Question: You are testing a 1 volt. d.c. source and have the following measurements obtained from the source every minute starting at time 0 To find
![п x[n] п x[п] 0 1.0 3 0.7 1 1.2 4 1.2 2 0.9 5 1.0](https://dsd5zvtm8ll6.cloudfront.net/si.question.images/images/question_images/1545/4/9/4/4145c1e5f8e676a11545476990232.jpg)
To find the average voltages for the first 5 min, i.e., to get rid of some of the noise in the data, you use the following averager
![y[n] = x[n] + x[n – 1] +x[n – 2]](https://dsd5zvtm8ll6.cloudfront.net/si.question.images/images/question_images/1545/4/9/4/4145c1e5f8e78da31545476990326.jpg)
(a) Use the input/output equation of the filter to compute and then plot the moving averages y[n] for 2 ¤ n ¤ 5.
(b) Find the impulse response h[n] of the averager and compute the output y[n] using the convolution sum. Do you obtain the same results as before?
(c) Suppose you use a median filter of length 3, so that at some sample nwe consider the samples values at n, n 1, and n 2, order these values and the one in the middle is the output ym[n]. Moving one sample we consider the next three values and find the new median. Compute and plot ym[n] for 2 ¤ n ¤ 5.
п x[n] п x[п] 0 1.0 3 0.7 1 1.2 4 1.2 2 0.9 5 1.0 y[n] = x[n] + x[n – 1] +x[n – 2]
Step by Step Solution
3.41 Rating (179 Votes )
There are 3 Steps involved in it
a Moving averages y2 09 12 13 103 y3 07 09 123 093 y4 12 07 093 093 y5 1 12 073 097 ... View full answer
Get step-by-step solutions from verified subject matter experts
