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# Graphical Convolution Algorithm

Module by: Richard Baraniuk. E-mail the author

Summary: This module discusses the Graphical Convolution Algorithm with the help of examples.

ct=fτgtτd τ c t τ f τ g t τ

## Step One

Plot fτ f τ and gτ g τ as functions of τ τ

## Step Two

Plot gtτ g t τ by reflecting gτ g τ over the 'y-axis' ( run time backwards) and then shifting right by t t.

## Step Three

For one value of ' t t' mutiply fτgtτ f τ g t τ and compute area underneath the curve to get ct c t . Area underneath

=fτgtτd τ =ct τ f τ g t τ c t
(1)

## Step Four

Repeat for all ' t t' to get ct c t for all t. Usually we will just have to consider several ranges of t.

## Remark

Since,

ct=fτgtτd τ =gτftτd τ c t τ f τ g t τ τ g τ f t τ
(2)
you can flip and shift either f or g. It is easier to flip and shift the 'simpler' of the two.

### Note:

Everyone is overwhelmed by convolution at first! Just practise and it will become second nature. Do examples 2.6 to 2.8 in Lathi!

## Example 1

Recall

Now compute output yt y t for a step input ft ut f t u t

### Solution

System is LTI with impulse response ht h t , so use convolution integral yt=fτhtτd τ y t τ f τ h t τ Since, fτ f τ is simpler, we rewrite it as hτftτd τ τ h τ f t τ

Plot things

#### Step 2

Do the flip and shift.

#### Step 3 & 4

Multiply and integrate.

##### Case 1

For, t<0 t 0

From the fact stated in the caption, hτftτd τ =yt=0 t:t<0 τ h τ f t τ y t 0 t t 0

##### Case 2

For t0 t 0

yt=hτftτd τ =0t1RCeτRCdτ =RCRCeτRC| τ =0t=1etRC y t τ h τ f t τ τ 0 t 1 R C τ R C τ 0 t R C R C τ R C 1 t R C
(3)

yt={0  if  t<01etRC  if  t0 y t 0 t 0 1 t R C t 0

#### Step 5

Do a reality check: As t tends to ∞ what happens? As t tends to −∞ what happens?

## Example 2

The input is ft=et f t t u t and the impulse response is ht=e(2t) h t 2 t u t . Compute the yt y t .

### Solution

We are given input and impulse response. So ride the convolution convoy! yt=fτhtτd τ y t τ f τ h t τ Both the functions are equally simple, so we flip and shift ht h t

#### Case 1

Again yt=0 y t 0 for all t<0 t 0

#### Case 2

For t0 t 0

yt=fτhtτd τ =0tfτhtτd τ =0teτe(2)(tτ)d τ =0te(2t)eτd τ =e(2t)0teτd τ =e(2t)eτ| τ =0t=e(2t)(et1) y t τ f τ h t τ τ 0 t f τ h t τ τ 0 t τ 2 t τ τ 0 t 2 t τ 2 t τ 0 t τ 2 t τ 0 t τ 2 t t 1
(4)
yt=ete(2t) t :t0 y t t 2 t t t 0

#### Combine Case 1 and 2

yt={0  if  t<0ete(2t)  if  t0=(ete(2t))ut y t 0 t 0 t 2 t t 0 t 2 t u t
(5)

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