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Second order system

A second order discrete system is described by

\begin{displaymath}y[n]+a_1y[n-1]+a_2y[n-2]=x[n] \end{displaymath}

The z-transform of this DE is

\begin{displaymath}Y(z)+a_1z^{-1}Y(z)+a_2z^{-2}Y(z)=(1+a_1z^{-1}+a_2z^{-2})Y(z)=X(z) \end{displaymath}

and the transfer function is

\begin{displaymath}H(z)=\frac{Y(z)}{X(z)}=\frac{1}{1+a_1z^{-1}+a_2z^{-2}}
=\frac{z^2}{z^2+a_1z+a_2}=\frac{z^2}{(z-z_{p_1})(z-z_{p_2})}
\end{displaymath}

which has two repeated zeros $z=0$ and a pair of complex conjugate poles

\begin{displaymath}z_{p_1,p_2}=-\frac{a_1}{2}\pm\frac{1}{2}\sqrt{a_1^2-4a_2}
=-\frac{a_1}{2}\pm\frac{j}{2}\sqrt{4a_2-a_1^2} \end{displaymath}

For convenience, we define

\begin{displaymath}a_1\stackrel{\triangle}{=}-2r\;cos\theta, \;\;\;\;\mbox{and}
...
...;\; a_2\stackrel{\triangle}{=}r^2\;\;\;\;(0\ge \theta \ge \pi) \end{displaymath}

and the transfer function becomes

\begin{displaymath}H(z)=\frac{z^2}{z^2-2r\;cos\theta z+r^2} \end{displaymath}

and the poles become

\begin{displaymath}
z_{p_1,p_2}=r(cos\theta \pm j\;sin\theta)=r\;e^{\pm j\theta} \end{displaymath}

For the system to be causal and stable, we have to have $\vert z_p\vert=r<1$. When $0<\theta <\pi$, $H(z)$ can be expanded
$\displaystyle H(z)$ $\textstyle =$ $\displaystyle \frac{z^2}{(z-z_{p_1})(z-z_{p_2})}=
\frac{1}{(1-z_{p_1}z^{-1})(1-z_{p_2}z^{-1})}$  
  $\textstyle =$ $\displaystyle \frac{1}{(1-r e^{j\theta}z^{-1})(1-r e^{-j\theta}z^{-1})}
= \frac{A}{1-r e^{j\theta}z^{-1}}+\frac{B}{1-r e^{-j\theta}z^{-1}}$  

where

\begin{displaymath}
A=\frac{e^{j\theta}}{2j\;sin\theta},\;\;\;\;B=\frac{-e^{-j\theta}}{2j\;sin\theta}
\end{displaymath}

The impulse response of the system can be found by inverse z-transform
$\displaystyle h[n]$ $\textstyle =$ $\displaystyle [A(re^{j\theta})^n+B(re^{-j\theta})^n]u[n]
=\frac{r^n}{2j\;sin\theta}[e^{j\theta(n+1)}-e^{-j\theta(n+1)}]u[n]$  
  $\textstyle =$ $\displaystyle r^n\frac{sin[(n+1)\theta]}{sin\theta}u[n]$  

When $\theta=0$, $H(z)$ has two repeated poles $z_p=r$ and becomes

\begin{displaymath}H(z)=\frac{1}{(1-rz^{-1})^2} \end{displaymath}

and the impulse response becomes

\begin{displaymath}h[n]=(n+1)r^n u[n] \end{displaymath}

When $\theta=\pi$, $H(z)$ has two repeated poles $z_p=-r$ and becomes

\begin{displaymath}H(z)=\frac{1}{(1+rz^{-1})^2} \end{displaymath}

and the impulse response becomes

\begin{displaymath}h[n]=(n+1)(-r)^n u[n] \end{displaymath}

To graphically evaluate the behavior of the system as a function of frequency $\omega$ in the z-plane, we let $z=e^{j\omega}$ in $H(z)$ and get the frequency response function of the system

\begin{displaymath}
H(e^{j\omega})=\frac{1}{(1-re^{j\theta}e^{-j\omega})(1-re^{-...
...j\theta})(e^{j\omega}-re^{-j\theta})}
=\frac{u^2}{v_1 \; v_2} \end{displaymath}

where $u$, $v_1$ and $v_2$ are three vectors in z-plane defined as

\begin{displaymath}u\stackrel{\triangle}{=}e^{j\omega}-0=e^{j\omega},
\;\;\;\;v_...
...},
\;\;\;\;v_2\stackrel{\triangle}{=}e^{j\omega}-re^{-j\theta}
\end{displaymath}

For any frequency $-\pi \le\omega\le\pi$ represented by a point $z=e^{j\omega}$ on the unit circle, the magnitude and phase angle of the frequency response function can be represented in the z-plane as

\begin{displaymath}\vert H(e^{j\omega})\vert=\frac{\vert u\vert^2}{\vert v_1\vert\vert v_2\vert} \end{displaymath}

and

\begin{displaymath}\angle{H(e^{j\omega})}=2\angle u-\angle v_1-\angle v_2 \end{displaymath}

When $\omega=\pm \pi$, $v_1$ and $v_2$ are maximized and thereby $\vert H(e^{j\omega})\vert$ is minimized. In particular, when $r$ is close to 1, $\vert v_1\vert=1-\vert r\vert$ is minimized when $\omega=\theta$, i.e., $\vert H(e^{j\omega})\vert$ has a peak.


next up previous
Next: System Algebra and Block Up: Evaluation of Fourier Transform Previous: First order system
Ruye Wang 2014-10-28