# Convolution Calculation (piecewise function)

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1. Mar 14, 2016

### roam

1. The problem statement, all variables and given/known data
Compute the convolution $(f*h)(t)$ where

$$f(t) = \left\{\begin{matrix}1, \ \ for \ \ |t|<1 \\ 0, \ \ \ \ otherwise \end{matrix}\right.$$

and

$$h(t) = \left\{\begin{matrix}2|t|-1, \ \ for \ \ |t|<1/2 \\ 0, \ \ \ \ otherwise \end{matrix}\right.$$

2. Relevant equations

Convolution integral: $(f*h)(t)= \int^\infty_{-\infty} f(\tau) h(t-\tau) d \tau$

3. The attempt at a solution

Here I first made a sketch of the two functions:

f(t) is non-zero on [-1, 1] and g(t) is non-zero on [-0.5, 0.5], so adding these two we find that (f ∗ g)(t) should be nonzero on [-1.5, 1.5]. The list of change points of (f ∗ g)(t) is {-1.5, -0.5, 0.5, 1.5}. For values t that are outside this range the convolution is zero.

So, do I need to calculate the convolution for 3 separate cases?

I mean $-1.5 \leq t \leq -0.5$, $-0.5 \leq t \leq 0.5$, and $0.5 \leq t \leq 1.5$?

Here is my attempt:

First case: we want to integrate from -1.5 to -0.5 using the formula given above ($(f*g)(t)=\int^{0.5}_{-1.5} f(\tau) h(t-\tau) d \tau$). $h(t)$ is zero. But for $f(t)$ the function is zero on the interval $(-1.5, -1)$, whereas it is equal to 1 from $(-1, -0.5)$, so what value do I need to be using for the function?

Any help is greatly appreciated.

2. Mar 14, 2016

### Samy_A

If $f$ is zero on an interval, the integral simply vanishes there. Then you get:
$(f*h)(t)=\displaystyle \int^{+ \infty}_{-\infty} f(\tau) h(t-\tau) d \tau = \int^{1}_{-1} h(t-\tau) d \tau$.

(EDIT: I had written $(f*g)(t)$ instead of $(f*h)(t)$, typo now corrected.)

Last edited: Mar 15, 2016
3. Mar 15, 2016

### Samy_A

To add to my previous post:

I would consider a substitution $\sigma =t- \tau$.
Also notice that both $f$ and $h$ are even functions, so their convolution will also be an even function.
Finally, when you distinguish between the different relevant cases for $t$ (ie $t\leq -1.5$, $-1.5 \leq t \leq -1$, $-0.5 \leq t \leq 0$), the results you obtain for each of the segments must be equal on their mutual endpoints.

EDIT: the relevant cases were meant to be $t\leq -1.5$, $-1.5 \leq t \leq -1$, $-1 \leq t \leq -0.5$, $-0.5 \leq t \leq 0$

Last edited: Mar 15, 2016
4. Mar 15, 2016

### roam

Thank you Samy for your response.

For the case $-1.5 \leq t \leq -0.5$, could you please explain how you got the integral limits to be $(-1,1)$ instead of $(-1.5, -0.5)$?

I think alternatively it is also possible to rewrite the integral limits like:

$$t+0.5 \leq \tau \leq t+1.5$$

In the interval $[-1.5, -0.5]$, $x+0.5$ is between -1 and 0. Whereas $x+1.5$ is between 0 and 1. So the integral can be rewritten:

$$(f*h)(t)=\int^{t+1.5}_{t+0.5} h(t- \tau) \ d \tau$$

I am not sure how to proceed from here. How exactly does a substitution help? (this is my first convolution problem of this type)

But shouldn't the three cases be the following?

(i) $-1.5 \leq t \leq -0.5$

(ii) $-0.5 \leq t \leq 0.5$

(iii) $0.5 \leq t \leq 1.5$

For values of t less than -1.5 or greater than 1.5, (f∗g)(t) is zero.

5. Mar 15, 2016

### Samy_A

I wasn't considering cases yet.

I simply started from the definition: $\displaystyle (f*h)(t)=\displaystyle \int^{+ \infty}_{-\infty} f(\tau) h(t-\tau) d \tau$.
Since $f$ is 0 outside of $[-1,1]$, the integration limits can be replaced by $-1$ and $+1$.
Since $f$ is 1 in $[-1,1]$, the $f(\tau)$ just becomes $1$, giving the result: $\displaystyle (f*h)(t)=\displaystyle \int^{+1}_{-1} h(t-\tau) d \tau$.
I made a mistake in listing the cases, sorry for that.

It really depends on how you want to tackle the integral. (There are probably a number of ways to do it, I'm not sure the one I suggest is the shortest.)

I suggested the substitution $\sigma =t-\tau$.
The integral then becomes $\displaystyle (f*h)(t)=\displaystyle -\int^{t+1}_{t-1} h(\sigma) d \sigma \ \ \ \ (1)$.
(There is no deep reason for this substitution. I think that dealing with $h(\sigma)$ is less error prone than dealing with $h(t-\tau)$ when treating the different cases.)
Notice that $h$ vanishes outside of $[-0.5,+0.5]$, and that the expression for $h$ is different for positive and negative $\sigma$. This will be important when treating the various cases.

Now for the cases.
a) For $t\leq -1.5$, the convolution is 0, as you already wrote.

b) For $-1.5 \leq t \leq -1$, the integration limits will be $-0.5 \to t+1 (\leq 0)$, so you can use the expression for $h$ as valid for negative $\sigma$.

c) For $-1 \leq t -0.5$, the integration limits will be $-0.5 \to t+1$, where $0 \leq t+1 \leq 0.5$. In this case the integral (1) will consist of two parts: the integral from $-0.5$ to $0$ and the integral from $0$ to $t+1$ (with different expressions for $h(\sigma)$).

d) For $-0.5 \leq t \leq 0$, the integration limits will be $-0.5 \to 0.5$, and the integral will also consist of two parts ($-0.5 \to 0, 0 \to 0.5$).
Since $h$ is even and the integration domain is symmetric around the y-axis, you can compute one of the integrals and multiply the result by 2.

e) Finally, for $t \geq 0$, make use of the fact that $f*h$ is an even function, and use (with the needed precautions) the results obtained for negative $t$.

Last edited: Mar 15, 2016
6. Mar 15, 2016

### roam

Thank you for the explanation. Yes $h$ is $-2t-1$ for negative σ. So, for instance for the case $-1.5 \leq t \leq -1$ this is what I found:

$$(f*h)(t) = -\int^{t+1}_{-0.5} h(\sigma) d \sigma = -\int^{t+1}_{-0.5} (-2t-1) \ dt = - \left( (t+1)^2 -(t+1)-0.25 \right)$$

Is this correct?

This function looks like an inverted parabola. This is the region of interest:

I just want to make sure it is done correctly so I can calculate the other integrals, and sketch $(f*h)(t)$ to make sure it is an even symmetric function.

7. Mar 15, 2016

### Samy_A

I find something different.
If I'm not mistaken, your function is equal to -0.5 in -1.5. That should be 0.

Your integral is the same as mine for that case, but the result is different.
When I computed the convolution this morning, I had a number of sign errors. It looks like that's what you have here too.

Last edited: Mar 15, 2016
8. Mar 15, 2016

### roam

Oops, I did the integral again and there was a sign error as you suggested:

$$- \int^{t+1}_{-0.5} (-2t-1).dt = -\Bigg[ \frac{-2 t^2}{2} -t \Bigg]^{t+1}_{-0.5}$$

$$= - \left( \Big[ -(t+1)^2 - (t+1) \Big] - \Big[ -(-0.5)^2 - (-0.5) \Big] \right) = (t+1)^2 + (t+1) + 0.25$$

This time it's a right way up parabola:

Is that what you have also?

9. Mar 15, 2016

### Samy_A

Yes, I have the same result. $(f*h)(t)=t²+3t+\frac{9}{4}$ for $t \in [-1.5, -1]$.
Notice that now the value in -1.5 is 0. The value of the first derivative in -1.5 is also 0. That is as expected, as $f*h$ is 0 below -1.5, and is a differentiable function.

10. Mar 16, 2016

### roam

Thank you so much for the explanation.

For case (c) I found:

$$(f*h)(t)= - \Big[ \int^0_{-0.5} f(t) h(\sigma) d \sigma + \int^{t+1}_{0} f(t) h(\sigma) d \sigma \Big] =- \Big[ \int^0_{-0.5} (-2t-1) d t + \int^{t+1}_{0} (2t-1) d t \Big]$$

$$= -(t+1)^2 + (t+1) + 0.25 = -t^2-t+\frac{1}{4}$$

This function is the inverted version of y=x2+3x+(9/4) we found for part (b).

For case (d) I found:

$$(f*h)(t)= - 2 \int^{0.5}_0 (2t-1)(f(t)=1) dt = 0.5$$

The graph in Matlab looks like this so far for $t \leq 0$ (parts (b), (c), and (d) are shown in magenta, blue and green respectively):

Since the $(f ∗ h)(t)$ is continuous and even I think the other half (in blue) should look like this:

Is this correct?

11. Mar 16, 2016

### Samy_A

Yes, I found the same results for cases c) and d).

As $f*h$ is even, you indeed can just take the mirror image of the graph for positive $t$.
To be clear, from $-1.5 \to -0.5$ the graph segments are parabolic, not straight lines (and likewise on the positive side).

12. Mar 17, 2016

### roam

I get it now. Thank you so much for your time and help.