How to calculate spectrum coefficient amplitude.

Thread Starter

lam58

Joined Jan 3, 2014
69
A question asks to calculate the spectrum amplitudes at harmonic frequencies which are closest to the frequencies of the sinusoid sample. The sinusoid has an amplitude of 220v with 50Hz freq and is sampled at 200Hz over 0.21s.

Not even sure how to start answering this question. Any advice?
 

shteii01

Joined Feb 19, 2010
4,644
Fourier series stuff.

This seems relevant: http://www.dspguide.com/ch13/4.htm

So you have a signal: x(t)=220*cos(2*pi*50*t)
I make a couple of assumptions:
1) The signal is even, that makes it cosine.
2) I am going to center the signal at t=0 seconds. This means that the signal is not from 0 seconds to 0.21 seconds. The way I set it up is that the signal is from -0.105 seconds to +0.105 seconds, total of 0.21 seconds, but centered on t=0.

Now just plug the numbers into the formula:
an=2*Amplitude*sin(pi*n*(k/T))/(n*pi)

Amplitude is known.

n is the harmonic, if you want to find amplitude of second harmonic, replace n with 2.

k is 0.21 seconds.

T is period of the sinusoid, you can calculate it using frequency of the sinusoid.




No offense mate, but if you are in school, then you should stop paying for your school because you are not getting anything for your money.
 

WBahn

Joined Mar 31, 2012
33,187
Is this signal that you are sampling coming from a mathematical function or is it, at least presumably, a real signal that has noise and harmonics in it?

Assuming the latter, you first have to determine what frequency is the closest harmonic to your fundamental.

Then you have to figure out how to single out that frequency from the data. What you are basically looking for is how to do a single frequency Fourier transform, which is closely related to the Fourier series.
 

Thread Starter

lam58

Joined Jan 3, 2014
69
Fourier series stuff.

This seems relevant: http://www.dspguide.com/ch13/4.htm

So you have a signal: x(t)=220*cos(2*pi*50*t)
I make a couple of assumptions:
1) The signal is even, that makes it cosine.
2) I am going to center the signal at t=0 seconds. This means that the signal is not from 0 seconds to 0.21 seconds. The way I set it up is that the signal is from -0.105 seconds to +0.105 seconds, total of 0.21 seconds, but centered on t=0.

Now just plug the numbers into the formula:
an=2*Amplitude*sin(pi*n*(k/T))/(n*pi)

Amplitude is known.

n is the harmonic, if you want to find amplitude of second harmonic, replace n with 2.

k is 0.21 seconds.

T is period of the sinusoid, you can calculate it using frequency of the sinusoid.




No offense mate, but if you are in school, then you should stop paying for your school because you are not getting anything for your money.
Really why do you say that?
 

shteii01

Joined Feb 19, 2010
4,644
Really why do you say that?
Because I have taken DSP class and I did not need to go look for help on the internets.

The amount of time you spend here looking for help tells me that you are having problems with how your instructor presents the material or you have an issue with the textbook for your class, or both.
 

Thread Starter

lam58

Joined Jan 3, 2014
69
Because I have taken DSP class and I did not need to go look for help on the internets.

The amount of time you spend here looking for help tells me that you are having problems with how your instructor presents the material or you have an issue with the textbook for your class, or both.
Just the way I learn, my tutors are very good at teaching, this is just the way I've always done it. The course I'm on is in a different country from where I come from so a lot of the content in the text books are going on the assumption that you've already learned previous material. Where as in my case I haven't. However I've been doing ok so far, I might be asking loads of questions on here and sound a bit ignorant at times but I soon pick it all up.

Better to ask on the internet when I can't get a hold of my tutor than to struggle and fail my exams when they come.
 
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