We know that a Fourier series formula for any signal $s(t)$ is given as

$$\frac {a_0} 2 + \sum \limits _{m=1} ^\infty (a_m \cos \frac {2 \pi m t} T + b_m \sin \frac {2 \pi m t} T)$$

Here,as we see from the formula ,except the DC component and fundamental frequency components there are all harmonics present at right side of the Fourier series formula.

Let us consider 4 sinusoidal periodic signal $$x(t),y(t),z(t) and g(t)$$ such that

$$x(t)=\cos \frac {1.2*2 \pi t} 8$$

$$y(t)=\cos \frac {1.4*2 \pi t} 8$$

$$z(t)= \cos \frac {1.6 \pi t} 8$$




  1. How would you apply Fourier series formula for a signal $$g(t)$$ which is nonharmonic?

  2. If there are fundamental frequency and harmonics present ,can anybody tell me what are they and their values?

  • 1
    \$\begingroup\$ I don't see any issue here at all. It's a composite signal g(t) comprising 3 sinewaves of different frequencies. Eventually g(t) will repeat so this limits the boundaries over the integration so what is the problem? \$\endgroup\$
    – Andy aka
    Commented Jun 24, 2015 at 11:59
  • \$\begingroup\$ @Andy aka as you can in the main Fourier series formula ,there are harmonic terms present but for $g(t)$ there are no harmonic frequency components. so how Fourier series formula is valid here? \$\endgroup\$ Commented Jun 24, 2015 at 12:02
  • \$\begingroup\$ @Andy aka could you tell me in case of g(t),what is fundamental frequency component here? also ,what are harmonics I.e.multiple of fundamental frequency components present here? \$\endgroup\$ Commented Jun 24, 2015 at 12:07
  • \$\begingroup\$ Model it as a harmonic sequence of cos(0.2*2*pi*t / 8) with all the harmonic weights set to 0, except for three corresponding to X,Y,Z. Job done. \$\endgroup\$
    – user16324
    Commented Jun 24, 2015 at 12:33
  • \$\begingroup\$ @Brian Drummond I really don't get you. Why are you avoiding X,Y and Z signals ?and representing signal g(t) by sum of other signals? \$\endgroup\$ Commented Jun 24, 2015 at 12:38

2 Answers 2


In general, Fourier coefficients are given by the formulas:

$$a_n = \frac{2}{T}\int_{x_0}^{x_0+T}g(x)\cos\left(\frac{2\pi nx}{T}\right)dx\\ b_n = \frac{2}{T}\int_{x_0}^{x_0+T}g(x)\sin\left(\frac{2\pi nx}{T}\right)dx$$

For any integrable \$g(x)\$ on the interval \$[x_0, x_0+T]\$. For periodic functions this interval can correspond to the period of the function, such that the Fourier series will be perfectly equal to the function itself. But the case of your example is very simple - it's just a sum of three cosines, so the \$a_n\$ terms can be calculated directly and there would be three of them. \$b_n\$ would equal to zero, as there is no sine terms.

  • \$\begingroup\$ Could you tell me in case of g(t),what is fundamental frequency component here? also ,what are harmonics I.e.multiple of fundamental frequency component present here? \$\endgroup\$ Commented Jun 24, 2015 at 12:10
  • \$\begingroup\$ The signal, as you can see is composed of three harmonic signals, I.e. three different frequencies. And you should be able to find them yourself. The fundamental one would be the lowest one. And there are (surprise!) two additional harmonics). As this question is suspected as a homework, I am not going to give you a complete solution. \$\endgroup\$
    – Eugene Sh.
    Commented Jun 24, 2015 at 12:14
  • \$\begingroup\$ 2.also how could you take $m$ as fractional value here? \$\endgroup\$ Commented Jun 24, 2015 at 12:15
  • \$\begingroup\$ @user3559780 if you refer to the numbers 1.2 1.4 and 1.6 - they are not the m. Hint: 1.2/8 = m_1/T, 1.4/8 = m_2/T, 1.6/8 = m_3/T \$\endgroup\$
    – Eugene Sh.
    Commented Jun 24, 2015 at 12:18
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    \$\begingroup\$ Actually in the equations above for m_3 it should be 0.8/8 = m_3/T, as there is no coefficient of 2 in the z(t). So yes, the idea is to find such a smallest T which will make ms non-fractional. 80 is not the smallest one. The smallest one you will get if reducing the fractions. It will be 40. Sorry, gotta go. Good luck. \$\endgroup\$
    – Eugene Sh.
    Commented Jun 24, 2015 at 12:39

To answer your recent comment: yes, in the formula for Fourier series, ANY of the \$a_m\$ or \$b_m\$ terms can be zero. The fundamental frequency has no special privileges.

Perhaps, to convince you, consider a composite signal:

\$g(\theta)=sin(2\theta)+sin(5\theta)\$, where \$\theta=2\pi ft=2\pi t/T\$

This does not have a component in \$sin(\theta)\$, i.e. the fundamental, yet \$g(\theta)\$ has been written as the sum of the 2nd and 5th harmonics of the fundamental. As \$\theta\$ goes from \$0\rightarrow 2\pi\$, \$sin(2\theta)\$ goes through 2 complete cycles and \$sin(5\pi)\$ goes through 5 complete cycles.

Looked at another way, \$g(\theta)\$ can be written:


and the lowest common multiple of \$0.5\$ and \$0.2\$ is \$1\$


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