signal very important

7/30/2019 Signal Very Important

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Basics in Systems and Circuits Theory

Michael E.Auer 01.11.2011 BSC04

Fourier Series Fourier Transform

Laplace Transform

Applications of Laplace Transform Z-Transform

BSC Modul 4: Advanced Circuit Analysis


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Fourier Series Fourier Transform Laplace Transform




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The Fourier series of a periodic functionf(t) is arepresentation that resolves f(t) into a dc component andan ac component comprising an infinite series of harmonicsinusoids.

Given a periodic function f(t) = f(t+nT)where n is aninteger and T is the period of the function.

where 0=2/T is called the fundamental frequency inradians per second.

ac

n

n

dc

tnbtnaatf

=

++=1

0000 )sincos()(

Trigonometric Fourier Series (1)


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)2()(and

21,0

10,1)( +=


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Three types of symmetry

1.Even Symmetry : a function f ( t ) if its plot issymmetrical about the vertical axis.

In this case,

)()( tftf =

0

)cos()(

4

)(2

2/

0 0

2/

00

==

=

n

T

n

T

b

dttntfTa

dttfT

a

Typical examples of even periodic function

Symmetry Considerations (1)


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2.Odd Symmetry : a function f ( t ) if its plot isanti-symmetrical about the vertical axis.

In this case,

)()( tftf =

=

=

2/

00

0

)sin()(4

0

T

n dttntf

T

b

a

Typical examples of odd periodic function



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3.Half-wave Symmetry : a function f ( t ) if

)()

2

( tfT

tf =

=

=

=

evenanfor,0

oddnfor,)sin()(4

evenanfor,0

oddnfor,)cos()(4

0

2/

00

2/

00

0

T

n

T

n

dttntfTb

dttntfTa

a

Typical examples of half-wave odd periodic functions



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Example 1

Find the Fourier series expansion of f(t) givenbelow.

=

=

1 2sin

2cos1

12)(

n

tnn

ntf

Ans:



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Example 2

Determine the Fourier series for the half-wavecosine function as shown below.

=

==1

2212,cos

14

2

1)(

k

knntn

tf

Ans:



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Steps for Applying Fourier Series

1.Express the excitation as a Fourier series.

2.Transform the circuit from the time domain to the

frequency domain.3.Find the response of the dc and ac components in

the Fourier series.

4.Add the individual dc and ac responses using the

superposition principle.

Circuit Applications (1)


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Example

Find the response v0( t ) of the circuit below whenthe voltage source vs( t ) is given by

( ) 12,sin12

2

1

)(1 =+=

=kntnntv n

s



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Amplitude spectrum of the output

voltage

Solution

Phasor of the circuit

For dc component, (n=0 or n=0), Vs = => Vo = 0

For nth harmonic,

In time domain,

s0 V25

2V

nj

nj

+=

)5

2tan(c425

4)(1

1

220

=

+

=k

ntnosn

tv

s22

1

0 V425

5/2tan4V,90

2V

n

n

nS

+

==



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Given:

The average power is

The rms value is

=

=+=+=

1

Im0mdc

1

0ndc )cos(II)(and)cos(VV)(mn

Vn tmtitntv

)(1

222

0

= ++= nnnrms baaF

= += 1nndcdc )cos(IV2

1

IVPn

nn

Average Power and RMS Values (1)


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Example

Determine the average power supplied to thecircuit shown below ifi( t )= 2+ 10cos(t + 10)+ 6cos(3t+ 35) A

Answer: 41.5W

Average Power and RMS Values (2)


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The exponential Fourier series of a periodic function f ( t ) describes the spectrum off ( t ) in terms of the amplitude andphase angle of ac components at positive and negativeharmonic.

The plots of magnitude and phase ofcnversus n0 are called

the complex amplitude spectrum and complex phasespectrum off ( t ) respectively.

==T tjn

n TdtetfT

c0

0 /2where,)(1 0

==

n

tjnn

oectf )(

Exponential Fourier Series (1)

B i i S d Ci i Th


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The complex frequency spectrum of the function

f(t)=et, 0


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Filter are an important component of electronics and communications

system.

This filtering process cannot be accomplished without the Fourier seriesexpansion of the input signal.

