CITSTUDENTS - · PDF fileSignals & Systems 10EC44 CITSTUDENTS.IN Page 1. ... 2nd edition, 1997. Indian Reprint ... Simon Haykin and Barry Van Veen “Signals and Systems”,

Signals & Systems 10EC44

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SUBJECT: SIGNALS & SYSTEMS

IA MARKS: 25 SUBJECT CODE: 10EC44

EXAM HOURS: 3 EXAM MARKS: 100

HOURS / WEEK: 4 TOTAL HOURS: 52

PART – A

UNIT 1:

Introduction: Definitions of a signal and a system, classification of signals, basic Operations on

signals, elementary signals, Systems viewed as Interconnections of operations, properties of systems.

07 Hours

UNIT 2:

Time-domain representations for LTI systems – 1: Convolution, impulse response representation,

Convolution Sum and Convolution Integral.

06 Hours

UNIT 3:

Time-domain representations for LTI systems – 2: properties of impulse response representation,

Differential and difference equation Representations, Block diagram representations.

07 Hours

UNIT 4:

Fourier representation for signals – 1: Introduction, Discrete time and continuous time Fourier series

(derivation of series excluded) and their properties .

06 Hours

PART – B

UNIT 5:

Fourier representation for signals – 2: Discrete and continuous Fourier transforms(derivations of

transforms are excluded) and their properties.

06 Hours

UNIT 6:

Applications of Fourier representations: Introduction, Frequency response of LTI systems, Fourier

transform representation of periodic signals, Fourier transform representation of discrete time

signals. 07 Hours

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UNIT 7:

Z-Transforms – 1: Introduction, Z – transform, properties of ROC, properties of Z – transforms,

inversion of Z – transforms.

07 Hours

UNIT 8:

Z-transforms – 2: Transform analysis of LTI Systems, unilateral Z Transform and its application to

solve difference equations. 06

Hours

TEXT BOOK

Simon Haykin and Barry Van Veen “Signals and Systems”, John Wiley & Sons, 2001.Reprint

2002

REFERENCE BOOKS :

1. Alan V Oppenheim, Alan S, Willsky and A Hamid Nawab, “Signals and Systems” Pearson

Education Asia / PHI, 2nd edition, 1997. Indian Reprint 2002

2. H. P Hsu, R. Ranjan, “Signals and Systems”, Scham‟s outlines, TMH, 2006

3. B. P. Lathi, “Linear Systems and Signals”, Oxford University Press, 2005

4. Ganesh Rao and Satish Tunga, “Signals and Systems”, Sanguine Technical Publishers, 2004

Question Paper Pattern: Student should answer FIVE full questions out of 8 questions to be set

each carrying 20 marks, selecting at least TWO questions from each part

Coverage in the Text:

UNIT 1: 1.1, 1.2, 1.4 to 1.8

UNIT 2: 2.1, 2.2 UNIT 3: 2.3, 2.4, 2.5

UNIT 4: 3.1, 3.2, 3.3, 3.6

UNIT 5: 3.4, 3.5, 3.6

UNIT 6: 4.1, 4.2, 4.3, 4.5, 4.6.

UNIT 7: 7.1, 7.2, 7.3, 7.4, 7.5

UNIT 8: 7.6 (Excluding „relating the transfer function and the State-Variable description,

determining the frequency response from poles and zeros) and 7.8 CITSTUDENTS.IN


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INDEX

SL.NO TOPIC PAGE NO.

PART A

UNIT – 1 INTRODUCTION

1.1 Definitions of Signal and system, classification of signals 5-24

1.4 Operation on signals:

1.5 Systems viewed as interconnections of operations

1.6 Properties of systems

UNIT – 2 TIME-DOMAIN REPRESENTATIONS FOR LTI SYSTEMS – 1

2.1 Convolution: concept and derivation 25-39

2.2 Impulse response representation

2.3 Convolution sum

2.5 Convolution Integral

UNIT – 3 TIME-DOMAIN REPRESENTATIONS FOR LTI SYSTEMS – 2

3.1 Properties of impulse response representation 40-55

3.3 Differential equation representation

3.5 Difference Equation representation

UNIT –4 FOURIER REPRESENTATION FOR SIGNALS – 1

4.1 Introduction 56-63

4.2 Discrete time fourier series

4.4 Properties of Fourier series

4.5 Properties of Fourier series

PART B

UNIT –5 FOURIER REPRESENTATION FOR SIGNALS – 2: 64-68

5.1 Introduction

5.1 Discrete and continuous fourier transforms

5.4 Properties of FT

5.5 Properties of FT

UNIT – 6 APPLICATIONS OF FOURIER REPRESENTATIONS


6.2 Frequency response of LTI systems

6.4 FT representation of periodic signals

6.6 FT representation of DT signals

UNIT – 7 Z-TRANSFORMS – 1


7.2 Z-Transform, Problems

7.3 Properties of ROC

7.5 Properties of Z-Transform

7.7 Inversion of Z-Transforms,Problems

UNIT – 8 Z-TRANSFORMS – 2

8.1 Transform analysis of LTI Systems 105-114

8.3 Unilateral Z- transforms

8.5 Application to solve Difference equations

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UNIT 1: Introduction Teaching hours: 7

Introduction: Definitions of a signal and a system, classification of signals, basic Operations on

signals, elementary signals, Systems viewed as Interconnections of operations, properties of systems.

TEXT BOOK


2002

REFERENCE BOOKS :






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Unit 1: Introduction

1.1.1 Signal definition

A signal is a function representing a physical quantity or variable, and typically it contains

information about the behaviour or nature of the phenomenon.

For instance, in a RC circuit the signal may represent the voltage across the capacitor or the

current flowing in the resistor. Mathematically, a signal is represented as a function of an

independent variable ‘t’. Usually ‘t’ represents time. Thus, a signal is denoted by x(t).

1.1.2 System definition

A system is a mathematical model of a physical process that relates the input (or excitation)

signal to the output (or response) signal.

Let x and y be the input and output signals, respectively, of a system. Then the system is

viewed as a transformation (or mapping) of x into y. This transformation is represented by the

mathematical notation

y= Tx -----------------------------------------(1.1)

where T is the operator representing some well-defined rule by which x is transformed into y.

Relationship (1.1) is depicted as shown in Fig. 1-1(a). Multiple input and/or output signals are

possible as shown in Fig. 1-1(b). We will restrict our attention for the most part in this text to the

single-input, single-output case.

1.1 System with single or multiple input and output signals

1.2 Classification of signals

Basically seven different classifications are there:

Continuous-Time and Discrete-Time Signals

Analog and Digital Signals Real

and Complex Signals Deterministic

and Random Signals Even and Odd

Signals

Periodic and Nonperiodic Signals

Energy and Power Signals

Continuous-Time and Discrete-Time Signals

A signal x(t) is a continuous-time signal if t is a continuous variable. If t is a discrete

variable, that is, x(t) is defined at discrete times, then x(t) is a discrete-time signal. Since a

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discrete-time signal is defined at discrete times, a discrete-time signal is often identified as a

sequence of numbers, denoted by {x,) or x[n], where n = integer. Illustrations of a continuous-

time signal x(t) and of a discrete-time signal x[n] are shown in Fig. 1-2.

1.2 Graphical representation of (a) continuous-time and (b) discrete-time signals

Analog and Digital Signals

If a continuous-time signal x(t) can take on any value in the continuous interval (a, b), where

a may be - ∞ and b may be +∞ then the continuous-time signal x(t) is called an analog signal. If a

discrete-time signal x[n] can take on only a finite number of distinct values, then we call this

signal a digital signal.

Real and Complex Signals

A signal x(t) is a real signal if its value is a real number, and a signal x(t) is a complex signal

if its value is a complex number. A general complex signal x(t) is a function of the form

x (t) = x1(t) + jx2 (t)--------------------------------1.2

where x1 (t) and x2 (t) are real signals and j = √-1

Note that in Eq. (1.2) ‘t’ represents either a continuous or a discrete variable.

Deterministic and Random Signals:

Deterministic signals are those signals whose values are completely specified for any given

time. Thus, a deterministic signal can be modelled by a known function of time ‘t’.

Random signals are those signals that take random values at any given time and must be

characterized statistically.

Even and Odd Signals

A signal x ( t ) or x[n] is referred to as an even signal if

x (- t) = x(t)

x [-n] = x [n] -------------(1.3)

A signal x ( t ) or x[n] is referred to as an odd signal if

x(-t) = - x(t)

x[- n] = - x[n]--------------(1.4)

Examples of even and odd signals are shown in Fig. 1.3.

