Z Transform

Definition

Recall that the Laplace Transform of a signal $x(t)$ is defined as:

$$X(s)\triangleq {\int }_{-\infty }^{+\infty }x(t)e^{-st}\mathrm{d}t$$

where $s$ is a complex variable. The Z-transform, on the other hand, operates on discrete-time signals. Consider a discrete-time signal (a sequence) denoted by $x[n]$, where $n$ is an integer. The Z-transform of $x[n]$ is defined as:

$$X(z)\triangleq \sum _{n=-\infty }^{\infty }x[n]z^{-n}$$

Here, $z$ is a complex variable, analogous to 's' in the Laplace transform. It should be noted that $X(z)$ is a function with the independent variable $z$, where $z$ is the complex variable. For convenience, the Z-transform is often expressed in the form of an operator $\mathcal{Z}\{x[n]\}$, and the transformation relationship between $x[n]$ and $X(z)$ is denoted as:

$$x[n]\stackrel{\mathcal{Z}}{\leftrightarrow }X(z)$$

Connection to Laplace Transform: You may notice the structural similarities between the Laplace and Z transforms. The integral has been replaced by a summation, the continuous time variable 't' is replaced by the discrete time variable 'n', and the complex exponential $e^{-st}$ is now represented as $z^{-n}$. This parallel structure is not by accident, and they are both examples of integral transforms.

The Role of $z$: The complex variable $z$ plays a similar role to 's' in the Laplace transform. We can express $z$ in polar form as $z=r{e}^{j\omega }$, where $r$ is the magnitude and $\omega$ is the angle (or frequency) of the complex variable. If $r=1$, then $z=e^{j\omega }$, and this is very reminiscent of the Fourier Transform!

When $|z|=1$

If we let $z=e^{j\Omega }$ (where $\Omega$ is the normalized digital frequency) we have:

$$X(e^{j\Omega })=\sum _{n=-\infty }^{\infty }x[n]e^{-j\Omega n}$$

This is the DTFT of the signal $x[n]$.

$$X(z){|}_{z=e^{j\Omega }}=\mathcal{F}\{x[n]\}$$

Similar to how the Laplace transform can be seen as a generalization of the Fourier Transform when we consider a complex variable, the Z-transform is a generalization of the Discrete-Time Fourier Transform.

Region of Convergence (ROC)

Similar to the Laplace transform, the Region of Convergence (ROC) is essential to define a unique Z-transform. The ROC is the set of values for the complex variable $z$ for which the Z-transform summation converges. Without specifying the ROC, the Z-transform is not complete or unique.

The ROC for the Z-transform is a region in the complex z-plane. Unlike the Laplace transform ROC, which is a vertical strip, the ROC for the Z-transform is typically a ring or a disk centered at the origin. It can be defined as:

• $|z|<r_1$ (inside a circle)
• $|z|>r_2$ (outside a circle)
• $r_1<|z|<r_2$ (an annulus / ring)
• Or even the entire z plane

Pole-Zero Plot

The Z-transform of a LTI system's impulse response, $h[n]$, results in the transfer function, $H(z)$, expressed as a ratio of two polynomials in the complex variable $z$:

$$H(z)=\frac{N(z)}{D(z)}$$

Where:

• $z$ is the complex frequency
• $N(z)$ is the numerator polynomial.
• $D(z)$ is the denominator polynomial. Poles and Zeros:

The concepts of poles and zeros apply to the Z-transform, analogous to the Laplace transform:

• Zeros: The values of z for which $N(z)=0$. When z equals a zero, the transfer function $H(z)$ evaluates to zero.
• Poles: The values of z for which $D(z)=0$. When z equals a pole, the transfer function $H(z)$ becomes infinite (or undefined).

Note

There can not be poles in the region of convergence.

Pole-Zero Plot: A pole-zero plot is a two-dimensional graph on the complex z-plane:

• The horizontal axis represents the real part of z.
• The vertical axis represents the imaginary part of z.
• Poles are marked with an "x".
• Zeros are marked with an "o".

