NCEES 519 Fourier Series

Professional Engineer & PE Exam Forum

Help Support Professional Engineer & PE Exam Forum:

This site may earn a commission from merchant affiliate links, including eBay, Amazon, and others.

schmidty99

Well-known member
Joined
Sep 14, 2007
Messages
128
Reaction score
0
Location
Mondovi, WI
Hey Guys back again. I'm having a a little trouble with this problem:

"A waveform having period T=2 is described by a constant value of 10 over the interval 0<t<1 and 0 over the interval of 1<t<2. This signal is repeated continually for all time "t." The first terms of the trigonometric Fourier series are most nearly:

(the answer is 5.0 + 6.4 sin (3.14t) + 2.1 sin (9.42t) + ... ) The given answer states that is has odd symmetry and half-wave symmetry.

My questions are as follows:

1. How does one figure out that is has odd and half-wave symmetry from what the question states? When I thought about it and drew it out I could tell it was a square wave, but that's about it.

2. Knowing the symmetry and tayloring my equations for the "b" coeffients I cannot come up with answer above. I end up with cosine terms.

3. With the information from the question only, I should be able to do the Fourier series equations and get the correct answer. In this case, all the "a" coefficents should go to zero and leave only "b" which would point to odd symmetry. Correct? To be honest I didn't actually try that....yet - its a lot of work.

The worst part is, I thought I had this figured out the last time I took the exam in Oct 2010, but working through it again, I'm just as confused as before. Any help would be appreciated. GrossBeckEE, got any gas left in the tank?

 
Ok, I think I have this figured out to the point of the integral which is:

An = 1/2 (Integral from 0 to 1) 10 cos (n*pi*t)2 + 1/2 (Integral from 1 to 2) 0 cos (n*pi*t)2 where n=1, 3...

Hope that makes sense. I'll be damned if I can get the answer to work out. HELP!

 
I don't have my book on me, but I'll look at it this weekend. To be perfectly honest I took the slow, but guaranteed, route with all things Fourier Series: I integrated over a single period of the waveform. Sure you can look for the symmetry and simplify things, but integrating is always a solid check on your work. I'll work it both ways, but I think you'll find that the peace of mind found from integration is worth the minute or two that you lose taking the long route.

 
1. How does one figure out that is has odd and half-wave symmetry from what the question states? When I thought about it and drew it out I could tell it was a square wave, but that's about it.
If you imagine the y-axis (x=0 line) to be the mirror, then don't move the mirror (left or right) but you are allowed to move the x-axis (y=0 line) to any other position. For this waveform, if you offset the x-axis up to y=5, then you can see that the waveform has odd symmetry. Hence FS will consist of sines only. Average value is 5. So, you are looking for something line y = 5 + sum of harmonic sines

To narrow down further, I think you have to do the integrations.

 
Thanks Guys. After some youtube videos and more reading, I think I understand the symmetry aspect of the problem. The trouble I'm having is with the integration. Like GB said, I'd rather take the longer route and do the integrations to make sure I'm correct. This is one of the problems I have to get correct on the exam to pass it. Kind of the low-hanging fruit aspect, the problem is I'm having trouble reaching it. :smileyballs:

 
Does this help -

http://mathworld.wolfram.com/FourierSeriesSquareWave.html

You may have already seen this.

For a square wave it might make sense to just memorize (or write down in your notes) the formulation of the terms - it's a pretty common type of problem. Integration is a universal method, but why do it if you don't have to?

 
Last edited by a moderator:
Thanks for the input benbo.

Well, yes and no. I understand the symmetry, the even and odd, etc. The diagram is helpful, but the integration confuses me. So in the final summation, is all we're really looking to do with the Fourier series is find coefficent that comes before summation sign? In the NCEES answer, I need to be able to calculate the coffecients of the series. Right now I can't come up with those answers...

 
Last edited by a moderator:
Thanks for the input benbo.
Well, yes and no. I understand the symmetry, the even and odd, etc. The diagram is helpful, but the integration confuses me. So in the final summation, is all we're really looking to do with the Fourier series is find coefficent that comes before summation sign? In the NCEES answer, I need to be able to calculate the coffecients of the series. Right now I can't come up with those answers...
I'm not sure what you're trying to find.

Here is what you say in your post -

the answer is 5.0 + 6.4 sin (3.14t) + 2.1 sin (9.42t) + ... ) The given answer states that is has odd symmetry and half-wave symmetry.
As you know, you don't need to do any integration to get the symmetry - you can tell that by looking at the figure.

So in the final summation, is all we're really looking to do with the Fourier series is find coefficent that comes before summation sign?
Maybe. As far as the coefficients, the answer you give here is the entire series. The coefficients are 6.4, 2.1, etc.

I guess the 5 also counts as a coefficient, although it is just a translation.

an, bn these are the coefficients, but since it's odd, all the an are zero.

For bn you just plug into the definition, but since it is from -T to T instead of -pi to pi you make a change of variables.