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1.3 Examples of even signals (a and b) and odd signals (c and d).

Any signal x(t) or x[n] can be expressed as a sum of two signals, one of which is even

and one of which is odd. That is,

Where,

-------(1.5)

-----(1.6)

Similarly for x[n],

Where,

-------(1.7)

--------(1.8)

Note that the product of two even signals or of two odd signals is an even signal and

that the product of an even signal and an odd signal is an odd signal.

Periodic and Nonperiodic Signals

A continuous-time signal x ( t ) is said to be periodic with period T if there is a positive

nonzero value of T for which

…………(1.9)

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An example of such a signal is given in Fig. 1-4(a). From Eq. (1.9) or Fig. 1-4(a) it follows

that

---------------------------(1.10)

for all t and any integer m. The fundamental period T, of x(t) is the smallest positive value of

T for which Eq. (1.9) holds. Note that this definition does not work for a constant

1.4 Examples of periodic signals.

signal x(t) (known as a dc signal). For a constant signal x(t) the fundamental period is

undefined since x(t) is periodic for any choice of T (and so there is no smallest positive

value). Any continuous-time signal which is not periodic is called a nonperiodic (or

aperiodic) signal.

Periodic discrete-time signals are defined analogously. A sequence (discrete-time

signal) x[n] is periodic with period N if there is a positive integer N for which

……….(1.11)

An example of such a sequence is given in Fig. 1-4(b). From Eq. (1.11) and Fig. 1-4(b) it

follows that

……………………..(1.12)

for all n and any integer m. The fundamental period No of x[n] is the smallest positive integer

N for which Eq.(1.11) holds. Any sequence which is not periodic is called a nonperiodic (or

aperiodic sequence.

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Note that a sequence obtained by uniform sampling of a periodic continuous-time signal may

not be periodic. Note also that the sum of two continuous-time periodic signals may not be

periodic but that the sum of two periodic sequences is always periodic.

Energy and Power Signals

Consider v(t) to be the voltage across a resistor R producing a current i(t). The

instantaneous power p(t) per ohm is defined as

…………(1.13)

Total energy E and average power P on a per-ohm basis are

……(1.14)

For an arbitrary continuous-time signal x(t), the normalized energy content E of x(t) is

defined as

…………………(1.15)

The normalized average power P of x(t) is defined as

(1.16)

Similarly, for a discrete-time signal x[n], the normalized energy content E of x[n] is

defined as

(1.17)

The normalized average power P of x[n] is defined as

(1.18)

Based on definitions (1.15) to (1.18), the following classes of signals are defined:

1. x(t) (or x[n]) is said to be an energy signal (or sequence) if and only if 0 < E < m, and

so P = 0.

2. x(t) (or x[n]) is said to be a power signal (or sequence) if and only if 0 < P < m, thus

implying that E = m.

3. Signals that satisfy neither property are referred to as neither energy signals nor power

signals.

Note that a periodic signal is a power signal if its energy content per period is finite, and

then the average power of this signal need only be calculated over a period

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1.3 Basic Operations on signals

The operations performed on signals can be broadly classified into two kinds

Operations on dependent variables

Operations on independent variables

Operations on dependent variables

The operations of the dependent variable can be classified into five types: amplitude scaling,

addition, multiplication, integration and differentiation.

Amplitude scaling Amplitude scaling of a signal x(t) given by equation 1.19, results in amplification of

x(t) if a >1, and attenuation if a <1.

y(t) =ax(t)……..(1.20)

1.5 Amplitude scaling of sinusoidal signal

Addition

The addition of signals is given by equation of 1.21. y(t) = x1(t) + x2 (t)……(1.21)

1.6 Example of the addition of a sinusoidal signal with a signal of constant amplitude

(positive constant)

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Physical significance of this operation is to add two signals like in the addition of the

background music along with the human audio. Another example is the undesired addition of

noise along with the desired audio signals.

Multiplication

The multiplication of signals is given by the simple equation of 1.22.

y(t) = x1(t).x2 (t)……..(1.22)

1.7 Example of multiplication of two signals

Differentiation

The differentiation of signals is given by the equation of 1.23 for the continuous.

…..1.23

The operation of differentiation gives the rate at which the signal changes with

respect to time, and can be computed using the following equation, with Δt being a

small interval of time.

….1.24

If a signal doesn‟t change with time, its derivative is zero, and if it changes at a fixed

rate with time, its derivative is constant. This is evident by the example given in

figure 1.8.

1.8 Differentiation of Sine - Cosine

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2 3 4 0 1 2 s t(time in cond!i} t (t•me '" seconas)

(11J

19

Integration of x(9

(l)

;

•

•

•

Signals &. S"'tems

Integrati>n

!OEC44

The integration of a si al x(, is given by equation 1.25

y( t ) = J x (r)d·r

->:• ...... 1.25

:•E

2 2

1.5 •••

1 0

• 1

Q

f •••

D D

D

Op eratintson indep endentv ar:iables

Time scaliltg

Time scaling operation is given by equation 1.26

y(f) = x(af) ...............1.26

This operation results in expansion in time for a<l and com pression intime for a>1. as

evident from the ex amp1es of figure 1.10.

••• 0.4

••• E K

0.0

M

-0.2

-1>.4

-1>.6

0 1 2 3 t tt•mc '" sccana,.;J

..,_. 4 0

2 • 4

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1.10 Examples of time scaling of a continuous time signal

An example of this operation is the compression or expansion of the time scale that results in

the „fast-forward’ or the „slow motion’ in a video, provided we have the entire video in some

stored form.

Time reflection

Time reflection is given by equation (1.27), and some examples are contained in fig1.11.

y(t) = x(−t) ………..1.27

(a)

(b) 1.11 Examples of time reflection of a continuous time signal

Time shifting

The equation representing time shifting is given by equation (1.28), and examples of this

operation are given in figure 1.12.

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0 0 \


y(t) x(t- tO)..............1.28

3,----------- ---, 3,----------- ---,

-3 4 -2 0 2 4 -2 0 2 4

t {time in seconds) t {time in seconds)

(a)

2 2

1

c:r <::. X

-1 -1

-2 -2 4 -2 0 2 4 4 -2 0 2 4

t (time in seconds) t (time in seconds]

(b)

1.12 Examples of time shift of a continuous time signal

Time shifting and scaling

The combined transformation of shifting and scaling is contained in equation (1.29),

along with examples in figure 1.13. Here, time shift has a higher precedence than time scale.

y(t) x(at -tO) .................1.29

t {time in seconds} t (time in seconds}

(a)

t {time in seconds)

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(b)

1.13 Examples of simultaneous time shifting and scaling. The signal has to be shifted first and then time scaled.

1.4 Elementary signals

Exponential signals:

The exponential signal given by equation (1.29), is a monotonically increasing function if a > 0, and is a decreasing function if a < 0.

……………………(1.29)

It can be seen that, for an exponential signal,

…………………..(1.30)

Hence, equation (1.30), shows that change in time by ±1/ a seconds, results in change in

magnitude by e±1 . The term 1/ a having units of time, is known as the time-constant. Let us

consider a decaying exponential signal

……………(1.31)

This signal has an initial value x(0) =1, and a final value x(∞) = 0 . The magnitude of this

signal at five times the time constant is,

………………….(1.32)

while at ten times the time constant, it is as low as,

……………(1.33) It can be seen that the value at ten times the time constant is almost zero, the final value of the signal. Hence, in most engineering applications, the exponential signal can be said to

have reached its final value in about ten times the time constant. If the time constant is 1

second, then final value is achieved in 10 seconds!! We have some examples of the

exponential signal in figure 1.14.

