Infinite Impulse Response Filters

Lecture by Prof. Brian L. Evans

Slides by Niranjan Damera-Venkata

Embedded Signal Processing Laboratory
Dept. of Electrical and Computer Engineering
The University of Texas at Austin
Austin, TX 78712-1084

http://signal.ece.utexas.edu

Introduction

Background
Bounded Input, Bounded Output (BIBO) stability of quarter-plane IIR filters
Not all output masks are recursively computable: Figure 1.

Figure 1: Two examples of not recursive computable masks
$\begin{figure}\centerline{ \begin{tabular}{c} \epsfxsize = 6cm \epsfbox{figure9.eps} \end{tabular}}\end{figure}$

**Figure 1:** Two examples of not recursive computable masks
$\begin{figure}\centerline{ \begin{tabular}{c} \epsfxsize = 6cm \epsfbox{figure9.eps} \end{tabular}}\end{figure}$

Background

2-D difference equations

$\begin{eqnarray*} y(n_{1},n_{2})&=&\sum_{r_{1}}\sum_{r_{2}}a(n_{1}-r_{1},n_{2}-... ...}\sum_{k_2 \ne n_2}b(n_{1}-k_{1},n_{2}-k_{2}) y(k_{1},k_{2})} \end{eqnarray*}$
Input and output masks
- is called the support of the input mask
- is called the support of the output mask
- Angle of support $\beta$ : minimum angle enclosing support of mask
- is computed as in Figure 2
  
  Figure 2: Recursive computation.
  $\begin{figure}\centerline{ \begin{tabular}{c} \epsfxsize = 15cm \epsfbox{figure1.eps} \end{tabular} } \end{figure}$
- What degrees of freedom exist for moving the output mask?
- What must the boundary conditions be?

**Figure 2:** Recursive computation.
$\begin{figure}\centerline{ \begin{tabular}{c} \epsfxsize = 15cm \epsfbox{figure1.eps} \end{tabular} } \end{figure}$

Recursive Computation

Consider a four-tap all-pole IIR filter to be passed over an $N \times N$ image in a raster scan fashion, e.g. in Floyd-Steinberg error diffusion halftoning.
What are the boundary conditions?
How many rows of the image do we need to keep in memory at one time?
What parallelism exists if any?
What is the tradeoff between the amount of parallelism exploited and memory?

Recursive Computability

Recursive computability: Computing the difference equation using known values of the shifted output samples and initial conditions.
- Support of output mask
- Initial conditions
  - Initial conditions must be 0, and must lie outside the support of the output sequence, for the filter to be Linear Shift Invariant (LSI).
- Order of recursion
Types of recursively computable masks
- Quarter-plane masks: supported in a quadrant in $\Re^2$
- Non-Symmetric Half Plane (NSHP): supported in a half-plane in $\Re^2$
Not all possible orderings are equivalent
- Amount of storage
- Degree of parallelism

Boundary Conditions

For a recursive system, how do you choose the boundary conditions.
- If the system is to be LTI, the initial conditions must be zero outside the support of the filter.
Example:
- Now assume a different set of boundary conditions: Figure 3.
  
  Figure 3: Boundary conditions (left) Response of to $\delta [n_1,n_2]$ (right)
  $\begin{figure}\centerline{ \begin{tabular}{cc} \epsfxsize = 4.5cm \epsfbox{f... ...2.eps}& \epsfxsize = 4.5cm \epsfbox{figure3.eps} \end{tabular} } \end{figure}$
- Response to $\delta [n_1,n_2]$ : Figure 3
- Response to $\delta [n_1-1,n_2-1]$ : Figure 4
  
  Figure 4: Response of to $\delta [n_1-1,n_2-1]$ (left) Another boundary condition (right)
  $\begin{figure}\centerline{ \begin{tabular}{cc} \epsfxsize = 4.5cm \epsfbox{f... ....eps}& \epsfxsize = 4.5cm \epsfbox{figure5.eps} \end{tabular} } \end{figure}$
- This is a shifted version of the above: consistent with support of the impulse response
- Let's begin guessing at some boundary conditions: Figure 4
- Calculated response to $\delta [n_1,n_2]$ : Figure 5
  
  Figure 5: Response of to $\delta [n_1,n_2]$ (left) Response of to $\delta [n_1-1,n_2-1]$ (right)
  $\begin{figure}\centerline{ \begin{tabular}{cc} \epsfxsize = 4.5cm \epsfbox{f... ....eps}& \epsfxsize = 4.5cm \epsfbox{figure7.eps} \end{tabular} } \end{figure}$
- Response to $\delta [n_1-1,n_2-1]$ : Figure 5
- Note: This is not a shifted version of the response $\delta [n_1,n_2]$ : These boundary conditions do not lead to an LTI system.

