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Thus, the superposition property (1.68) is satisfied and the system is linear. in .NET framework
Thus, the superposition property (1.68) is satisfied and the system is linear. Scan QR Code 2d Barcode In Visual Studio .NET Using Barcode Control SDK for VS .NET Control to generate, create, read, scan barcode image in VS .NET applications. Print QR Code JIS X 0510 In .NET Using Barcode creator for VS .NET Control to generate, create QR Code 2d barcode image in .NET applications. SIGNALS AND SYSTEMS
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UCC  12 Generator In Java Using Barcode creator for Android Control to generate, create UCC.EAN  128 image in Android applications. Paint GS1128 In None Using Barcode creation for Office Excel Control to generate, create GS1 128 image in Microsoft Excel applications. Hence, the system is not timeinvariant. (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. 138) and the system is not B I B 0 stable. 139. A system has the inputoutput relation given by
where k,,is a positive integer. Is the system timeinvariant
Hence, the system is not timeinvariant 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 timevarying. 1.40. Consider the system whose inputoutput 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. 139 is known to be timeinvariant. 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. 139 it is seen that
x 3 [ n ]= x l [ n ] + x , [ n
 21 CHAP. 1) SIGNALS AND SYSTEMS
Fig. 139 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. 140. From Figs. 139 and 140 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. 190 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 discretetime 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 complexvalued 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 continuoustime 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 inputoutput 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 inputoutput 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

