Discovering the Type of Waveform from a Fourier Series | Homework Help

AI Thread Summary
The discussion revolves around determining the type of waveform generated by a Fourier series expression. The first waveform presented, V(t) = 2/π(sin(ωt) + 1/2sin(2ωt) + ...), suggests a sawtooth waveform due to its use of odd harmonics. The second waveform, 5sin(ωt) + 5sin(2ωt) + ..., raises concerns as it appears to continuously increase voltage without a valid waveform representation. Participants clarify that Fourier series typically decrease the amplitude of higher frequency components, and the peak amplitude should be factored into the calculations for coefficients. The conversation emphasizes the importance of understanding how amplitude and frequency interact in Fourier series to accurately identify waveforms.
suv79
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Homework Statement



what type of waveform would this make ?

Homework Equations



V(t)=2/π(sin(ωt)+1/2sin(2ωt)+1/3sin(3ωt)+1/4sin(4ωt)+...)

5sin(ωt)+5sin(2ωt)+5sin(3ωt)+5sin(4ωt)...

The Attempt at a Solution

 

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all that I have read about Fourier series, it uses odd number for 'n' to make an sawtooth waveform,

5sin(ωt)+5sin(2ωt)+5sin(3ωt)+5sin(4ωt)... this is not add together as it is increasing the voltage ?
 
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Where did you come up with

suv79 said:
5sin(ωt)+5sin(2ωt)+5sin(3ωt)+5sin(4ωt)
 
that is from the input, see the 2nd attached.
i don't understand what waveform the input will be making.
 
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Those inputs I don't think are valid, cause you are right as they stand they'd keep adding.
Generally with Fourier series as the frequency of the component goes up the scalar goes down.
Take a look at http://en.wikipedia.org/wiki/Fourier_series
example 1 is actually the sawtooth waveform. You'll notice that is not 5 as you have in your example but rather 2*(-1)^{n+1}\over(\pi*n)
 
the 5 is Peak amplitude.
 
Well then it would be multiplied by 5. The peak amplitude is generally taken into account when you do the integral to calculate Bn & An
 
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