fourier and laplace examples

[PG1995]

Thank you very much, Steve. It's really very kind of you.

Q1:
I don't know if I'm just going to make a fool of myself but it looks like you have used a different expression for Ck from the one we originally had in the problem under discussion. Please look here and I think you can notice the expression you used and the answer reached by you differ from the ones we originally had.

Q2:
Could you please also help me with this problem?

Thanks a lot for your time and patience.

Regards
PG

 

[steveB]

For Q1 above, perhaps I did the wrong problem. I worked out the calculation and conversion for Q4 from the first post. In any event, the two problems are very similar. The limit is 0 to 1 versus 0 to pi and the function is exp(-t) versus exp(-t/2), but all the math is the same and you can easily do the other case now.

For Q2, I think your statement is reasonably accurate. I hate to say anything is impossible because we can never be sure that there is not a viewpoint to make "regular" integration sensible. However, I think other mathematical techniques are typically used to evaluate these problematic non-converging (or ambiguous) integrals. One can approach the problem using complex integration techniques, or using transform properties, as you mentioned. Also, this issue helps motivate the use of Laplace Transforms. Such integrals can become convergent over some of the complex s-plane, even when s=jw gives non-convergent forms.

 

[PG1995]

Thank you very much.

So, you did Q4 from the first post. That's good because there is a problem with my solution for Q4; I discovered it after comparing it with yours. Please have a look here. Thanks.

Best wishes
PG

 

[steveB]

Thank you very much.

So, you did Q4 from the first post. That's good because there is a problem with my solution for Q4; I discovered it after comparing it with yours. Please have a look here. Thanks.

Best wishes
PG

So, I think you made one mistake, which is you have an extra factor of 2, which seems to be a mistake in calculating the period. The period is pi, so the factor in front should be 1/T=1/pi.

The other issue is not a mistake, but just a matter of finding the simpler form. You can notice that exp(i2n∏)=1 for any integer n.

I believe our formulas will match if you correct the mistake and simplify the form.

 

[PG1995]

So, I think you make one mistake, which is you have an extra factor of 2, which seems to be a mistake in calculating the period. The period is pi, so the factor in front should be 1/T=1/pi.

The other issue is not a mistake, but just a matter of finding the simpler form. You can notice that exp(i2n∏)=1 for any integer n.

I believe our formulas will match if you correct the mistake and simplify the form.

Hi

I think there is some other error. I'm using a little different formula. "L" is half of a period in the given formula. Instead of "-4", I should have "-2" as your solution has. Thanks.

 

[steveB]

It's hard to judge that formula without the full context of the definition of L and the inverse transform. However, it looks incorrect to me.

In your work you showed 1/(2L) in front that then you determined that L=pi/2. This seems to correctly put a 1/pi in front of the integral. Hence, I originally thought you just made a slip of the pen.

Anyway, just be careful of which definition you use. I find these myriad forms very confusing myself, which is another reason i use one, and only one, form for all of my work. I would not even trust someone else's published conversion formulas, but would instead directly derive them myself if I needed them.

 

[PG1995]

Thank you.

I have traced the error. Your comment, "However, it looks incorrect to me", was very helpful and it was good that I was able to track the error otherwise I could have made mistake in an exam. You were right the formula was wrong. The e-book has wrong version of the formula but I also have the paperback which gives correct formula.

Now we need to return to the original problem we were discussing yesterday and also some days ago. Please have a look here. Thanks.

Regards
PG

 

[PG1995]

Hi

I have a feeling that you haven't checked my previous post after it was updated. If you have, and don't have time to address that query at the moment then no problem. Take your time and please help me when you are free. Many thanks.

Regards
PG

 

[steveB]

I have time to address it because I can just do a cut and paste from the other problem. As I mentioned, the two problems are very similar. I think you have a couple of simple algebra mistakes in your work. I'll let you track them down using the following guide.

I took the Ck definition and determined the following formula.

[latex] c_k= \int_0^1 e^{-t\cdot(1+j\cdot 2\cdot\pi\cdot k)}\cdot dt=\frac{(1-e^{-1})}{(1+j\cdot 2\cdot\pi\cdot k)},\ \ {\rm for \ all} \ k[/latex]

Again, the standard trick when you have a complex number in the denominator is to multiply both the numerator and the denominator by the complex conjugate of the denominator. This leads to the following form.

