Chapter 8 Discrete-Time Signals And Systems

EE 422G Notes: Chapter 8 Instructor: Zhang

Chapter 8 Discrete-Time Signals and Systems

8-1 Introduction

Most “real” signals and natural (physical) processes: continuous – time

A : System Design Problem

How the computer sees “ the rest”? an equivalent

(Physical Process + Sensor +A/D + D/A )=> discrete-time system

The Equivalent discrete-time system

e Modeled by a discrete-time model

System Design (Design of the computer control Algorithm):

Based on discrete-time model description.

e Needs for discrete-time system analysis and design tool:

Z-Transform (Similar position as Laplace Transform for continuous-time system.)

B. How does the computer understand the progress and behaviors of the

process being monitored and controlled? By sampling the output of the

continuous-time system!

How can we ensure that the sampled signal is a sufficient representation of

its continuous-time origin. i.e., how fast we have to sample?

A question we must answer before z-transform based analysis!

C. Two basic parts of the chapter

Part one : Theoretical frame work for determining how fast we have to sample.

Part two : z-transform

Part one: How fast

8.2A Analog-to-Digital Conversion

1.
Sample Operation

Needs to Know:

(1) Sampling period: T

(2) x(t) is sampled at t=nT

(3) What do we mean by x(n)

(4) Sampling function: p(t)

(5) Sampled signal x s = x(t)p(t)

2. Mathematical Description of Sampling Process

Sampled signal : x s (t) = x(t)p(t)

Objective: Derivation of x s (t)’s Fourier Series Expression (Time Domain)

Derivation :

Sampling function: A Periodical function,

(thus can be expressed using Fourier series), with

period T on fundamental frequency

With Fourier series coefficients:

3. Spectrum of sampled signal

Objective: Find the spectrum of the sampled signal x s (t).

Derivation :

Take Fourier Transform for

4. Spectral Characteristic of ‘Real Signal’

Most ‘real’ signals: continuous with time

e Highest frequency f h can be found

e X(f) = 0 if

5. How ‘sampling process’ modifies the spectrum

Consider a

No spectrum modification

6. How fast we have to sample in order to keep the spectrum:

Condition

Should we consider ? Of course not!

( implies that the real process

changes faster than the sampling rate.)

Consider only =>

Answer :

Sampling rate: at least twice as the highest

frequency of the “original process”

Sampling Theorem: ………….

7. What about if r ? 0 ? (show Figure 8-4)

8. Practical sampling rate:

8-2B Data Reconstruction

1. What’s Data Reconstruction?

Original x(t) t 3 0 (anytime)

Its samples x s (t) t = 0, T, 2T, … (Discrete time)

Can we tell x(t) between sampled points ( nT < t < (n+1)T ) based on x s (t)?

Data Reconstruction problem!

2. Data Reconstruction Method

What’s a filter? A system which processes the input to generate an output.

It could be an algorithm (mathematical equation/operation set) or

circuit/analog computer, depending on the form of x s (t) (digital

number or analogy signal .)

Let’s see how a filter works!

Output

Weighted sum of the ‘should be point x(kT)

and its surrounding points

|h(0)| should > h( t ) " t 1 0

and |h( t )| decreases as t ? ￥

What is ? (In addition to being an algorithm)

Let’s see:

Consider x s (t)*h(t) :

Reconstruction Filter: with h(t) as impulse response!

Output of the reconstruction Filter (y(t)): Convolution of x s (t) and h( t ) !

3. Design of Reconstruction Filter: Ideal case

Assumption : f s >2f h (x s (t) was generated at a frequency higher than the

Nyquist rate).

1/2 f s > f h f h : highest frequency of the original signal

Ideal Filter

Question : why do we need this low-pass filter to reconstruct x(t) from x s (t)?

answer : x s (t) contains frequencies higher than , but x(t)does not!

Question : Will any spectrum (other than x(t)’s introduced by sampling operation

remains after the filter?

Answer: No. , has ensured that no overlapping between x(t)’s

frequencies and the undesired frequencies in x s (t) introduced by

sampling!

Implementation of Ideal Reconstruction Filter

(Given the Impulse response of the filter)

Inverse Fourier transform =>

Characteristic of the Ideal Reconstruction Filter: Non causal!

Output at t ( y(t) ) must be generated using x s ( t ) t > t

=> Not good for real-time application!

How to reconstruction x(t) from nT < t < nT + T ?

