301
Signals EECE & Systems
Prof. Mark Fowler
Note Set #11
C-T Systems: Computing Convolution
Reading Assignment: Section 2.6 of Kamen and Heck
1/20
Course Flow Diagram
The 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).
New Signal
Models
Ch. 1 Intro
C-T Signal Model
Functions on Real Line
System Properties
LTI
Causal
Etc
D-T Signal Model
Functions on Integers
New Signal
Model
Powerful
Analysis Tool
Ch. 3: CT Fourier
Signal Models
Ch. 5: CT Fourier
System Models
Ch. 6 & 8: Laplace
Models for CT
Signals & Systems
Fourier Series
Periodic Signals
Fourier Transform (CTFT)
Non-Periodic Signals
Frequency Response
Based on Fourier Transform
Transfer Function
New System Model
New System Model
Ch. 2 Diff Eqs
C-T System Model
Differential Equations
D-T Signal Model
Difference Equations
Ch. 2 Convolution
Zero-State Response
C-T System Model
Convolution Integral
Zero-Input Response
Characteristic Eq.
D-T System Model
Convolution Sum
Ch. 4: DT Fourier
Signal Models
DTFT
(for Hand Analysis)
DFT & FFT
(for Computer Analysis)
Ch. 5: DT Fourier
System Models
Freq. Response for DT
Based on DTFT
New System Model
New System Model
Ch. 7: Z Trans.
Models for DT
Signals & Systems
Transfer Function
New System
Model2/20
C-T convolution properties
Many of these are the same as for DT convolution.
We only discuss the new ones here.
1. Derivative Property:
See the next slide for the others
d
[ x (t ) v (t )] = x (t ) v (t )
dt
= x (t ) v (t )
derivative
2. Integration Property Let y(t) = x(t)*h(t), then
t x( )d h(t ) = x(t ) t h( )d
y( )d =
t
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 might
have on the physical interface conditions.
3/20
Convolution Properties
These are things you can exploit to make it easier to solve convolution problems
1.Commutativity
x (t ) h (t ) = h (t ) x (t )
You can choose which signal to flip
2. Associativity
x (t ) (v (t ) w(t )) = ( x (t ) v (t )) w(t )
Can change order sometimes one order is easier than another
3. Distributivity
OR
x (t ) ( h1 (t ) + h2 (t )) = x (t ) h1 (t ) + x (t ) h2 (t )
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)
4. Convolution with impulses
x (t ) (t ) = x (t )
4/20
Computing CT Convolution
-For D-T systems, convolution is something we do for analysis and for
implementation (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.
5/20
Steps for Graphical Convolution x(t)*h(t)
y (t ) = x ( )h(t )d
1.
Re-Write the signals as functions of : x() and h()
2.
Flip just one of the signals around t = 0 to get either x(-) 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 of h(-): call it L,0
b. Find the right-hand-edge -value of h(-): call it R,0
4.
Shift h(-) by an arbitrary value of t to get h(t - ) and get its edges
a.
Find the left-hand-edge -value of h(t - ) as a function of t: call it L,t
Important: It will always be L,t = t + L,0
b.
Note: I use for
what the book
uses ... It is not
a big deal as they
are just dummy
variables!!!
Find the right-hand-edge -value of h(t - ) as a function of t: call it R,t
Important: It will always be R,t = t + R,0
Note: If the signal you flipped is NOT finite duration,
one or both of L,t and R,t will be infinite (L,t = and/or
R,t= )
6/20
Steps Continued
5.
Find Regions of -Overlap
a. What you are trying to do here is find intervals of t over which the product
x() h(t - ) has a single mathematical form in terms
b. of In each region find: Interval of t that makes the identified overlap happen
c. Working examples is the best way to learn how this is done
Tips:
6.
Regions should be contiguous with no gaps!!!
Dont worry about < vs. etc.
For Each Region: Form the Product x() h(t - ) and Integrate
a. Form product x() h(t - )
b. Find the Limits of Integration by finding the interval of over which the
product is nonzero
i.
Found by seeing where the edges of x() and h(t - ) lie
ii. Recall that the edges of h(t - ) are L,t and R,t , which often depend on
the value of t
So the limits of integration may depend on t
c. Integrate the product x() h(t - ) over the limits found in 6b
i.
The result is generally a function of t, 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 output y(t)
7/20
Steps Continued
7.
