Signals and Systems - Electrical Engineering

8.3 Discrete-Time Systems 499

to signals we can then see that the output of a causal LTI discrete-time system can be written in terms
of the convolution sum as

y[n]=

∑∞

k=−∞

x[k]h[n−k]=

∑∞

k= 0

x[k]h[n−k]=

∑n

k= 0

x[k]h[n−k]

where we first used the causality of the input (x[k]=0 fork<0) and then that of the system (i.e.,
h[n−k]=0 whenevern−k<0 ork>n). According to this equation the output depends on inputs
{x[0],...,x[n]}, which are past and present values of the input.

nExample 8.25

So far we have considered the convolution sum as a way of computing the outputy[n] of an LTI

system with impulse responseh[n] for a given inputx[n]. But it actually can be used to find either

of these three variables given the other two. The problem is then calleddeconvolution. Assume

the inputx[n] and the outputy[n] of a causal LTI system are given. Find equations to compute

recursively the impulse responseh[n] of the system. Consider finding the impulse responseh[n]

of a causal LTI system with inputx[n]=u[n] and outputy[n]=δ[n]. Use the MATLAB function

deconvto findh[n].

Solution

If the system is causal and LTI, the inputx[n] and the outputy[n] are connected by the convolu-

tion sum

y[n]=

∑n

m= 0

h[n−m]x[m]=h[n]x[0]+

∑n

m= 1

h[n−m]x[m]

To findh[n] from givenx[n] andy[n], under the condition thatx[0]6=0, the above equation can

be rewritten as

h[n]=

1

x[0]

[

y[n]−

∑n

m= 1

h[n−m]x[m]

]

so that the impulse response of the causal LTI can be found recursively as follows:

h[0]=

1

x[0]

y[0]

h[1]=

1

x[0]

(

y[1]−h[0]x[1]

)

h[2]=

1

x[0]

(

y[2]−h[0]x[2]−h[1]x[1]

)

..

.