This explains it better-

http://mathworld.wolfram.com/FourierSeries.html

But like I said, on the test you don't need to know how to integratre everything. It's better if you do, but in a pinch you can write down all the common types of functions and their coefficients/series. Remember, you just need to pick out the right answer.

 
Last edited by a moderator:
Thanks for the input benbo.
Well, yes and no. I understand the symmetry, the even and odd, etc. The diagram is helpful, but the integration confuses me. So in the final summation, is all we're really looking to do with the Fourier series is find coefficent that comes before summation sign? In the NCEES answer, I need to be able to calculate the coffecients of the series. Right now I can't come up with those answers...
I'm not sure what you're trying to find.

Here is what you say in your post -

the answer is 5.0 + 6.4 sin (3.14t) + 2.1 sin (9.42t) + ... ) The given answer states that is has odd symmetry and half-wave symmetry.
As you know, you don't need to do any integration to get the symmetry - you can tell that by looking at the figure.

So in the final summation, is all we're really looking to do with the Fourier series is find coefficent that comes before summation sign?
Maybe. As far as the coefficients, the answer you give here is the entire series. The coefficients are 6.4, 2.1, etc.

I guess the 5 also counts as a coefficient, although it is just a translation.

an, bn these are the coefficients, but since it's odd, all the an are zero.

For bn you just plug into the definition, but since it is from -T to T instead of -pi to pi you make a change of variables.

This explains it better-

http://mathworld.wolfram.com/FourierSeries.html

But like I said, on the test you don't need to know how to integratre everything. It's better if you do, but in a pinch you can write down all the common types of functions and their coefficients/series. Remember, you just need to pick out the right answer.

I see what you are saying. Over the last couple days, I've learned enough that I understand I should have sine terms in my answer. Then the problem I have is calculating the numbers 6.4, (3.14t), 2.1, (9.42t). I can't "educated guess" the right answer without calculating those terms. (I don't think so anyway.)

Its getting to the point where I'm probably over thinking the whole problem. Maybe this will help: When I set up the bn equation and calculate, all my sine terms change to cosine per intgration. So, its impossible for me to get the NCEES answer because it has all sine terms. What am I doing wrong? What part am I missing?

 
Last edited by a moderator:
Its getting to the point where I'm probably over thinking the whole problem. Maybe this will help: When I set up the bn equation and calculate, all my sine terms change to cosine per intgration. So, its impossible for me to get the NCEES answer because it has all sine terms. What am I doing wrong? What part am I missing?
Yeah, maybe. I think more acurately you are confusing two things that you need to keep separate. I'm not sure it's entirely true to say this, but for simplicity sake - The SINE in the terms is not the SINE you use to calculate the coefficients. They are two separate things. Separate it in your mind.

IGNORE MY EARLIER POST HERE - I have to go back and look at the formula again, because I think I am I am calculating the integral incorrectly. But the part about the coefficients and the terms being separate still stands. Just the fact that you may wind up with a cosine doesn't matter because you will be caluclating a definite integral. But I'm not sure getting a cosine here is correct.

But once you do get the coefficients, the SINE comes back. You just put it here, along with the n and the t..

 
Last edited by a moderator:
^^^

schmidty, I may be wrong but I think the problem is that to calculate the coefficients you need to calculate the integral of the Heaviside Step function times sine(n*pi*x). This may requires integration by parts, giving you a sin squared term somewhere.

Although there may be a trick to doing it. I'm pretty rusty at this, but when I get home I'll give it a shot if I have time. Or GB may answer it. I'd be curious to see if I'm overlooking something.

That said, I still think that you don't want to be doing this on the test if you don't have to. If you can write out some standard formats for Fourier series of things like sawtooth, imulse, step function, etc and include them in your notes I think you will be better off.

 
Last edited by a moderator:
Yeah, maybe. I think more acurately you are confusing two things that you need to keep separate. I'm not sure it's entirely true to say this, but for simplicity sake - The SINE in the terms is not the SINE you use to calculate the coefficients. They are two separate things. Separate it in your mind.

I was starting to think along those lines, that would be the only way it would work. If that's how it works, then I have the 95% of the problem figured out. Maybe GB can show me the math.

i would agree with your last statement, I'll put together some notes on to deal with the various waveforms. Thanks!

 
Yeah, maybe. I think more acurately you are confusing two things that you need to keep separate. I'm not sure it's entirely true to say this, but for simplicity sake - The SINE in the terms is not the SINE you use to calculate the coefficients. They are two separate things. Separate it in your mind.
I was starting to think along those lines, that would be the only way it would work. If that's how it works, then I have the 95% of the problem figured out. Maybe GB can show me the math.

i would agree with your last statement, I'll put together some notes on to deal with the various waveforms. Thanks!
Yeah - I think you already knew what I wrote in my post, so I wasn't much help. I'm having a little trouble figuring out how they came up with the coefficients myself (although they seem to generally match what's inthe standard format), but I'm sure I must be overcomplicating it. But now I can't help but think about it until you, me or somebody else figures it out.

