Thus, the superposition property (1.68) is satisfied and the system is linear. in .NET framework

Make QR Code in .NET framework Thus, the superposition property (1.68) is satisfied and the system is linear.

Thus, the superposition property (1.68) is satisfied and the system is linear.
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[CHAP. 1
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Fig. 1-38
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( d ) Let y , [ n ] be the response to x l [ n ] = x [ n - n o ] . Then
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{ x [ - n o ] ) = n x [ n - n,] n n -n,)x[n -no] # ~ , ( n ]
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Hence, the system is not time-invariant. (el Let x [ n ] = u [ n l . Then y [ n l = nu[nl. Thus, the bounded unit step sequence produces an output sequence that grows without bound (Fig. 1-38) and the system is not B I B 0 stable.
139. A system has the input-output relation given by
where k,,is a positive integer. Is the system time-invariant
Hence, the system is not time-invariant unless k, = 1. Note that the system described by Eq. (1.114) is called a compressor. It creates the output sequence by selecting every koth sample of the input sequence. Thus, it is obvious that this system is time-varying.
1.40. Consider the system whose input-output relation is given by the linear equation y=ax+b
(1.115)
where x and y are the input and output of the system, respectively, and a and b are constants. Is this system linear
If b + 0, then the system is not linear because x system is linear.
implies y
=b #
0. If b
= 0,
then the
1.41. The system represented by T in Fig. 1-39 is known to be time-invariant. When the inputs to the system are xl[n], x,[n], and x,[n], the outputs of the system are yl[n], y,[n], and y , [ n ] as shown. Determine whether the system is linear.
From Fig. 1-39 it is seen that
x 3 [ n ]= x l [ n ] + x , [ n
- 21
CHAP. 1)
SIGNALS AND SYSTEMS
Fig. 1-39
Thus, if T is linear, then
T{xJnI) = T{x,[nl} + T(x,[n - 21) = y , [ n I + y , [ n - 21 which is shown in Fig. 1-40. From Figs. 1-39 and 1-40 we see that y 3 b I +y,[nI + y J n - 21
Hence, the system is not linear.
- 2 - 1 0 1 2 3 4
-2 - 1
Fig. 1-90
1.42. Give an example of a system that satisfies the condition of additivity (1.66) but not the condition of homogeneity ( 1.67).
SIGNALS AND SYSTEMS
[CHAP. 1
Consider a discrete-time system represented by an operator T such that
y [ n ] = T ( x [ n ] )= x * [ n ] (1.116)
where x * [ n ] is the complex conjugate of x [ n ] . Then
T { x l [ n I + x 2 [ n I } = { x ~ [ n+I ~ 2 [ n I } = x T [ n I + x [ n I = ~ l [ n + ~ 2 [ n I * I
Next, if a is any arbitrary complex-valued constant, then
T { a x [ n ] } = { a x [ n ] )* = cu*x*[n] = a * y [ n ] # a y [ n ]
Thus, the system is additive but not homogeneous.
Show that the causality for a continuous-time linear system is equivalent to the following statement: For any time t o and any input x ( t ) with x ( t ) = 0 for t r r,,, the output y ( t ) is zero for t I o . t Find a nonlinear system that is causal but does not satisfy this condition. Find a nonlinear system that satisfies this condition but is not causal.
Since the system is linear, if x ( t ) = 0 for all I , then y ( t ) = 0 for all t . Thus, if the system is causal, then x ( r ) = 0 for t _< t o implies that y ( t ) = 0 for 1 I I,. This is the necessary condition. That this condition is also sufficient is shown as follows: let x , ( t ) and x,(t) be two inputs of the system and let y , ( t ) and y,(t) be the corresponding outputs. If x , ( t ) = x , ( t ) for t I t o , or x ( r ) = x , ( t ) - x 2 ( t ) = 0 for I s t , , then y , ( t ) = y 2 ( t ) for I s t , , or y ( t ) = y , ( ~ ) - y 2 ( t ) = 0for t s t,. Consider the system with the input-output relation
y(t) =x(t)
This system is nonlinear (Prob. 1.40) and causal since the value of y ( t ) depends on only the present value of x ( t ) . But with x ( t ) = 0 for I II,, y ( t ) = I for t s t,. Consider the system with the input-output relation
y(t) =x(t)x(t y ( t ) at time I depends on the value of x ( t t I implies that y ( t ) = 0 for I I I,. t,
+ 1)
It is obvious that this system is nonlinear (see Prob. 1.35) and noncausal since the value of + I ) of the input at time I + 1. Yet x ( t ) = 0 for
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