# unit impulse function, unit step function, unit ramp function

Discussion in 'Mathematics and Physics' started by PG1995, Oct 4, 2012.

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Hi

Regards
PG

2. ### steveBWell-Known MemberMost Helpful Member

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For part A, you just failed to consider the limits of integration carefully. If you look carefully, you will see that the first term is a delta function that is not inside the range of integration.

For part B, you would be correct if the delta function is a standard Dirac delta function. However, I'm confused by the notation. What does the subscript 2 mean in

???

3. ### PG1995Active Member

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Thank you, Steve.

I'm sorry. I see it now. I thought the delta function had been shifted toward the right. But the shift is toward negative x-axis.

It's a standard Dirac function because in the book the author hasn't introduced Kronecker function so far. Please let me know what to do. Thank you.

Regards
PG

Last edited: Oct 17, 2012

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5. ### steveBWell-Known MemberMost Helpful Member

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I dont' know what to do. If I assume it is a standard Dirac delta function, then I get 0, as you do. However, why would they use the subscript 2 for the δ symbol? Either their answer is wrong, or we are both making the same mistake or the 2 means something and the book answer is correct.

I'm not sure which it is.

6. ### PG1995Active Member

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Thank you.

You are right. The book does have something different to tell. Mostly I use electronic copy of the though I also happen to have the hard copy. There are few differences between the two. This is what I found in the book. I hope it settles the confusion. It's actually a rect(t) function. But I'm still at loss to solve the question! Thank you.

Regards
PG

Last edited: Oct 17, 2012
7. ### steveBWell-Known MemberMost Helpful Member

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OK, good job finding that definition. If we assume this definition is correct, then I still get an answer of zero, not one. So, either the book is wrong, or we still have the wrong definition of
.

Personally, I don't worry about weird questions like this. If this is required homework, then just write up your interpretation of what you think the question asks and be clear about the uncertainty of the definition. Then give your best attempt at the solution. Point out it doesn't agree with the book answer.

I would say don't worry about this at all, but perhaps you will later get a test question related to this. So, best to get to the bottom of this by consulting your teacher. If your teacher can't make sense of it, then skip it and you won't be missing anything important as far as I can tell.

Last edited: Oct 17, 2012
8. ### PG1995Active Member

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If possible, could you please tell me how you found answer to be still zero? I suspect integration interval lies outside the limits of rectangular pulse function. I found new limits for rectangular function to be: -1/3 < t < 1/3; but the integration interval is (1/2, 5/2). Or, perhaps I have it all wrong. It could be that you are using substitution approach.

Best wishes
PG

9. ### steveBWell-Known MemberMost Helpful Member

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I just worked it out quickly in my head, so I hope I didn't make a mistake. However, I saw it exactly as you say above.

We also have to consider that the total integrated area of that pulse is only 1/3, so there should be no way to get an area of 1. So, that's why I say that either the book is wrong, or we have the wrong definition for

10. ### PG1995Active Member

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Well, I hope the book is wrong!

Thanks.

11. ### PG1995Active Member

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Hi

Regards
PG

PS: I have just realized that the solution part is not easily readable. The handwriting is not readily legible and moreover the capture is not clear. If you like I can update it. Thanks.

Last edited: Oct 19, 2012
12. ### steveBWell-Known MemberMost Helpful Member

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First of all, I'm surprised you would get 4 for question B, but maybe this is just a typo. Perhaps question A is 4 and question B is not 4. Surely, question B can not be 4 because the signal is scaled by the constant A. Anything scaled by "A" would either have to have energy 0, infinity or a constant multiplied by A^2.

For your actual question C, you just need to be careful drawing the waveform, and then be careful to square it correctly. When you square real numbers you will never get a negative number. Once you correctly draw a plot of the signal squared, the answer will be obvious.

