrac 2{1/24000}\tan (\dfrac {\pi/6}2)=12861.56 $$

$\Omega_s=\dfrac 2T\tan (\dfrac {W_s}2)=\dfrac 2{1/24000}\tan (\dfrac {\pi/4}2)=19882.25 $

C) Calculation of order of filter.

The order of filter can be given as,

$N=\dfrac 12 \dfrac {\log\Bigg[\dfrac {10^{0.1A_S}-1}{10^{0.1A_p}-1}\Bigg]}{\log \Bigg(\dfrac {\Omega_s}{\Omega_p}\Bigg)} \\ N=\dfrac 12 \dfrac {\log\Bigg[\dfrac {10^{0.1(40)}-1}{10^{0.1(1)}-1}\Bigg]}{\log \Bigg(\dfrac {19882.25}{12861.56}\Bigg)} \\ N=\dfrac 12.\dfrac {4.5867}{0.1891}\\ N=12.080 \approx 13$

d) Calculation of cut off frequency

$\Omega_c=\dfrac 2T\tan (\dfrac {W_c}2) \\ \dfrac {\Omega_c T}2=\tan\dfrac {W_c}2\space \space \space \space \space \therefore 2\tan^{-1}(\dfrac {\Omega_cT}2)$

But

$\Omega_c=\dfrac {\Omega_p}{(10^{A_p/1}-1)}=\dfrac {12861.25}{(10^{0.1}-1)^{1/2\times 12.080}} \\ =\dfrac {12861.25}{0.9457} \\ =13599.7145 rad/sec \\ W_c=2\tan^{-1} (\dfrac {\Omega_cT}2) \\ =2\tan^{-1} (\dfrac {13599.7145\times 1/24000}2)\\ =2\tan^{-1}(0.2833)\\ =0.552 rad $