• Acknowledgment: Multiple slides are from internet and Prof. Mark Fowler in Binghamton University.
Convolution Examples
2
ττ )dx(t)h(t=h(t)x(t)=f(t) −∫∞
∞-
*
Convolution Example
5
( ) ( ) ( ) ( ) ( )
( ) ( )
( )
Let d where
0 1 0 01 1 2 0 1
and 2 2 3 0 10 3
Find .
y t f t g t f g t
t tt t t
f t g tt t
ty t
τ τ τ∞
−∞
= ∗ = −
< − < − ≤ < ≤ < = = ≤ < ≤ ≤
∫
1
g( ) g(- ) τ
0 0 -1
τ
Band 1: Sliding Function Method
0 -1
g(- ) τ
Band 2: Sliding Function Method
Band 3: Sliding Function Method
Band 4: Sliding Function Method
Band 5: Sliding Function Method
Band 6: Sliding Function Method
Convolution Example “Table view”
h(-m)
h(1-m)
Discrete-Time Convolution Example: “Sliding Tape View”
D-T Convolution Examples
( ) ]4[][][][21][ −−== nununhnunx
n
-3 -2 -1 1 2 3 4 5 6 7 8 9
][ih
i…
-3 -2 -1 1 2 3 4 5 6 7 8 9
][ix
i…
-3 -2 -1 1 2 3 4 5 6 7 8 9
]0[ ih −
i
Choose to flip and slide h[n] This shows h[n-i] for n = 0
…
For n < 0 h[n-i]x(i) = 0 ∀i
00][ <= nforny⇒
][][ ixih −
i
For n = 0
⇒ y[0] = 1
-3 -2 -1 1 2 3 4 5 6 7 8 9
][ix
i
Notice that for n = 0, n = 1, …, n = 3
The general result is:
( ) 3,2,1,021][
0==∑
=
nfornyn
i
i
( ))(
2112
111
SumGeometric
n
−
−=
+
( ) 3,2,1,02112][
1=⎥⎦
⎤⎢⎣⎡ −=
+nforny
n
-2 -1 1 2 3 4 5 6 7 8
…
]1[][ ihinh −=−
i
⇒ y(1) = 1 + ½ = 3/2
For n = 1
-3 -2 -1 1 2 3 4 5 6 7 8 9
][ix
i
n = 4 case
-3 -2 -1 1 2 3 4 5 6 7 8 9
]4[][ ihinh
…
−=−
i
Now for n = 4, n = 5, …
13 =−= ni 4== nin = 5 case
-3 -2 -1 1 2 3 4 5 6 7 8 9
…
]5[][ ihinh −=−
i
23 =−= ni 5== ni
Notice that: for n = 4, 5, 6, …
( )( ) ( )
!2
112
12
1
,...6,5,421][
133
simplifythen
nforny
nn
in
ni
−
−=
==
+−
−=∑
Then we can write out the solution as:
( )[ ]( ) ( )[ ]⎪
⎪⎩
⎪⎪⎨
⎧
=−
=−
<
=
+−
+
...,6,5,4,2/12/12
3,2,1,0,2/112
0,0
][
13
1
n
n
n
ny
nn
n
-3 -2 -1 1 2 3 4 5 6 7
…
][nh
n1
2
2. Same Impulse Response: h[n]
n
i
etc
( ) ][2cos][ nunnx π=
][ix
][ ih −
i
Again y[n] = 0 for n < 0:
00101]5[01010]4[00101]3[
0101]2[101]1[
1]0[
=+++−==++−==+−+=
=−+==+=
=
yyyyyy
Notice: y[n] = 0 ∀ n = 2, 3, 4, 5, …!
So suppose we had a desired part of our signal as:
But say we “receive” our desired pulse signal with an “interfering” sinusoid:
( ) [ ]nunnpnx 2cos][][ π+=
][ˆ][ npny =][nh][nx
From above we know that system “zeros out” (or suppresses) the sinusoid…
We also know that the system will “pass” the pulses, although their edges will be smoothed.
9-point pulse keeps repeating
][np 9 points0.25
n