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timc{&cc)

er-------------------------,

sr--------------------------,

2

2 - . 0-=::::;,======! t:imo {'S>QC}

(a)

o.s '

o.s

! 0.4

2

2 ====- 1==== 0 -! (b)

(c) (rl)

Fig 1o14 The continuous time exponential signal (a) e-t, (b) et, (c) e-ltl, and (d) eltl

The sinusoidal signal:

The sinusoidal continuous time periodic signal is given by equation 1.34, and examples are

given in figure 1015

The different parameters are:

Angular frequency co 2nfin radians,

Frequency fin Hertz, (cycles per second)

Amplitude A in Volts (or Amperes)

Period Tin seconds

The complex exponential:

We now represent the complex exponential using the Euler's identity (equation (1.35)),

=( 1 000000000 oooooo(l.35)

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to represent sinusoidal signals. We have the complex exponential signal given by

equation (1.36)

………(1.36) Since sine and cosine signals are periodic, the complex exponential is also periodic with

the same period as sine or cosine. From equation (1.36), we can see that the real periodic

sinusoidal signals can be expressed as:

………………..(1.37)

Let us consider the signal x(t) given by equation (1.38). The sketch of this is given in fig 1.15

……………………..(1.38)

The unit impulse:

The unit impulse usually represented as δ (t) , also known as the dirac delta function, is

given by,

…….(1.38) From equation (1.38), it can be seen that the impulse exists only at t = 0 , such that its area is

1. This is a function which cannot be practically generated. Figure 1.16, has the plot of the

impulse function

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The unit step:

The unit step function, usually represented as u(t) , is given by,

……………….(1.39)

Fig 1.17 Plot of the unit step function along with a few of its transformations

The unit ramp:

The unit ramp function, usually represented as r(t) , is given by,

…………….(1.40)

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"-"' "

_,

" '-''

<?

"" "_,

(a)

"d'

"' 1.15

1

'"'

-1 -<:L5 0 ( $<1>'<;:- )

(b)

4

3

2

(<)

0

-2 0

t( sec f

(d)

Fig U8 Plot of the unit ramp function along with a few of its transformations

The signum function:

The signum function, usually represented as sgn(t) , is given by

1 t

0 t-

1 t

000 000000000 000000000 0000000000(1.41)

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2r--------------,

t 0

!(sec l

2 2

. . I 1

0 0

4 . -1 r .

_,

0 1 2 !( ...)

- _, 0 1 2

Fig 1.19

{c)

Plot of the unit signum function along with a few of its transformations

1.5 System viewed as interconnection of operation:

This article is dealt in detail again in chapter 2/3. This article basically deals with system

connected in series or paralleL Further these systems are connected with adders/subtractor,

multipliers etc.

1.6 Properties of system:

In this article discrete systems are taken into account. The same explanation stands for

continuous time systems also.

The discrete time system:

The discrete time system is a device which accepts a discrete time signal as its input,

transforms it to another desirable discrete time signal at its output as shown in figure 1.20

input Ui rl!te time

systt:ltt xlnl

output

y[n]

Fig 1.20 DTsystem

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-1

J

"

' ..,

= l

Stability

A system is stable if 'bounded input results in a bounded output'. This condition, denoted

by BIBO, can be represented by:

.......(1.42)

Hence, a finite input should produce a finite output, if the system is stable. Some examples of

stable and unstable systems are given in figure 1.21

Nlah!!l" 'Y'tLm

2 2

1

k .-=·.

il. .s l

·'

·2 0 5 10 15 0

n

10 15

8 8;------------------,

A.

! 2 ...;, ..

5 10

n

Fig 1.21

15

Examples for system stability

Memory

The system is memory-less if its instantaneous output depends only on the current input.

In memory-less systems, the output does not depend on the previous or the future input.

Examples of memory less systems:

r["] =

I I= + + +..... CIT

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Causality:

A system is causal, if its output at any instant depends on the current and past values of

input. The output of a causal system does not depend on the future values of input. This

can be represented as:

y[n] F x[m] for m n

For a causal system, the output should occur only after the input is applied, hence,

x[n] 0 for n 0 implies y[n] 0 for n

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All physical systems are causal (examples in figure 7.5). Non-causal systems do not exist.

This classification of a system may seem redundant. But, it is not so. This is because,

sometimes, it may be necessary to design systems for given specifications. When a system

design problem is attempted, it becomes necessary to test the causality of the system, which

if not satisfied, cannot be realized by any means. Hypothetical examples of non-causal

systems are given in figure below.

Invertibility:

A system is invertible if,

Linearity:

The system is a device which accepts a signal, transforms it to another desirable signal, and is

available at its output. We give the signal to the system, because the output is s

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Time invariance:

A system is time invariant, if its output depends on the input applied, and not on the time of

application of the input. Hence, time invariant systems, give delayed outputs for delayed

inputs.

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UNIT 2: Time-domain representations for LTI systems – 1 Teaching hours: 6

Time-domain representations for LTI systems – 1: Convolution, impulse response representation,

Convolution Sum and Convolution Integral.

TEXT BOOK


2002

REFERENCE BOOKS :






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UNIT 2

Time-domain representations for LTI systems – 1

2.1 Introduction: The Linear time invariant (LTI) system:

Systems which satisfy the condition of linearity as well as time invariance are known as linear time

invariant systems. Throughout the rest of the course we shall be dealing with LTI systems. If the

output of the system is known for a particular input, it is possible to obtain the output for a number

of other inputs. We shall see through examples, the procedure to compute the output from a given

input-output relation, for LTI systems.

Example – I:

2.1.1 Convolution:

A continuous time system as shown below, accepts a continuous time signal x(t) and gives out

a transformed continuous time signal y(t).

Figure 1: The continuous time system

Some of the different methods of representing the continuous time system are:

i) Differential equation

ii) Block diagram

iii) Impulse response

iv) Frequency response

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v) Laplace-transform

vi) Pole-zero plot

It is possible to switch from one form of representation to another, and each of the representations

is complete. Moreover, from each of the above representations, it is possible to obtain the system

properties using parameters as: stability, causality, linearity, invertibility etc. We now attempt to

develop the convolution integral.

2.2 Impulse Response

The impulse response of a continuous time system is defined as the output of the system when its

input is an unit impulse, δ (t ) . Usually the impulse response is denoted by h(t ) .

Figure 2: The impulse response of a continuous time system

2.3 Convolution Sum: We now attempt to obtain the output of a digital system for an arbitrary input x[n], from

the knowledge of the system impulse response h[n].

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LTI L

x[n] = L"'

i'l= "' l.l

An input impulse re punse cori'*''Poodillj! output

x[u] y[nJ

bful

x[m].5[n-m] ylnl= mlhln-ml <;ystem 'h <!G

input

h[u]

"utput

y[n]=x[n]•h[n]

y[n] = x[n] h[n]

Methods of evaluating the convolution sum:

Given the system impulse response h[n], and the input x[n], the system output y[n], is

given by the convolution sum:

=

Problem:

To obtain the digital system output y[n], given the system impulse response h[n], and the

system input x[n] as:

I, -I 31

]= 1 ' l.

-l 4 7

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- -

r - l

1. Evaluation as the weighted sum of individual responses

The convolution sum of equation( ... ), can be equivalently represented as:

y[n] D D ..... D OCJ[D l:fz[n D l]D IX[O]h[n]D Dx[l]h[n D l]D ...... .

input x[n] impulse response: h[n]

8 .............;........ 8 ""'\"'

6 6

4 4

c: 2 :E

2 II 2 4 2 II 2 4 6 n n

h[n+1] x[·1].h[n+11

111 111

.. 5

.5.

.c

5

+c

..;::

-5 -5

·2 0 2 ·2 0 2 4 6 n n

h[nl x[O].h[n] 111 111

li

-.c

li 'i'

.., T c: .c

' 1 0 0 0 0 0 .,; 0 0 0 0 0 '

-5 ...................... ................ ··.··········· -5 ........................... : .......................

-2 0 2 4 6 -2 0 2 4 s n n

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-"'

'

h[n-1] 10

-• 5

. I:

x(1].h(n-1]

···0· .. '1': ...

0!:! ··0·· C>

. 'i' . ..·i\1·· ··0·····

·2 0 2 4 n

h[n-:2] 10

-6

s ·2 0 2 4 s n

x(2].h(n-2]

10

6 ''"'"'""''""' '"' ''"'"'"'"' <:"

.E.

6 ''"'"'"'"'"'

-c • T .c J::. 0 0 0

! 0

-6

-6·

0 0 0

1 0

-:2 0 2 4 n

Convolution as matrix multiplication:

Given

=

B -2 0 2 4 B

n

X

]

1: = (L+M-1)

+

0 0 0 0

+I] h 0 ' X

0 1

+ 0 = =

+

0 0

0 0 0 0

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Evaluation using graphical representation:

Another method of computing the convolution is through the direct computation of each value of the

output y[n]. This method is based on evaluation of the convolution sum for a single value of n, and

varying n over all possible values.

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-

=

IS lwo

4: I.