For example,

(a) Input and output spectra of a lowpass filter, (b) the lowpass filter passes

only the dc component when c


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(a) Input and output spectra of a bandpass filter, (b) the bandpass filter

passes only the dc component when


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Fourier Series

Fourier Transform Laplace Transform





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It is an integral transformation of f(t) from the time domainto the frequency domain F()

F() is a complex function; its magnitude is called theamplitude spectrum, while its phase is called the phase

spectrum.Given a function f(t), its Fourier transform denoted by F(),is defined by

= )()( dtetfF

tj

Definition of Fourier Transform (1)



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Example 1:

Determine the Fourier transform of a singlerectangular pulse of wide and height A, as shownbelow.


2sin

2

2

2/

2/

)(

2/2/

2/

2/

cA

j

eeA

ej

A

dtAeF

jj

tj

tj

=

=

=

=

Solution:

Amplitude spectrum of the

rectangular pulse



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Example 2:

Obtain the Fourier transform of theswitched-on exponential function asshown.


Solution:

ja

dte

dteedtetfF

etuetf

tja

tjjattj

at

at

+=

=

==

==

+

1

)()(

Hence,

0t,0

0t,)()(

)(



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[ ] )()()()( 22112211 FaFatfatfaF +=+

Linearity:

IfF1() and F2() are, respectively, the FourierTransforms off1(t) and f2(t)

Example:

[ ] ( ) ( )[ ] [ ])()(21

)sin( 00000

+==

jeFeFjtF

tjtj

Properties of Fourier Transform (1)



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[ ] constantais,)(1

)( aa

Fa

atfF

=

Time Scaling:

IfF() is the Fourier Transforms off(t), then

If |a|>1, frequency compression, or time expansion

If |a|


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y y


[ ] )()(0

0

FettfFtj

=

Time Shifting:

IfF() is the Fourier Transforms off( t ) , then

Example:

[ ]

jetueF

j

t

+=

1

)2(

2

)2(




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y y


)()( 00

= FetfF tj

Frequency Shifting (Amplitude Modulation):IfF() is the Fourier Transforms off( t ) , then

Example:

[ ] )(21

)(2

1)cos()( 000 ++= FFttfF




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)()( sFjtudt

df

F =

Time Differentiation:IfF() is the Fourier Transforms off( t ) , then theFourier Transform of its derivative is

Example:

( ) jatuedtd

Fat

+=

1)(




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)()0(

)(

)(

Fj

F

dttfF

t

=

Time Integration:IfF() is the Fourier Transforms off( t ) , then theFourier Transform of its integral is

Example:

[ ] )(1

)( +=

jtuF




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Fourier transforms can be applied to circuits with non-sinusoidalexcitation in exactly the same way as phasor techniques beingapplied to circuits with sinusoidal excitations.

By transforming the functions for the circuit elements into thefrequency domain and take the Fourier transforms of theexcitations, conventional circuit analysis techniques could beapplied to determine unknown response in frequency domain.

Finally, apply the inverse Fourier transform to obtain theresponse in the time domain.

Y() = H()X()

Circuit Application (1)



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Example:

Find v0( t ) in the circuit shown below for

vi( t ) =2e -3tu(t )

Circuit Application (2)

Solution:

)()(4.0)(givesansformFourier trinversetheTaking

)5.0)(3(

1)(V

Hence,

21

1

)(V

)(V)(iscircuittheoffunctiontransferThe

3

2)(VissignalinputtheofansformFourier trThe

35.0

0

0

i

0

i

tueetv

jj

jH

j

tt =

++

=

+==

+

=



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Laplace Transform





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It is an integral transformation of f(t) from the timedomain to the complex frequency domain F(s)

Given a function f(t), its Laplace transform denoted byF(s), is defined by

Where the parameter s is a complex number

[ ] ==

0)()()( dtetftfLsF st

Definition of Laplace Transform

js += , real numbers



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When one says "the Laplace transform" without qualification,the unilateral or one-sided transform is normally intended.The Laplace transform can be alternatively defined as thebilateral Laplace transformor two-sided Laplace transform byextending the limits of integration to be the entire real axis. If

that is done the common unilateral transform simply becomesa special case of the bilateral transform.

