**IIR Design** - Convert the band-pass analog filter with
transfer function:

_{}

which has a resonance (peak frequency response) at W = 4 to a digital filter using the bi-linear
transformation.

The digital filter is to have a resonance at w = p/2

A. What is the desired mapping?

Td = ½ to map W =
4 to
w
= p/2 using: _{} or _{}

The mapping is therefore: _{}

B. What is the resulting H(z)?

The resulting H(z) is:

_{}

C. Where are the poles and zeros of this filter?

Factoring the numerator gives the zeros and
factoring the denominator yields the poles.
The poles are approximately at 0.987e^{±j}^{p/2 }and
the zeros are at -1 and 0.95

D. The Frequency Response of the digital filter (at a 1 kHz sample rate) is:

**FIR Design** - Starting with
the ideal low-pass filter function with cutoff W_{c} = p/4 ,
develop an FIR low pass filter using a Hamming window that has the following
characteristics:

w_{c}
= p/4 (cut off frequency)

M = 31 (Filter length)

A. What is the FIR filter pulse response? Sketch it below.

The pulse response is the product of sin(x)/x (the Inverse Fourier Transform of the ideal filter shifted to center at n=16) and the Hamming raised cosine window function.

B. What is the approximate transition width of the resulting filter?

The transition width for the rectangular window would be 1/T in normalized frequency, with the Hamming Window the Transition width is doubled to 2/T in normalized frequency.

C. Sketch the FIR filter frequency response below.

This is a low pass filter. The positive frequencies start at index 1 and go to index 16. The highest negative frequency is at index 17. DC (f=0) is at index 0 and 32.