• 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