Calculating Harmonics from FFT of sin(x) Function

In summary, the Fourier Cosine Transform of cos(x) for 0<x<a and 0 everywhere else can be represented as F(\omega)=\displaystyle\frac{1}{\sqrt{2 \pi}}[\frac{\sin a (1-\omega)}{1-\omega}+\frac{\sin a (1+\omega)}{1+\omega}]. When plotted, the amplitude spectrum of F(\omega) against (\omega) is a continuous function. However, since the function is non-periodic, it does not have distinct harmonics. Therefore, it is not possible to determine the number of harmonics in a more complicated function, such as an amplitude modulation signal, when using the FFT method.
  • #1
bugatti79
794
1
Hi Folks,

The Fourier Cosine Transform of [tex]cos(x) [/tex] for [tex]0<x<a [/tex] and 0 everywhere else is given as

[tex]F(\omega)=\displaystyle\frac{1}{\sqrt{2 \pi}}[\frac{\sin a (1-\omega)}{1-\omega}+\frac{\sin a (1+\omega)}{1+\omega}][/tex]

I can plot this and we get a continuous amlitude spectrum of [tex]F(\omega)[/tex] against [tex](\omega)[/tex]

but how do I extract/obtain the harmonic of this function which we know has just one harmonic. How do i extract this mathematically and/or from the graph say?

Thanks
 
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  • #2
bugatti79 said:
Hi Folks,

The Fourier Cosine Transform of [tex]cos(x) [/tex] for [tex]0<x<a [/tex] and 0 everywhere else is given as

[tex]F(\omega)=\displaystyle\frac{1}{\sqrt{2 \pi}}[\frac{\sin a (1-\omega)}{1-\omega}+\frac{\sin a (1+\omega)}{1+\omega}][/tex]

I can plot this and we get a continuous amlitude spectrum of [tex]F(\omega)[/tex] against [tex](\omega)[/tex]

but how do I extract/obtain the harmonic of this function which we know has just one harmonic. How do i extract this mathematically and/or from the graph say?

Thanks

The function would have only one harmonic if it were defined as $\cos x$ for $- \infty< x < + \infty$ ... but it is not so ...

Kind regards

$\chi$ $\sigma$
 
  • #3
chisigma said:
The function would have only one harmonic if it were defined as $\cos x$ for $- \infty< x < + \infty$ ... but it is not so ...

Kind regards

$\chi$ $\sigma$

Is there not one harmonic over 1 period?

In either case, is it possible to determine how many harmonics there are? The reason I ask is because i want to find the harmonics of a more complicated function like an amplitude modulation signal when i do the fft

$$A_c[1+A_1\cos(\omega_1 t+\phi_1)]\cos(\omega_c+\phi_c)$$

thanks
 
  • #4
bugatti79 said:
Is there not one harmonic over 1 period?

In either case, is it possible to determine how many harmonics there are? The reason I ask is because i want to find the harmonics of a more complicated function like an amplitude modulation signal when i do the fft

$$A_c[1+A_1\cos(\omega_1 t+\phi_1)]\cos(\omega_c+\phi_c)$$

thanks

... there is a little detail... the function is non periodic... and that means that its spectrum doesn't contain 'harmonics' but is a continuos function...

Kind regards

$\chi$ $\sigma$
 

FAQ: Calculating Harmonics from FFT of sin(x) Function

How do you calculate harmonics from an FFT of a sin(x) function?

The FFT (Fast Fourier Transform) of a sin(x) function can be used to calculate the harmonics present in the signal. This is done by taking the magnitude of the FFT output and identifying the peaks, which correspond to the frequencies of the harmonics. The amplitude of each peak represents the strength of the harmonic.

What is the purpose of calculating harmonics from an FFT of a sin(x) function?

Calculating the harmonics from an FFT of a sin(x) function allows us to analyze the frequency components of the signal. This can be useful in various applications such as signal processing, audio engineering, and vibration analysis.

How many harmonics can be calculated from an FFT of a sin(x) function?

The number of harmonics that can be calculated depends on the length of the signal and the sampling rate. According to the Nyquist-Shannon sampling theorem, the maximum frequency that can be accurately represented is half of the sampling rate. Therefore, the number of harmonics will be limited by the highest frequency present in the signal.

Can harmonics be present in a sin(x) function?

Yes, harmonics can be present in a sin(x) function. In fact, any periodic function can have harmonics. A sin(x) function is a pure tone at a single frequency, but it can be decomposed into a series of harmonics with different frequencies and amplitudes.

How does the amplitude of harmonics in a sin(x) function change with frequency?

The amplitude of harmonics in a sin(x) function decreases as the frequency increases. This is known as the harmonic decay or roll-off. The rate of decay depends on the shape and duration of the signal, as well as any external factors such as noise or interference.

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