Answer :

for example t = nT + 0.5T

l points before t = nT + 0.5T (k=n-l+1,…,n)

l points after t = nT + 0.5T (k=n+1,…,n+l)

Part Two

8-3A The z-Transform

1. Definition

For Laplace transform, we are given a function x(t),

For z-Transform, we are given a sampling sequence: x(0) , x(T), x(2T), …

· Definition: z-transform of a given sequence x(0) , x(T), x(2T), …

· Why do we define such a transform?

x(t)

If we want to compute this Laplace transform by computer

On the other hand

· Relationship between z- and s-plane

Basic Relationship :

(1) (note )

l.h .p. (s- plane) ? inside the unit circle (z- plane)

(2)

s: r.h.p. ? z: outside the unit circle

(3)

s: j w axis. ? z : unit circle

(4) s = 0 ()

z = 1

2. Basic z-Transform pairs

· Example 8-4: z-transform of unit pulse :

Solution :

· Example 8-5 z-Transform of unit step sequence u(n):

Solution :

· How to understand?

Step function u(t) :

Does give the same spectrum if T ? 0 ?

T ? 0 :

e z-Transform gives the same spectrum as Laplace transform if the sampling rate ?￥

· Example 8-6 : z-transform of unit exponential sequence

Solution:

· Is this result reasonable?

Why? Because

· Example 8-6 B

· Summary: Basic z-transform pairs

Would these be sufficient? No!

3. Extended z – transform pairs

4. Find z-transform using symbolic tool box

Example 8-7

Solution:

Analysis:

(1) n : odd =>

(2) n : even =>

Very complex!

Using Symbolic ToolBox

syms a n z % Declare symbolic

xn = a^n*cos(n*pi/2); % Define x(n)

xz = ztrans (xn, n, z); % Determine X(z)

xz (enter)

xz =

z^2/(a^2+z^2)

MatLab: always in terms of z instead of z -1.

8-3B Properties of z - transform

1. Linearity

2. Initial Value

why?

3. Final value

Why?

But,

8-3C Inverse z-Transform

Two Basic Methods:

(1) Express X(z) into “Definition Form”

(very simple, use long division or MatLab:

n = 8

X = dimpulse(num,den, n) (enter)

gives the first n terms)

(2) Express X(z) into partial-fraction from

- -

partial-fraction each term has an inverse transform

expansion

what Terms?

What about if you have ?

What about if you have?

Let’s see:

Can we now find A and B? What is the inverse z-transform of

What to do if you have ?

Important: before doing partial-fraction expansion, make sure the z-transform is in proper rational function of !

Example 8.9

Solution :

Heaviside’s Expansion Method:

(1)

(2) è

Example 8-9B MatLab Method

(1) Find partial-fraction expansion

b = 1;

a = [1 –1.2 0.2];

[r, p, k] = residuez(b,a);

(2) Directly Find Inverse Transform

syms n, z; % Declare symbolic

xz = 1/(1-1.2*z^(-1)+0.2*z^(-2)); % define X(z)

xn = iztrans(xz,z,n); % compute x(n)

xn ?

xn = 5/4-(1/4)*(1/5)^n è x(nT) = 1.25-0.25(0.2)n

Example 8-10

Solution :

Question: Define (or )

any relationship between

and ?

n = 0 5 + 1.25 - 6.25 = 0 1.25 - 0.25 = 1

n = 1 0 + 1.25 - 6.25*0.2 = 0 1.25 - 0.25*0.2 = 1.2

n = 2 0 + 1.25 - 6.25*0.22 = 1 1.25 - 0.25*0.22 = 1.24

n = 3 0 + 1.25 - 6.25*0.23 =1.2 1.25 - 0.25*0.23 = 1.248

Why? 6.25*0.2*0.2=0.25 =>

y(n+2) = x(n)!

Does always imply has two-step-delay

than ? Yes!

z -1 : Delay operator! (Must Assume X(nT)(the sequence to be z^(-1) processed)=0 for n<0)

8-3D Delay operator : z -k ( k steps ) ( k > 0 )

We want to establish the relationship between Z(x(nT-kT)) and Z(x(nT)) !

Let’s see what’s :

(1)? Yes!

(2)

Question : If , what’s ? Answer:

8-4 Difference Equation and Discrete-Time Systems

Continuous-Time System: Differential Equation, Laplace Transform

Discrete-Time System: Difference Equation, z-Transform

Properties of Continuous-Time Systems

Properties of Discrete-Time Systems

8-4A Properties of Discrete-Time Systems

System : Processes input to generate output

How to process : system-dependent

General symbolic notation for Discrete-Time System:

y( nT ) = H [ x(nT) ]

- -

what does this operator or

notation tell us? Processor

1. Shift-Invariant System

(Time-Invariant Systems for continuous-time or general)

An example of time-varying system

The “processing algorithm” which maps input to output changes!