Last edited: Oct 19, 2012
13. ### PG1995Active Member

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Hi

Here you can see the solutions for parts (a), (b), and (c). Sorry about the writing, I was really doing it quickly. I have reworked the part (c). I still think there is no problem with my solutions, and in my opinion the answer for part (c) should be "10". Please help me with this. Thank you.

Regards
PG

PS: I have figured out the solution for the part (c). I was thinking about the problem in wrong way.

Last edited: Oct 20, 2012
14. ### PG1995Active Member

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even and odd parts of a function

Hi

Could you please help me with this query? Please note that my query is only related to even part of the function. Thank you.

Regards
PG

Last edited: Oct 20, 2012
15. ### steveBWell-Known MemberMost Helpful Member

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I'm confused because you say the plot on the left is your answer for the even part of the function, but you are showing only g(t) and g(-t). You would need to add these and divide by two to get the actual even function g_e(t). This is NOT shown in your plot.

Almost looks like you had one of my "senior moments", but you are too young for that.

16. ### PG1995Active Member

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Thank you.

Yes, I realized that what you are saying once I was done with it. But I don't think adding the shown g(t) and g(-t) and dividing the sum by "2" would give you the result shown in the book's answer.

You are not that much "senior"! You have to wait some more years for that. But I do have quite a few crazy moments everyday.

Best wishes
PG

17. ### steveBWell-Known MemberMost Helpful Member

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Hmmmm .... It looks to me that it does give the correct answer, at least in general shape. I'm just checking by eye, and my "almost senior" eyes are not as good as they used to be, but check it again and see if it doesn't add up correctly.

This reminds me of one of those comparison puzzles where you are supposed to look at two pictures and see what is different between them. The only thing I see wrong with your sketch of the plot is that the peak value of g(t) appears to be nearer to x=1.5, rather than x=1. So maybe just sketch it a little better, and then add it up.

EDIT: looking again, it appears that you plotted g(t)/2 and g(-t)/2 rather than g(t) and g(-t)

Last edited: Oct 20, 2012
18. ### PG1995Active Member

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I'm sorry but it still doesn't make sense to me. g(t)_even = {g(t) + g(-t)}/2 = g(t)/2 + g(-t)/2. I have looked quite carefully (at least I have tried to!) but nothing adds up to give the form shown in the book's answer. I have placed the graphs side by side for easy comparison. Thanks a lot.

Regards
PG

19. ### steveBWell-Known MemberMost Helpful Member

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Have you tried adding the two functions in your plot? You are plotting g(t)/2 and g(-t)/2. So, just add them together and the shape will match.

As I mentioned above, the value of "x" where g(t)/2 and g(-t)/2 is maximum seems a little off in your plot. It should be at x=1 and x=-1, respectively, but your plots seems more like x=1.4 and x=-1.4.

Last edited: Oct 20, 2012
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20. ### PG1995Active Member

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Thank you very much.

It was all craziness!

Best regards
PG

21. ### MrAlWell-Known MemberMost Helpful Member

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Hi PG and Steve,

As the attachment shows, we can do this in a few relatively simple steps.

Starting with the original function (blue) if we take the right side and divide by 2 and reflect that we get the green function just to the right following the dark arrow.
Now if we SUBTRACT that green curve from the blue one, we get the red curve below the first green curve (follow the dark arrow again).
From the resulting red graph, we can see that this is not yet a true odd function, but it's easy to see what we have to do to make it so. All we have to do is add 1/2 to all points (shift it up by 1/2). This brings us to the next red graph (following the dark arrow again), where we see it is now a true odd function.
But since we had to add 1/2 to the red part of the (now) composite function, that means we have to subtract 1/2 from the other part of the function (the green part). Doing this we get the final green function shown above the final red function (follow dark arrow again).
So we are left with one odd function and one even function.

BTW, is the book wrong again with the first set, where we have a single pulse? The two composite functions they have shown as the presumed solution dont add up to that single pulse function, they add up to twice that.

Last edited: Oct 20, 2012