S! y[lj

-m[

is two

6: 2.

=

WhiCh is

are

10: I]

is I.

-m[

is two

ll: IS ,.

12:

=

two

13: all y[n], are

x[m]

10.--------.,

5' ····· '' ''

1'1[-1-m] J[-1J=H> 10.--------., 10.--------.

5• ..!.. .c;

5 'X

-5 "'""' ""' ..i .......

-6 0 5 m

-5 ......................, ......... .

-6 0 5 m

,, '" ,.: '" '" '"" '".

(.l 5 m

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T

...........

9

"-G"

x[m] h[-m) y[OJ=£1.5+2.5)

10 10 10

5· e =..!...

6 .. -;;: 5

.,.....

..15 ' ...e;

0 5 -5 0 5 ...e; 0 5 m m m

K[m] h[1-rn) y[1]=j-3-3.75+0.8}

10 10 10

5 ,.. '"' ''" '" "'. 5

:z .§.

5 ...

-5 ""' -6 -6

-6 I) -6 0 5 -5 0 5

m m m

10 10 10

6 6 6

••

-5. -6 -6 .. . . ... ... . ............ -6 -5 0 -6 0 5

m m m

x[m] h(S..m] y[3)=(2.4-1.876) 10 10 10

...... 5 5 E 5

O·&Ge, 00 e 00600 00·

>< -6 -5 -6

-6 0 5 -6 0 5 -5 0 5

m m m

x[mJ h[4-m] y[4]==(3.75)

10 10 10

6 ..... 6 5

f • 6

-5 -6 -6

-6 -6 0 5 -6 0 6 m m m

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Output: y[nl-•1·1]. h[n+1}+x[OJ.h[nJ+x[1J. In-1)+x[2J.nIr>-2)

10

8

G

4- " " '!

2

2 ---- ,----7----t ---- 2---- --_J -- --

n

Evaluation from direct convolution sum:

While small length, finite duration sequences can be convolved by any of the above three methods,

when the sequences to be convolved are of infinite length, the convolution is easier performed by

direct use of the 'convolution sum' of equation(. .. ).

m 0 smce:

m O

0 =

0

<n =

m 11

l m '5c 11

Example: A system has impulse response h[n] D D exp(D O u[n]. Obtain the unit step

response.

Solution:

=L: ( n-m

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II

2 -2

f

-

- 1

- -" -

-

c

2

=m'o n-

=I: (

(

{1- rl)

(1- )

input:x[n] impulse response: h[n] ouput:y[n]

- .c..... '"E' 2

-:I ::1

m '""'"''''""''""''

c

c 'i' 1 .

-II ci ......

.......

-II

.!!..

c • "' >< c .c

.=.... :;.,

-1 0 5 10 15 5 10 15 0 5 10 15

n n n

input:x[n] impulse response: h[n] ouput:y[n]

.... ::1

2 c.......

'.E.=' 2 ..... 2

ll:

J::

'*

II <I c >< II

>( c 1:

-1 -1 :>. -1

0 5 10 0 5 10 0 5 10 n n n

-.I.I.:.. 1

::1

II

.....:.. ><

-1

input:x[nJ

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impulse response; h[n]

2.--.,-----.

ouput: y[n)

0 5 10 n

10 0 10 n n

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- :::1 -

-

..:: 10 15 $()

input:x[n]

-impulse response: h(n] oupu1: y[n]

-c 2,..-- -...,

2 1::: '2

2

::::1

!:!' 1

'2' t " f 1

<:j :

fa.. l >(

1 ."I-I'. "-' .

,.I.I..,

E. !>< -1'-------'

0 5 10 n

1:

.c: -1

!::

• '>: -1 0 5 10

n n input:x[n)

n

impuh>e response: h(n]

0 5 10 15 20

n input:x[n)

n

impulse response; h[n]

9 11

-2· ------ ------ ---- ------- ------ ------ n

ouput: y[n]

n

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Evaluation from Z-transforms:

Another method of computing the convolution of two sequences is through use of Z-transforms. This

method will be discussed later while doing Z-transforms. This approach converts convolution to

multiplication in the transformed domain.

input xfnl

impulse response

h[nJ

output yLnJ

.1(11] I 7

d/[Z] JtnJ z rLXJ

t'[X ]=XIX JH[XJ

Evaluation from Discrete Time Fourier transform (DTFT):

It is possible to compute the convolution of two sequences by transforming them to the frequency

domain through application of the Discrete Fourier Transform. This approach also converts the

convolution operator to multiplication. Since efficient algorithms for DFT computation exist, this

method is often used during software implementation of the convolution operator.

. bml)·si.· u.o;ing. Disoett•-time Fmtrit'1' Transform (DTFT)

i11put x[11]

huJmlse l'e!>Jlom>ll

uulpul y[n]

- hfnl

l.'[n] l.Y!F! I(PJ] /)TPT JI[K] Jtn] DTF'J' Y[K]

YIKI=XIKIHIKI

Evaluation from block diagram representation:

While small length, finite duration sequences can be convolved by any of the above three methods,

when the sequences to be convolved are of infinite length, the convolution is easier performed by

direct use of the 'convolution sum' .

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2.4 Convolution Integral:

We now attempt to obtain the output of a continuous time/Analog digital system for an arbitrary

input x(t), from the knowledge of the system impulse response h(t), and the properties of the impulse

response of an LTI system.

Methods of evaluating the convolution integral: (Same as Convolution sum)

Given the system impulse response h(t), and the input x(t), the system output y(t), is given by the

convolution integral:

Some of the different methods of evaluating the convolution integral are: Graphical representation,

Mathematical equation, Laplace-transforms, Fourier Transform, Differential equation, Block

diagram representation, and finally by going to the digital domain.

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UNIT 3: Time-domain representations for LTI systems – 2 Teaching hours: 7

Time-domain representations for LTI systems – 2: properties of impulse response representation,

Differential and difference equation Representations, Block diagram representations.

TEXT BOOK


2002

REFERENCE BOOKS :






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UNIT 3:

Time-domain representations for LTI systems – 2

3.1 Properties of impulse response representation: Impulse Response

Def. Linear system: system that satisfies superposition theorem.

For any system, we can define its impulse response as:

h(t) y(t) whenx(t) (t)

For linear time invariant system, the output can be modeled as the convolution of the impulse

response of the system with the input.

y(t)

x(t) * h(t)

x( )h(t )d

For casual system, it can be modeled as convolution integral.

y(t) x( )h(t )d0

3.2 Differential equation representation:

General form of differential equation is

where ak and bk are coefficients, x(.) is input and y(.) is output and order of differential or

difference equation is (M,N).

Example of Differential equation • Consider the RLC circuit as shown in figure below. Let x(t) be the input voltage source and

y(t) be the output current. Then summing up the voltage drops around the loop gives

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-)p

T)dt

R L

x(l) c

3.3 Solving differential equation:

A wide variety of continuous time systems are described the linear differential equations:

• Just as before, in order to solve the equation for y(t), we need the ICs. In this case, the ICs are

given by specifying the value of y and its derivatives 1 through N -1 at t0-

• Note: the ICs are given at t0- to allow for impulses and other discontinuities at t0.

• Systems described in this way are

• linear time-invariant (LTI): easy to verify by inspection

• Causal: the value of the output at timet depends only on the output and the input at times 0 S

tSt

• As in the case of discrete-time system, the solution y(t) can be decomposed into y(t)

Yh(t)+yp(t), where homogeneous solution or zero-input response (ZIR), yh(t) satisfies

the equation

• The zero state resoo1nse (ZSR) or

Nl

+2 i=O

\Vlth ) - !)

1ll

=2 t 0 i=O

) = ... = I)

) = 0.

Homogeneous solution (ZIR) for CT

• standard method for u"'"""'"

'"'""IS all terms unut·v "'"' the input Ia zero.

N

2 i=O

(t) 0, t 0

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and solution is of the form

N

rJ = 2: i=i

lV

and <J:re solved

Hmnogeneous solution (ZIR) for DT

N

2: k=D

is

wlwrP n are the roots

····k =0

and are solved

Example 1 (ZIR)

• of

t) + + + d

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is

t) Ct

• 2] = l

Example 2 (ZIR)

. find solution.

