Signals and Systems - Electrical Engineering

8.3 Discrete-Time Systems 483

y[n−2]=ay[n−3]+bx[n−2], and so on, we obtain

y[n]=a(ay[n−2]+bx[n−1])+bx[n]

=a(a(ay[n−3]+bx[n−2]))+abx[n−1]+bx[n]

=···

=···a^3 bx[n−3]+a^2 bx[n−2]+abx[n−1]+bx[n]

until we reachx[0]. The solution can be written as

y[n]=

∑n

k= 0

bakx[n−k] (8.30)

which we will see in the next section is the convolution sum of the impulse response of the system
and the input.

To see that Equation (8.30) is actually the solution of the given difference equation, we need to
show that when replacing the above expression fory[n] in the right term of the difference equation
we obtain the left termy[n]. Indeed, we have that

ay[n−1]+bx[n]=a

[n− 1

∑

k= 0

bakx[n− 1 −k]

]

+bx[n]

=

∑n

m= 1

bamx[n−m]+bx[n]=

∑n

m= 0

bamx[n−m]=y[n]

where the dummy variable in the sum was changed tom=k+1, so that the limits of the sum-
mation becamem=1 whenk=0, andm=nwhenk=n−1. The final equation is identical
toy[n].

To establish if the system represented by the difference equation is linear, we use the solution
Eq. (8.30) with inputx[n]=αx 1 [n]+βx 2 [n], where the outputs{yi[n], i=1, 2}correspond to
inputs{xi[n],i=1, 2}, andαandβare constants. The output forx[n] is

∑n

k= 0

bakx[n−k]=

∑n

k= 0

bak

(

αx 1 [n−k]+βx 2 [n−k]

)

=α

∑n

k= 0

bakx 1 [n−k]+β

∑n

k= 0

bakx 2 [n−k]=αy 1 [n]+βy 2 [n]

So the system is linear.