We know that Fourier series for periodic signal \$y(t)\$ is given by

$$ y(t) = \sum\limits_{m=0}^{+\infty} a_m \cos(w_m t) + \sum\limits_{m=0}^{+\infty}b_m \sin(w_m t). \quad$$

y(t) is sinusoidal periodic signal with period T

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  1. How would you decompose the given signal into simple sinusoid?

  2. If possible can anybody show me waveforms of fundamental and harmonics components for the given signal y(t)?

  • \$\begingroup\$ Throw it into some math tools fft function? If you are interested into the dirty math details behind such a function, you might be better asking that on a math related stack. \$\endgroup\$
    – PlasmaHH
    Commented Jul 8, 2015 at 14:22
  • \$\begingroup\$ This is a pure math problem. \$\endgroup\$
    – MathieuL
    Commented Jul 8, 2015 at 14:23
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    \$\begingroup\$ If the signal is periodic, then in theory you can decompose it into a Fourier series. What do you get if you run it through a Fourier transform? \$\endgroup\$ Commented Jul 8, 2015 at 14:32
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    \$\begingroup\$ It's clearly periodic therefore it can be expressed as a Fourier series. I don't know what you mean by 'other simple sinusoidal signals'. \$\endgroup\$
    – Chu
    Commented Jul 8, 2015 at 16:28
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    \$\begingroup\$ ...express given signal using sum of fundamental and harmonicsexpress given signal using sum of fundamental and harmonics...[is there] any other way than Fourier series? A sum of fundamental and harmonics is a Fourier series. You are asking whether there is any way to express the signal as a Fourier series without expressing it as a Fourier series. Doesn't make sense. \$\endgroup\$ Commented Jul 8, 2015 at 17:08

2 Answers 2


Take the shortest repeating interval of the waveform - that seems to be one-quarter of what you have drawn - that is the time period of the fundamental frequency and, to find the amplitude of that fundamental, multiply that section of signal by a sine wave and a cosine wave of the same time period.

Then integrate (over the time period) the two multiplied waveforms to get two numbers. Divide those two numbers by the time period and you get the a and b coefficients that pertain to the fundamental signal. Well actually you get the RMS values so multiply them by 1.4142 to get the true a and b coefficients.

Repeat for the 2nd harmonic and keep going up in harmonics until you are satisfied there is no appreciable signal energy left to worth considering.

You can do it in excel if you have sample values for the repeating signal. If all you have is a picture then you are out of luck.


Can anybody express given signal using sum of fundamental and harmonics?

The equation you gave is an expression of the signal in terms of the fundamental and harmonics.

\$a_n\cos\left(\omega t\right) + b_n\sin\left(\omega t\right)\$ is the \$n\$th harmonic. The fundamental is the same as the 1st harmonic.

The coefficients \$a_n\$ and \$b_n\$ can be calculated by formulas given in the Wikipedia article on the Fourier series.

You can easily plot an individual harmonic using the form

$$A \sin\left(\omega{}t +\phi\right) = a\cos\left(\omega{}t\right)+b\sin\left(\omega{}t\right)$$

with \$A=\sqrt{a^2+b^2}\$ and \$\phi=\tan^{-1}\dfrac{a}{b}\$.


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