# How to read the filter IIR design numericals specifications?

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

$$A_p$$=$$\sqrt\frac{1}{\epsilon^2+1}$$

$$A_s$$=$$\sqrt\frac{1}{\lambda^2+1}$$

$$\epsilon$$=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)

$$\epsilon=\sqrt{10^{\alpha_p/10}-1}$$

$$\lambda=\sqrt{10^{\alpha_s/10}-1}$$

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

$$F_p$$=1.2KHz

$$F_s$$=2.5KHz

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-:

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 $$\log_{10}(\delta_p)$$

$$\omega_s$$=0.45$$\pi$$

$$\alpha_s$$=-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

$$\omega_s$$=1.3

$$\omega_c$$=1

Bandwidth=?=1

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.