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From Discrete-Time Fourier Transform to Z-Transform

Forward Z-Transform

The Fourier transform of a discrete signal $x[n]$ is defined as:

\begin{displaymath}X(e^{j\omega})={\cal F}[x[n]]=\sum_{n=-\infty}^\infty x[n]e^{-j2\pi f n}
=\sum_{n=-\infty}^\infty x[n]e^{-j\omega n} \end{displaymath}

provided $x[n]$ is absolutely summable:

\begin{displaymath}\sum_{n=-\infty}^\infty \vert x[n] \vert<\infty \end{displaymath}

Obviously some signals may not satisfy this condition and their Fourier transform do not exist. To overcome this difficulty, we can multiply the given $x[n]$ by an exponential function $e^{-\sigma n}$ so that $x[n]$ may be forced to be summable for certain values of the real parameter $\sigma$. Now the discrete time Fourier transform becomes:
$\displaystyle {\cal F}[x[n]e^{-\sigma n}]$ $\textstyle =$ $\displaystyle \sum_{n=-\infty}^\infty [x[n]e^{-\sigma n}]\;e^{-j\omega n}
=\sum_{n=-\infty}^\infty x[n]e^{-(\sigma +j\omega) n}$  
  $\textstyle =$ $\displaystyle \sum_{n=-\infty}^\infty x[n]e^{-s n}=\sum_{n=-\infty}^\infty x[n] z^{-n}

The result of this summation is a function of a complex variable defined as:

\begin{displaymath}z=e^s=e^{\sigma+j\omega} \end{displaymath}

This is the forward Z-transform of the discrete signal $x[n]$:

\begin{displaymath}X(z)={\cal Z}[x[n]]=\sum_{n=-\infty}^\infty x[n] z^{-n} \end{displaymath}

Inverse Z-Transform

Given the Z transform $X(z)$, the original time signal can be obtained by the inverse Z transform, which can be derived from the corresponding Fourier transform. As shown above, we have:

\begin{displaymath}X(z)=X(e^{\sigma+j\omega})=\sum_{n=-\infty}^\infty [x[n] e^{-\sigma n}]
e^{-j\omega n} ={\cal F}[x[n] e^{-\sigma n}]

Now $x[n]e^{-\sigma n}$ can be obtained by the inverse Fourier transform:

\begin{displaymath}x[n] e^{-\sigma n}={\cal F}^{-1}[X(e^{\sigma+j\omega})]
...2\pi}\int_0^{2\pi} X(e^{\sigma+j\omega}) e^{j\omega n} d\omega

Multiplying both sides by $e^{\sigma n}$, we get:

x[n]=\frac{1}{2\pi}\int_0^{2\pi} X(e^{\sigma+j\omega}) e^{(\... n} d\omega
=\frac{1}{2\pi}\int_0^{2\pi} X(z) z^n d\omega

To represent the inverse transform in terms of $z$ (instead of $\omega$), we note

\begin{displaymath}dz=d(e^{\sigma+j\omega})=e^\sigma j\;e^{j\omega} d\omega=jz\;d\omega,
\;\;\;\;\;\mbox{i.e.,}\;\;\;\;\;\;d\omega=\frac{dz}{jz} \end{displaymath}

and the inverse Z transform can be obtained as:

\begin{displaymath}x[n]={\cal Z}^{-1}[X(z)]=\frac{1}{j2\pi}\oint X(z) z^{n-1} dz \end{displaymath}

Note that the integral with respect to $\omega$ from $0$ to $2\pi$ becomes an integral with respect to $z=e^{\sigma+j\omega}$ in the complex z-plane, along a circle with a fixed radius $e^\sigma$ and a varying angle $\omega$ from $0$ to $2\pi$. Now we have the z-transform pair:

\begin{displaymath}X(z)={\cal Z}[x[n]]=\sum_{n=-\infty}^\infty x[n] z^{-n} \end{displaymath}

\begin{displaymath}x[n]={\cal Z}^{-1}[X(z)]=\frac{1}{2\pi j}\oint X(z)z^{n-1} dz \end{displaymath}

The forward and inverse z-transform pair can also be represented as

\begin{displaymath}x[n] \stackrel{ {\cal Z} }{\longleftrightarrow} X(z) \end{displaymath}

In particular, if we let $\sigma=0$, i.e., $z=e^{j\omega}$, then the Z transform becomes the discrete-time Fourier transform:

\begin{displaymath}X(z)\bigg\vert _{z=e^{j\omega}}=\sum_{n=-\infty}^\infty x[n] e^{-j\omega n}=X(e^j\omega) \end{displaymath}

This is the reason why sometimes the discrete Fourier spectrum is expressed as a function of $e^{j\omega}$.


Different from the discrete-time Fourier transform which converts a 1-D signal $x[n]$ in time domain to a 1-D complex spectrum $X(e^{j\omega})$ in frequency domain, the Z transform $X(s)$ converts the 1D signal $x[n]$ to a complex function defined over a 2-D complex plane, called z-plane, represented in polar form by radius $\vert z\vert=\vert e^{\sigma+j\omega}\vert=e^\sigma$ and angle $\angle z=\angle(e^{\sigma+j\omega})=\omega$.

In particular, when this 2D function $X(z)=X(e^{\sigma+j\omega})$ is evaluated along the unit circle $\vert z\vert=e^0=1$ corresponding to $\sigma=0$, it becomes a 1D periodic function $X(e^{j\omega})$, the discrete Fourier transform of $x[n]$. Graphically, the periodic spectrum of the signal can be found as the cross section of the 2D function $X(z)=X(e^{\sigma+j\omega})$ along the unit circle $\vert z\vert=e^0$.

Transfer Function of LTI system

The output $y[n]$ of a discrete LTI system with input $x[n]$ can be found by convolution

\begin{displaymath}y[n]={\cal O}[x[n]]=h[n]*x[n]=\sum_{k=-\infty}^\infty h[k]x[n-k] \end{displaymath}

where $h[n]$ is the impulse response function of the system. In particular, if the input is a complex exponential

\begin{displaymath}x[n]=z^n=(e^s)^n=e^{sn} \end{displaymath}

then the output $y[n]$ can be found to be:

\begin{displaymath}y[n]={\cal O}[z^n]=\sum_{k=-\infty}^\infty h[k]z^{n-k}
=z^n \sum_{k=-\infty}^\infty h[k]z^{-k}=H(z) z^n \end{displaymath}

This is the eigenequation with the complex exponential $x[n]=z^n=e^{sn}$ being the eigenfunction of any discrete LTI system, corresponding to its eigenvalue defined as:

\begin{displaymath}H(z)\stackrel{\triangle}{=}\sum_{k=-\infty}^\infty h[k] z^{-k} \end{displaymath}

which is the z-transform of its impulse response $h[n]$, called the transfer function of the LTI system. In particular, when $\sigma=0$, i.e., $z=e^s=e^{j\omega}$, the transfer function $H(z)$ becomes the frequency response function, the Fourier transform of the impulse response:

\begin{displaymath}H(e^{j\omega})\stackrel{\triangle}{=}\sum_{n=-\infty}^\infty h[n] e^{-j\omega n} \end{displaymath}

next up previous
Next: Conformal Mapping between S-Plane Up: Z_Transform Previous: Z_Transform
Ruye Wang 2014-10-28