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First order system

The first order discrete system is described by

\begin{displaymath}y[n]-ay[n-1]=x[n] \end{displaymath}

The impulse response $h[n]$ can be found by solving the following

\begin{displaymath}h[n]-ah[n-1]=\delta[n] \end{displaymath}

to be

\begin{displaymath}h[n]=a^n u[n] \end{displaymath}

Alternatively, we can take z-transform of the DE and get

\begin{displaymath}Y(z)-az^{-1}Y(z)=(1-az^{-1})Y(z)=X(z) \end{displaymath}

and the transfer function of the system (assumed causal)

\begin{displaymath}H(z)=\frac{Y(z)}{X(z)}=\frac{1}{1-az^{-1}} \;\;\;\;\;\;\vert z\vert>\vert a\vert \end{displaymath}

$H(z)$ has a zero at $z=0$ and a pole $z_p=a$ and its ROC is the region $\vert z\vert > \vert a\vert$ outside the pole. If $\vert a\vert<1$, then the unit circle $\vert z\vert=r=1$ can be included in the ROC, the Fourier transform exists and the system is stable. The impulse response (unit sample response) of the system is

\begin{displaymath}h[n]={\cal Z}^{-1}[H(z)]=a^nu[n] \end{displaymath}

Although $h[n]$ seems to have a form different from the typical impulse response in continuous case $h(t)=e^{-t/\tau}u(t)$, they are essentially the same as $h(t)$ can be rewritten as

\begin{displaymath}h(t)=e^{-t/\tau}u(t)=(e^{-1/\tau})^t u(t)=a^t u(t) \end{displaymath}

where

\begin{displaymath}a \stackrel{\triangle}{=}e^{-1/\tau} \end{displaymath}

Letting $z=e^{j\omega}$ in $H(z)$, we get the frequency response function of the system

\begin{displaymath}H(e^{j\omega})=\frac{1}{1-ae^{-j\omega}}
=\frac{e^{j\omega}}{e^{j\omega}-a}=\frac{e^{j\omega}-0}{e^{j\omega}-a}
=\frac{u}{v} \end{displaymath}

where $u$ and $v$ are two vectors in z-plane defined as

\begin{displaymath}u\stackrel{\triangle}{=}e^{j\omega}-0=e^{j\omega},\;\;\;\;
v\stackrel{\triangle}{=}e^{j\omega}-a \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}{\vert v\vert} \end{displaymath}

and

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

which can be evaluated graphically in the z-plane as the frequency $\omega$ changes in the range $-\pi \le\omega\le\pi$. If we assume $0<a< 1$, then when $\omega=0$, the denominator reaches its minimum of $1-a$, and $\vert H(e^{j\omega})\vert$ is maximized to be $1/(1-a)$; and when $\omega=\pi$, the denominator reaches its maximum of $\vert(-1-a)\vert=1+a$, and $\vert H(e^{j\omega})\vert$ is minimized to be $1/(1+a)$. The phase angle of $H(e^{j\omega})=\angle u-\angle v$ is zero when $\omega=0$ or $\omega=\pi$, and is negative for $0<\omega<\pi$ and positive for $-\pi<\omega<0$.


next up previous
Next: Second order system Up: Evaluation of Fourier Transform Previous: Evaluation of Fourier Transform
Ruye Wang 2014-10-28