Pinpoint DSP CT : Z-transform and Filter

Questions:

(a)

Advantages of Z-transform over DTFT system:

• $\delta(n), u(n)$ can’t be analyzed by DTFT but Z.

• The transient response of a system due to initial condition or due to changing inputs cannot be computed by using DTFT but Z.

• The Z-transform might exist anywhere in the Z-plane; the DTFT can only exist on the unit circle.

• Z transform allows for the discrete time system analysis in the frequency domain. This facilitates the analysis of system characteristics such as stability, causality, linearity and time and frequency response.

• Z transform allows the representations of signals and systems in the frequency domain.

• It becomes possible to analyze frequency content of signals determine spectral properties, apply filtering and modulation.

• Useful for determining the stability of DT system.

• It has convolution property that simplifies the analysis of linear time invariant system.

• Simplifies mathematical operation.

The Transfer Function

The transfer function $H(z)$ of a DT linear system is defined as the ration of the Z-transform of the output signal $Y(z)$ to the Z-transform of the input signal $X(z)$

$H(z) = \frac{Y(z)}{X(z)} = \frac{\sum_{n=-\infin}^{+-\infin} b_n z^{-n}}{\sum_{m=-\infin}^{+-\infin} a_m z^{-m}}$

$=\frac{b_0 + b_1z^{-1}+...+b_mz^{-m} }{a_0 + a_1z^{-1}+...+a_nz^{-n} }$

Where, $b_0, b_1, ...$ are input coefficient and $a_0, a_1 ...$ are output.

if $a_o = 1$, then this called the standard form of transfer function.

If we can convert, $H(z) = G\frac{(z-z_1)(z-z_2)... (z-z_n)}{(z-p_1)(z-p_2)...(z-p_n)}$

Then, $z_1, z_2 ...$ are called zeros and $p_1, p_2 ....$ are called poles.

Distinguish between Zeros and Poles

(b) Transfer function representation : $X(z) = \frac{N(z)}{D(z)} = \frac{b_0 + b_1 z^{-1}+ ....+ b_mz^{-m}}{1 + a_1 z^{-1}+ ....+ a_nz^{-n}} = \frac{b_0 + b_1 z^{-1}+ ....+ b_mz^{-m}}{a_0 + a_1 z^{-1}+ ....+ a_nz^{-n}}$ [$a_0 = 1$ means normalized] .$z=e^{j\omega}$

Difference equation representation : Transfer function → Inverse Z-transform → $y(n)$

$H(z) = \frac{1+z^{-1}+z^{-2}}{1+0.5z^{-1}+0.75z^{-2}}$, [Transfer function, $H(z) = \frac{Y(z)}{X(z)} = ..$ ]

$X(z)+z^{-1}X(z) + z^{-2}X(z) = Y(z)+0.5z^{-1}Y(z)+0.75z^{-2}Y(z)$ [cross product]

⇒ $x[n] + x[n-1]+x[n-2] = y[n] + 0.5y[n-1]+0.75y[n-2]$ [Inverse Z-function]

⇒ $y[n] = x[n] + x[n-1]+x[n-2] + 0.5y[n-1]+0.75y[n-2]]$

(c)

Idea : $X(z) = \frac{N(z)}{D(z)} = \frac{b_0 + b_1 z^{-1}+ ....+ b_mz^{-m}}{a_0 + a_1 z^{-1}+ ....+ a_nz^{-n}}$, make given $X(z)$ this format and compare with this ( $a_0 =1$ ) where $b_0, ...., b_n$ = input and $a_0+...a_n =$ output coefficient.

$X(z) = \frac{2 + z + 3z^2}{0.5z^3 \left( 10z^{-3} + 18z^{-2} + 1 \right)} = \frac{4z^{-3} + 2z^{-2} + 6z^{-1}}{10z^{-3} + 18z^{-2} + 1}$ [Normalized]

Input coefficient : 4, 2, 6

Output Coefficient : 1, 0, 18, 10

(a) Check the answer above.

(b) Transfer function representation:

$H(z) = \frac{z^{-2} + z^{-1}}{1 - 0.9z^{-1} + 0.81 z^{-2}}$

$H(z) = \frac{e^{j\omega+1}}{e^{2j\omega} + 0.9 e^{j\omega} + 0/81}$

Difference equation representation

$H(z) = \frac{z^{-2} + z^{-1}}{1 - 0.9z^{-1} + 0.81 z^{-2}} = \frac{Y(z)}{X(z)}$

⇒ $z^{-2}X(z)+z^{-1}X(z)=Y(z) + 0.9 z^{-1} Y(z)+0.82z^{-2} Y(z)$

⇒ $x[n-2]+x[n-1] = y[n]+0.9y[n-1]+0.81y[n-2]$

(c) $X(z) = \frac{3 + z^{-2}+2z^{-4}}{z^{-2}+4z^{-3}+3z^{-4}}$

Input Coefficients: 3, 0, 1, 0,2

Output Coefficients: 0, 0, 1, 4, 3

• Given difference equation, find transfer function representation.

