First of all I want to let you know about notations that I am using-:

$\alpha_p$=attenuation factor of passband(in dB)

$\alpha_s$=attenuation factor of stopband (in dB)

$A_p$=Gain at passband edge frequency

$A_s$=Gain at stopband edge frequency



$\epsilon$=passband frequency parameter,

$\lambda$=stopband frequency paramter

$\Omega_c$=cut off frequency(analog)

$\Omega_p$=passband frequency(analog)

$\Omega_s$=stopband frequency(analog)

$\omega_c$ $\omega_s$ $\omega_p$ are respectively cutoff,stopband and passband frequency(digital)



I have asked lots of related questions in this post but the point I am trying to ask is how to understand digital filter design specifications from given word problem-:

1) Design a digital low pass butterworth filter by applying bilinear transformation techniques for the given specifications-:

Passband peak to peak ripple $\leq$ 1dB

Passband edge frequency=1.2 kHz

Stopband attenuation $\geq$ 40 dB

Stopband edge frequency=2.5 KHz

Sample rate=8 KHz

My solution-:

$\alpha_p$=1 dB

$\alpha_s$=40 dB



Hence $\omega_p=2\pi F_p T$

Here $T=\frac{1}{8 \cdot 10^3}$

2) Design a low pass discrete IIR filter by bilinear transformation method to an approximate butterworth filter having the specifications having specifications as below-:

pass band edge frequency $\omega_p$=0.22 $\pi$ radians

stop band edge frequency $\omega_s$=0.54 $\pi$ radians

Passband ripple $\delta_p$=0.11

Stopband ripple $\delta_s$=0.22

My solution-:

$\alpha_p$=-20 $\log_{10}(\delta_p)$

$\alpha_s$=-20 $\log_{10}(\delta_s)$

I am confused in this formula because of this image-: enter image description here

what's the pass band ripple and stop band attenuation of a digital filter?

pass band edge frequency $\omega_p$=0.22 $\pi$ radians

stop band edge frequency $\omega_s$=0.54 $\pi$ radians

3) Design a digital low pass filter with the following specifications-:

i) Pass-band magnitude constant to 0.7 dB below the frequency of 0.15$\pi$

ii) Stop band attenuation at least 14 dB for the frequencies between 0.6$\pi$ to $\pi$

Use butterworth approximation as a prototype and use impulse invariance method to obtain the digital filter.

My solution-:

$\alpha_p$=0.7 dB

$\alpha_s$=14 dB

$\omega_p$=0.15 $\pi$

$\omega_s$=0.6 $\pi$

4) Design a low pass digital filter by bilinear trasnformation method to an approximate butterworth filter if passband edge frequency is $0.25$ $\pi$ radians and maximum deviation of 1 dB below o dB gain in the passband The maximum gain of -15 dB and frequency is $0.45\pi$ radians in stopband. Consider sampling frequency 1 Hz.

$\omega_p$=0.25 $\pi$

$\delta_p$=1 dB

$\alpha_p$=-20 $\log_{10}(\delta_p)$


$\alpha_s$=-15 dB

5) Design a digital low pass filter of 1 rad/sec bandwidth and 2 sec sampling interval using bilinear transformation which meets the following specifications-:

a) Acceptable passband ripple of 2dB

b) Cut off frequency of 1 rad/sec

c) Stopband attenuation of 20dB or more beyond the frequency of 1.3 rad/sec

My solution-:

$\alpha_p$=2 dB

$\alpha_s$=20 dB




I have no idea how we calculate $\omega_p$ from this. We need to find order of filter and as you know

$N \geq \frac{\log(\lambda/\epsilon)}{\log(\omega_s/\omega_p)}$

So we need $\omega_s$ and $\omega_p$. But I have no idea how to calculate $\omega_p$ from given specifications.



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