• to is 11-

1] =

solution for N 1 is

• p = 0, 11e!lCe =P

Example 3 (ZIR)

• Consider the RC described by -'c RC t)

•

solution

t) t) = 0

=q

is root =0

• This

•

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3.4 Difference equation representation:

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• n=O

N 111

L: L: k=l k=O

depends on /Cs on

• at n= 1

After

N M"

L: 1 L: l k] k=I k=O

I

I I k=l k=fl

ou ICs au

• at n = 2

l]- I

I k=l

-k]

Example of Difference equation

• An exillll!Pi<"

1]

•

• initial conditions

Initial Conditions

Initial Conditions summarise all the information about the systems past that is needed to

determine the future outputs.

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• In disrrete rasP, for an

1].. . . 1

• inHhd conditions for n-[}l"()pr differ<ential are the values

of the first /1/ derivatives of the output

d y(t) f)

Solving difference equation

• an example of differ<encP

0. L2 .. with I]= 0 Then

-1] = . n=

+

x[O] 1

)+x

and so on

as a surn of tvvo tenns:

=( I 1] + ll = 1.

• term(- 's but not on input

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• not on the

ICs • This is true for ARMA

svstc'rn outpur at time n is a sum

rmtm ,, at tirne !1.

• LM" 'u" an ARMA (N,l\1) systentJ

n 0. L

11 + 1] •...

• +

are

• not on

• on the input. hut not on the s

• is

sv<;IPtcrt delf>mnlrted by the

only

input to

the input only

n] is the output of thE> svst<•m

to

• uut<'u the zet·o·immt respm·tse

of to the that it is fietc>lr-

thl" only)

• is called the zero·state response (ZSR) usually as

ular solution the filter to the it is determined

the with the set to

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0.5 rI

+5

Step response" of a system

• Consi,cler the uutpw det ot+np<>sliion +

l'vf) fiIter

N

=-2.: i=i

with the 1}' "

M

-iJ+:L i=O

n=O,L2... ,

• The of an ARIVIA filter at lime 11 is the sum of the ZIR aud the

ZSR at time n.

Example of difference e<Jttation

• A is described

2] 0.0675x[n 0.134Hx[n l

• the =L

0.1

3.5 Block Diagram representation:

-Ll -1 +0.41

0.675x-n 2]

1 -0.41 -+-

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• A block is an interconnection that

act on

• This method is more detailed representation of the system than

res;ocms:e or

• represent

•

the input-omput behavior of a sv ;tel:n

de:scribc;s how are rm·lp,·prl

block

Scalar multiplication: t) = cx(t) or . where cis a

+ t} or

•

• = x[n]. where cis a

lar Multipl tion

ll= +

l= +

Addition

• AddiHon: or

Time shift for

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X["I- 8 -"''"' = '''" II

Integration and timeshifting

s

IJ

s

Ll I

Exan1plf" 1

• Lo,nslclt·r rhe SV'>tern described rhe block

• Lo,nsicl<•r the within the dashed box

• The input

one to

is linw sh Ifred

The

1 to 1] and tim<' shifted

icano,ns art> GHTic·<l out and

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---.--- :!: ---!: -----.- vfnl

,,,

1.] 1: 2: DhY•ct rorm I

-1]+ -2].

• VVrhe .v_n_ in tenus of

• PutIhe value of

- 1] + ..... lJ ·l

-2]

·····2]= ..... I] c ..... 2]

• The block di<1g1carn represents an LTI system

Exacmple 2

• Consider the sv,;tern described by the block and its eli fference

is

Exarnp]l:' 3

(1 1] (1 2]

• Consider the system described by the block

is + ( 1 1] ( 1 1]

x]n]--.........- s-

(b)

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l)

• Block diagram repr<es•enta>tic>n is not"""'!'"« direct form IIstructure of

Example l

• can the order without tiW inplH UUC<PU

"'v''" input are relatr·d

= al +

• The acts as an 10 the second

1+ T[n 2]

• systeln is not unique

Continuous Urne

set

be an and

r vvhere (t) is the n·fold

• Revvdre in tenns of an iniU::_d condition on the as

(r)

J{o"

(() , II

• If \Ve assutne zero then differentiation and

ie.

d (r)=

dr

• if N ,\1 and intet,r<ile /If times, we

of the svstecm

k)

•

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f I

f I

Di

r)

J

,

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UNIT 4: Fourier representation for signals – 1 Teaching hours: 6

Fourier representation for signals – 1: Introduction, Discrete time and continuous time Fourier series

(derivation of series excluded) and their properties . TEXT BOOK


2002

REFERENCE BOOKS :






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UNIT 4

Fourier representation for signals – 1

4.1 Introduction:

Fourier series has long provided one of the principal methods of analysis for mathematical

physics, engineering, and signal processing. It has spurred generalizations and applications that

continue to develop right up to the present. While the original theory of Fourier series applies to

periodic functions occurring in wave motion, such as with light and sound, its generalizations often

relate to wider settings, such as the time-frequency analysis underlying the recent theories of wavelet

analysis and local trigonometric analysis.

• In 1807, Jean Baptiste Joseph Fourier Submitted a paper of using trigonometric series to represent

“any” periodic signal.

• But Lagrange rejected it!

• In 1822, Fourier published a book “The Analytical Theory of Heat” Fourier‟s main contributions:

Studied vibration, heat diffusion, etc. and found that a series of harmonically related sinusoids is

useful in representing the temperature distribution through a body.

• He also claimed that “any” periodic signal could be represented by Fourier series. These arguments

were still imprecise and it remained for P.L.Dirichlet in 1829 to provide precise conditions under

which a periodic signal could be represented by a FS.

• He however obtained a representation for aperiodic signals i.e., Fourier integral or transform

• Fourier did not actually contribute to the mathematical theory of Fourier series.

• Hence out of this long history what emerged is a powerful and cohesive framework for the analysis

of continuous- time and discrete-time signals and systems and an extraordinarily broad array of

existing and potential application.

The Response of LTI Systems to Complex Exponentials:

We have seen in previous chapters how advantageous it is in LTI systems to represent signals as a

linear combinations of basic signals having the following properties.

Key Properties: for Input to LTI System

1. To represent signals as linear combinations of basic signals.

2. Set of basic signals used to construct a broad class of signals.

3. The response of an LTI system to each signal should be simple enough in structure.

4. It then provides us with a convenient representation for the response of the system.

5. Response is then a linear combination of basic signal.

Eigenfunctions and Values :

• One of the reasons the Fourier series is so important is that it represents a signal in terms of eigenfunctions of LTI systems.

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• When I put a complex exponential function like x(t) = ejωt through a linear time-invariant system,

the output is y(t) = H(s)x(t) = H(s) ejωt where H(s) is a complex constant (it does not depend on

time).

• The LTI system scales the complex exponential ejωt .

Historical background

There are antecedents to the notion of Fourier series in the work of Euler and D. Bernoulli on

vibrating strings, but the theory of Fourier series truly began with the profound work of Fourier on

heat conduction at the beginning of the century. In [5], Fourier deals with the problem of describing

the evolution of the temperature of a thin wire of length X. He proposed that the initial temperature

could be expanded in a series of sine functions:

The following relationships can be readily established, and will be used in subsequent sections for

derivation of useful formulas for the unknown Fourier coefficients, in both time and frequency

domains.

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T T

T T

T


f sin(kw0 t)dt = f cos(kw0 t)dt (1)

0 0

=0

f sin 2 (kw t)dt = f cos

2 (kw t)dt (2)

0 0

0 0

T

= 2

T

f cos(kw0 t) sin(gw0 t)dt = 0 (3) 0

f sin(kw0 t)sin(gw0 t)dt = 0 (4)

0

T

where

f cos(kw0 t)cos(gw0t)dt = 0 0

Wo = 27(

[=!._ T

(5) (6) (7)

where f and T represents the frequency (in cycles/time) and period (in seconds) respectively. Also,

k and g are integers.

A periodic function f(t) with a period T should satisfy the following equation

f(t + T) = f(t) (8)

Example 1

Prove that

f sin(kw0 t) = 0 0

for Wo = 27(

[=!._ T

and k is an integer.

Solution

Let T

A= f sin(kw0 t)dt (9) 0

1 = -(- )[cos(kw0 t)]

kw0

A= (2)[cos(kw0 T)- cos(O)]

kw0

= (2)[cos(k27r) -1]

kw0

(10)

=0

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2

0

T

2

Example 2

Prove that

sin 2

0

(k w t ) T

0 2

for w0 2f

f 1 T

and k is an integer.