Significance of Pole Locations

• Stability: For a discrete-time LTI system to be stable, all poles of its transfer function must lie strictly inside the unit circle on the complex z-plane (i.e., their magnitude must be less than 1, $|z|<1$).
• Poles outside the unit circle will lead to an unstable system.
• Poles on the unit circle represent a marginally stable system, which can lead to oscillations.
• Transient Response: Pole locations influence the transient behavior of the discrete-time system, like rise time, overshoot, and settling time. Poles closer to the unit circle lead to more oscillatory behavior.
• Frequency Response: The magnitude of the transfer function, $|H(e^{j\Omega })|$, at different frequencies $\Omega$ is related to the distance from each pole to the unit circle. Poles closer to the unit circle have a larger impact on the frequency response. Significance of Zero Locations
• Zeros affect the system's response by causing attenuation of specific frequency components. They effectively "block" frequencies.
• Zeros can influence the shape of the frequency response curve.

Note

In general, for a LTI system to be stable (BIBO stable), the ROC of its Z-transform must necessarily include the unit circle.

Inverse Z-Transform

Similar to the inverse Laplace transform, the inverse Z-transform recovers the discrete-time sequence $x[n]$ from its Z-transform $X(z)$. There are a variety of methods to find the inverse Z-transform, including partial fraction expansion:

If the rational function $X(z)$ can be expressed as $N(z)/D(z)$, we can decompose $X(z)$ into simpler fractions, each with a known inverse Z-transform. The inverse transform is of the form $a^n$ if the ROC is outside the pole, and $-a^n$ if the ROC is inside the pole.

• Causal Systems: If the system is known to be causal, then the ROC is always outside of the rightmost pole, and $x[n]$ will be right-sided (the signal will only be non-zero for $n\ge 0$).
• Anti-Causal Systems: If the system is anti-causal, then ROC will be inside of the leftmost pole, and the signal will be left-sided (the signal will only be non-zero for $n<0$).

Example Let's say we have $H(z)=1/((z-1/2)(z-1/3))$

1. Partial Fraction Expansion: $H(z)=A/(z-1/2)+B/(z-1/3)$ Solving for A and B yields $A=6$ and $B=-6$, so that $H(z)=6/(z-1/2)-6/(z-1/3)$. However, these do not have readily known inverse z-transforms. It is easier to work with terms with $z^{-1}$ in them. We can use $\frac{z}{z-a}\to a^{n}u[n]$, so let us reformulate $H(z)$ as: $H(z)=\frac{6z^{-1}}{1-\frac{1}{2}{z}^{-1}}-\frac{6z^{-1}}{1-\frac{1}{3}{z}^{-1}}=\frac{12z^{-1}}{2-z^{-1}}-\frac{18z^{-1}}{3-z^{-1}}$. By multiplying by $\frac{z}{z}$ we have $H(z)=\frac{12z}{2z-1}-\frac{18z}{3z-1}=\frac{6}{1-\frac{1}{2}{z}^{-1}}-\frac{6}{1-\frac{1}{3}{z}^{-1}}$.
2. Poles and Possible ROCs:
   • Poles at $z=1/2$ and $z=1/3$. These poles divide the z-plane in to three regions
     • Region 1 : $|z|<1/3$.
     • Region 2 : $1/3<|z|<1/2$.
     • Region 3 : $|z|>1/2$.
3. Inverse Transforms based on ROC:
   • If ROC is |z| > 1/2 (Causal System):
     • $h[n]=6(\frac{1}{2}{)}^{n}u[n]-6(\frac{1}{3}{)}^{n}u[n]$
   • If ROC is 1/3 < |z| < 1/2 (Non-causal System):
     • $h[n]=-6(\frac{1}{2}{)}^{n}u[-n-1]-6(\frac{1}{3}{)}^{n}u[n]$
   • If ROC is |z| < 1/3 (Anti-causal System)
     • $h[n]=-6(\frac{1}{2}{)}^{n}u[-n-1]+6(\frac{1}{3}{)}^{n}u[-n-1]$