EE424 Quiz 5 Solutions

1. Comb and Notch filters

a) Describe one application that a notch filter would be used for and state why you want a notch filter. (2 points)

b) Find the spectrum of the echo filter with the linear difference equation:

Sketch the spectrum magnitude for D=10. (3 points)

c) A reverberator has the transfer function:

Find the pole and zero locations of this filter? Plot the pole-zero diagram and sketch the spectrum magnitude for D=5. (5 points) (Hint: Look for pole-zero cancellation)

Solution:

a) Notch filters remove signal harmonics. Two examples of this are 60 Hz power harmonics and strong radar signals.

b) The matlab program q5p1.m generates the figures for parts 1b and 1c.

c) zeros are at:

The poles cancel the zeros at k = multiple of 4. The spectrum is shown below. Peaks in the filter occur where the poles cancel out the zeros at every 2*pi*k/5. In this case the pole-zero representation gives more insight into the system than the transfer function.

2. Noise Reduction Ratio

a) Describe what a noise reduction filter does. Give an example of an application of this filter. (2 points)

b) For an application of a signal with white noise, we need a noise reduction ratio of 30 dB. How long of an averaging FIR filter is needed to acheive this ratio? What is it's settling time and bandwidth?(3 points)
c) For the same filter as in part b, Design a first order IIR smoother filter that achieves the same NRR of 30 dB as in part b. What is its settling time and bandwidth? (3 points)
d) Compare the results of parts b and c, which filter would you use and why?(2 points)

Solutions:

a) Filter reduces noise in the system so the signal can be seen. Common applications are: ECG, speech signals, communications systems, etc.

b) For a noise reduction of 35 dB, the NRR should be -35 dB (this means that the filtered noise is 35 dB down from the unfiltered noise.

NRR_dB=-35=20*log10(NRR)

NRR=10^(-35/20)=0.0178

For an FIR filter:

NRR=1/N => N=57

This gives a settling time of 57 and a cutoff frequency of w_c=pi/N=.0551 radians.

c) a = (1-NRR)/(1+NRR) = 0.9651

Settling time is:

n_eff=ln(eps)/ln(a)=ln(0.01)/ln(0.9651)=129.5 which rounds to 130 samples.

Bandwidth is w_c= 1 - a = 0.0349

d) I would use the FIR filter since it has the same NRR and a shorter settling time. It also has a slightly larger bandwidth which may be useful for capturing more signal information. The figure below shows the frequency responses of both filters. Note how the IIR filter appears to be shaped like an envelope for the FIR filter. This result can be generated using q5p2.m