laplacetransform…transfer function

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  • 7/30/2019 LaplaceTransformTransfer Function

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    EECE 301

    Signals & Systems

    Prof. Mark Fowler

    Note Set #29

    C-T Systems: Laplace Transform Transfer Function Reading Assignment: Section 6.5 of Kamen and Heck

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    Ch. 1 Intro

    C-T Signal Model

    Functions on Real Line

    D-T Signal Model

    Functions on Integers

    System Properties

    LTICausal

    Etc

    Ch. 2 Diff EqsC-T System Model

    Differential Equations

    D-T Signal ModelDifference Equations

    Zero-State Response

    Zero-Input Response

    Characteristic Eq.

    Ch. 2 Convolution

    C-T System Model

    Convolution Integral

    D-T Signal Model

    Convolution Sum

    Ch. 3: CT FourierSignalModels

    Fourier Series

    Periodic Signals

    Fourier Transform (CTFT)

    Non-Periodic Signals

    New System Model

    New Signal

    Models

    Ch. 5: CT FourierSystem Models

    Frequency Response

    Based on Fourier Transform

    New System Model

    Ch. 4: DT Fourier

    SignalModels

    DTFT

    (for Hand Analysis)DFT & FFT

    (for Computer Analysis)

    New Signal

    Model

    Powerful

    Analysis Tool

    Ch. 6 & 8: LaplaceModels for CT

    Signals & Systems

    Transfer Function

    New System Model

    Ch. 7: Z Trans.

    Models for DT

    Signals & Systems

    Transfer Function

    New System

    Model

    Ch. 5: DT Fourier

    System Models

    Freq. Response for DT

    Based on DTFT

    New System Model

    Course Flow DiagramThe arrows here show conceptual flow between ideas. Note the parallel structure between

    the pink blocks (C-T Freq. Analysis) and the blue blocks (D-T Freq. Analysis).

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    6.5 Transfer FunctionWeve seen that the system outputs LT is:

    )()(

    )(

    )(

    )()( sX

    sA

    sB

    sA

    sCsY +=

    Part due to ICs

    So, if the system is in zero-state then we only get the second term:

    )()(

    )()( sXsA

    sBsY =

    )()()( sXsHsY =

    Define:)(

    )()(

    sA

    sBsH

    =

    System effect in zero-state case is completely set by the transfer function

    Part due to Input

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    Note: If the system is described by a linear, constant coefficient differential

    equation, we can getH(s) by inspection!! Lets see how

    )()()()()( 01012

    sXbssXbsYassYasYs +=++

    )()(

    )()()(

    01012

    2

    txbdt

    tdxbtya

    dt

    tdya

    dt

    tyd+=++

    To illustrateTake the LT of a Diff. Eq. under the zero-state case:

    Solve for Y(s) and identify theH(s):)()(

    )(01

    2

    01 sXasas

    bsbsY

    sH44 344 21

    =

    ++

    +=

    With zero ICs we have that each higher derivative

    corresponds to just another power ofs.

    We can then apply this idea to get the Transfer Function

    )()0()0()()( )1(21)( txxsxssXstx nnnnn L&Recall:

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    )()()()()( 01012

    2

    txbdt

    tdxbtyadt

    tdyadt

    tyd +=++

    01

    2

    01)(

    asas

    bsbsH

    ++

    +=

    So now it is possible to directly identify the TFH(s) from the Diff. Eq.:

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    But, we have also seen that for the zero-state case the system output is:

    )(*)()()()(0

    txthdtxhtyt

    ==

    ButWe have an LT property for convolution that says:

    )(*)()()()()( txthtysXsHsY ==

    where: { })()( thsH L=

    { }ResponseImpulseFunctionTransfer L=

    So, we have two ways to getH(s)

    - Inspect the Diff. Eq. and identify the transfer functionH(s)

    - Take the LT of the impulse response h(t)

    This gives an easy way to get the impulse response from a Diff. Eq.:- Identify the H(s) from he Diff. Eq. and then find the ILT of that

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    Recall: If the ROC ofH(s) includes thejaxis, then

    js

    sHH=

    = )()(

    This is the connection between

    Note that the transfer function does essentially the same thing that the frequency

    response does

    )(*)()()()()( txthtysXsHsY ==

    )(*)()()()()( txthtyXHY ==

    Recall that the LT is a generalized, more-powerful version of the FT this

    result just says that we can do the same thing withH(s) that we did withH(),

    but we can do it for a larger class of systems

    There are some systems for which we can use either method those are the

    ones for which the ROC ofH(s) includes thejaxis.

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    Poles and Zeros of a system

    Given a system with Transfer Function:

    01

    1

    1

    011

    1

    ...