Assemble the output from the output time-sections for all the regions
a. Note: you do NOT add the sections together
b. You define the output piecewise
c. Finally, if possible, look for a way to write the output in a simpler form
8/20
Example: Graphically Convolve Two Signals
y (t ) =
h() x(t )d
=
x()h(t )d
Convolve these two signals:
By Properties of
By Properties of
Convolution
Convolution
these two forms are
these two forms are
Equal
Equal
This is why we can
This is why we can
flip either signal
flip either signal
x(t)
2
2
t
h(t)
3
1
t
9/20
Step #1: Write as Function of
x()
2
2
h()
3
1
Step #2: Flip h() to get h(-)
x()
2
0
2
h(-)
3
1
0
Usually Easier
to Flip the
Shorter Signal
10/20
Step #3: Find Edges of Flipped Signal
x()
2
0
2
h(-)
3
1
L,0 = 1
0
R,0 = 0
11/20
Motivating Step #4: Shift by t to get h(t-) & Its Edges
Just looking at 2 arbitrary t values
In Each Case We Get
L,t = t + L,0
R,t = t + R,0
For tt= 2
For = 2
For tt= -2
For = -2
h(t-) =h(2-)
3 h(t-) =h(-2-)
3
3
2
1
2
L,t = t + L,0
R,t = t + R,0
L,t = t + L,0
R,t = t + R,0
L,t = t 1
R,t = t + 0
L,t = t 1
R,t = t + 0
L,-2 = -2 1
R ,-2 = 2+0
L,2 = 2 1
R,2 = 2+0
12/20
Doing Step #4: Shift by t to get h(t-) & Its Edges
For Arbitrary Shift by tt
For Arbitrary Shift by
h(t )
3
t1
t
L,t = t + L,0
R,t = t + R,0
L,t = t 1
R,t = t + 0
13/20
Step #5: Find Regions of -Overlap
2
x()
2
Region II
Region
h(t-)
No -Overlap
No -Overlap
tt< 0
<0
3
L,t R,t= t
= t -1
Want R,t< 0
2
t<0
x()
2
h(t-)
Partial -Overlap
Partial -Overlap
3
t -1
Want L,t 0
t-1 0
t1
t
Region II
Region II
0 tt 1
0 1
Want R,t 0
t0
14/20
Step #5 (Continued): Find Regions of -Overlap
x()
2
Region III
Region III
2
h(t-)
Total -Overlap
Total -Overlap
3
1 < tt 2
1< 2
t -1
Want L,t > 0
t-1 > 0
t
Want R,t 2
t>1
t2
x()
2
Region IV
Region IV
2
h(t-)
Partial -Overlap
Partial -Overlap
3
2 < tt 3
2< 3
t -1
Want L,t 2
t-1 2
t3
t
Want R,t > 2
t>2
15/20
Step #5 (Continued): Find Regions of -Overlap
2
x()
2
h(t-)
No -Overlap
No -Overlap
tt> 3
>3
3
t -1
Want L,t > 2
Region V
Region V
t
t-1 > 2
t>3
16/20
Step #6: Form Product & Integrate For Each Region
2
h(t-)
3
t -1
2
Region I: tt< 0
Region I: < 0
y (t ) =
t
x()h(t )d
=
h(t-) x() = 0
x()
0 d = 0
y (t ) = 0 for all t < 0
Region II: 0 tt 1
Region II: 0 1
2
h(t-) 3
t -1
With 0 integrand
the limits dont
matter!!!
2
0
y (t ) =
t
x( )h(t )d
t
t
= 6d = [6 ]0 = 6t 6 0 = 6t
h(t-) x()
6
0
y (t ) = 6t for 0 t 1
0t
17/20
Step #6 (Continued): Form Product & Integrate For Each Region
x()
h(t-)
Region III: 1 < tt 2
Region III: 1 < 2
2
2
3
y (t ) =
t -1
t
x( )h(t )d
h(t-) x()
t
=
6
6d = [6 ]t 1 = 6t 6(t 1) = 6
t
t 1
t -1
x()
h(t-)
t
y (t ) = 6 for all t such that : 1 < t 2
Region IV: 2 < tt 3
Region IV: 2 < 3
2
2
3
y (t ) =
t -1
t
x( )h(t )d
2
h(t-) x()
=
6
6d = [6 ]t 1 = 6 2 6(t 1) = 6t + 18
2
t 1
y (t ) = 6t + 18 for 2 < t 3
t -1 2
18/20
Step #6 (Continued): Form Product & Integrate For Each Region
x()
2
3
Region V: tt> 3
Region V: > 3
h(t-)
2
t -1
t
y (t ) =
x()h(t )d
=
h(t-) x() = 0
0 d = 0
y (t ) = 0 for all t > 3
19/20
Step #7: Assemble Output Signal
Region II
Region
tt< 0
<0
Region II
Region II
0 tt 1
0 1
Region III
Region III
1 < tt 2
1< 2
Region IV
Region IV
2 < tt 3
2< 3
Region V
Region V
tt> 3
>3
y (t ) = 0
y (t ) = 6t
y (t ) = 6
y (t ) = 6t + 18
y (t ) = 0
y (t )
6
0
1
2
3
t
20/20