Solution

Let T

2

Recall

B sin 0

(k w0 t)dt (11)

Thus,

sin 2 ( )

1 c os(2 ) 2

(12)

B 1

1

cos(2kw t)dt

(13)

o 2

1

0

T

1 1 t sin(2kw0 t)

2 2 2kw0 0

T B

1 sin(2kw T ) 0

(14)

0 4k w0

T 1

sin(2k * 2 )

2 4k w

T 2

Example 3

Prove that

sin( gw0 t) c os(k w0 t) 0 0

for w0 2f

f 1 T

and k and g are integers.

Solution

Let T

C sin(gw0 t) c os(k w0 t)dt 0

(15)

Recall that

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Hence,

sin( ) sin( ) cos( ) sin( ) cos( )

T

C sing k w0 t sin(kw0 t) cos(gw0 t)dt 0

T T

sin( g k )w0 t dt sin(kw0 t) cos(gw0 t)dt 0 0

(16)

(17)

(18)

From Equation (1), T

[sin(g k )w0 t]dt 0 0

then

T

C 0 sin(kw0 t) cos(gw0 t)dt 0

(19)

Adding Equations (15), (19),

T T

2C sin(gw0 t) cos(kw0 t)dt sin(kw0 t) cos(gw0 t)dt 0 0

T T

singw0 t (kw0 t)dt sin( g k )w0 t dt 0 0

(20)

2C 0 , since the right side of the above equation is zero (see Equation 1). Thus,

T

C sin(gw0 t) cos(k w0 t)dt 0 o

0

Example 4

Prove that T

sin(k w0 t) sin( gw0 t)dt 0 0

(21)

for w0 2f

f 1 T

k, g integers

Solution

Since

Let

T

D sin(k w0 t) sin(gw0 t)dt 0

(22)

or

Thus,

cos( ) cos( ) cos( ) sin( ) sin( )

sin( ) sin( ) cos( ) cos( ) cos( )

T T

D cos(kw0 t) cos(gw0 t)dt cos(k g )w0 t dt 0 0

(23)

From Equation (1)

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T

T

T

T T


then

f co{(k + g)w0 t]dt = 0 0

D = f cos(kw t) cos(gw t)dt- 0 (24)

0 0

0

Adding Equations (23), (26)

2D = f sin(kw0 t)sin(gw

0 t) + f cos(kw

0 t)cos(gw

0 t)dt

0 0

= f co{kw0 t- gw0 t]dt (25)

0

T

= fco{(k-g)w 0 t]dt

0

2D = 0, since the right side of the above equation is zero (see Equation 1). Thus, T

D = f sin(kw0 t)sin(gw0 t)dt = 0 0

(26)

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UNIT 5: Fourier representation for signals – 2 Teaching hours: 6

Fourier representation for signals – 2: Discrete and continuous Fourier transforms(derivations of

transforms are excluded) and their properties. TEXT BOOK


2002

REFERENCE BOOKS :






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S;,;nalo & Systems lOEC44

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Po,;e 64

l w

UNIT5

Fourier representation for signals -2

5.1 Introduction:

Fourier Representation for four Signal Classes

I Fourier Representation Types I

I I

I Periodic Signals I ! Aperiodic Signals I

I I I I

!Continuous Timt:l

FS

r Discrete Time I ronlinuuus Timt: I[Discrete time J

l DTF< T DTFS

5.2 The Fomier transfonn

5.2.1 From Discrete Fourier Series to Fourier Transform:

Letx I1lJ be a nonperiodic sequence of finite duration. That is, for some positive

integer N,

x(n] = 0

Such a sequence is shown in Fig. l(a). Letx,Jtz] be a periodic sequence formed by

repeatingx Itz]with fundamental period No as shown in Fig. 6-l(b). If we let No-, m, we

have

lim x N [ n] = x[ n] No-+"" o

The discrete Fourier series of xNoltz] is given by

X No( n] - L ckejkfl on

k< No>

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Properties of the Fourier transform

Periodicity

As a consequence of Eq. (6.41), in the discrete-time case we have to consider values of R(radians) only over the range0 < Ω < 2π or π < Ω < π, while in the continuous-time case we

have to consider values of 0 (radians/second) over the entire range –∞ < ω < ∞.

Linearity:

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Signals &Systems lOECH

-

Time Shifting:

Freguency Shifting:

Com ugation:

x"'(n] <->X*( -D)

Time Reversal:

x[ -n] x( -0)

Time Scaling:

Duality:

The duality property of a continuous-time Fourier transform is expressed as

X(t).-.21Tx( -w) There is no discrete-time counterpart of this propety. Howellet', there is a duality between

the discrete-time Fourier transform and the continuous-time Fourier series. Let

x[ n J +-> X(fl)

X(!))= L: x[ n)e -i0n

X(fi+2,.) =X(O)

Since 0 is a continuous va riable, letting n =Iand n = -k

X(t ) = E x(-kJe' 4 '

k- _ ...,

Since X(t) is periodic with period To= 2 1t and the fundamental frequency= 2x!I' 0 = 1 ,

Equation indicates that the Fourier series coefficients of X( t) will be x [ - k] . This duality

relationship is denoted by

where FS denotes the Fourier series and c, are its Fourier coefficients.

CITSTUDENTS.IN



Differentiation in Frequency:

Differencing:

The sequence x[n] -x[n – 1] is called the first difference sequence. Equation is easily obtained

from the linearity property and the time-shifting property .

Accumulation:

Note that accumulation is the discrete-time counterpart of integration. The impulse term on the

right-hand side of Eq. (6.57) reflects the dc or average value that can result from the accumulation.

Convolution:

As in the case of the z-transform, this convolution property plays an important role in the

study of discrete-time LTI systems.

Multiplication:

where @ denotes the periodic convolution defined by

The multiplication property (6.59) is the dual property of Eq. (6.58).

Parseval's Relations:

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1 -X +

J

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UNIT 6: Applications of Fourier representations Teaching hours: 7 Applications of Fourier representations: Introduction, Frequency response of LTI systems, Fourier

transform representation of periodic signals, Fourier transform representation of discrete time

signals.

TEXT BOOK


2002

REFERENCE BOOKS :






CITSTUDENTS.IN



1

-.

I

UNIT 6: Applications of Fourier representations

6.1 Introduction:

• can Ill

• lUI

a simple fonn.

• Fourier representation is to the

Then, with +oo

y(t) = akH(eP<wo)eP<wot

IS neimrlu•

coefficients

Hence in

the tleClUellCYfeSJ)Oll:Se

Example:

fonn

k:::-ro

fimdamental fi·e<Juency 2n,

+3

X

1 . e

+jw

at mo = 2n . We v<"'""'· +3

y(t) = k=-:-1 with - akH(jk2rr),so that

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La.

the respolttse ts

a

a d:!s•cret.e-IJ:me svs·rem 1s

IN

=M

I aky[n- k]

k=O k=O

bkx[n-

To ommn

!)7FT'

y[n- k] .,...... ...,. e N

(e-iw)kY(ejw)

=LN

a.(e-jw)"x(ei"') k=O k=O

• eq11taiH>n as

• tre<lUeJ:lcy respm11se 1s

Differential Equation Descriptions

d +4 +

wheJre. x(t) = (1 + e-t)u(t)

FT. d

=3 +

d d

-y(t) + 4 dty(t) + Sy(t) = 3 dt +x(t)

+ 4(jw) Y(jw) = +

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1 sinceu(t) --- rro(w)+-.

)W

Fr 1

jw+1

=

Aml [(jw)2 + 4(jw) + Y(jw) = (3jw + l)X(jw)

we R= IP=1I Q=6

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we

Readjusting

we know that,

e cos

rr Y(2) = ... 8(w)

J

e sin J

Readjusting the last term,we get

1 =-

5

we

Differential Equation Description

• Find

the differential »m•,.lictn

d2 d d dt

2 y(t) + 3 dty(t) + 2y(t) = 2 dtx(t) +x(t)

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we

Differential Equation Descriptions

2

2

inverse FT <

Diffe1·ence Equation

lm,.or constant coefficient diflference eqtJcatl\)11

y[n] = 1.3433y[n- 1]- 0.9025y[n- 2] + x[n]

- 1.4142x[n- 1] x[n- 2]

Find

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1- 1.3433e

Ex: If the unit impulse response of an LTI :sy,stel l find the response of

the svs:tetn to an input defined

Soln:

, where j'l,a < 1 and a* f!

