Reference no: EM132723985
Lab Exercise: Filter Design
In this section, we will use PeZ to place the poles and zeros of H(z) to make a filter with a desirable frequency response. Filter design is a process that selects the coefficients {ak} and {bk} to accomplish a
Figure 1: Magnitude response of two unknown filters. Frequency axis is normalized. Use PeZ to help you find the filter coefficients that will match these frequency responses as closely as possible. (a) Second-order FIR filter. (b) Second-order IIR filter.
given task. The task here is to create a filter that has a very narrow "notch." This filter would be useful for removing one frequency component while leaving others undisturbed. The notch filter can be synthesized from the cascade of two simpler filters shown in Fig. 1.
(a) Start the process by using PeZ to design each of the filters given in Fig. 1. (You will have to determine the locations of the poles and zeros from the plots in Fig. 1.) Both filters are second-order. Make sure that you enter the poles and zeros precisely. PeZ will do the conversion between between root locations and polynomial coefficients, but you could also do this with the MATLAB commands roots and poly. You can check your results by also calculating the filter coefficients by hand (see the next section on polynomials with complex coefficients). Record the coefficients of your filters in the table provided.
Note: Use PeZ or freqz() to verify that the frequency response of each filter is correct.
(b) Now use PeZ to connect the filters together in a cascade. Place the poles and zeros, and then view the frequency response. Determine the filter coefficients for the overall cascaded filter H(z).
(c) Use freqz() to determine the frequency response of the cascade of the two filters that you "de- signed" in part (e). Plot the magnitude of the overall frequency response of the cascade system for, and print a copy of the plot for your lab report.
2 Explain briefly why the frequency response magnitude has a notch, and explain why the gain (i.e., |H(ejω)|) at ω = 0 and ω = Π is the same.
Exercise - complex poles and zeros
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