Question 3.16: Zero-Input Response of a Second-Order System with Complex Ro......
Zero-Input Response of a Second-Order System with Complex Roots
Consider a second-order difference equation with complex-conjugate roots:
\left(E^2-1.56 E+0.81\right) y[n]=(E+3) x[n]
Determine the zero-input response y_0[n] if the initial conditions are y_0[-1]=2 \text { and } y_0[-2]=1 \text {. }
The characteristic polynomial is \left(\gamma^2-1.56 \gamma+0.81\right)=(\gamma-0.78-j 0.45)(\gamma-0.78+j 0.45). The characteristic roots are 0.78 \pm j 0.45 \text {; that is, } 0.9 e^{ \pm j(\pi / 6)}. We could immediately write the solution as
y_0[n]=c(0.9)^n e^{j \pi n / 6}+c^*(0.9)^n e^{-j \pi n / 6}
Setting n=−1 and −2 and using the initial conditions y_0[-1]=2 \text { and } y_0[-2]=1 \text {, } we find c=1.1550-j 0.2025=1.1726 e^{-j 0.1735} \text { and } c^*=1.1550+j 0.2025=1.1726 e^{j 0.1735} .
Script File
>> gamma = roots([1 -1.56 0.81]);
>> c = inv([gamma(1)^(-1) gamma(2)^(-1);gamma(1)^(-2) gamma(2)^(-2)])*[2;1]
c = 1.1550 - 0.2025i
1.1550 + 0.2025i
Alternately, we could also find the unknown coefficient by using the real form of the solution, as given in Eq. (3.25). In the present case, the roots are 0.9 e^{ \pm j(\pi / 6)} Hence, |γ| = 0.9 and β = π/6, and the zero-input response, according to Eq. (3.25), is given by
y_0[n]=\frac{c}{2}|\gamma|^n\left[e^{j(\beta n+\theta)}+e^{-j(\beta n+\theta)}\right]=c|\gamma|^n \cos (\beta n+\theta) (3.25)
y_0[n]=c(0.9)^n \cos \left(\frac{\pi}{6} n+\theta\right)
To determine the constants c and θ, we set n=−1 and −2 in this equation and substitute the initial conditions y_0[-1]=2 \text { and } y_0[-2]=1 to obtain
2=\frac{c}{0.9} \cos \left(-\frac{\pi}{6}+\theta\right)=\frac{c}{0.9}\left[\frac{\sqrt{3}}{2} \cos \theta+\frac{1}{2} \sin \theta\right]
1=\frac{c}{(0.9)^2} \cos \left(-\frac{\pi}{3}+\theta\right)=\frac{c}{0.81}\left[\frac{1}{2} \cos \theta+\frac{\sqrt{3}}{2} \sin \theta\right]
or
\frac{\sqrt{3}}{1.8} c \cos \theta+\frac{1}{1.8} c \sin \theta=2
\frac{1}{1.62} c \cos \theta+\frac{\sqrt{3}}{1.62} c \sin \theta=1
These are two simultaneous equations in two unknowns ccos θ and csin θ. Solution of these equations yields
c \cos \theta=2.308
c \sin \theta=-0.397
Dividing csin θ by ccos θ yields
\tan \theta=\frac{-0.397}{2.308}=\frac{-0.172}{1}
\theta=\tan ^{-1}(-0.172)=-0.17 ~\mathrm{rad}
Substituting θ =−0.17 radian in ccos θ = 2.308 yields c = 2.34 and
y_0[n]=2.34(0.9)^n \cos \left(\frac{\pi}{6} n-0.17\right) \quad n \geq 0
Observe that here we have used radian units for both β and θ. We also could have used the degree unit, although this practice is not recommended. The important consideration is to be consistent and to use the same units for both β and θ.