= h.[n] * Taking DTFT on both sides of the equation, we get

1 1

Y(eJw) 1 1 A

=-----:,-X .

" 1 ue-Jw 1 pe-Jw

Bare constants to

1- ae

=V

A= a

B= _.:

performing partial we get a- {3' a- {3

u[n]

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• Sampling theory

• Reconstruction theory

){

IIIC

-

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L 1

Sampling: Spatial Domain

ure we can see

Hence, Sa:mp:ling 1s multiplication

saulpln:lg 1s detem1im d FT

cmTespm1ds to

\Ve

2rr

n=-OCI

1 +=

Jilt--

n=-= 00

r I n=-=

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Aliasing : an example

cammt be

l

-

WS

tlme

whirhis

reconstructed

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Sampling below the Nyguist rate

X

*

'

- -

Reconstruction below the Nvguist rate

*

-

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FT of sampled signal for different sampling frequency

{b) Spectwun1 of sampled signdt =3W

• •

•

• DTFT !be sa1npled signal is obtailltedfrom ""Vu'1

relationship Q= IS

x[n] -or_"'-X(ew)- w-

• This rhe nule >endellll v•ri•h1!"" unpl:Jes that m=m, cmcre:srxnu'ls to

Q=21t

•

• pu:m1 v o: integeL

• repn:senl x[n]as a

•

• as the impulse ;;a1npled CT

=7+I=-oo

•

8(jw)- X(j(w- kws)

11:=-oo

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as a sa1np.led ve1.·sio>n

we may express y[n]

[n] = x[qn] = x(nqr)

• y]n] is T•- ·r Hence

FT

=

• Hence sul1sti.tuting T,

1

l =

ro.,'= +oo

I

=- q

k=-co

• \Ve hRve as a fiu <'tinn

• . Let us as a proper

k m -=l+-, q q

k l m

q

= J ! oo q-i

=-L

m 0 q-1

m

q

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UNIT 7: Z-Transforms – 1 Teaching hours: 7

Z-Transforms – 1: Introduction, Z – transform, properties of ROC, properties of Z – transforms,

inversion of Z – transforms.

TEXT BOOK


2002

REFERENCE BOOKS :






.

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UNIT 7

Z-Transforms – 1

7.1 Introduction to z-transform:

The z-transform is a transform for sequences. Just like the Laplace transform takes a function

of t and replaces it with another function of an auxiliary variable s. The z-transform takes a sequence

and replaces it with a function of an auxiliary variable, z. The reason for doing this is that it makes

difference equations easier to solve, again, this is very like what happens with the Laplace transform,

where taking the Laplace transform makes it easier to solve differential equations. A difference

equation is an equation which tells you what the k+2th term in a sequence is in terms of the k+1th

and kth terms, for example. Difference equations arise in numerical treatments of differential

equations, in discrete time sampling and when studying systems that are intrinsically discrete, such

as population models in ecology and epidemiology and mathematical modelling of mylinated nerves.

Generalizes the complex sinusoidal representations of DTFT to more

generalized representation using complex exponential signals

• It is the discrete time counterpart of Laplace transform

The z-Plane

• Complex number z = re

jis represented as a location in a complex plane (z-plane)

7.2 The z-transform:

• Let z = re j be a complex number with magnitude r and a

• The signal x[n] = zn is a complex exponential and x[n] = rn cos( n)+ jrn sin( n)

• The real part of x[n] is exponentially damped cosine

• The imaginary part of x[n] is exponentially damped sine

• Apply x[n] to an LTI system with impulse response h[n], Then

y[n] = H{x[n]} = h[n] ∗ x[n]

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You can see that when you do the z-transform it sums up all the sequence, and so the individual

terms affect the dependence on z, but the resulting function is just a function of z, it has no k in it. It

will become clearer later why we might do this.

• This has the form of an eigen relation, where zn is the eigen function and H(z) is the eigen value.

• The action of an LTI system is equivalent to multiplication of the input by the complex number

H(z).

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• If H(z) = IH(z! then the systern output is

= H(z)

_vfnj = IH(

• Revvriting x[ n]

• If \VC contpare and • "\VC see that the systcrn rnodifics

the arnpHtude of the .input by jl /{ refil) I and

shifts the phase by reii2)

DTFT and the z-transform

• Put zin WE'

n

• see that correspm1ds to fl

• inverse must

• can write

1

2n:

The z-transfon:n contd..

•

h[n] H(

)

• VVe ciln convert this equation .inlu an inLegral over L puHing rein= L

• Integration is over !l, V\!e n1ay consider rasa constant

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• w have

jreludn = Jzlln

1 dQ=

• Consider lilnits on integral

- fl varies from -n: to n:

- z traverses a circle of radius r in a countcrdodnvise direction

• vVe can write as knJ = 2 J .f H(z)z" 1 dz

where § i:s integration around the circle of raUJu:s

clock wise direction

• The z transfbnn of any signal is

X(z) = I,

Il=----"""

• The iurerse z-rrtmsfi_)nn of is_

= 1 in a counter

• lnver<>e z-transform t>xpresses

<"XIJotwntiails z"

as a weighted pP!"]JOSition of com-

• The wc:IL'hts are dz

• requires the know'le<lL!e of eom]>lexvariable

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Example 1:

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Properties of Region of Convergence:

• ROC Is related to characteristics of x:n]

• ROC can be .identified frorn and lhnited knovvledge of

• The relationship bet\<veen R()C and characteristic& of the

find invet'se z-transfnnTl

is used to

Property 1

R()C. can not contain any poles

• ROC is the set of all z for vvhich z-transfonn converges

• X(z) n1ust be finite for all z

• If pis a

the pole

then H( = = and z-transforrn does not converge at

• Pole can not lie in the HOC

Property Z

The RC)C for a finite duration includes entire L-plane except L = 0

or/and z= "->(_j

• Let x[nJ be nonzero on the interval n1 : n n2. The z-transform is

r;z

X(zl = L, x[n]z n

n=n;

The RtJC for a finite duration

oJhutd L = <:><?

includes entire z-plane except z = 0

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• Ifa

hence ROC can not ind ttcle z = 0

will have a term z !

• Ifa is non-causal (nt 0) then

powers of z hence can not include z= 06

ROC for a duratiou si!;n<li includes entire

• If nz 0 then the ROC will include z = 0

• If nt 2 0

• shows the only "!''"uwhose ROC is entire z-plane is

where c is a constant

=coin],

Filrlte duration signals

< 2:

tudes

• is

tirne parts

• Let

---- 1

I L iz; 11

n=----QC

I+ = L '-1nJ zl u

n-0

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Properties of Z – transform:

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P2: Titnfl' rmrersaJ

•

z .... I

'

if is of the form a b then the

a 1 b or 1 zl I

with ROC

Then with ROC

Proof: Th:ne reversal

• n

!l

I= n, then

I

P3: Tim" shift

• shift of n0 in th<e time domain co•rnest>onds to of

no in the z-dornain

Then x[n llo] z

with

with ROC z=Oor lzl==

P3: Titne shift. 110 0

z no introduces a of order 110 at z = 0

• can not include z = 0. even if does include z = 0

• If has a zero of at least order n0 at z = 0 that cancels all of the

new then ROC can include z 0

P3: Tbue :shift, 11

0 0

• Ifthese

of

are not canceled

can not include

zeros at in then the ROC

= 0<1-

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Proof: Time sbift

• z n

I= n- !10 , then

P4: Multij)lkatiou by o;£1

If with ROC Rx

Then with ROC

•

• If is a b then the new ROC is

• If contains a pole d, ie. the factor is in the denontlnator

thc•n u J h:J_s a factor in the denorninator and thus a at

•If contains a zero c, then has a Zf'fCJ Ei.t flC

• This indicates that the and zeros of have their radii changed

• Their ang1es are changPd by

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P6: Dilf<>rentialion in the z domain

• n in the lime dmlla'in eorrPSJlOitds to

with resne<':r to zand multiplication of the result -z in the z domain

If ROC

• ROC remains

Proof: Differentiation in the z domain

• We

L.<

wllh -z

= L.- !!=----""""

with

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Example 1

the z l'ransform properties 10 determine

ztransform

• x[n] (i

• zI

1 2

c[n] =

Use the z transform DI'OIJ>ertlc:s to detennine the z transform

• x[n] =

!1 = 4

= (bn] 4

=( 4

4

Example 2

the z transform properties to determine the z transform

• where a is real and +ve

to determine the z transform

• , where a is real and +ve

•

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L

Inverse Z transform:

Three different methods are:

1. Partial fraction method

2. Power series method

3. Long division method

4.

