9.5 One-Sided Z-Transform Inverse 555You can then see that the first term in Equation (9.38) is a convolution sum, and the second is the
impulse response multiplied byay[−1]—that is,y[n]=yzs[n]+yzi[n]=
∑∞
k= 0h[k]x[n−k]+ay[−1]h[n]
nnExample 9.17
Consider a discrete-time system represented by a second-order difference equation with constant
coefficientsy[n]−a 1 y[n−1]−a 2 y[n−2]=x[n]+b 1 x[n−1]+b 2 x[n−2] n≥ 0wherex[n] is the input,y[n] is the output, and the initial conditions arey[−1] andy[−2]. Use the
Z-transform to obtain the complete solution.SolutionApplying the one-sided Z-transform to the two sides of the difference equation, we haveZ(y[n]−a 1 y[n−1]−a 2 y[n−2])=Z(x[n]+b 1 x[n−1]+b 2 x[n−2])Y(z)−a 1 (z−^1 Y(z)+y[−1])−a 2 (z−^2 Y(z)+y[−1]z−^1 +y[−2])=X(z)( 1 +b 1 z−^1 +b 2 z−^2 )where we used the linearity and the time-shift properties of the Z-transform. It was also assumed
that the input is causal,x[n]=0 forn<0, so that the Z-transforms ofx[n−1] andx[n−2] are
simplyz−^1 X(z)andz−^2 X(z). Rearranging the above equation, we haveY(z)( 1 −a 1 z−^1 −a 2 z−^2 )=(y[−1](a 1 +a 2 z−^1 )+a 2 y[−2])+X(z)( 1 +b 1 z−^1 +b 2 z−^2 )and solving forY(z), we haveY(z)=X(z)( 1 +b 1 z−^1 +b 2 z−^2 )
1 −a 1 z−^1 −a 2 z−^2+
y[−1](a 1 +a 2 z−^1 )+a 2 y[−2])
1 −a 1 z−^1 −a 2 z−^2where again the first term is the Z-transform of the zero-state response, due to the input only, and
the second term is the Z-transform of the zero-input response, which is due to the initial conditions
alone. The inverse Z-transform ofY(z)will give us the complete response. nRemarksAs we saw in Chapter 8, if either the initial conditions are not zero or the input is not causal,
the system is not linear time invariant (LTI). However, the time-shift property allows us to find the complete
response in that case. We can think of two inputs applied to the system: one due to the initial conditions
and the other due to the regular input. By using superposition, we obtain the zero-state and the zero-input
responses, which add to the total response.