barcode lib ssrs Because u ( - k ) = 0 fork > 0 and u(-(n in Software

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Because u ( - k ) = 0 fork > 0 and u(-(n
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- 1) = 0 fork < n
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+ I, the convolution sum becomes
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With the change of variables m = -k, and using the series formulas given in Table I -I, we have
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Let us check this answer for a few values of n using graphical convolution. Time-reversing x ( k ) , we see that h(k) and x ( - k ) do not overlap for any k and, thus, y ( 0 ) = 0 . In fact, it is not until we shift x ( - k ) to the left by two that there is any overlap. With x ( - 2 - k ) and h ( k ) overlapping at one point, and the product being equal to it follows that y ( - 2 ) = Evaluating the expression above for y ( n ) above at index n = -2, we obtain the same result. For n = -3, the sequences x ( - 3 - k ) and h ( k ) overlap at two points, and the sum of the products gives y(-3) = f $ = $, which, again, is the same as the expression above.
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If the response of a linear shift-invariant system to a unit step (i.e., the step response) is
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find the unit sample response, h(n).
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In this problem, we begin by noting that S(n) = u ( n ) - u(n - 1) Therefore, the unit sample response, h ( n ) ,is related to the step response, s ( n ) , as follows:
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Thus, given s(n), we have h ( n ) = s ( n ) - s(n - I)
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= n(;)"u(n) - (n =
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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 shift-invariant 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
n-cc
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 finite-length 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,
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