What do we mean by a time-invariant system?

Shift-invariant systems:

Physical:

Mathematic:

Assume x(nT): x(0), x(T), … has generated

y(nT): y(0), y(T), …

For example:

has

If we apply as input

look at if

generated

Question: Is this system shift-invariant? Yes!

Question: Is this example telling us ? Yes!

Question: Is

always true for different systems?

No! only for time-invariant systems!

Shift-invariant system: if true for any n 0 .

2. Causal and noncausal systems

Physical Description: A system is causal or nonanticipatory if the system’s response to an input does not depend on future values of the input.

Mathematical Description:

Causal system:

Why? Although x1(nT) may not be the same as x2(nT) for n > n 0 , such difference does not affect the output determined by input up to n = n 0 .

3. Linear System

Linear System

Linear Systems: can be modeled as

response of the shift-invariant linear system at t=kT to an impulse input applied at t=0. (Or the response at to an impulse input applied at )

Causal systems:

Linear+causal+

Example: Given

x(0) = 1, x(T) = 2, x(2T) = 2, x(3T) = 1, …

h(0) = 3, h(T) = 2, h(2T) = 1, h(3T) = 0, …

MatLab:

x = [1 2 2 1 1];

h = [3 2 1];

y = conv(x,h);

3 8 11 9 7 3 1

Example 8-13:

Can you write a program (algorithm) to calculate y(nT) = x(nT)*h(nT) ?

Example 8-13: Symbolic Tool Box

syms n z % Declare Symbolic

xn =(1/2)^n % x(n)

hn = (1/3)^n % h(n)

xz = ztrans(xn, n, z) % z-transform of x(n)

hz = ztrans(hn, n, z) % z-transform of h(n)

yz = xz*hz % multiply, not convolution

yn = iztrans (yz, z, n); % Do you know why?

yn (enter)

yn = 3*(1/2)^n-2*(1/3)^n % y(nT)=3(1/2)n – 2(1/3)n

* Analytic solution of convolution

i.e.

Example:

Find x(nT)*h(nT)

Solution:

4. Stable system

Consider linear shift-invariant systems only.

· Definition of BIBO stable:

for all bounded x(nT).

· Derivation of Criterion

x(kT) bounded =>

Criterion:

· How to use this criterion: A

h: h(0), h(1), … h(N), 0, 0, …

(causal)

causal + Limited N => stable

Why!

for any fixed n in ,

for example,

In general

Conclusion: limited terms of h => stable!

Example : stable?

What about ?

· How to use this criterion: B

If we know

Z(h(nT)) = H(z)

h(0) h(1) h(2)

Why? |0.2| < 1 !

What about

Not BIBO stable!

In general, deg(num) < deg(den)

(poles inside the unit circle!)

· Example 8-14:

Solution:

Stable

8-4B Difference Equations

1. Difference Equations

Problem: determine the output of the system at the present time :

t = nT y(nT)

What information to use:

(1) input: current input u(nT)

previous input u(kT) (k < n)

future input u(kT) (k > n)

causal system : no future input!

Previous input

We do not need all of them è use u(n-1), … , u(n-m)

(2) output: previous output (its history): y(kT) (k < n) ? Yes.

future output y(kT) (k >n) ? No, no future output

previous outputs

We do not need all of them! y(n-1), …. , y(n-r) would be sufficient!

Mathematical Equation

y(nT) : depends on

linear system

weights:

Larger weight: more important in determining y(nT)

Would the weights be the same? No!

(r, m): system’s order

different systems: different order and weights (parameters)

2. z-transfer function

Different Equation

z-transform =>

z-transfer function

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

Why H(z) is the z-transform of impulse response h(nT) ?

8-4C Steady-State Frequency Response of a Linear Discrete-Time System

x(t)’s spectrum

x(nT)’s spectrum

y(t)’s spectrum

y(nT)’s spectrum

System’s frequency response

What is Y(z)/X(z) ? H(z) = Y(z)/X(z)

System frequency response

· Property of frequency response

T: sampling period

: sampling frequency

Frequency Response H: periodic function with period

è when the frequency increase by , the system’s frequency

response does not change.

Example: Input 1: T = 1 second

Input 2 :

Generate the same output amplitude?

· Normalized Frequency

: frequency period

Frequency Response in terms of r (argument)

· Amplitude Response or

Phase Response or

Question: what are their physical meaning?

Example 8-15: y(nT) = x(nT) + x(nT-2T)

Solution :

· Comment: z-transform: good for analysis

difference equation: computer program

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