$y[n] = x[n] + x[n-1]+x[n-2] + 0.5y[n-1]+0.75y[n-2]]$

⇒ $Y(z) = X(z) + z^{-1} X(z) + z^{-2} X(z) + 0.5 z^{-1} Y(z) + 0.75 z^{-2} Y(z)$

⇒ $\frac{Y(z)}{X(z)} = ....$

⇒ $H(z) = \frac{Y(z)}{X(z)} = ....$

(a) The set of $Z$ for which $X(z)$ is converges (gives finite value)/the set of points in $Z$-plane for which $X(z)$ is converges is called Region Of Converges.

Properties:

(b) $x(n) = \begin{cases} (-\frac{1}{2}); n \le 1 \\ 0 ; n \ge 0 \end{cases}$

$X(z) = \sum_{n=-\infin}^{1} (-0.5)^{n} z^{-n} =\sum_{n=-\infin}^{-1} (-0.5 z^{-1})^{n} = \sum_{n=1}^{\infin} (-0.5^{-1} z)^{n}$

If $0 < |-0.5^{-1} z| < 1,$ then

$X(z) = \frac{1}{1 - (-0.5^{-1}z)} = \frac{0.5}{0.5 - z}$

Condition for converges,

$|-0.5^{-1} z| < 1$ ⇒ $|z| < 2$

• Differences between Z-transform and Laplace Transform

Digital Filters

A digital filter is a system that performs mathematical operations on a sampled discrete time signal to reduce or enhance certain aspects of that signal.

Types : FIR, IIR

Basic element to design Digital Filter

• Adder :

• Multiplexer

• Delay Element

IIR (Infinite Duration Impulse Response) : If the impulse response exists infinitely, it is an IIR Filter.

• Frequency Response of IIR:

  $H(z) = \frac{B(z)}{A(z)} = \frac{\sum_{m=0}^{M} b_m z^{-m}}{1 + \sum_{n=1}^{N} a_n z^{-n}}$, $[a_0 = 1]$

• Difference equation of IIR Filter

  $y(n) = \sum_{m=0}^{M}b_mx(n-m) - \sum_{m=1}^{N}a_my(n-m)$

• $N-th$order Direct form (I)

  $y(n) = b_0x(n)+b_1x(n-1)...+b_N(n-N) - (a_1x(n-1)+...+ a_Nx(n-N))$

• $N-th$ order direct form (II)

  $y(n) = b_0x(n)+b_1x(n-1)...+b_N(n-N) - (a_1x(n-1)+...+ a_Nx(n-N))$

  

• Cascade form :

  In this form the frequency response, $H(z)$ is factored into smaller second section, called bi-quads. The frequency response is then represented as a product of these bi-quads section.

  $H(z) = \frac{B(z)}{A(z)} = b_0\frac{1 + \frac{b_1}{b_0} z^{-1}+ ....+ \frac{b_N}{b_0} z^{-N}}{1 + a_1 z^{-1}+ ....+ a_Nz^{-N}}$

  $=b_0\prod_{k=1}^{k} \frac{1+B_{k,1}z^{-1}+B_{k,2}z^{-2}}{1+A_{k,1}z^{-1}+A_{k,2}z^{-2}}$ [for second-order]

  $k = \frac{N}{2}$ [Always]

Cascade form for $N-th$ order

$K=\frac{N}{2}$

FIR (Finite Duration Impulse Response): The filters have a finite impulse response function which has finite length of time.

• Transfer Function

  $M : \text{length of filter}$

  $N : \text{Order of filter} = M- 1$

  $H(z) = \sum_{n=0}^{M-1} b_nZ^{-n}$

  $h(n) = \begin{cases} b_n, 0 \le n \le M-1 \\ 0, \text{otherwise} \end{cases}$

• Difference equation

  $y[n] = \sum_{m=0}^{M-1}b_mx(n-m)$

• Direct form I if $M=M$

  Order of filter, $N = M-1$

  $N-th$ order, $y(n) = b_0x(n) + b_1x(n-1)+...+b_Nx(n-N)$

  

• Cascade FIR

  For second order section,

  $H(z) = H(z) = \sum_{n=0}^{M-1} b_nZ^{-n} = b_0 + b_1z^{-1}+....+b_{M-1}z^{-M+1}$

  $= b_0 (1+\frac{b_1}{b_0} z^{-1}+ .... +\frac{b_{M-1}}{b_0} z^{-M+1})$

  $=b_0 \prod_{k=1}^{k} (1 + B_{k, 1}z^{-1}+ B_{k, 2}z^{-2})$

  $K = \lfloor\frac{M}{2}\rfloor$

  Design cascade form of FIR where filter length is $M$

  $K = \lfloor\frac{M}{2}\rfloor$

• Properties of Linear-phase FIR filter

Let, $h(n), 0 < n < M-1$, be the impulse response of length M. The frequency response,

$H(z) = \sum_{n=0}^{M-1} h(n)z^{-n}$

We know, $H(e^{jw}) = \sum_{n=0}^{M-1} h(n) e^{-jwn}, -\pi \le w \le \pi$

Where, $\angle H(e^{jw} ) = -\alpha w$ (A system has linear phase if its phase response θ(ω) = ∠H(e^jω) = −cω for all ω and any constant c.)

where, $\alpha$ is a constant phase delay, $h(n)$ must be symmetric, if $h(n) = h(M-1-n)$

$0 \le n \le M-1$ with $\alpha = \frac{M-1}{2}$ if M is odd.

Anti symmetry : $h(n) = -h(M-1-n)$

Linear phase system fall into one of 4 categories: