# Convolution in signal processing

L

#### Lily Bepant

Jan 1, 1970
0
Hello,
I'm stuck in a convolution question.

x(t) = u(t) - 2u(t-2) + u(t-5)
h(t) = e^(2t) u(1-t)
y(t) = x(t) * h(t)

(I'm using capital T as tao and Int as integral from minus infinity to
plus infinity)

y(t) = Int ( x(t-T) h(T) ) dT

= Int ( u(t-T)-2u(t-T-2)+u(t-T-5)[e^(2T)u(1-T)] )dT

= Int ( [e^(2T)u(1-T)] u(t-T) ) - Int ....
------------------------------------------

I'm going to set integral limits due to the interval that the step
functions are nonzero. So:

1-T > 0 -> T < 1
t-T > 0 -> T < t

But the T is smaller than both 1 and t...

so.. What do I have to take as integral limits? I'm stuck here.
I know that it's basic calculus but, because it's about signal
processing, I posted it here.

Lily

A

#### Activ8

Jan 1, 1970
0
Hello,
I'm stuck in a convolution question.

x(t) = u(t) - 2u(t-2) + u(t-5)
h(t) = e^(2t) u(1-t)
y(t) = x(t) * h(t)

(I'm using capital T as tao and Int as integral from minus infinity to
plus infinity)

y(t) = Int ( x(t-T) h(T) ) dT

= Int ( u(t-T)-2u(t-T-2)+u(t-T-5)[e^(2T)u(1-T)] )dT

= Int ( [e^(2T)u(1-T)] u(t-T) ) - Int ....

Are you confusing concolution with the sampling function and step
with impulse?
1-T > 0 -> T < 1
t-T > 0 -> T < t

How can that be when t starts at zero and goes forever?
But the T is smaller than both 1 and t...

so.. What do I have to take as integral limits? I'm stuck here.
I know that it's basic calculus but, because it's about signal
processing, I posted it here.

Lily

The T I'm thinking of should be the sampling interval so it should
be smaller than t_max from -oo to oo or 0 to t_max ... isn't that
the idea behind dT, and infinitely small increment?

So integrate y(t) = f(t)u(t-T)dT from -oo to oo (or wherever you get
non zero values) for each t

J

#### Jem Berkes

Jan 1, 1970
0
Hello,
I'm stuck in a convolution question.

x(t) = u(t) - 2u(t-2) + u(t-5)
h(t) = e^(2t) u(1-t)
y(t) = x(t) * h(t)
...

so.. What do I have to take as integral limits? I'm stuck here.
I know that it's basic calculus but, because it's about signal
processing, I posted it here.

Why integrate? Can't you solve this entirely in the frequency domain? Use
Laplace transform/identities to get X(s), H(s)

Y(s) = X(s) x H(s)

Inverse transform to find y(t) in time domain.

B

#### Ben Moss

Jan 1, 1970
0
Linear Systems, I'm guessing? If so, you'll love it once you learn to
solve these with Laplace transforms (like Jem said). It makes life
easy.

It's easiest to solve convolution problems like these graphically. So
I'll try to do some ascii art. h(t) looks like this (somewhat):
h(t)
| ,
| /|
|.' |
.'| |
======='--+----------> t
| t=1
|
|
|
t=0

x(t) looks like this:

x(t)
|
|
| 1 2 3 4 5
|__
| |
----------+--+-----------+--------> t
| |___________|
|
|
|

It is easiest to break this up into two parts. Convolution is linear
and time-invariant so it is correct to do this.
Say x1(t)=u(t)-u(t-1) and x2(t)=-u(t-1)+u(t-5), so x1(t) is the upper
square and x2(t) is the lower rectangle, and x1(t) + x2(t) = x(t)

Like you said, y(t) = Int ( x(t-T) h(T) ) dT
This is also equivalent to y(t) = Int ( x(T) h(t - T) ) dT which is
easier to handle, IMO.

So, h(t) needs to be mirrored horizontally so we can see h(t-T).

h(t-T)
, |
|\ |
| .|
| |.
------+---+--========
T=t-1 |
|
|
|
T=t

Overlay this over x1(t), and this will be the case for t=0:

h(0-T)--> , |
|\ |_____
| .| | <----x1(t)
| |. |
------+---+--=+====== T
-1 |
|
|
|
T

Ok, so you are integrating from -oo to oo and multiplying these two
functions together. x1(t) is zero for everywhere less than 0 and
everywhere greater than 1. So when you multiply 0 by h(t), you get
zero. This confines the bounds to 0 and 1, but only when t is less
than 1. (Otherwise, the two functions would not completely overlap).

This makes the very first case for x1(t), and it works for t <= 1:

y1(t) = Int ( x1(T) h(t - T) ) dT from T=0 to T=1

and since x1(t) is 1 from 0 to 1, you can take that term out of the
equation, and now it is
y1(t) = Int ( e^(2t - 2T) ) dT = e^(2t) * Int( e^(-2T)) dT
= e^(2t) * e^(2T)/(-2) evaluated from 0 to 1 for T
= e^(2t) * (e^(-2) - 1 ) / (-2) ONLY FOR t<=1 !!

(BTW, * denotes multiplication there, not convolution)

Now the second case for x1(t) comes in when t>1 and t<=2:
Same intetgral, but different limits. Now the limits are from t to 1.
If you slide h(t-T) over in the picture, you can see why this is the
case, because that is the only time the two functions overlap.

y1(t) = Int (e^(2t - 2T) ) dT = e^(2t) * e^(2T)/(-2) evaluated from t
to 1 for T
= e^(2t) * (e^(-2) - e^(-2t) ) /(-2) ONLY FOR t>1, t<=2 !!!

To get the total y(t), you have to continue in the same manner for
x2(t) and then add y1(t) + y2(t) to get y(t). Hopefully you see how
this works. Draw it out on a piece of paper, and these get kind of
fun to do. You have to consider each case to find the bounds for the
integral.

I realize the ascii art isn't the best, so if you want I could write
out the solution and scan it in for you. It would be much easier to
see.
Hope that helps,
-Ben

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