# Analytic solution of a Convolution Integral

1. Feb 13, 2015

### Legend101

1. The problem statement, all variables and given/known data
The question is in the attached image . My problem starts when dealing with the limits of integration . I need an analytic procedure of solving such problems without involving graphical method . The equations of the graphs of h(t) and x(t) are easily derived .

2. Relevant equations
See attached image:
The convolution integral of 2 functions is defined as :
[itеx] y(t) = \int_{-infinity}^{+infinity}h(\tau}*x(t-\tau)d\tau [\itеx]

3. The attempt at a solution
See attached image

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Last edited: Feb 13, 2015
2. Feb 13, 2015

### Staff: Mentor

The integral in your second attachment (both attachments have image.jpg as their filename) is not difficult to write using LaTeX.
$$y(t) = \int_{t - 4}^{t - 2} h(\tau) d\tau$$
The LaTeX script that produces this integral is y(t) = \int_{t - 4}^{t - 2}h(\tau}d\tau. Use two \$ symbols at each end. More info here: https://www.physicsforums.com/help/latexhelp/

3. Feb 13, 2015

### RUber

$h(\tau)$ is a linear function, and should be easy to integrate if you break it into two pieces.

4. Feb 13, 2015

### Legend101

The problem , as i mentioned , is in the intervals of integration . We should consider to which interval t belongs . The overlapping confuses me

5. Feb 13, 2015

### RUber

h(t) is non-zero for 1<t<3, so your integral will be zero for t<3, it will grow on 3<t<5, then shrink down again on 5<t<7 and be zero from that point on.
So from the outset, restrict t to between 3 and 7 and see where that gets you.

6. Feb 13, 2015

### Legend101

I seriously didn't understand . How did you know that the integral will be zero for t<3 and grow in....?
How am i supposed to know the different intervals of t for integration ?

7. Feb 13, 2015

### RUber

$f(t) = \int_{t-4}^{t-2} h(\tau) d\tau = \int_{2}^{4} h(t-\tau) d\tau$
If the intersection: $[t-4, t-2] \cap [1,3]$ is empty, then the function is zero, since the integrand is zero over the entire integral.
This intersection is maximized when the two ranges match exactly, i.e. t=5.
Other than that, you are evaluating a portion of the area under the tent of h.

8. Feb 13, 2015

### RUber

Are you able to write the integral of h as a function of x?
something like $f(x) = \left\{ \begin{array} {l l } \displaystyle \int_1^x g(t) dt & 1<x\leq 2 \\ \displaystyle \int_1^2 g(t) dt + \int_2^x k(t) dt & 2< x\leq 3 \end{array} \right.$

9. Feb 13, 2015

### Ray Vickson

Always draw a picture! Look at the feasible region in $(\tau,t)$-space.

10. Feb 13, 2015

### vela

Staff Emeritus
I have to admit I'm always puzzled by students when they insist on a non-graphical method of analyzing a problem. Analyzing the problem graphically is probably one of the most powerful tools you have and much less error-prone than trying to apply a non-intuitive, non-graphical method.

11. Feb 13, 2015

### Legend101

Unfortunately , i'm not . It seems I need to do further readings :(

12. Feb 13, 2015

### RUber

In your attached photo, you already have the piecewise definition of h as:
$\left\{ \begin{array} {l l} t-1 & 1<t\leq 2 \\ 3-t & 2<t\leq 3 \\ 0 & \text{Otherwise} \end{array} \right.$
What happens if you take the integral of this for t from 0 to 2? From 1 to 3? From -2 to 0? From 2 to 4?
If you can answer these questions, you should know enough to complete this problem.

13. Feb 14, 2015

### Legend101

But the integral is from t-4 to t-2 . The intervals of t that we should consider are not the same as h(tau) 's intervals of definition . For now , i obtained 7 cases of t

14. Feb 14, 2015

### RUber

What if you made a shift and said x= t-4 and x+2 = t-2?
Either way, you will get an integral that depends on some variable which can be evaluated as a function that has maximum of 1 and is zero for all but a small interval of length 4.
7 cases? Could you show what you have so far?

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