Signals and Systems - Electrical Engineering

8.3 Discrete-Time Systems 485

Solution

For an initial conditiony[−1]=−2 andx[n]=u[n], we get recursively

y[0]=0.5y[−1]+x[0]+x[−1]= 0

y[1]=0.5y[0]+x[1]+x[0]= 2

y[2]=0.5y[1]+x[2]+x[1]= 3

...

Let us then double the input (i.e.,x[n]= 2 u[n]) and call the responsey 1 [n]. As the initial condition
remains the same (i.e.,y 1 [−1]=−2), we get

y 1 [0]=0.5y 1 [−1]+x[0]+x[−1]= 1

y 1 [1]=0.5y 1 [0]+x[1]+x[0]=4.5

y 1 [2]=0.5y 1 [1]+x[2]+x[1]=6.25

...

It is clear that they 1 [n] is not 2y[n]. Due to the initial condition not being zero, the system is
nonlinear.

If the initial condition is set to zero, and the inputx[n]=u[n], the response is

y[0]=0.5y[−1]+x[0]+x[−1]= 1

y[1]=0.5y[0]+x[1]+x[0]=2.5

y[2]=0.5y[1]+x[2]+x[1]=3.25

...

If we double the input (i.e.,x[n]= 2 u[n]) and call the responsey 1 [n],y 1 [−1]=0, we obtain

y 1 [0]=0.5y 1 [−1]+x[0]+x[−1]= 2

y 1 [1]=0.5y 1 [0]+x[1]+x[0]= 5

y 1 [2]=0.5y 1 [1]+x[2]+x[1]=6.5

...