Analytic solution of a Convolution Integral

Legend101
Messages
15
Reaction score
0

Homework Statement


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 .
image.jpg


Homework 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]

The Attempt at a Solution


See attached image
 

Attachments

  • image.jpg
    image.jpg
    29.4 KB · Views: 641
  • image.jpg
    image.jpg
    26.1 KB · Views: 675
Last edited:
Physics news on Phys.org
Legend101 said:

Homework Statement


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 .View attachment 79013

Homework Equations


See attached image

The Attempt at a Solution


See attached image
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/
 
##h(\tau) ## is a linear function, and should be easy to integrate if you break it into two pieces.
 
RUber said:
##h(\tau) ## is a linear function, and should be easy to integrate if you break it into two pieces.
The problem , as i mentioned , is in the intervals of integration . We should consider to which interval t belongs . The overlapping confuses me
 
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.
 
RUber said:
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.
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 ?
 
##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.
 
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. ##
 
Legend101 said:
The problem , as i mentioned , is in the intervals of integration . We should consider to which interval t belongs . The overlapping confuses me

Always draw a picture! Look at the feasible region in ##(\tau,t)##-space.
 
  • #10
Legend101 said:

Homework Statement


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.
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
RUber said:
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. ##
Unfortunately , I'm not . It seems I need to do further readings :(
 
  • #12
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
RUber said:
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.
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
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?
 

Similar threads

Replies
1
Views
1K
Replies
11
Views
7K
Replies
19
Views
29K
Replies
4
Views
2K
Replies
14
Views
2K
Replies
23
Views
3K
Replies
16
Views
13K
Back
Top