[latex] c_k=\frac{(1-e^{-1})\cdot (1-j\cdot 2\cdot\pi\cdot k)}{(1+4\cdot\pi^2 \cdot k^2)} \approx \frac{0.632\cdot (1-j\cdot 2\cdot\pi\cdot k)}{(1+4\cdot\pi^2 \cdot k^2)},\ \ {\rm for\ all} \ k[/latex]

From here it is easy to see the form of the "a" and "b" coefficients for the rectangular cosine/sine form, because the conversions are the following.

[latex] c_k=\frac{1}{2}\cdot (a_k-j\cdot b_k),\ \ {\rm for} \ \ k=1,2,3 ...[/latex]
[latex] c_{-k}=\frac{1}{2}\cdot (a_k+j\cdot b_k),\ \ {\rm for} \ \ k=1,2,3 ...[/latex]

and the answers given before were as follows.

[latex] a_k=0.632\cdot \frac{2}{1+4\cdot \pi^2\cdot k^2}[/latex]
[latex] b_k=0.632\cdot \frac{4\cdot\pi\cdot k}{1+4\cdot \pi^2\cdot k^2}[/latex]

The zero coefficient can also be handled directly, but I've seen two forms of Fourier series with either ao or ao/2. As long as you are consistent with definitions, conversions and usage, the alternate forms will work out correctly.

 

[PG1995]

Thank you.

I have time to address it because I can just do a cut and paste from the other problem.

But I'm afraid this time it looks like there is also a problem with your solution because your final answer also differs from the one given by the book. Please have a look here. I'm extremely sorry if I'm being silly. Thank you.

Regards
PG

 

[steveB]

Even thought the form of the two problems you presented resulting in a nice simplification with exp(j2pi n)=1, you don't need such luck to solve these problems. When you ended up with exp(jpi n)=(-1)^n, you should have still been able to proceed. In general, you might have a complex number to deal with, and you should still be able to proceed. All you do is separate the real and imaginary parts.

Let's make this specific with your problem, even though you arrived at this point after making a mistake. Here is where you left off, but I am substituting c=0.73576 in to make it neater.

[latex] \frac{2-c\cdot (-1)^n-2\cdot j\cdot n\cdot \pi +j\cdot c \cdot n\cdot \pi\cdot (-1)^n}{4+4\cdotn^2 \cdot \pi^2}[/latex]

Now separate the numerator into a simple complex number with the real part and the imaginary part clearly visible.

[latex] \frac{(2-c\cdot (-1)^n)-j\cdot (2\cdot n\cdot \pi + c \cdot n\cdot \pi\cdot (-1)^n)}{4+4\cdotn^2 \cdot \pi^2}[/latex]

Since the denominator is real, you can make the final step easily.

[latex] \frac{2-c\cdot (-1)^n}{4+4\cdotn^2 \cdot \pi^2}+j\cdot \frac{ 2\cdot n\cdot \pi + c \cdot n\cdot \pi\cdot (-1)^n}{4+4\cdotn^2 \cdot \pi^2}[/latex]

You can now clearly identify the real and imaginary parts and associate them with the ak and bk coefficients.

There can be cases where the exp function part yields a complex number and not 1 or (-1)^n. You should still have no trouble because exp(ix)=cos(x)+j sin(x), and you can still do the math to separate the real and imaginary parts.

Work out some more complex examples to make sure you can do the math quickly and efficiently without confusing. For example try exp(-t) from t=1.48 to t=4.5. Here you will not get the nice simplifications.

 

[steveB]

Thank you.





But I'm afraid this time it looks like there is also a problem with your solution because your final answer also differs from the one given by the book. Please have a look here. I'm extremely sorry if I'm being silly. Thank you.

Regards
PG

No, you are not being silly. You correctly identified a mistake I made. So i failed to notice that the function they chose is zero from 1<t<2. I thought it was like the other problem and i treated the period as if it was 1, but the period is clearly 2. Sorry, my mistake.

So, your work is reflecting the correct problem. I didn't check all the details, but fortunately I already addressed the issue in my previous post. You should now be able to recheck your work and then proceed with the (-1)^n terms that do seem to be there.

 

[PG1995]

Thank a lot.

I was able to find a_n, b_n and a_0.

Let's make this specific with your problem, even though you arrived at this point after making a mistake. Here is where you left off, but I am substituting c=0.73576 in to make it neater.

[latex] \frac{2-c\cdot (-1)^n-2\cdot j\cdot n\cdot \pi +j\cdot c \cdot n\cdot \pi\cdot (-1)^n}{4+4\cdotn^2 \cdot \pi^2}[/latex]

I have been through it but sorry I couldn't find that mistake. I previously knew that that my answer and book's answer differ by a minus sign (but then a couple of days ago you said that the book's answer might have a typo). Now I believe you have checked my work carefully and found that book's answer is correct and there is a mistake in mine. If possible, let me know where that mistake occurred. Thanks a lot.

