Question

Let \(\left\{ {{y_k}} \right\}\) be the sequence produced by sampling the continuous signal \({\bf{2cos}}\frac{{\pi t}}{{\bf{4}}} + {\bf{cos}}\frac{{{\bf{3}}\pi t}}{{\bf{4}}}\) at \(t = {\bf{0}},\,{\bf{1}},\,{\bf{2}},\,......\)as shown in the figure. The values of \({y_k}\), beginning with \(k = {\bf{0}}\), are

3, .7, 0, \( - .{\bf{7}}\), \( - {\bf{3}}\), \( - .{\bf{7}}\), 0, .7, 3, .7, 0

Where .7 is an abbreviation for \(\frac{{\sqrt {\bf{2}} }}{{\bf{2}}}\).

a. Compute the output signal \(\left\{ {{z_k}} \right\}\) when \(\left\{ {{y_k}} \right\}\) is fed into the filter in Example 3.

b. Explain how and why the output in part (a) is related to the calculations in Example 3.

Exercises 23 and 24 refer to a difference equation of the form \({y_{k + {\bf{1}}}} - a{y_k} = b\), for suitable contain a and b.

Short answer

a. 1.4, 0, \( - 1.4\), \( - 1.99\), \( - 1.4\), 0, 1.4, 1.99, 1.4

b. The output values are slightly higher than in Example 3.

Step-by-step solution

Write the output signal from Example 3

The output signal to produce the sequence is

\({z_k} = .35{y_{k + 2}} + 0.5{y_{k + 1}} + .35{y_k}\).

Find the values of \({z_k}\)

\(k\)	\({y_k}\)	\({z_k} = .35{y_{k + 2}} + 0.5{y_{k + 1}} + .35{y_k}\)
0	3
1	0.7	1.4
2	0	0
3	\( - 0.7\)	\( - 1.4\)
4	\( - 3\)	\( - 1.99\)
5	\( - 0.7\)	\( - 1.4\)
6	0	0
7	0.7	1.4
8	3	1.99
9	0.7	1.4
10	0

Compare the output with that in Example 3

In Example 3, the maximum output value is 0.7, and the minimum output value is \( - 1\). But here, the output values are slightly higher.