The bilateral Laplace transform is defined as follows:

Bilateral Laplace Transform

[ ]

== )()()( dtetftfLsF st

http://en.wikipedia.org/wiki/Two-sided_Laplace_transformhttp://en.wikipedia.org/wiki/Two-sided_Laplace_transform


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Determine the Laplace transform of each of the following functions shown:

Examples of Laplace Transforms (1)

a) The Laplace Transform of unit step, u(t) is given by

[ ]

=== 01

1)()( sdtesFtuLst



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b) The Laplace Transform of exponential function, e-atu(t), a>0is given by

[ ]

+===

0

1)()(

sdteesFtuL

stt


c) The Laplace Transform of impulse function, (t) is given by

[ ]

===0

1)()()( dtetsFtuL st



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ssF

1)( =

+=

ssF

1)( 1)( =sF



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Table of Selected Laplace Transforms (1)



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[ ] )()()()( 22112211 sFasFatfatfaL +=+

Linearity:IfF1(s) and F2(s) are, respectively, the LaplaceTransforms off1( t ) and f2( t )

Example:

[ ] ( )22

)(

2

1)()cos(

+

=

+=

s

stueeLtutL

tjtj

Properties of Laplace Transform (1)



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[ ] )(1)(a

sFa

atfL =

Scaling:

IfF(s) is the Laplace Transforms off( t ) , then

Example:

[ ] 224

2)()2sin(

+

=s

tutL




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[ ] )()()( sFeatuatfL as=

Time Shift:

IfF(s) is the Laplace Transforms off( t ) , then

Example:

[ ] 22)())(cos(

+=

sseatuatL as




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)()()( asFtutfeL at +=

Frequency Shift:

IfF(s) is the Laplace Transforms off(t), then

Example: [ ]22)(

)()cos(

++

+=

as

astuteL

at




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)0()()( =

fssFtu

dt

dfL

Time Differentiation:

IfF(s) is the Laplace Transforms off(t), then theLaplace Transform of its derivative is


Time Integration:

IfF(s) is the Laplace Transforms off( t ) , then theLaplace Transform of its integral is

)(1

)(0

sFs

dttfLt

=



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)(lim)0( ssFfs

=

Initial and Final Values:

The initial-value and final-value properties allow usto find f(0) and f() off(t) directly from its Laplacetransform F(s).

Initial-value theorem

)(lim)(0

ssFfs= Final-value theorem




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In principle we could recover f(t) from F(s) via

But, this formula isnt really useful.

sesFj

tf

j

j

std)(

2

1)(

+

=

The Inverse Laplace Transform (1)

= xF

1



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Suppose F(s) has the general form of

The finding the inverse Laplace transform ofF(s) involves two steps:

1.Decompose F(s) into simple terms using partialfraction expansion.

2.Find the inverse of each term by matching entriesin Laplace Transform Table.

)(

)()(

sD

sNsF =


numerator polynomial

denominator polynomial



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ExampleFind the inverse Laplace transform of

Solution:4

6

1

53)(

2

+

+

+

=

sss

sF

0t),()2sin(353(

4

6

1

53)(

2

111

+=

++

+

=

tute

sL

sL

sLtf

t




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The Laplace transform is useful in solving linearintegro-differential equations.

Each term in the integro-differential equation istransformed into s-domain.

Initial conditions are automatically taken intoaccount.

The resulting algebraic equation in the s-domain canthen be solved easily.

The solution is then converted back to time domain.

Application to Integro-differential Equations (1)



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Example:

Use the Laplace transform to solve the differentialequation

Given: v(0) = 1; v(0) = -2

)(2)(8)(6)(2

2

tutvdt

tdvdt

tvd =++




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Solution:Taking the Laplace transform of each term in the givendifferential equation and obtain

[ ] [ ]

)()21(4

1)(

Transform,LaplaceinverseBy the

42)(

2424)()86(

havewe,2)0(';1)0(ngSubstituti

2

)(8)0()(6)0(')0()(

42

41

21

412

2

2

tueetv

ssssV

s

ss

sssVss

vv

ssVvssVvsvsVs

tt ++=

++

++=

++=++=++

===++




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Laplace Transform





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Steps in Applying the Laplace Transform:1.Transform the circuit from the time domain to

the s-domain

2.Solve the circuit using nodal analysis, mesh

analysis, source transformation, superposition, orany circuit analysis technique with which we arefamiliar

3.Take the inverse transform of the solution and

thus obtain the solution in the time domain.

