laplacetransform…transfer function
TRANSCRIPT
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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: