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Consider a square wave f(x) of length 1. Over the range [0,1), this can be written as
Consider this mathematical question intuitively: Can a discontinuous function, like the square wave, be expressed as asum, even an infinite one, of continuous signals? One should at least be suspicious, and in fact, it can't be thusexpressed.
The extraneous peaks in the square wave's Fourier series never disappear; they are termed Gibb's phenomenon after the American physicist Josiah Willard Gibbs. They occur whenever the signal isdiscontinuous, and will always be present whenever the signal has jumps.
The Square wave is the standard example, but other important signals are also useful to analyze, and these are included here.
This signal is relatively self-explanatory: the time-varying portion of the Fourier Coefficient is taken out, and we are left simply with a constant function over all time.
With this signal, only a specific frequency of time-varying Coefficient is chosen (given that the Fourier Series equation includes a sine wave, this is intuitive), and all others are filtered out, and this single time-varying coefficient will exactly match the desired signal.
link: http://yoder-3.institute.rose-hulman.edu/visible3/chapters/03spect/demosLV/fseries/index.htm
To summarize, a great deal of variety exists among the common Fourier Transforms. A summary table is provided here with the essential information.
Description | Time Domain Signal for $t\in [0,1)$ | Frequency Domain Signal |
Constant Waveform | $x\left(t\right)=1$ | ${c}_{k}=\left\{\begin{array}{cc}1& k=0\\ 0& k\ne 0\end{array}\right)$ |
Sinusoid Waveform | $x\left(t\right)=sin\left(\pi t\right)$ | ${c}_{k}=\left\{\begin{array}{cc}1/2& k=\pm 1\\ 0& k\ne \pm 1\end{array}\right)$ |
Square Waveform | $x\left(t\right)=\left\{\begin{array}{cc}1& t\le 1/2\\ -1& t>1/2\end{array}\right)$ | ${c}_{k}=\left\{\begin{array}{cc}4/\pi k& \mathrm{k}\mathrm{odd}\\ 0& \mathrm{k}\mathrm{even}\end{array}\right)$ |
Triangle Waveform | $x\left(t\right)=\left\{\begin{array}{cc}t& t\le 1/2\\ 1-t& t>1/2\end{array}\right)$ | ${c}_{k}=\left\{\begin{array}{cc}-8\mathrm{Sin(k\pi )/2)}/{\left(\pi k\right)}^{2}& \mathrm{k}\mathrm{odd}\\ 0& \mathrm{k}\mathrm{even}\end{array}\right)$ |
Sawtooth Waveform | $x\left(t\right)=t/2$ | ${c}_{k}=\left\{\begin{array}{cc}0.5& k=0\\ \mathrm{-1}/\pi k& k\ne 0\end{array}\right)$ |
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