# Convolution of 2 summed signals

#### AlexMak

Joined Jan 2, 2018
32
Hello , I have an issue with my homework.
I have the signal in the attachment and I must convolute M(t)*M(t) where * - convolution then the result I must convolute again with M(t) and then again with M(t). So I would have to do 3 convolutions. I can't figure out the first one I know it should look like a triangle ( a derivated constant ) but cant see how to start. Could I get some help ?

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#### WBahn

Joined Mar 31, 2012
25,757
Where does the "summed signals" come into this?

What is the definition of a(t) * b(t)?

What is the mathematical definition of M(t)?

What do you get when you plug M(t) into the definition of M(t) * M(t)?

You need to show your best attempt.

#### AlexMak

Joined Jan 2, 2018
32
There were 2 summed signals that resulted the M(t) ( my bad I missed that ). So I got that out of the way. The convolution definition would be a(t) * b(t) = ∫a(t-τ)b(t)dτ = ∫a(t)b(t-τ)dτ from -inf to +inf . M(t) is 1 when t is within (-0.5;+0.5) i think and 0 outside.

What do you get when you plug M(t) into the definition of M(t) * M(t)?
This I don't understand. it would be something like this ∫M(t-τ)M(t)dτ from -inf to +inf but I don't get how to solve it.

#### AlexMak

Joined Jan 2, 2018
32
Hello, I'm back, I succesfully convoluted the first signal ( y(t) = x(t)*x(t) ) but now I'm having trouble at the 2nd one ( z(t) = y(t)*x(t) ). I used the laplace method and got here ( attachment ) and I can't seem to be able to make a represent the signal.

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#### WBahn

Joined Mar 31, 2012
25,757
The convolution definition would be a(t) * b(t) = ∫a(t-τ)b(t)dτ = ∫a(t)b(t-τ)dτ from -inf to +inf .
Close, but not quite. Remember, tau is the variable of integration that you want to go away and you want t from one function to survive.

a(t) * b(t) = ∫a(t-τ)b(τ)dτ = ∫a(τ)b(t-τ)dτ from -inf to +inf

M(t) is 1 when t is within (-0.5;+0.5) i think and 0 outside.
Okay.

This I don't understand. it would be something like this ∫M(t-τ)M(t)dτ from -inf to +inf but I don't get how to solve it.
You solve it by plugging M(τ) and M(t-τ) into the definition for convolution and performing the integration. That equation is more than just a pretty picture!

Just like any time you integrate a piecewise defined function, you set up separate integrals for the different segments of its definition.

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#### AlexMak

Joined Jan 2, 2018
32
You solve it by plugging M(τ) and M(t-τ) into the definition for convolution and performing the integration. That equation is more than just a pretty picture!

Just like any time you integrate a piecewise defined function, you set up separate integrals for the different segments of its definition.
I did it, the first convolution is the one in the attachments but I'm stuck at the 2nd one. I made it with the laplace method and I can't represent the signal with the given expression as you can see in my 2nd reply.

This is the expression : 2z(t) = (t-1.5)²u(t-1.5) - (t-0.5)²u(t-0.5) - (t+0.5)²u(t+0.5) + (t+1.5)²u(t+1.5)
How do I represent this signal ?

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#### WBahn

Joined Mar 31, 2012
25,757
I did it, the first convolution is the one in the attachments but I'm stuck at the 2nd one. I made it with the laplace method and I can't represent the signal with the given expression as you can see in my 2nd reply.

This is the expression : 2z(t) = (t-1.5)²u(t-1.5) - (t-0.5)²u(t-0.5) - (t+0.5)²u(t+0.5) + (t+1.5)²u(t+1.5)
How do I represent this signal ?
If you want to use the Laplace transform, then remember that the transform of the sum is the sum of the transforms.

But can (and should be able to) evaluate the convolution directly, too.

Since you got the first one via Laplace, I'll show you how to start setting it up directly.

$$M(t) = \left\{ \begin{array}{rcrcl} 0 & \mbox{for} & & -0.5 & \lt t \\ 1 & \mbox{for} & -0.5 \leq & t & \leq +0.5 \\ 0 & \mbox{for} & & +0.5 & \lt t \end{array}$$

Notice that since M(τ) is 0 outside of the range from -0.5 to +0.5, the convolution integrand is also zero outside this region. This, by itself, limits the range of integration to

$$M(t)*M(t) = \int_{-\infty}^{+\infty} M(\tau) M(t-\tau) d \tau \; M(t)*M(t) = \int_{-0.5}^{+0.5} M(\tau) M(t-\tau) d \tau$$

But that only reflects the limits imposed by M(τ). We also have to consider the limits imposed by M(t-τ).

