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[CHAP. 1
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Case 3 6 5 n 5 10. For 6 5 n 5 10, all of the nonzero values of x ( n sum. and
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- k ) are within the limits of the
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Case 4 1 I 5 n 5 15. When n is in the range I 1 5 n 5 15, the sequences h ( k ) and x ( n - k ) overlap for n - 5 5 k 5 10. Therefore,
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Case 5 n > 15. Finally, For n > 15, there is again no overlap between h ( k ) and x ( n - k ) , and the product h ( k ) x ( n - k ) is equal to zero for all k. Therefore, y ( n ) = 0 for n 15.
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In summary, for the convolution we have
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The correlation of two sequences is an operation defined by the relation
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Note that we use a star * to denote correlation and an asterisk to denote convolution.
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( a ) Find the correlation between the sequence x ( n ) = u ( n ) - u(n - 6 ) and h(n) = u(n - 2 ) - u(n - 5).
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( b ) Find the correlation of x ( n ) = crnu(n)with itself (i.e., h(n) = x(n)). This is known as the autocora relation of x ( n ) . Assume that l1 < 1.
(a) If we compare the expression for the correlation of x ( n ) and h ( n ) with the convolution
we see that the only difference is that, in the case of convolution, h ( k ) is time-reversed prior to shifting by n , whereas for correlation h ( k ) is shifted without time-reversal. Therefore, with a graphical approach to compute the correlation, we simply need to plot x ( k ) and h ( k ) , shift h ( k ) by n (to the left if n > 0 and to the right if n < 0), multiply the two sequences x ( k ) and h ( n k ) , and sum the products. Shown in the figure below is a plot o F x ( k ) and h ( k ) .
CHAP. I]
SIGNALS AND SYSTEMS
Denoting the correlation by r,h(n), it is clear that for n = O the correlation is equal to 3. In fact, this will be the value of r r h ( n )for - I 5 n 5 2. For n = 3, s ( k ) and h(3 k ) only overlap at two points, and r r h ( 3 = 2. ) Similarly, because x ( k ) and h(4 k ) only overlap at one point, r,y,1(4) I . Finally, r r h ( n = 0 for n > 4. = ) Proceeding in a similar fashion for n < 0 , we find that r,,,(-2) = 2, and r-,,,(-3) = I . The correlation is shown in the figure below.
~.vII(~)
(h) Let r,(n) denote the autocorrelation of .x(n),and note that the autocorrelation is the convolution of x ( n ) with x(-n):
In addition observe that r , ( n ) is an even function of n:
Therefore, it is only necessary to find the values of r , ( 1 1 ) for n 1 0.For n 2 0 , we have
Using the symmetry of r , ( n ) , we have, for n < 0,
Combining these two results together, we finally have
Difference Equations 1.37
Consider a system described by the difference equation
Find the response of this system to the input
with initial conditions y ( - I ) = 0.75 and y(-2) = 0.25.
The first step in solving this difference equation is to find the particular solution. With x ( n ) = (OS)"u(n), assume we a solution of the form y,(n) = C l ( 0 . 5 ) " n 20 Substituting this solution into the difference equation, we have
SIGNALS AND SYSTEMS Dividing by (0.5)". CI = 2 c I - 4 C I + o . 5 + 1 which gives
[CHAP. 1
c, = ;
The next step is to find the homogeneous solution. The characteristic equation is z2-z+l=O
Therefore, the form of the homogeneous solution is yh(n) = and the total solution becomes y(n) = (0.5)"+' ein"l> ~~~-1nnl3
+ A , e ~ n n l > +2e-in+
n z 0
The constants A1 and A2 must now be found so that the total solution satisfies the given initial conditions, y(-1) = 0.75 and y(-2) = 0.25. Because the solution given in Eq. (1.25) is only applicable for n > 0, we must derive an equivalent set of initial conditions for y(0) and y(l). Evaluating the difference equation for n = 0 and n = I, we have y(0) = y(- 1) - y(-2) and y(l) = y(0) - y(-1)
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