Regards
PG

 

[steveB]

I Wrote that you had a mistake before you showed me that I was the one who made the mistake. I have not checked the details of your work to know if it is correct. So far I've focused on making sure you have correct use of the tools/techniques, which it seems you have.

The simplest way for you to check your answer is to calculate a0, an and bn using the sine/cosine integral forms. Then compare the answers and make sure they agree.

 

[PG1995]

Thank you, Steve.

The simplest way for you to check your answer is to calculate a0, an and bn using the sine/cosine integral forms. Then compare the answers and make sure they agree.

Yes, you are correct. But that problem was not easy to solve using 'regular' integration formulas as we discussed in posts #2 and #3. But it's okay. The good thing is that now I have better understanding of the conversion.

For the last few days I have been observing one thing that there are few math problems, especially integrations ones, which need some kind of special software weapon to solve them. For example, my calculator, TI-89, really becomes useless. Perhaps, I need a computer algebra system such as Maple which I can get for free from my school. Have you used it? I would like to use a program which is easy to learn. Please recommend me the right program if you know of any. And I have Matlab installed and it has symbolic toolbox but I can't make heads or tails of it. Thanks a lot.

Best wishes
PG

 

[steveB]

For the last few days I have been observing one thing that there are few math problems, especially integrations ones, which need some kind of special software weapon to solve them. For example, my calculator, TI-89, really becomes useless. Perhaps, I need a computer algebra system such as Maple which I can get for free from my school. Have you used it? I would like to use a program which is easy to learn. Please recommend me the right program if you know of any. And I have Matlab installed and it has symbolic toolbox but I can't make heads or tails of it. Thanks a lot.

First, let's remember that it is the responsibility of every engineer, doing real work, to double-, triple and even quadruple-check their work to make sure every assumption, formulation and calculation is correct from an engineer's point of view (which differs from the mathematician's and scientist's point of view). It doesn't matter how difficult or easy this checking might be. It MUST be done for ethical reasons because, quite literally, life, property and jobs are at risk when an engineer makes a mistake.

In line with this, we are fortunate to have better and better tools as time goes on, and these symbolic processors are very good for saving time in the double-check process. You should never rely on them alone because you might misuse them or they might have bugs in them. In fact, I have found errors in answers given by symbolic processors more than once. I think MrAl mentioned doing calculations by three different ways, and I agree with this for any serious work. Non-critical (homework, test, self-learning, fun problem solving) work can be calculated by two ways to save time. And, you would be crazy not to use symbolic processors as one of your methods.

I used Maple briefly years for basic things, and I'm sure it is easy to learn and use. Similarly, I've heard good things about Mathematica. Personally, I use Matlab and Maxima mostly. Maxima is freely available, which is what got me to try it. It is very easy to use if you get a version with a menu driven front end on it. If you have the Matlab symbolic toolbox, a quick way to use it is to run the "mupad" tool, which brings up a menu driven window that is very easy to use.

 

[PG1995]

Hi

I was going through the material and thought that I should update some of the information. This relates to the post #11 where you said this:

Second, the Fourier transform (i) is correct, but the versions (ii) and (iii) appear to me to be the Fourier Sine Transform and the Fourier Cosine Transform, which are not quite the same thing. Formulas (ii) and (iii) are certainly related and can be used to construct the full transform, but I believe you have misinterpreted their meaning.

This text tells more about the cosine and sine forms of Fourier transform and when they are applicable. Also notice that the formulas I used previously while solving the problem are little different from the ones given in that text. Thank you.

Regards
PG

 

[PG1995]

Hi

Problem #1:
Could you please help me with this query? And I'm sorry for the poor quality image. If you can't read it well, then please let me know. Here is the conversion formula (in green highlight).

Problem #2:
Could you please help me with these queries? Thank you.

Regards
PG

 

[steveB]

For problem one, remember that 0/0 is an indeterminate case. In words, this means that you can't determine the angle with that formula. This turns out to make physical sense, because a vector with components of all zero values does not have a well defined angle.

 

[PG1995]

For problem one, remember that 0/0 is an indeterminate case. In words, this means that you can't determine the angle with that formula. This turns out to make physical sense, because a vector with components of all zero values does not have a well defined angle.

Can't we do this limit (x->0){arctan(x/x)}? The limit evaluates to 45 degrees. If I can't do this then how can I sketch the spectrum?

If possible then please give these queries a look? Thanks a lot.

Regards
PG