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7/31/2019 Convolution Graphical
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EECE 301
Signals & Systems
Prof. Mark Fowler
Note Set #11
C-T Systems: Computing Convolution Reading Assignment: Section 2.6 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 System 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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)()(
)()()]()([
tvtx
tvtxtvtx
dt
d
=
=
C-T convolution properties
Many of these are the same as for DT convolution.
We only discuss the new ones here.
See the next slide for the others
derivative
=
=
ttt
dhtxthdxdy )()()()()(
The properties of convolution help perform analysis and design tasks that
involve convolution. For example, the associative property says that (in
theory) we can interchange to order of two linear systems in practice,
before we can switch the order we need to check what impact that mighthave on the physical interface conditions.
1. Derivative Property:
2. Integration Property Let y(t) =x(t)*h(t), then
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Convolution Properties
These are things you can exploit to make it easier to solve convolution problems
1.Commutativity
You can choose which signal to flip)()()()( txththtx =
2. Associativity
Can change order sometimes one order is easier than another
)())()(())()(()( twtvtxtwtvtx =
3. Distributivity
may be easier to split complicated system h[n] into sum of simple ones
we can split complicated input into sum of simple ones
(nothing more than linearity)
OR
4. Convolution with impulses
)()()( = txttx
)()()()())()(()( 2121 thtxthtxththtx +=+
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Computing CT Convolution
-For D-T systems, convolution is something we do for analysis and forimplementation (either via H/W or S/W).
-For C-T systems, we do convolution for analysis
nature does convolution for implementation.
If we are analyzing a given system (e.g., a circuit) we may need to compute a
convolution to determine how it behaves in response to various different input
signals
If we are designing a system (e.g., a circuit) we may need to be able to visualize
how convolution works in order to choose the correct type of system impulse
response to make the system work the way we want it to.
Well learn how to perform Graphical Convolution, which is nothing more
than steps that help you use graphical insight to evaluate the convolution integral.
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Steps for Graphical Convolutionx(t)*h(t)
1. Re-Write the signals as functions of : x() and h()
2. Flipjust one of the signals around t= 0 to get eitherx(-) or h(-)a. It is usually best to flip the signal with shorter duration
b. For notational purposes here: well flip h() to get h(-)
3. Find Edges of the flipped signal
a. Find the left-hand-edge -value ofh(-): call it L,0
b. Find the right-hand-edge -value ofh(-): call it R,0
4. Shift h(-) by an arbitrary value oftto get h(t- ) and get its edgesa. Find the left-hand-edge -value ofh(t- ) as a function of t: call it L,t
Important: It will always beL,t
= t +L,0
b. Find the right-hand-edge -value ofh(t- ) as a function of t: call it R,t Important: It will always be R,t = t + R,0
Note: I use for
what the book
uses ... It is not
a big deal as they
are just dummy
variables!!!
dthxty )()()( =
Note: If the signal you flipped is NOT finite duration,one or both ofL,tand R,twill be infinite (L,t= and/or R,t= )
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5. Find Regions of -Overlap
a. What you are trying to do here is find intervals oftover which the product
x() h(t-
) has a single mathematical form in terms of
b. In each region find: Interval oftthat makes the identified overlap happen
c. Working examples is the best way to learn how this is done
Tips: Regions should be contiguous with no gaps!!!
Dont worry about < vs. etc.