Give it some thought and see if you can figure out how to do that.

Hint #1: The limits will involve 't', which is how the convolution result ends up being a function of 't'.

Hint #2: You will end up with different integrals based on the value of 't'.

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#### AlexMak

Joined Jan 2, 2018
32
If you want to use the Laplace transform, then remember that the transform of the sum is the sum of the transforms.

But can (and should be able to) evaluate the convolution directly, too.

Since you got the first one via Laplace, I'll show you how to start setting it up directly.

$$M(t) = \left\{ \begin{array}{rcrcl} 0 & \mbox{for} & & -0.5 & \lt t \\ 1 & \mbox{for} & -0.5 \leq & t & \leq +0.5 \\ 0 & \mbox{for} & & +0.5 & \lt t \end{array}$$

Notice that since M(τ) is 0 outside of the range from -0.5 to +0.5, the convolution integrand is also zero outside this region. This, by itself, limits the range of integration to

$$M(t)*M(t) = \int_{-\infty}^{+\infty} M(\tau) M(t-\tau) d \tau \; M(t)*M(t) = \int_{-0.5}^{+0.5} M(\tau) M(t-\tau) d \tau$$

But that only reflects the limits imposed by M(τ). We also have to consider the limits imposed by M(t-τ).

Give it some thought and see if you can figure out how to do that.

Hint #1: The limits will involve 't', which is how the convolution result ends up being a function of 't'.

Hint #2: You will end up with different integrals based on the value of 't'.
I can't see the code properly I think it bugged out.

#### WBahn

Joined Mar 31, 2012
25,757
I can't see the code properly I think it bugged out.
What browser are you using? New versions of Chrome seems to have problems rendering the MimeTex code. Try FireFox or IE.

#### AlexMak

Joined Jan 2, 2018
32
Alright, I see what you're saying, but the thing is I've done the first. I'm stuck on the 2nd one right now and I want to do it with laplace aswell.

#### WBahn

Joined Mar 31, 2012
25,757
Alright, I see what you're saying, but the thing is I've done the first. I'm stuck on the 2nd one right now and I want to do it with laplace aswell.
If you did the first one by Laplace, then you already have the Laplace transforms you need to do the second one, don't you?

#### AlexMak

Joined Jan 2, 2018
32
I do, I attached my work in the attachments. The M(t) in this case is x(t) for me ( preference ). So what I did I used the derivate of y(t) to find out y(t). Then I used the property that in laplace transformation the convolution is a product. Therefore, with all the calculus I get to this ecuation 2z(t) = (t-1.5)²u(t-1.5) - (t-0.5)²u(t-0.5) - (t+0.5)²u(t+0.5) + (t+1.5)²u(t+1.5) and from this I have to figure out how to sketch a signal but I can't do it... I tried

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#### WBahn

Joined Mar 31, 2012
25,757
"I used the derivate of y(t) to find out y(t)." Huh????

To use the normal notation, you have m(t) given as being 1 from -0.5 to +0.5 and zero elsewhere, right?

m(t) = 1 for -0.5 <= t <= +0.5; 0 elsewhere.

Right?

Now, for some reason that I don't follow, you seem to want

f(t) = [ [m(t)*m(t)] * m(t) ] * m(t)

Since convolution is both commutative and associative, this can be rewritten as

f(t) = [m(t)*m(t)] * [m(t)*m(t)]

If

m_2(t) = m(t)*m(t)

then

f(t) = m_2(t)*m_2(t)

Right?

So what is

M(s) = L{m(t)}

And what is L(s) in terms of M(s)?

EDIT: Replaced F{} with L{} for clarification.

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#### AlexMak

Joined Jan 2, 2018
32
So M(t) = x(t) in my picture. y(t) = x(t) * x(t) and z(t) = y(t) * x(t)
Z(s) = Y(s) x X(s)

and z(t) = (inverse laplace ) (Z(s)) right.

M(s) = F{m(t)} is laplace of m(t) ?

Is there something wrong ?

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#### WBahn

Joined Mar 31, 2012
25,757
So M(t) = x(t) in my picture. y(t) = x(t) * x(t) and z(t) = y(t) * x(t)
Z(s) = Y(s) x X(s)

and z(t) = (inverse laplace ) (Z(s)) right.

M(s) = F{m(t)} is laplace of m(t) ?
Yes. I meant to use L and not F. So

M(s) = L{m(t)}

Is there something wrong ?
Your y(t) is the first convolution and your z(t) is just the second convolution. You said that you needed a third convolution.

