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30 Chapter 2 DISCRETE-TIME SIGNALS AND SYSTEMS
c. x(n) =cos(0.04πn)+0.2w(n), 0 ≤ n ≤ 50.
>> n = [0:50]; x = cos(0.04*pi*n)+0.2*randn(size(n));
>> subplot(2,2,2); stem(n,x); title(’Sequence in Problem 2.1c’)
>> xlabel(’n’); ylabel(’x(n)’);
The plot of the sequence is shown in Figure 2.1c.
d. ˜x(n) ={..., 5, 4, 3, 2, 1, 5, 4, 3, 2, 1, 5, 4, 3, 2, 1, ...}; −10 ≤ n ≤ 9.
↑
Note that over the given interval, the sequence ˜x (n) has four periods.
>> n = [-10:9]; x = [5,4,3,2,1];
>> xtilde = x’ * ones(1,4); xtilde = (xtilde(:))’;
>> subplot(2,2,4); stem(n,xtilde); title(’Sequence in Problem 2.1d’)
>> xlabel(’n’); ylabel(’xtilde(n)’);
The plot of the sequence is shown in Figure 2.1d.
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□ EXAMPLE 2.2 Let x(n) = {1, 2, 3, 4, 5, 6, 7, 6, 5, 4, 3, 2, 1}. Determine and plot the following
↑
sequences.
a. x 1(n) =2x(n − 5) − 3x(n +4)
b. x 2(n) =x(3 − n)+x(n) x(n − 2)
Solution
The sequence x(n) isnonzero over −2 ≤ n ≤ 10. Hence
>> n = -2:10; x = [1:7,6:-1:1];
will generate x(n).
a. x 1(n) =2x(n − 5) − 3x(n + 4).
The first part is obtained by shifting x(n) by5and the second part by
shifting x(n) by−4. This shifting and the addition can be easily done using
the sigshift and the sigadd functions.
>> [x11,n11] = sigshift(x,n,5); [x12,n12] = sigshift(x,n,-4);
>> [x1,n1] = sigadd(2*x11,n11,-3*x12,n12);
>> subplot(2,1,1); stem(n1,x1); title(’Sequence in Example 2.2a’)
>> xlabel(’n’); ylabel(’x1(n)’);
The plot of x 1(n) isshown in Figure 2.2a.
b. x 2(n) =x(3 − n)+x(n) x(n − 2).
The first term can be written as x(−(n − 3)). Hence it is obtained by first
folding x(n) and then shifting the result by 3. The second part is a multiplication
of x(n) and x(n−2), both of which have the same length but different
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