    ...)(

    asasas

    bsbsbsbsH

    N

    N

    N

    MM

    MM

    ++++++++=

    )(sB

    )(sA

    We can factorB(s) andA(s): (Recall:A(s) = characteristic polynomial)

    ))...()((

    ))...()(()(

    21

    21

    N

    MM

    pspsps

    zszszsbsH

    =

    Assume any common factors inB(s) andA(s) have been cancelled out

    So we know thatH(s) is completely described by the Diff. Eq. Therefore

    we should expect that we can tell a lot about a system by looking at the

    structure of the transfer functionH(s) This structure is captured in the idea

    of Poles and Zeros

    Note: MisHizs

    ...,,2,10)( ===

    NisHips

    ...,,2,1)( ===

    { } )"(ofpoles"calledare sHpi

    Note: pi are the roots of the char. polynomial

    { } )"(ofzeros"calledare sHzi

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    Note that knowing the sets { } { }Nii

    M

    iipz

    11&

    ==

    tells us whatH(s) is: (up to the multiplicative scale factor bM)-bMis like a gain (i.e. amplification)

    Pole-Zero Plot

    This gives us a graphical view of the systems behavior

    jplane-s

    1

    1

    Pole-Zero Plot for thisH(s)

    )1)(1)(4(

    )3)(3(2

    8106

    20122)(

    23

    2

    jsjss

    jsjs

    sss

    sssH

    ++++

    +++=

    +++

    ++=

    Real coefficients complex conjugate pairs

    Example:

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    From the Pole-Zero Plots we can Visualize the TF function on the s-plane:

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    As the pole moves closer to thej

    axis it has a stronger effect on the

    frequency responseH(). Poles

    close to thejaxis will yield

    sharper and taller bumps in the

    frequency response.

    By being able to visualize what |H(s)|will look like based on where the poles

    and zeros are, an engineer gains the

    ability to know what kind of transfer

    function is needed to achieve a desiredfrequency response then through

    accumulated knowledge of electronic

    circuits (requires experience

    accumulated AFTER graduation) the

    engineer can devise a circuit that will

    achieve the desired effect.

    Can also look at a pole-zero plot and see the effects on Freq. Resp.

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    Differential

    Equation

    Differential

    Equation

    Transfer

    Function

    Transfer

    Function

    FrequencyResponse

    FrequencyResponse

    Impulse

    Response

    ImpulseResponse

    Fourier Transform

    Laplace Transform

    Inspection in Practice

    Time Domain Freq Domain

    Continuous-Time System Relationships

    h(t)

    011

    1

    0111)(asasas

    bsbsbsbsHN

    NN

    MMMM

    ++++++++=

    L

    L

    jssHH == |)()(

    s = j

    )()()()(

    )()()()(

    011

    1

    1

    011

    1

    1

    txbdt

    tdxb

    dt

    txdb

    dt

    txdb

    tyadt

    tdya

    dt

    tyda

    dt

    tyd

    M

    M

    MM

    M

    M

    N

    N

    NN

    N

    ++++

    =++++

    L

    L

    Char.

    Modes

    LT in Theory

    Pole-Zero

    Diagram

    Pole-Zero

    Diagram

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    1. From the differential equation you can get:

    a. Transfer function, then the impulse response, the pole-zeroplot, and if allowable you can get the frequency response

    2. From the impulse response you can get:

    a. Transfer function, then the Diff. Eq., the pole-zero plot, andif allowable you can get the frequency response

    3. From the Transfer Function you can get:

    a. Diff. Eq., the impulse response, the pole-zero plot, and ifallowable you can get the frequency response

    4. From the Frequency Response you can get:

    a. Transfer function, then the Diff. Eq., the pole-zero plot, andthe impulse response

    5. From the Pole-Zero Plot you can get:

    a. (up to a scaling factor) Transfer function, then the Diff. Eq.,the impulse response, and possible the Frequency Response

    In practice you may need to start your work in any spot on this diagram

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    Similarly, we can get the transfer function using the s-domain impedances:

    sLsZsC

    sZRsZ LCR === )(1

    )()(

    Recall: We get the frequency response from a circuit by using frequency

    dependent impedances

    LjZCj

    ZRZ LCR

    === )(1

    )()(

    and then doing circuit analysis.

    In Practice we often get the transfer function from a circuit Heres an example:

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    Example 6.37 (Assume zero-state)

    )(tx )(ty

    +

    )(sY

    +

    )(sX

    Now we treat everything here as if we have a

    DC Circuit which leads to simple algebraic

    manipulation rather than differential equations!

    For this circuit the easiest approach is to use the Voltage Divider

    )()/1(

    /1

    )( sXCsLsR

    Cs

    sY

    ++=

    )()/1()/(

    /1

    2 sXLCsLRs

    LC

    ++=

    )(sH

    A standard form:

    a ratio of two polynomials in s

    unity coefficient on the highest power in the denominator

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    Once you start including linear amplifiers with gain > 1 this may not be true

    If you include non-linear devices the system becomes non-linear

    But, you may be able to linearize the system over a small operating range

    E.g. A transistor can be used to build a (nearly) linear amplifier even though

    the transistor is itself a non-linear device

    Some comments: