understanding convolution

[PG1995]

Hi

Q1:
Please note that I have very basic knowledge of signal and system theory. I have no knowledge of Fourier transforms etc. I'm only after intuitive understanding of the convolution if that's possible. I have been through different webpages but they don't really make much sense to me but out of all the ones I have been to **broken link removed** one was the best.

Q2:
Please help me with **broken link removed** query. Please note that in the attachment I have labelled the query as "Q1", just ignore the "1". Here you can see the complete PDF which I have used for the attachment.

Q3:
I have been told that when an impulse function is transformed into frequency domain (by the way, which I haven't studied so far) it contains the components of all frequencies with the same magnitude. What does this really mean in 'easy' words?

Thanks a lot for the help.

Regards
PG

Update Q1: I think the concept of convolution comes into play whenever we are considering the effect of some previous input values with the effect of present input values. For instance, suppose you are lecturing in some big hall where significant echo is produced and some individuals are listening to you. Everything you speak can be considered an input. The persons who are listening, listening is output, to you are not only hearing what you have just said but there are hearing the combined effect of what you said some moments ago, i.e. echo, and what you have just said. Please note that an echo in itself is a combination of echos. The effect of what you have just said would be more prominent than anything else and effect of each recent echo would be more prominent than the previous echos.

Update Q3:

Mathematically, how the impulse is described depends on whether the system is modeled in discrete or continuous time. The impulse can be modeled as a Dirac delta function for continuous-time systems, or as the Kronecker delta for discrete-time systems. The Dirac delta represents the limiting case of a pulse made very short in time while maintaining its area or integral (thus giving an infinitely high peak). While this is impossible in any real system, it is a useful idealisation. In Fourier analysis theory, such an impulse comprises equal portions of all possible excitation frequencies, which makes it a convenient test probe.

From Wikipedia article on impulse response.

 

[MrAl]

Hi,

A simple way to look at convolution of two functions is that one function is like an amplifier circuit or filter circuit and the other is the input, and the output is the input function modified by the filter function.
A few simple examples:
An amplifier with a gain of 2:
Vout=Vin*2
where that '2' is the amplification factor, and what we did to get Vout was we convolved Vin with that 2 and got Vin *2.
In fact, in the frequency domain convolution is just multiplication, that's why we get Vin*2 because we multiply Vin times 2.
So if we have two functions in the frequency domain F1 and F2 the output is:
Output=F1*F2
If F1 was the input and F2 the filter function then so be it, but F1 and F2 could be two filters wired together one after the other:
Vin-->F1--->F2---Vout
In this case we would actually have to convolve the input with F1 and F2 also:
Vout=Vin*F1*F2
and it's as simple as that.
So if we had two amplifiers where the second one amplified the output of the first, and the first had a gain of 2 and the second a gain of 3 we would get:
Vout=Vin*2*3=Vin*6
and that has to be as simple as we can get.

In the time domain you have to use the integral because we can not multiply in the time domain, but in the frequency domain we just multiply.
[Note in the above examples you could multiply in the time domain too, but that's not generally the case where the functions are time dependent so we either use the integral or transform to the frequency domain and multiply, then transform back to the time domain]

The frequency content of a signal is what it is. Im sure you have read that a square wave has components that are all odd frequencies and each amplitude is 1/N, well other signals have other components and that's just because that is what happens when we move from the time domain to the frequency domain...it takes a certain number of frequencies at the correct amplitudes to add up to the same signal that we have in the time domain. So adding all the components in the frequency domain gives you the signal in the time domain.

 

[steveB]

Q2 Response.

My first comment is that I'm curious why you are concerned about this question. Is it just a point of curiosity and for learning how to be mathematically correct? Or, do you feel this issue will impact the answer?

My view is that the distinction is not too important, since the limits of integration and the final answer are not affected by how you define the regions.

However, in order to answer the question, one has to be very careful about definitions. How are the functions defined and what is the definition of "overlap". The function definition matters because when you have steps and flat pulses, you need to define where the endpoints of function pieces go (left or right side). The definition of "overlap" matters because one has to question whether functions that have only one nonzero point in common are overlapping. To me the function definitions and the definition of overlap are arbitrary. However, perhaps mathematicians have standards and definitions already in place. However, if they do, I don't know what they are.

 

[MrAl]

Hello,

The first rule is to include each point only once. I think you've already done this. So
0<=x<=1 for a region A and then 1<=x<=2 for region B would not be correct because the point x=1 had been included twice.
This would have to be corrected as either:
0<=x<=1 and 1<x<=2
or
0<=x<1 and 1<=x<=2
but i think you got that already.

The second rule is that if the infinitesimal area is the same for both ways of defining the region then it doesnt matter which way we do it.
For example, say region A above is a ramp from y=0 to y=1, and region B is a pulse of amplitude 1. This wave is just a ramp up to 1 from x=0 to 1 and then a pulse from x=1 to 2. Note some ambiguity already. Is the first region 0<=x<=1 and the second 1<x<=2 or is the first 0<=x<1 and the second 1<=x<=2 ?
If we look at the point where they join, x=1, we see that the infinitesimal area y*dx is the same for both the ramp and the pulse, so it doesnt matter which way we define it as long as we also observe rule 1 above.
There is something to be said about consistency though, in that if we define the right hand side as x<=a and the left hand side of the next region as a<x then we should not swap and turn to x<b for the right and x<=b for the left for any of the remaining regions.

The exception to this is when we intend to discretize the region along x. Depending on the step size, choosing the borders may have a big impact on the result. With the ramp and pulse example, choosing x=1 to be part of the ramp will lead to a result that is too small, and that's because part of the pulse will get included with the ramp and the ramp has less area per unit x than the pulse does. The remedy then might be to exclude x=1 from both regions, and define a third region that handles the region close to x=1 in a totally different manner. This is a side point though and doesnt matter as much for the subject of the paper.