#### AlexMak

Joined Jan 2, 2018
32
I need z(t) , I cant figure it out.

#### MrAl

Joined Jun 17, 2014
7,591
I do, I attached my work in the attachments. The M(t) in this case is x(t) for me ( preference ). So what I did I used the derivate of y(t) to find out y(t). Then I used the property that in laplace transformation the convolution is a product. Therefore, with all the calculus I get to this ecuation 2z(t) = (t-1.5)²u(t-1.5) - (t-0.5)²u(t-0.5) - (t+0.5)²u(t+0.5) + (t+1.5)²u(t+1.5) and from this I have to figure out how to sketch a signal but I can't do it... I tried
Hi,

Those images are hard to see because they are so badly copied. That makes it hard to follow your work.

You said you also want to do this with Laplace right? So are you just trying to find:
((M*M)*M)*M (?)

where M is the Laplace of m(t).

So it looks like multiplication in the frequency domain gets you a bunch of exponentials and then after expanding you should be able to make sense of it i think.

Here's the image cleared up a little...

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#### AlexMak

Joined Jan 2, 2018
32
Hi,

Those images are hard to see because they are so badly copied. That makes it hard to follow your work.

You said you also want to do this with Laplace right? So are you just trying to find:
((M*M)*M)*M (?)

where M is the Laplace of m(t).

So it looks like multiplication in the frequency domain gets you a bunch of exponentials and then after expanding you should be able to make sense of it i think.

Here's the image cleared up a little...
No, I need (M*M)*M where M is the Laplace of m(t). I will ned ((M*M)*M)*M after this, which would be a third convolution.
After calculating the bunch of exponentials and making the inverse laplace I get this 2z(t) = (t-1.5)²u(t-1.5) - (t-0.5)²u(t-0.5) - (t+0.5)²u(t+0.5) + (t+1.5)²u(t+1.5) which I can't represent as a signal.. and this is what I need help on atm

#### MrAl

Joined Jun 17, 2014
7,591
No, I need (M*M)*M where M is the Laplace of m(t). I will ned ((M*M)*M)*M after this, which would be a third convolution.
After calculating the bunch of exponentials and making the inverse laplace I get this 2z(t) = (t-1.5)²u(t-1.5) - (t-0.5)²u(t-0.5) - (t+0.5)²u(t+0.5) + (t+1.5)²u(t+1.5) which I can't represent as a signal.. and this is what I need help on atm
Hi,

What do you mean that you cant represent it as a signal? That is the signal, at least if you got it correct.

The first convolution will result in an upside down triangle, possibly shifted or delayed or both. I think you got this one already.
The second convolution will result in a blip, which looks sort of like a rounded top triangle but also with curved sides.
The third will result in another rounded top and curved side triangle, possibly sharper or flatter.

I think maybe all you need to do is graph your results. You know when the signal turns 'on' based on your u(...) functions, and you know your time functions from the factors you get for those on/off functions. I think that is the only way it may make sense to you because sometimes it is harder to visualize the results without actually graphing.

If this still doesnt make sense, i'll graph a different example for a pulse 2<=t and t<=3 (a pulse that is 'on' from 2 to 3 seconds).

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#### AlexMak

Joined Jan 2, 2018
32
Hi,

What do you mean that you cant represent it as a signal? That is the signal, at least if you got it correct.

The first convolution will result in an upside down triangle, possibly shifted or delayed or both. I think you got this one already.
The second convolution will result in a blip, which looks sort of like a rounded top triangle but also with curved sides.
The third will result in another rounded top and curved side triangle, possibly sharper.

I think maybe all you need to do is graph your results. You know when the signal turns 'on' based on your u(...) functions, and you know your time functions from the factors you get for those on/off functions. I think that is the only way it may make sense to you because sometimes it is harder to visualize the results without actually graphing.

If this still doesnt make sense, i'll graph a different example for a pulse 2<=t and t<=3 (a pulse that is 'on' from 2 to 3 seconds).
Here, I moved the actual question here https://forum.allaboutcircuits.com/threads/issues-representing-signal.144007/ since the one in this post was replied. The problem is , if I take the results that I got and try to represent the signal I will get different results from what you're saying ( -t^2 + 3*t -2.25 is the first part of my signal which would be concave parabola and my graphic wouldnt look correct if i'd do that )

I did the convolution in matlab. And it should look like : first portion a convex parabola then a concave one then another convex one to end it right ?