Z Transform for Filters

The Z-transform can be used to investigate the properties of discrete time LTI systems, such as FIR and IIR filters. This is achieved by applying the transform to the impulse resonse or filter coefficients, similar to the Laplace transform on the impulse response of a continuous time system.

Stability¶

Most importantly, the z transform allows us to determine if a system is stable. BIBO (bounded-input, bounded-output) stability of a system is given when for any bounded input $x[n]$, the output $y[n]$ is also bounded.

• A causal system is BIBO stable, if its transfer function has no poles outside unit circle.
  • An anti-causal system is BIBO stable, if its transfer function has no poles inside unit circle.
• Another way to express this is that the unit circle must be part of the ROC.
  • This also means a DFT exists for $x[n]$.

Transfer Function¶

In general, the transfer function of a system describes the relation between input and output. We know that for a system with the impulse response $h[n]$:

$$ y[n] = x[n] * h[n] $$

Like for Fourier an Laplace, this translates to a multiplication in the z-domain:

$$ Y(z) = X(z) H(z) $$

And the transfer function $H(z)$:

$$ H(z) = \frac{Y(z)}{X(z)} = \frac{\mathcal Z\{y[n]\}}{\mathcal Z\{x[n]\}} $$

Difference Equation¶

We know the difference equation of a digital filter can be expressed with two sums - one for forward and one for recursive coefficients:

$$ y[n] = \sum\limits_{i=0}^{N} b_i x[n-i] - \sum\limits_{i=1}^{N} a_i y[n-i] $$

Transform¶

The transfer function of the filter can be obtained via the z-transform:

$$\begin{eqnarray} Y(z) & = & X(z) \sum\limits_{i=0}^{i=N} b_i z^{-i} - Y(z) \sum\limits_{i=1}^{i=N} a_i z^{-i} \\ \end{eqnarray} $$

$$\begin{eqnarray} Y(z) + Y(z) \sum\limits_{i=1}^{i=N} a_i z^{-i} & = & X(z) \sum\limits_{i=0}^{i=N} b_i z^{-i} \\ \end{eqnarray} $$

$$\begin{eqnarray} H(z) = \frac{ X(z) \sum\limits_{i=0}^{i=N} b_i z^{-i}}{1 + Y(z) \sum\limits_{i=1}^{i=N} a_i z^{-i}} \\ \end{eqnarray} $$

$$ H(z) = \frac{\sum\limits_{n=0}^{n=N} b_n z^{-n}}{1 +\sum\limits_{n=1}^{n=N} a_n z^{-n}} $$

First Order FIR Example¶

We are considering an IIR filter with the following difference equation and coefficients:

$$\begin{eqnarray} y[n] & = & b_0 x[n] + b_1 x[n-1] - a_1 y[n-1] \\ & = & x[n] + 0.5 x[n-1] - 0.75 y[n-1] \end{eqnarray}$$

Taking th Z-transform of both sides:

$$ Y(Z) = X(z) b_0 + X(z) b_1 z^{-1} - Y(z) a_1 z^{-1} $$

$$ Y(z)\left(1 + a_1 z^{-1}\right) = X(z) \left( b_0 + b_1 z^{-1}\right) $$

The z-transform of this filter is:

$$\begin{eqnarray} H(z) &=& \frac{b_0 + b_1 z^{-1}}{1+a_1 z^{-1}}\\ &=& \frac{1 + 0.5 z^{-1}}{1 + 0.75 z^{-1}} \end{eqnarray}$$

We are now expanding the fraction with $z^{1} $ to get rid of negative exponents:

$$\begin{eqnarray} H(z) &=& \frac{z + 0.5 }{z + 0.75} \end{eqnarray}$$

We now have a fraction with two polynoms. To analyse the filter, we are observing:

• Poles: when the denominator becomes $0$
• Zeros: when the numerator becomes $0$

Zero at $\mathbf{z=-0.5}$

Pole at $\mathbf{z=-0.75}$

In the Z-Plane¶

The Z-Plane is the time-discrete counterpart to the S-Plane for the Laplace transform. It shows the complex variable $z$ with its real and imaginary component.

The poles and zeros for the above filter are plotted below, giving direct information on the stability of the filter:

• The pole is inside the unit cirlce.
• The unit circle is inside the ROC.
• The system is stable.

Relation to the Amplitude Response¶

The relation between the pole-zero plot and the magnitude response can be established on the unit circle - in this visualization from $0 \leq \omega \leq \frac{\omega_s}{2}$:

• 1: take product of the lengths of all vectors from zeros to point
• 2: devide by product of the lengths of all vectors from poles to point

In this example we have only one pole and one zero. We can take the following extreme values:

$$\begin{eqnarray} H(\omega=0) &=& 1.5/1.75 = 0.86 \\ H(\omega=f_s/2) &=& 0.5/0.25 = 2 \end{eqnarray}$$

With a lower gain for $\omega=0$ than for $H(\omega=f_s/2)$, the filter shows high-pass character. Its magnitude response is shown below:

Surface Plot¶

A more detailed visual representation for results of the Z-transform can be delivered with a surface plot. It shows the magnitude of $H(z)$ in the z-plane.

• The red peak represents the pole.
• The blue dip represents the zero.

This visualization shows how the poles and zeros have a certain gravity and shape the contour of the surface.