Take a look at this.

http://www.facstaff.bucknell.edu/mastascu/...ml#Coefficients

It shows the integrals that produce the coefficients as cosines, which is what you are getting and what makes sense to me. I'm not sure where they got the sin^^2 in the other thing I linked to.

In other words, I think you are probaably correct in the cosine thing you just need to evaluate the definite integral. And select the proper limits of integration if you make a change of variables.

 
Last edited by a moderator:
^^^schmidty, I may be wrong but I think the problem is that to calculate the coefficients you need to calculate the integral of the Heaviside Step function times sine(n*pi*x). This may requires integration by parts, giving you a sin squared term somewhere.

Although there may be a trick to doing it. I'm pretty rusty at this, but when I get home I'll give it a shot if I have time. Or GB may answer it. I'd be curious to see if I'm overlooking something.

That said, I still think that you don't want to be doing this on the test if you don't have to. If you can write out some standard formats for Fourier series of things like sawtooth, imulse, step function, etc and include them in your notes I think you will be better off.

Ok. I worked the problem again using cosines and I came up with everything except the 3.14t and 9.42t. I have pi/2 and 3*pi/2 in the terms. I need to get rid of the 1/2 and I'll have it. Grrr...

 
Last edited by a moderator:
^^^schmidty, I may be wrong but I think the problem is that to calculate the coefficients you need to calculate the integral of the Heaviside Step function times sine(n*pi*x). This may requires integration by parts, giving you a sin squared term somewhere.

Although there may be a trick to doing it. I'm pretty rusty at this, but when I get home I'll give it a shot if I have time. Or GB may answer it. I'd be curious to see if I'm overlooking something.

That said, I still think that you don't want to be doing this on the test if you don't have to. If you can write out some standard formats for Fourier series of things like sawtooth, imulse, step function, etc and include them in your notes I think you will be better off.

Ok. I worked the problem again using cosines and I came up with everything except the 3.14t and 9.42t. I have pi/2 and 3*pi/2 in the terms. I need to get rid of the 1/2 and I'll have it. Grrr...
THe 3.14t and the 9.23t should be n*pi*t with n = 1 and 3 (the odd coefficients for bn = n=1, n=3, etc. You get that by substituing for nwt in the sine. Of course, 9.23 is not 3*pi so maybe back to the drawing board for me. It seems like it should be 9.42t

You should not have a 2 in the denominator because it cancels out when you convert the angular frequency in the numerator = you would have w = 2pi*t/T, where T=2. THe 2s cancel out.

To calculate the coefficients did you make a substition u= nwt = 2*n*pi*t/T = 2*n*pi*t/2 = n*pi*t

Then du = n*pi*dt so dt = du/(n*pi)

Then you just integrate udu from 0 to 4*pi. At least that's how I think I finally solved it once I figured out it wasn't that darned complicated. THere may even be an easier way.

 
Last edited by a moderator:
Ok. I believe you, but I don't see anywhere in the Fourier formulas I have seen where the angular frequency "w" comes into play. I know w=2*pi*f or (2*pi)/T. How would I have known that? Am I just not recognizing it? I mean, what you're saying makes sense, but I just don't see it....

 
Last edited by a moderator:
Ok. I believe you, but I don't see anywhere in the Fourier formulas I have seen where the angular frequency "w" comes into play. I know w=2*pi*f or (2*pi)/T. How would I have known that? Am I just not recognizing it? I mean, what you're saying makes sense, but I just don't see it....
Sorry to confuse again. 2*pi*f is the angular frequency, but you don't necessarily need to know that.

Even if you start with 2*n*pi*t/T the 2s should still cancel out - the 2 in the numerator from the formula should cancel out fromthe 2 in the denominator from the period T=2). Doesn't it?.

 
Last edited by a moderator:
Ok. I believe you, but I don't see anywhere in the Fourier formulas I have seen where the angular frequency "w" comes into play. I know w=2*pi*f or (2*pi)/T. How would I have known that? Am I just not recognizing it? I mean, what you're saying makes sense, but I just don't see it....
Sorry to confuse again. 2*pi*f is the angular frequency, but you don't necessarily need to know that.

Even if you start with 2*n*pi*t/T the 2s should still cancel out - the 2 in the numerator from the formula should cancel out fromthe 2 in the denominator from the period T=2). Doesn't it?.
It doesn't start out with a 2 in the numerator, that's why I can't get it to work out. ????

 
Attached is the long form method of integrating this problem. It is more complex than "eye-balling" the symmetry, but I got into the habit of integrating everything in place of actual thinking. For an 8 hour exam and 80 problems, this leaves 6 minutes per problem, which is more than enough time to integrate, especially if you skip things. E.g., why calculate ao? All answers have the same ao, so why bother. Anyway, the answer comes out to B via this method. In any case, if you learn to eye-ball these problems as well then you are pretty much guaranteed that your answer is correct if the integration matches what your eyes are telling you.

Lastly, be VERY wary of the stupid radian scale shenanigans. Chapter 10, section 6 explains the Fourier series fairly well, but I wish they would just skip putting it into radian scale and do the examples the long way. I hope this helps.

Fourier_Series.gif

 
Back
Top