# Convolution

Discussion in 'Homework Help' started by salmanshaheen_88, Jun 26, 2010.

Mar 5, 2009
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help me guyz

2. ### Georacer Moderator

Nov 25, 2009
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1,266
You keep asking for answers without making any attempts on your own on solving the problems. This isn't the way this thread works.

3. ### steveb Senior Member

Jul 3, 2008
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Just plug your functions into the convolution integral and out pops the answer. What could be easier?

4. ### salmanshaheen_88 Thread Starter Active Member

Mar 5, 2009
88
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@ georacer: if you don't know whether I have attempted or not then keep shut, no need of your cheep comments

5. ### JoeJester AAC Fanatic!

Apr 26, 2005
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1,211
You failed to post your attempts, so that indicates to me that you have done nothing to solve the problem.

Why should the members bother to go through an analysis when you chose not to abide by the forum rules.

6. ### salmanshaheen_88 Thread Starter Active Member

Mar 5, 2009
88
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Don't forget that this is HOMEWORK HELP SECTION, i you don' want to help then stay away from the forum and sell omelet

Apr 5, 2008
15,799
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Hello salmanshaheen_88,

Georacer and JoeJester are right.
You did not show any attemp on you solution.

Without that we can not see where you got stuck.

You must have seen this in the regulations:

Bertus

Jun 4, 2010
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9. ### salmanshaheen_88 Thread Starter Active Member

Mar 5, 2009
88
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ok my question is that how I will pass x(t) through h(t) as x(t) has magnitude greater than h(t) ???

10. ### steveb Senior Member

Jul 3, 2008
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469
The idea of "passing through" is not a physical one, but instead is the idea of multiplying, shifting/sliding and adding/integrating. To see this graphically, go back to the page that BenjaminSweet recommended and use your mouse to draw your signals as best you can. Make sure that you make the heights, widths and shapes of the waveforms as close as possible to your example. Then use your mouse to slide one signal by the other, and you will see a graphical answer that is almost the answer to your problem. It will display all of the regions, shapes and kinks of the real answer.

Once you can visualize what is happening, then you can write out the integral equation which does this exact calculation. The integrals are trivially easy, but the hard part is to figure out how to break the integral into pieces that are valid over a resticted time period. You will end up with a solution that has 5 pieces, I think. To help you along, I'll identify the regions where you will get different functional answers.

t < -1
-1 < t < 1
1 < t < 3
3 < t < 5
t > 5

Did you notice that the graphical answer displayed these 5 regions?

Try writing out the integrals for each of these 5 cases. If you can't do it completely or with confidence, post what you are able to do, and then we can step you through the process.

11. ### salmanshaheen_88 Thread Starter Active Member

Mar 5, 2009
88
1
my professor told me that in convolution we pass one signal from another, in fact convolution means multiplication. but my question about this problem is that how it is possible to pass a signal with high amplitude through a low amplitude signal

12. ### steveb Senior Member

Jul 3, 2008
2,433
469
That is the exact question I tried to answer. Did you try what I suggested? Did you go back to the page and create your own signal with a high amplitude? If you do that, you will see how it is possible to pass a signal with a high amplitude by (not through) a low amplitude signal.

Convolution does not in fact mean multiplication, although it does correspond to multiplication in the frequency domain, but that's a totally different concept. In the time domain, convolution is a process of multiplying, shifting and adding for discrete time signals, and it is a process of multiplying, sliding and integrating for continuous time signals. Note that one of the signals is flipped on the time axis (like a mirror image) for the sliding/integrating process.

Last edited: Jun 28, 2010