Circuit Element Models (1)



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Assume zero initial condition forthe inductor and capacitor,

Resistor : V(s)=RI(s)

Inductor: V(s)=sLI(s)

Capacitor: V(s) = I(s)/sCThe impedance in the s-domain

is defined as Z(s) = V(s)/I(s)

The admittance in the s-domainis defined as Y(s) = I(s)/V(s)

Time-domain and s-domain representations of passive

elements under zero initial conditions.




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Non-zero initial condition for the

inductor and capacitor,

Resistor : V(s)=RI(s)

Inductor: V(s)=sLI(s) + LI(0)

Capacitor: V(s) = I(s)/sC + v(0)/s


I d E l


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Introductory Example

Charging of a capacitor v

V(0) = 0


Ci i El M d l E l (1)


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Example 1:

Find v0(t) in the circuit shown below, assuming zero initialconditions.

Circuit Element Models Examples (1)


Ci it El t M d l E l (2)


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Solution:

Transform the circuit from the time domain to the s-domain:

s

L

tu

3sC1F

31

ssH1

s

1)(

=

=

Apply mesh analysis, on solving for V0(s):

220 )2()4(

2

2

3)(V ++= ss 0V,)2sin(2

3)( 40 =

ttetv t

Inverse transform





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Example 2:

Determine v0(t) in the circuit shown below, assuming zeroinitial conditions.

V)()21(8:Ans 22 tutee tt





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Example 3:

Find v0(t) in the circuit shown below. Assume v0(0)=5V.

V)()1510()(v:Ans 20 tueettt +=





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Example 4:

The switch shown below has been in position bfor a longtime. It is moved to position aat t=0. Determine v(t) for t > 0.

where0,t,I)IV()(v:Ans 0/

00 RCReRtt =>+=



Circuit Analysis


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Circuit analysis is relatively easy to do in the s-domain.By transforming a complicated set of mathematicalrelationships in the time domain into the s-domain where weconvert operators (derivatives and integrals) into simple

multipliers of s and 1/s.This allow us to use algebra to set up and solve the circuitequations.

In this case, all the circuit theorems and relationships

developed for dc circuits are perfectly valid in the s-domain.

Circuit Analysis


Circuit Analysis Example (1)


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Example:

Consider the circuit below. Find thevalue of the voltage across thecapacitor assuming that the value

of vs(t)=10u(t) V and assume thatat t=0, -1A flows through theinductor and +5V is across thecapacitor.



Ci it A l i E l (2)


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Solution:Transform the circuit from time-domain (a) into s-domain (b) using LaplaceTransform. On rearranging the terms, we have

By taking the inverse transform, we get

2

30

1

35V1

+

+

=ss

V)()3035()(v 21 tueettt =





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Example:

The initial energy in the circuit below is zero at t=0. Assume that vs=5u(t) V.(a) Find V0(s) using the Thevenin theorem. (b) Apply the initial- and final-value theorem to find v0(0) an v0(). (c) Obtain v0(t).

Ans: (a) V0(s) = 4(s+0.25)/(s(s+0.3)) (b) 4,3.333V, (c) (3.333+0.6667e-0.3t)u(t) V.



Transfer Functions


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The transfer function H(s) is the ratio of the output responseY(s) to the input response X(s), assuming all the initialconditions are zero.

h ( t ) i s t h e i m p u l se r e sp o n se f u n c t i on .

Four types of gain:

1. H(s) = voltage gain = V0(s)/Vi(s)

2. H(s) = Current gain = I0(s)/Ii(s)

3. H(s) = Impedance = V(s)/I(s)

4. H(s) = Admittance = I(s)/V(s)

)(

)()(

sX

sYsH =

Transfer Functions


Transfer Functions Example (1)


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Example:The output of a linear system is y(t)=10e-tcos4twhen the input is x(t)=e-tu(t).

Find the transfer function of the system and its impulse response.