 

[PG1995]

Thanks a lot, MrAl, Steve.

My first comment is that I'm curious why you are concerned about this question. Is it just a point of curiosity and for learning how to be mathematically correct?

I was just curious. But I believe sometimes one shouldn't be too curious. Curiosity killed the cat!

Best wishes
PG

 

[steveB]

Thanks a lot, MrAl, Steve.



I was just curious. But I believe sometimes one shouldn't be too curious. Curiosity killed the cat!

Best wishes
PG

Killed many people too, but also it leads to many great accomplishments. Notice that I was curious if you were curious. Just can't get away from that curiosity thing, can we!

Anyway, did we answer adequately enough to help?

 

[PG1995]

Killed many people too, but also it leads to many great accomplishments.

In my opinion, when one is enrolled in an academic institution then the former case is more true than the latter.

Anyway, did we answer adequately enough to help?

Of course, you guys did!

 

[PG1995]

Hi

Please help me with **broken link removed** query. Thank you very much.

You can see the original PDF here.

Best wishes
PG

 

[steveB]

The question you are asking here is a bit strange in the sense that you are asking a very specific question, but the failure to understand the specific point is an indication that you don't understand the method itself. In order to understand why the limits are what they are, you need to understand the operations being done and the reasons for creating the separate regions.

Anyway, to answer the specific question, region 4 is the range where the narrower pulse is sliding out of the range (region 3) where it is completely inside the wider pulse, as it is sliding along, being dragged by the parameter "t". Hence, the upper limit is 2, because the wider pulse is fixed and ends at the point tau=2.

I'm not sure if this answer will make sense, since it relies on a full understanding of the method.

 

[PG1995]

Thank you, Steve.

Kindly help me with **broken link removed** related problem. Thanks.

Best wishes
PG

 

[steveB]

It seems that you have it understood, but I don't know why you chose those limits for region 3. Regions 2 and 4 are more difficult and you got those right.

A key point to notice is that x1 is nonzero only in the range of -0.5<t<0.5. Hence, no limits should go beyond these endpoints. In region 3, the limits are these endpoints because x2 is wider and in sliding past x1, but the narrower pulse in x1 forces the limits to be fixed at -0.5 and +0.5.

The pictures shown are very helpful, but you need to do more than make the pictures. You need to visualize the motion of once function moving past the other function. The parameter "t" is the thing that "drags" the moving function and the other function is fixed. Just imagine what's happening as if it is an animation. When you imagine regions 3, you should see x1 fixed and narrow, with x2 moving and wider. Hence, the limits do not change in this region.

 

[PG1995]

Thank you, Steve.

It seems that you have it understood, but I don't know why you chose those limits for region 3. Regions 2 and 4 are more difficult and you got those right.

I don't understand the method yet. I was taking help from **broken link removed**. I believe doing more examples might help me. Thanks a lot for always being there to help me. Much obliged.

With best wishes
PG

 

[PG1995]

Hi

Just did **broken link removed**. Looks fine though I still don't exactly understand it! If you have time, please give it a look, and kindly let me know if there are any errors. Thanks a lot.

Regards
PG

 

[steveB]

It seems ok. I'm not sure why you don't exactly understand it. Are you able to visualize all of the steps.

1. Flip one signal around to get x(-tau)
2. Shift that signal by t to get x(t-tau)
3. consider the product with the other function x(t-tau)*h(tau)
4. consider regions where integral of this product can be simplified

The last step is the hard part because different functions can be simplified in different ways. The key is visualizing the movement of the shifted function by the fixed function. Then, it is easier to identify regions where simplifications are possible.

 

[PG1995]

Thank you.

Don't you think the **broken link removed** I used for the Region 1 should be -∞ to -0.5, and for the Region 5 it should be 3.5 to ∞? I know limits don't really matter for these regions but still choosing wrong limits shows that I was visualizing it wrongly. Thank you for the patience.

Regards
PG

 

[steveB]

It's not clear what the limits mean in regions 1 and 5. Strictly, the limits are always -inf to +inf, but we cut the limits down when the integrand is clearly zero over some of this range.

However, I noticed that what you wrote for region 5 is incorrect. You are correct that there is no overlap of the pulses and that the answer is zero, but you wrote the integrand as 8, when it is zero. Sorry, I didn't notice this earlier. It's probably just a slip of the pen anyway.

 

[fermat7]

The convolution product return the area of the overlapping region of the product of the two signals, it is function of the amount of translation time of one of the two signals, it is well illustrated here by StevB's post.

 

[PG1995]

discrete-time convolution

Hi

Could you please help me with **broken link removed**? Thank you very much.

Regards
PG

 

[steveB]

Why do you care about our preference? If you are doing hand calculations, use the method you prefer when given a choice, and use the method you are told to use, when you are not given a choice.

I think a more relevant question is whether one method is more efficient (numerically). As I've mentioned, you will never need to do these calculations by hand in the real world. Hence, personal preference is completely irrelevant. However, if you don't use Matlab to do the calculations for you, you may decide to program one of these methods yourself using a a high level computer language.

If you do program it, then you will care about whether one method is easier to program, or whether one method is numerically more efficient (i.e. will run in less computer time). If I were going to program a convolution algorithm, these are the two main questions I would ask. However, since I'm never going to do that, I won't bother to figure out the answers to these questions.

 

[PG1995]

Hi

I'm so sorry to ask you this again but don't you think **broken link removed** I used for the Region 2 should be "-0.5" instead of "0.5"? Kindly let me know. Thanks a lot.

Regards
PG