Partial fraction method: .;-jl n c a se:;;oJfTL T1I ·onJmonly encountered form

• If N then use long division method and express in the form

MN

X(z)

"\Vhere B(z) no"\v has the order one less than the denorninator polyno

rnial and usP partial fraction rnethod to find z-transfonn

• The inverse z-transforrn of the tenns in the sunnnation are obtained

frorn the transforn1 and time shift property

1 o[n]

• If ..¥'(z) is expressed a.-.; ratio of polynontiaJs in z instead of z- 1 then

convert into the polynoxnial of z---

• Convert the denoniinator into product of first-order tenns

where dk are the poles of X(z)

For distinct poles

• For all distinct poles. the X(z) can be written as

• Depending on ROC, the irt\lerse z-transfonn associated \Vith each tenn

is then deterntined using the appropriate transfonn pair

• We get

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,'-z- 1

,

w-ith ROC z> UH

11 -_-,A-ck

,-•

-Ax(dk) ll[-n-1] co 1 1

with ROC z <:, dk

• For each tenn the relationship between the RCJC ;:1ssociated vvHh .,¥(z)

and each pole detennines Yvhether the right-sided or left sided inverse

lrcmsfonn is selected

For Repeated poles,

• If pole d, is repeated r tirnes, then there are r tern1:s in the partiaJ

fraclion expansion a:ssucinted with that pule

A;1 A;2 A1,

1 - d,z 1 • ( 1 - d,z ·1 ) 2 • · · · ' ( 1 - d;z 'Y

• Here also, the ROC of X(z) detennines vvhether the right or left sided

inverse Lran funn i:s chosen,

•4 (n+l) ... (n+m-l)

( "[] / A d;!un

1 •

> d; ( n1 - f J 1. ' - ( l- d;z - )m ·

• If the ROC is of the fonn ;z; < d1, the left-sided inverse z-transfonn is

chosen. ie.

-A (n+1) .. (n+m-1)( )n.[-n-

l} z

=--;cA--.-=

(m-1)! d- u '

with ROC (l-

Deciding ROC

• The ROC of X( is the intersection of the ROCs associated vvlth the

individual tcnns in the partial fraction expansion.

• order to chose the correct inverse z-transform. we nu1st infer the

ROC of each wrm from the ROC of

• corn paring the JocaHon of each wllh the ROC of

• sided inverse transform: if the of has the

radius grea1ter than that of the associated \Vith the tenn

• Clros:e the left sided inverse transform: if the ROC of has the

radius less than that of the associated \Vith the tenn

Partial fraction method

•

in arf' real then the t'XJ"'n"''urr coettr

to complex

poles will be complex con-

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• Here we use unique inverse trans

form

•

·If inverse transfmm is

cbosen

• If the S1!511<et is stable. then t is absolutely sllmmable and has DTFT

• circle in the z-plane, ie, = I

• The inverse z-transform is determined the and the

unit circle

• Ifthe is inside the unit circle then the JgJH-:siclled inverse z-transfon11

is chosen

• Ifthe is outside the unit circle then the left-sided inverse z-transform

is chosen

Power series expansion method

• as a oowr·r series in z 1 or z as

tion

• The v"''"t"' of the are coefficient as:so•c <Hect

wHhz"

• limited to one

• Signals with ROCs of the form a or

• If the is Cl, then eXDH)SS as a po\ver series in z ·· 1 and

we

• If is zl as a uctwer series in z we

left

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Long division method:

•

1

2

• is: use rmcthod to write as a nnwr•r series

in z , since ROC indicates that is

• We

2+ l+z + 1

• Compare with z-tnmsfonn

= 2: n

Jt=

• We

+ -l]+B[n-t

3]+--

z

• 'vVe

=-2-8z-

• \Ve can vvTite as

= -2o(n]- +1 - +

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• Find

ROC zexceot

•

• can \¥rite as

• VVe can ".-vrite as

x[n] = { f'W

n > 0 or n is odd

other'.-vise

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NIT 8: Z-Transforms – 2 Teaching hours: 6

Z-transforms – 2: Transform analysis of LTI Systems, unilateral Z Transform and its application to

solve difference equations. TEXT BOOK


2002

REFERENCE BOOKS :






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Page 106


UNIT 8

Z-Transforms – 2

8.1 Transform analysis of LTI systems:

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Page 107

• Find the

to find inverse

l' '

)

wilh 1

3 ROC 1

• \Ve can write as

1

•

'With l

• need to find inverse z,transfonn to find

\Vrite as

2 2 with ROC 1

h n] = 1 + 1

Relation between transfer function and difference equation

• The can from

scription anLTI

• know N M

2 2 k] k=O k=O

• know the function H( z) an eigen the

outpm of

• k in difference equation,

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Page 108

we

• Vve can sollve for H(z)

M

z':L k=O

H(z)

• transfer function described by a difference "' '"'"u'u is a ratio of

is tPrm•"l as a rational "'"n<t<>r function.

• - k] in the

• The coefficient of

•

in the denominator polynomial is the coefficient

- k] in the

us to

a rational

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Page 109

Transfer function:

• The poles and zeros of a rational function

system

into LTI

the numerator

• If and

is the

are zeros ami

• assumes are no zeros at z 0

• The tfh order at z = 0 occurs

• The order zero at z = 0 occurs when

• we can as

where b =

• In the example we had first order at z= 0

• The zeros and gain factor b uniquely determine transfer func-

• The poles are the roots of characteristic equation CIT

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8.2 Unilateral Z- transfonns:

•

• The

time at wtttcll the

1l 0

then times

Advantages

• do not

with initial conditions

Unilateral z-transforn1

is defined as

X(z) n

• IBIS

n20

llj 2

Properties of unilateral Z transform:

• the Pquation desr:ri ption of an LTI

N lcf

L. A']= L. k] k O kdl

• write as

and k

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• sarne are both unilateral and bilateral z

the time shift un)D•ertv

• '""''m'"''tv for unilateral z-tran<>form: -r unllateral z-transfonn of Jc,

=L n=O

z rr = L, r1=0

1] x[n

-1 + L m=O

• unilateral z transform of is

(rn+l

l+zt2:, zm n1=0

1 z l

• oneunit time z 1

constant 1]

is

+' x[-k+

'

k+l-':- z k X(z) for k 0

to

x[n+

for k 0

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" I

8.3 Application to solve difference equations

Solving Differential equations using initial conditions:

• \Ne N-----1 N

=L. L. HJ=0k=m+1

• \Ve have assurned that r"tin] is causal and

-k+ m

• The terrn depend!s on the ;V initial conditions l],y[-

• is zero if all the initial conditions are zero

• The input ami

• to input

--X(zl

• The the

• is the polynomial, the of natural eSJlotJse are roots

tlw of the tracnster function CITSTUDENTS.IN


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Page 113

• The form of natural response cleoenclls

which are the roots of the characteristic ey:m•wJu

First orde1·recursive systmn

•

y[n] 1]

where p 1- 100, and r is the interest rate per period in and

"-"vfnJ is the balance after the or \Vithdra\vt.d of

•

• Assume hank account has em initial balance of $10,000/- aml earrl'> 69<,

Interest month of the

second year. the owner withdraws $100 per month from the account

at the

of each month.

month.

•

- 1] -r.-Z !

) = X(z)

• we

( 1- pz 1 ) +pv[-1]

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Page 114

}

• consists of two terms

one

initial the

• The initial balance of $10,000 at &tart of the first month is initial

l ' !I

• y[n

monttL

account at start n+

• \Ne have p 1+ 1

• Since owner withdraws $100 per month at the start of month 13

(n 11)

• l0011[n 11]. we

• 'We

y ) _ -lOOz 11 L005(10.000)

tz -(1-zl)(l-LOOSzl)+ 1-L005z 1

• After a partial fraction expansion \Ve get

20.OOOz 11 20, OOOz 11 10,050 7

(z)= 1-zl +1-1.005z 1 +1-1.005z 1

• Monthly account balance is obtained by inverse z-transfornting

vVe get

- 11] - 20. 000( !.005 11 - 11]

+ 10. L005)"u[n]

• The last tcrrn HL 050( 1.005 is the natural response '¥vHh the ini-

tial balance

• The account balance

• The natural balance

• The forced response

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