文档介绍：§3-1 Convolution Representation of Linear Time-invariant Discrete-Time Systems
Chapter 3 Convolution Representation
§3-2 Convolution of Discrete-Time Signals
Problems
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§3-1 Convolution Representation of Linear Time-invariant Discrete-Time Systems
Consider a single-input single-output discrete-time system with input x[n] and output y[n]. Throughout this section it is assumed that y[n] is the output response resulting from x[n] with no initial energy in the system prior to the application of x[n]. It is also assumed that the system under consideration is causal, linear, and time-invariant, but the system is not necessarily finite dimensional.
Chapter 3 Convolution Representation
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Unit-Pulse Response
Let h[n] denote the output response of the system when the input x[n] is equal to the unit pulse δ[n] with no initial energy in the system at time n=0. (Recall that δ[0]=1 and δ[n]=0 for all n0.) The response h[n] is called the unit-pulse response of the system. Note that since δ[n]=0 for n=1, 2, …, by causality the unit-pulse response must be zero for all integers n<0 (since in a causal system there can be no response before an input is applied). An example of the form of the unit-pulse response is illustrated in Fig. 3-1.
Fig. 3-1
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Example
Consider the finite-dimensional discrete-time system given by the input/output difference equation
y[n] +ay[n1] bx[n2] ()
where a and b are constants. The unit-pulse response for this system can puted by solving Eq.() with initial condition y[1]=0 and with input x[n]=δ[n]. From the discussion in Section , the solution to Eq.() can be expressed in the form
()
But sinceδ[n]=0 for all n0 and δ[0]=1, Eq. () reduces to
y[n] =(–a)n b , n =1, 2, …()
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Hence the unit-pulse response h[n] for this system is given by
Returning to the general case, consider a causal linear time-invariant discrete-time system with input x[n], ouput y[n], and unit-pulse response h[n]. Since the system is time in