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barcode lib ssrs Because u (  k ) = 0 fork > 0 and u((n in Software
Because u (  k ) = 0 fork > 0 and u((n USS Code 128 Scanner In None Using Barcode Control SDK for Software Control to generate, create, read, scan barcode image in Software applications. Code128 Creator In None Using Barcode printer for Software Control to generate, create Code 128 image in Software applications.  1) = 0 fork < n
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Recognize Code 128B In .NET Using Barcode reader for .NET framework Control to read, scan read, scan image in VS .NET applications. Painting Bar Code In None Using Barcode generation for Microsoft Word Control to generate, create barcode image in Microsoft Word applications. I)(;) u(n  I ) [.(;In
 2(n  ~ ) ( ; ) " ] u ( I ) n
= ( 2  n ) ( i ) " u ( n I ) Prove the commutative property of convolution
Proving the commutative property is straightforward and only involves a simple manipulation of the convolution sum. With the convolution of x ( n ) with h ( n ) given by with the substitution 1 = n
 k , we have
and the commutative property is established.
CHAP. 1 1
SIGNALS AND SYSTEMS
Prove the distributive property of convolution
To prove the distributive property, we have
Therefore, and the property is established.
h(n) = 3(;)"u(n)  2(;)"'u(n) be the unit sample response of a linear shiftinvariant system. If the input to this system is a unit step, x(n) = n z O
else
find limn,, With
y ( n ) where y ( n ) = h ( n ) x ( n ) . y(n) = h ( n ) * x ( n ) = h(k)x(n  k ) k=w
if x ( n ) is a unit step, Therefore, Evaluating the sum, we have
ncc
lim y(n) = h(k) k=m
Convolve
with a ramp
The convolution of x ( n ) with h ( n ) is
[ ( 0 . 9 ) ~ u ( k ) ] [  k)u(n  k ) ] (n
k=03 SIGNALS AND SYSTEMS Because u(k) is zero fork < 0, and u(n  k) is zero fork > n, this sum may be rewritten as follows: [CHAP. 1
Using the series given in Table 1 1, we have
which may be simplified to y (n) = [Ion
 90 + 90(0.9)"]u(n) Perform the convolution
y(n) = x ( n ) * 0 ) when h(n) = (;)"u(n) x ( n ) = ( i ) " [ u ( n ) u(n  101)l  With we begin by substituting x(n) and h(n) into the convolution sum
To evaluate this sum, which depends on n, we consider three cases. First, for n c 0, the sum is equal to zero because u(n  k ) = 0 for 0 5 k 5 100. Therefore, Second, note that for 0 5 n 5 100, the step u(n  k) is only equal to 1 fork 5 n. Therefore, I (f)" 1(f1) ,I+ l
= 3(;)"[1  (f)"+'] CHAP. 11
SIGNALS AND SYSTEMS
Finally, for n 2 100, note that u ( n  k ) is equal to I for all k in the range 0 5 k 5 100. Therefore, =(f)" In summary, we have
1  ($O' = 3(!)"[l
(f) ] Let h(n) be a truncated exponential
and x ( n ) a discrete pulse of the form x(n) = Find the convolution y ( n ) = h(n) * x ( n ) . To find the convolution of these two finitelength sequences, we need to evaluate the sum
O s n s 5 else
To evaluate this sum, it will be useful to make a plot of h ( k ) and x ( n  k ) as a function of k as shown in the following figure: Note that the amount of overlap between h ( k ) and x ( n  k ) depends on the value of n . For example, if n < 0, there is no overlap, whereas for 0 5 n 5 5 , the two sequences overlap for 0 5 k 5 n . Therefore, in the following, we consider five separate cases. Case 1 n = 0. When n c 0, there is no overlap between h ( k ) and x ( n  k ) . Therefore, the product h(k)x(n  k) = 0 for all k , and y ( n ) = 0. Case 2 0 _< n 5 5 . For this case, the product h ( k ) x ( n  k ) is nonzero only fork in the range 0 5 k 5 n . Therefore,

