Pass band ripple: ≤1dB Pass band Edge: 4KHz

Stop band Attenuation: ≥40dB; Stop band Edge: 6KHz

Sampling Rate: 24KHz. Use bilinear transformation

Step-1: Identification of specification of filter

$A_p=1dB ; A_s=40dB ; F_PB=4KHz ; F_SB=6KHz; F_S=24KHz ; T_S=41.66 µsec$

Now, we have

$F_{SB}=\frac{F_{SB}}{F_S} =0.25$

$ω_S=2πF_{SB}=1.57$

And

$F_PB=\frac{F_{PB}}{F_S} =0.166$

$ω_P=2πF_{PB}=1.043$

Now,

$(-A)_p (dB)=20log⁡(A_p)$

$-1=20 log⁡(A_p)$

$A_p=0.891$

And

$(-A)_s (dB)=20log⁡(A_s)$

$-40=20log(A_s)$

$∴A_s=0.01$

$Ω_P=\frac{2}{T} tan(\frac{ω_P}{2})=27.583 K rad/sec$

$Ω_S=\frac{2}{T} tan(\frac{ω_S}{2})=47.969 K rad/sec$

Step-2: Order of filter

Order of filter is given by

$N≥\frac{1}{2}(19.08)$

$∴N=9.54≅10$

Step-3: Cut of frequency

$Ωc=\frac{Ω_P}{(\frac{1}{(A_P^2 )-1)^{\frac{1}{2N}}}}$

$Ωc=29.506 K rad/sec$

Step-4: Calculation of poles

$P_k=Ωce^{(j(N+2k+1)^{\frac{π}{2N}}}$

When k=0

$∴P_o=-4615+j29142$

$When k=1; P_1=-13395+j26290$

$When k=2; P_2=-20863+j20863$

$When =3 ; P_3=-26290+j13395$

$When =4 ; P_4=-29142+j4615$

$When =5 ; P_5=-29142-j4615$

$When =6 ; P_6=-26290-j13395$

$When k=7; P_7=-20863-j20863$

$When =8 ; P_8=-13395-j26290$

$When =9 ; P_9=-4615-j29142$

Step-5: Calculation of Transfer Function

$H(s)=\frac{(Ωc)^N}{((S-P_0)(S-P_1)(S-P_2)(S-P_3)(S-P_4)(S-P_5)(S-P_6)(S-P_7)(S-P_8)(S-P_9))}$