Solution:

Transform y(t) and x(t) into s-domain and apply H(s)=Y(s)/X(s), we get

Apply inverse transform for H(s), we get

16)1(

44010

16)1(

)1(10

)(

)()(

22

2

++=

++

+==

ss

s

sX

sYsH

)()4sin(40)(10)( tutetth t=

Transfer Functions Example (1)


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Laplace Transform Applications of Laplace Transform

Z-Transform



Introduction


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In continuous systems Laplace transforms play a unique role. They

allow system and circuit designers to analyze systems and predictperformance, and to think in different terms - like frequencyresponses - to help understand linear continuous systems.

Z-transforms play the role in sampled systems that Laplacetransforms play in continuous systems.

In continuous systems, inputs and outputs are related by differentialequations and Laplace transform techniques are used to solvethose differential equations.

In sampled systems, inputs and outputs are related by difference

equations and Z-transform techniques are used to solve thosedifferential equations.


Fourier, Laplace and Z-Transforms


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, p

For right-sided signals (zero-valued for negative time index) theLaplace transform is a generalization of the Fourier transform ofa continuous-time signal, and the z-transform is a generalizationof the Fourier transform of a discrete-time signal.

Fourier Transform

Laplace Transform Z-Transform

generalization

continuous-time signals discrete-time signals



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The Z-transform converts a discretetime-domain signal, which isa sequence of real or complex numbers, into a complexfrequency-domain representation.It can be considered as a discrete-time equivalent of the Laplacetransform.

There are numerous sampled systems that look like the one shown below.

discrete-time signals


Definition of the Z-Transform
http://en.wikipedia.org/wiki/Discrete_mathematicshttp://en.wikipedia.org/wiki/Time-domainhttp://en.wikipedia.org/wiki/Sequencehttp://en.wikipedia.org/wiki/Real_numberhttp://en.wikipedia.org/wiki/Complex_numberhttp://en.wikipedia.org/wiki/Frequency-domainhttp://en.wikipedia.org/wiki/Laplace_transformhttp://en.wikipedia.org/wiki/Laplace_transformhttp://en.wikipedia.org/wiki/Laplace_transformhttp://en.wikipedia.org/wiki/Laplace_transformhttp://en.wikipedia.org/wiki/Frequency-domainhttp://en.wikipedia.org/wiki/Frequency-domainhttp://en.wikipedia.org/wiki/Frequency-domainhttp://en.wikipedia.org/wiki/Complex_numberhttp://en.wikipedia.org/wiki/Real_numberhttp://en.wikipedia.org/wiki/Sequencehttp://en.wikipedia.org/wiki/Time-domainhttp://en.wikipedia.org/wiki/Time-domainhttp://en.wikipedia.org/wiki/Time-domainhttp://en.wikipedia.org/wiki/Discrete_mathematics


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Let us assume that we have a sequence, yk.

The subscript "k" indicates a sampled time interval and that ykis thevalue of y(t) at the kth sample instant.

ykcould be generated from a sample of a time function.For example: yk= y(kT), where y(t) is a continuous time function, and T is thesampling interval.

We will focus on the index variable k, rather than the exact time kT, in all that wedo in the following.

[ ] kk

kk zyyZ

==

0


Z-Transform Example


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k

kayy = 0

p

Given the following sampled signal:

[ ]az

z

z

az

azaaZ

k

kk

kkk

=

=

==

=

=

1

11

00

We get the Z-Transform for y0 = 1


Z-Transform of Unit Impulse and Unit Step


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Given the following sampled signal Dk:

Dkis zero for k>0, so all those terms are zero.Dkis one for k = 0, so that

[ ] 1=k

DZ

[ ] 1...1 321

=+++=

z

zzzzuZ

k

Given the following sampled signal uk:

ukis one for all k.


More Complex Example of Z-Transform


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Given the following sampled signal fk:

( ) )sin(bkTekTff akTk

==

[ ]k

k

akT

k

k

kk zbkTezffZ

=

=

==

)sin(00

Finally:

[ ]

+

=*

2

1

cz

z

cz

z

j

fZ k jbTaTec+=where


S- and Z-Plane Presentation


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Inverse Z-Transform


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The inverse z-transform can be obtained using one of two methods:

a) the inspection method,b) the partial fraction method.

In the inspection method each simple term of a polynomial in z,H(z), is substituted by its time-domain equivalent.

For the more complicated functions of z, the partial fraction methodis used to describe the polynomial in terms of simpler terms, andthen each simple term is substituted by its time-domain equivalentterm.

signal very important

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