lab7 report

# lab7 report - Results and Explanation: 3.1 Deconvolution...

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R e s u l t s a n d E x p l a n a t i o n : 3.1 Deconvolution Experiment for 1-D Filters a) In this part of the lab, it was asked to use the firfilt() function to implement this filter w[n] = x[n] − 0.9x[n − 1] on the input signal xx = 256*(rem(0:100,50)<10) which was given in MATLAB coding. Then the firfilt() function was used for the convolution operation and both the input and output signal were plotted by using subplot and the both of the signals were indexed by starting to plot them from n+1. The coding of this operation can be seen in Appendix 3.1 part a. Figure Stem plot of xx the input signal that is on the top and yy the output signal on the bottom after FIR filtering (b) The length of the signal can be determined by using the hint given in the question and if we use the relation the length of the x[n] is N1 and the length of the filter signal is N2 the relation (N1+N2)/2 would give us the length of the filtered signal. . Also the length of the input size can be determined by N2 -1. Since this is a convolution operation it could be seen better if the dconvdemo function was used because that allows us to actually move the h[n] through the input signal and shows how a averaging filter works better but since it is a demo the output that it shows to us is limited and that why the computation is done in MATLAB and the stem plot in MATLAB is used to view the output to get better and more accurate results. 3.1.1 Restoration Filter a) In this part of the lab the main purpose is to use the restoration filter provided with the specification r=0.9 and M=22 and using a vector mm to make the signal 22 units long. The code for this part of the lab can be seen in Appendix part b. However, when the output signal is not the exact replica of the original signal and this could be seen using the subplot graphing function of MATLAB. b) In this part it is asked from us to analyze the two stem plots which should be the same graph since the restoration filter is used but it is not exactly the same and could be seen on the graph below.

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Figure The graph on the top is the result after the restoration filter and the one on the bottom is the original signal c) In order to find the error difference between x[n] and y[n], which should have been identical, y[n] is substracted from the x[n] and then the absolute value of it is taken since a negative error doesn’t mean anything and they multiplied by a 100 for the percent error and the graph below is obtained when plotted. EDU>> stem(n,ans(n+1)) Figure Error difference between the original signal and the restoration filter output signal 3.1.2 Worst-Case Error
a) In this part worst-case error is evaluated by using the max function of the ans that was obtained in the previous section. Basically what max function does is to find the max difference between points and the answer obtained by the function is 22.6891 as it can be seen from below, which also confirms that the graph above is correct. EDU>> max(ans)

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## This note was uploaded on 11/14/2011 for the course ECE 2140 taught by Professor Carrol during the Spring '11 term at GWU.

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lab7 report - Results and Explanation: 3.1 Deconvolution...

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