6. For Each Region: Form the Productx( )h(t - ) and Integrate
a. Form productx() h(t- )
b. Find the Limits of Integration by finding the interval of over which theproduct is nonzeroi. Found by seeing where the edges ofx() and h(t- ) lieii. Recall that the edges ofh(t- ) are L,tand R,t, which often depend on
the value oft
So the limits of integration may depend on t
c. Integrate the productx() h(t- ) over the limits found in 6bi. The result is generally a function oft, but is only valid for the interval
of t found for the current region
ii. Think of the result as a time-section of the outputy(t)
Steps Continued
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7. Assemble the output from the output time-sections for all the regions
a. Note: you do NOT add the sections togetherb. You define the output piecewise
c. Finally, if possible, look for a way to write the output in a simpler form
Steps Continued
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x(t)
t2
2
h(t)
t1
3
Example: Graphically Convolve Two Signals
=
=
dthx
dtxhty
)()(
)()()(By Properties of
Convolution
these two forms are
Equal
This is why we can
flip either signal
By Properties of
Convolution
these two forms areEqual
This is why we can
flip either signalConvolve these two signals:
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x()
2
2
h(-)3
Step #2: Fliph( ) to geth(- )
1
x()
2
2
h()
1
3
Step #1: Write as Function of
Usually Easier
to Flip theShorter Signal
0
0
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x()
2
2
h(-)3
Step #3: Find Edges of Flipped Signal
0
1 0
R,0 = 0L,0 = 1
i i S #4 S if h( ) &
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Motivating Step #4: Shift byt to geth(t- ) & Its Edges
L,t = t + L,0
R,t =t + R,0
h(t-) =h(-2-)
3
3
Fort = -2Fort = -2
2
R,t = t + R,0
R,t = t + 0
R ,-2 = 2+0
L,t= t + L,0
L,t= t 1
L,-2 = -2 1
Just looking at 2 arbitrary tvalues
In Each Case We Get
3
1
h(t-) =h(2-)
Fort = 2Fort = 2
2
R,t = t + R,0
R,t = t + 0
R,2 = 2+0
L,t= t + L,0
L,t= t 1
L,2 = 2 1
D i St #4 Shift b t t h( ) & It Ed
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Doing Step #4: Shift byt to geth(t- ) & Its Edges
3
t 1
h(t )
ForArbitrary Shift bytForArbitrary Shift byt
t
R,t = t + R,0
R,t = t + 0
L,t= t + L,0
L,t= t 1
Step #5: Find Regions of Overlap
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h(t-)3
L,t=t -1
R,t=t
Step #5: Find Regions of -Overlap
x()
2
2
Region I
No -Overlap
t < 0
Region I
No -Overlap
t < 0
h(t-)3
t -1 t
x()
2
2
Region II
Partial -Overlap
0 t 1
Region II
Partial -Overlap
0 t 1
Want R,t 0 t 0Want L,t 0 t-1 0 t 1
Want R,t< 0 t< 0
Step #5 (Continued): Find Regions of Overlap
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Step #5 (Continued): Find Regions of -Overlap
h(t-)3
t -1 t
x()
2
2
Region IIITotal -Overlap
1 0 t > 1
h(t-)3
t -1 t
x()
2
2
Region IV
Partial -Overlap
2 2Want L,t 2 t-1 2 t 3
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h(t-)3
t -1 t
x()
2
2 Region V
No -Overlap
t > 3
Region V
No -Overlap
t > 3
Want L,t> 2 t-1 > 2 t > 3
Step #5 (Continued): Find Regions of -Overlap
Step #6: Form Product & Integrate For Each Region
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Step #6: Form Product & Integrate For Each Region
Region I:t < 0Region I:t < 0
Region II: 0 t 1Region II: 0 t 1
h(t-)3
t -1 t
2
2
h(t-)x() = 0
0allfor0)(
00
)()()(
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Step #6 (Continued): Form Product & Integrate For Each Region
h(t-) 3
t -1 t
x()
2
2
Region III: 1
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Region V:t > 3Region V:t > 3
Step #6 (Continued): Form Product & Integrate For Each Region
h(t-)3
t -1 t
2
2
h(t-)x() = 0
3allfor0)(
00
)()()(
>=
==
=
tty
d
dthxty
x()
Step #7: Assemble Output Signal
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Step #7: Assemble Output Signal
Region It < 0
Region I
t < 0
0)( =ty
Region II0 t 1
Region II
0 t 1
tty 6)( =
Region III1
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