**Figure 3:** Boundary conditions (left) Response of to $\delta [n_1,n_2]$ (right)
$\begin{figure}\centerline{ \begin{tabular}{cc} \epsfxsize = 4.5cm \epsfbox{f... ...2.eps}& \epsfxsize = 4.5cm \epsfbox{figure3.eps} \end{tabular} } \end{figure}$

**Figure 4:** Response of to $\delta [n_1-1,n_2-1]$ (left) Another boundary condition (right)
$\begin{figure}\centerline{ \begin{tabular}{cc} \epsfxsize = 4.5cm \epsfbox{f... ....eps}& \epsfxsize = 4.5cm \epsfbox{figure5.eps} \end{tabular} } \end{figure}$

**Figure 5:** Response of to $\delta [n_1,n_2]$ (left) Response of to $\delta [n_1-1,n_2-1]$ (right)
$\begin{figure}\centerline{ \begin{tabular}{cc} \epsfxsize = 4.5cm \epsfbox{f... ....eps}& \epsfxsize = 4.5cm \epsfbox{figure7.eps} \end{tabular} } \end{figure}$

The Transfer Function

$\begin{eqnarray*} H(z_1,z_2) & = & \sum_{n_1}\sum_{n_2} h[n_1,n_2] z_1^{-n_1} z_2^{-n_2} \end{eqnarray*}$

The z-transform will not converge for all values of
If it converges for $z_1=e^{j\omega_1}$ , $z_2=e^{j\omega_2}$ , then the Discrete-Time Fourier Transform exists and the system is stable.

$\vert z_1\vert=1$ and $\vert z_2\vert=1$ $\Rightarrow$ Unit Bicircle

Properties of the z-transform

Separable Signals: $v[n_1] w[n_2] \Longleftrightarrow V(z_1) W(z_2)$
Linearity
Shift
Convolution
Linear mapping (look familiar?)

$\begin{eqnarray*} x[n_1,n_2]&=&\left\{ \begin{array}{cc} w[m_1,m_2]&n_1=Im_1+J... ...neq & 0\\ X(z_1,z_2)& = & W(z_1^I z_2^K, z_1^J z_2^L); \end{eqnarray*}$

Z-transform of Linear Mappings

Linear mapping

$\begin{eqnarray*} x[n_1,n_2]&=&\left\{ \begin{array}{cc} w[m_1,m_2]&n_1=Im_1+J... ... \neq & 0\\ X(z_1,z_2)& = & W(z_1^I z_2^K, z_1^J z_2^L) \end{eqnarray*}$
Notice the regular insertion of zeros
This is upsampling by upsampling matrix

$\begin{displaymath} {\bf R} = \left[ \matrix{ I & J \cr K & L } \right] \end{displaymath}$

where $\det {\bf R} = I L - J K \ne 0$
Frequency response $X({\bf\omega})$ of an upsampler to input $W({\bf\omega})$

$\begin{displaymath} X_u({\bf\omega}) = X({\bf R}^t {\bf\omega}) \end{displaymath}$
Let $z_1=e^{j\omega_1}$ and $z_2=e^{j\omega_2}$

$\begin{eqnarray*} X(z_1,z_2) = W(z_1^I z_2^K, z_1^J z_2^L) \\ X(e^{j \ome... ...a_2}) \\ {\bf R}^t = \left[ \matrix{ I & K \cr J & L } \right] \end{eqnarray*}$

Inverse z-transform

$\begin{eqnarray*} x[n_1,n_2]&=&\frac{1}{(2\pi j)^2} \oint_{c_2}\oint_{c_1}X(z_1,... ...{-1}-bz_2^{-1}} \quad\mbox{$\vert a\vert+\vert b\vert\le1$}\quad \end{eqnarray*}$

Choose ROC that includes the unit bicircle.

$\begin{eqnarray*} x[n_1,n_2]&=&\frac{1}{(2\pi j)^2} \oint_{c_2}\oint_{c_1}\frac{... ...{n_2}\oint_{c_1}\frac{z_1^{n_1}}{z_1-\frac{az_2}{z_2-b}}dz_1dz_2 \end{eqnarray*}$
The inner integral (with respect to ) is the inverse -transform of a first-order system with a pole at $\displaystyle\frac{az_2}{z_2-b}$

$\begin{eqnarray*} x[n_1,n_2]&=&\frac{1}{2\pi j}\oint_{c_2}\frac{z_2^{n_2}}{z_2-b... ...}}dz_2\\ &=&\frac{(n_1+n_2)!}{n_1!n_2!}a^{n_1}b^{n_2}u[n_1,n_2] \end{eqnarray*}$
What does this tell us about other examples? very little
Consider an example:

$\begin{eqnarray*} H(z_1,z_2)&=&\frac{1}{1-az_1^{-1}-bz_2^{-1}-cz_1^{-1}z_2^{-1}}... ...1},z_2)&=&1-ae^{-j\omega_1}-bz_2^{-1}-ce^{-j\omega_1}z_2^{-1}\\ \end{eqnarray*}$

Setting $B(e^{j\omega_1},z_2)=0$ and solving gives

$\begin{eqnarray*} z_2=\frac{b+ce^{-j\omega_1}}{1-ae^{-j\omega_1}} \end{eqnarray*}$

This is a bilinear transformation which must map circles into circles.

Figure 6: Bilinear transform
$\begin{figure}\centerline{ \begin{tabular}{c} \epsfxsize = 6cm \epsfbox{figure8.eps} \end{tabular} }\end{figure}$

Point A $\omega_1=\pi$ $z_2=\frac{b-c}{1+a}$

Point B $\omega_1=0$ $z_2=\frac{b+c}{1-a}$

$\begin{eqnarray*} \left\vert \frac{b-c}{1+a} \right\vert <1 & \left\vert \displaystyle\frac{b+c}{1-a} \right\vert <1 \end{eqnarray*}$

From the other part of the locus

$\begin{eqnarray*} \left\vert \frac{a-c}{1+b} \right\vert <1 & \left\vert \displaystyle\frac{a+b}{1-b} \right\vert <1 \end{eqnarray*}$