Finding the amplitude of the spectral component of a wave

Thread Starter

BRetonDP

Joined Apr 11, 2014
5
Hello, AAC community!

I am just beginning my Power Electronics course and the problem posted in the attachment is part of my first homework.

I do not understand what is the "spectral component" of the wave, and what steps I should follow to find it. I've tried googling what the spectral component of a wave is, but haven't really found instructions to solve a problem as this one. Can anyone give me a hand and help me answer the problem, lead me in the right direction or show me the reading material to figure this problem out?

I am currently a study abroad student and am not taking all of the courses I should be taking back in my home university. Had I stayed, I would be taking my Signals class in hand with the Power Electronics course, so maybe I am stuck because of the lack of such information.

Thanks in advance!
 

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Roderick Young

Joined Feb 22, 2015
408
If this is a Power Electronics course, I'm guessing it's ok to just use a formula? Or do they want you to take a Fourier transform of the waveform as part of the work?

That looks like a triangle wave at 10 Hz. If you look here you'll see that a triangle wave can be broken down into component frequencies (pure sine waves) that are odd harmonics of the fundamental. That means if your waveform is at 10 Hz, it will have sine wave components in it of frequency 10 Hz, 30 Hz, 50 Hz, 70 Hz, 90 Hz, and so forth (multiply the fundamental by 1, 3, 5, ... odd numbers). These frequencies (also called harmonics) are what they mean by "spectral components."

The amplitude of the first harmonic (lowest frequency) is \(8/{\pi^2}\) of the amplitude of the triangle wave. You can derive that my doing the Fourier integral, but for practical purposes, you can just use the value 0.8 or 0.81. In your case, that would be 80 or 81 volts.

The 3rd harmonic has an amplitude of \(1/{3^2}\) of the first (fundamental) harmonic, and in general, the nth harmonic has an amplitude of \(1/{n^2}\) of the fundamental, while remembering that there are no even harmonics. So the amplitude of the harmonic at 20 Hz is zero.

But definitely look at that link above on triangular waves.
 

Papabravo

Joined Feb 24, 2006
21,159
In this context "spectral" is an adjective form of the noun spectrum. So a spectral component is asking for how much energy is contained in a small range of frequencies. Think about the instrument called a spectrum analyzer that displays signal energy as a function of frequency. Mathematically you compute this by the use of a Fourier Transform. Have you studied Fourier Transforms?

https://en.wikipedia.org/wiki/Fourier_transform
 

RBR1317

Joined Nov 13, 2010
713
Attached is a table of common Fourier series taken from the chapter "Signal Analysis in the Frequency Domain" of the textbook Linear Circuits by R. Scott, 1960. It may be helpful.
 

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Thread Starter

BRetonDP

Joined Apr 11, 2014
5
If this is a Power Electronics course, I'm guessing it's ok to just use a formula? Or do they want you to take a Fourier transform of the waveform as part of the work?

That looks like a triangle wave at 10 Hz. If you look here you'll see that a triangle wave can be broken down into component frequencies (pure sine waves) that are odd harmonics of the fundamental. That means if your waveform is at 10 Hz, it will have sine wave components in it of frequency 10 Hz, 30 Hz, 50 Hz, 70 Hz, 90 Hz, and so forth (multiply the fundamental by 1, 3, 5, ... odd numbers). These frequencies (also called harmonics) are what they mean by "spectral components."

The amplitude of the first harmonic (lowest frequency) is \(8/{\pi^2}\) of the amplitude of the triangle wave. You can derive that my doing the Fourier integral, but for practical purposes, you can just use the value 0.8 or 0.81. In your case, that would be 80 or 81 volts.

The 3rd harmonic has an amplitude of \(1/{3^2}\) of the first (fundamental) harmonic, and in general, the nth harmonic has an amplitude of \(1/{n^2}\) of the fundamental, while remembering that there are no even harmonics. So the amplitude of the harmonic at 20 Hz is zero.

But definitely look at that link above on triangular waves.
I see! That makes a lot of sense! I am looking into videos regarding the Fourier series and I am beginning to understand :)
With the information you have given, that means that the amplitude of the component at 50 Hz is 3.2422, then? I'll have than in mind and try to elaborate on the answer with the information I have been given in the posts below.

What does the textbook say?
I am using "Power Electronics" by Daniel W. Hart (2011) and sadly I cannot seem to find the topic. My guess is that the professor's intention is for students to review the theory from Signals and Systems or other class.

In this context "spectral" is an adjective form of the noun spectrum. So a spectral component is asking for how much energy is contained in a small range of frequencies. Think about the instrument called a spectrum analyzer that displays signal energy as a function of frequency. Mathematically you compute this by the use of a Fourier Transform. Have you studied Fourier Transforms?

https://en.wikipedia.org/wiki/Fourier_transform
There lies the problem, I've never studied Fourier. In my home university I would've taken my signals class side-by-side with the power course, but this is not an option here. I'm trying to study Signals on my own with online courses right now, though.

Attached is a table of common Fourier series taken from the chapter "Signal Analysis in the Frequency Domain" of the textbook Linear Circuits by R. Scott, 1960. It may be helpful.
Thanks! Quick question, though: my signal is odd, right? And the triangular wave displayed in your attachment is even. Is the only thing that changes the use of sine instead of the cosines?
 

Roderick Young

Joined Feb 22, 2015
408
Power electronics is never going to be precise to that many digits, but yes, I believe your answer is about right. The 5th harmonic should have 4% the amplitude of the fundamental, so 80 * 4% = 3.2 volts. Good! Keep it up.
 
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