Using Laurent Series of a function $f(z)$ around a point $a \in \mathbb{C}$ $$f(z) = \sum^{\infty}_{n=-\infty} c_n(z-a)^n \ \ \ \ (1)$$ where $$c_n = \frac{1} {2\pi i}\int\limits_{\gamma}\frac {f(z)} {(z-a)^{n+1}} dz \ \ \ \ (2)$$ where $\gamma$ is a closed curve around $a$.

And choosing $\gamma$ such that $z$ can be parameterized as $z=a+e^{it}\ t \in [-\pi,\pi] $ in order to obtain Fourier Series

$$ f(t) = \sum^{\infty}_{n=-\infty} c_ne^{int} \ \ \ \ (3) $$ where $$ c_n = \frac{1} {2\pi}\int_{-\pi}^{\pi} f(t)e^{-int}dt \ \ \ \ (4) $$

we can "intuitively" prove Fourier Series by just using introductory complex calculus without any advanced mathematical background in Hilbert-Banach Spaces, Functional Analysis etc.

My question is, how can we derive other forms of Fourier Series (also called Fourier Transforms) such as Discrete Fourier Transform (DFT) and Continuous Time Fourier Transform (CTFT) by a similar manner? The problem here is that DFT equations are not closed contour integrals anymore, just discrete finite sums. I could not wrap my head around how Laurent Series may still work in this discrete case. For the CTFT case we have to let integral limits in $(4)$ to go infinity, which corresponds to infinitely many contour integrals in $(2)$. This seems to be divergent, if I am not terribly mistaken?

  • $\begingroup$ a continuous time Fourier transform is the limit of a Fourier series when the period of the periodic function becomes infinitely large; is there more to say? $\endgroup$ Jan 8, 2020 at 19:53
  • $\begingroup$ letting the "periodicity" of n to go infinity with redefining, say, n = n'/w and letting w to go infinity may change the nature of equation (4). Since int = in't/w after our new definition, letting w go to infinity will cause t/w to converge to 0 (remember, t is finite by definition). This means we do not describe a circular closed contour in (2), Laurent series will not work. If we change the definition of t to eliminate this issue by redefining t within an infinite interval, then we obtain a fraction of infinities t/w = inf/inf. This may yield anything. $\endgroup$
    – user740171
    Jan 8, 2020 at 20:10

1 Answer 1


Continuous time Fourier transform and Laurent series:

I recall equations (1) and (2), for convenience set $a=0$, and substitute $z=e^{it/T}$. The function $g(t)=f(e^{it/T})$, with $t\in(-\pi T,\pi T)$, is periodic with period $2\pi T$, given by the Laurent series $$g(t)=\sum_{n=-\infty}^\infty c_n e^{int/T},$$ with coefficients $$c_n=\frac{1}{2\pi T}\int_{-\pi T}^{\pi T}g(t)e^{-int/T}\,dt.$$ For $T\gg t$ the sum over $n$ may be approximated by an integral over $\omega=n/T$ with coefficients $C(\omega)=Tc_{n=\omega T}$, giving $$g(t)=\int_{-\infty}^\infty C(\omega) e^{i\omega t}d\omega,$$ $$C(\omega) =\frac{1}{2\pi}\int_{-\infty}^\infty g(t)e^{-i\omega t}\,dt.$$

In this way the Fourier integral can be obtained as the limit of the Laurent series when the periodicity of the function tends to infinity. Notice that the exponent $nt/T$ cannot be set to zero because, even though $t/T\ll 1$, the product $nt/T=\omega t$ need not be small.

Discrete Fourier transform and Laurent series:

We now start from a discrete time signal $x_n$ and construct the Z-transform $$X(z)=\sum_{n=-\infty}^\infty x_{n}z^{-n}=\sum_{n=-\infty}^\infty x_{-n}z^n.$$ This is a Laurent series centered at $a=0$, with inversion formula $$x_{n}=\frac{1}{2\pi i}\oint X(z)z^{n-1}\,dz.$$ For the discrete Fourier transform one has only $N$ distinct values of $x_n$, the set $\{x_0,x_1,x_2,\ldots x_{N-1}\}$. We extend this set periodically by $x_{n+N}=x_n$. The Z-transform is a series of $N$ Dirac delta functions with coefficients $X_k$, $$X(z=e^{2\pi iq/N})=\sum_{k=0}^{N-1}X_k\delta(q-k),\;\;X_k=\sum_{n=0}^{N-1}x_n e^{-2\pi i kn/N}.$$ The delta function converts the integral in the inversion formula into a sum, $$x_n=\frac{1}{N}\sum_{k=0}^{N-1}X_k e^{2\pi ikn/N}.$$

  • $\begingroup$ n is a countable number, the sum has countably infinite elements in it. However an integral is performed on real numbers which are uncountably infinite. How can we do that transition from countably infinite numbers to uncountably infinite numbers ? Is it just an approximation ? $\endgroup$
    – user740171
    Jan 8, 2020 at 21:54
  • $\begingroup$ Isn’t that what we do when we approximate an integral by a Riemann sum? The approximation becomes more and more accurate as the discretization interval t/T becomes smaller and smaller. $\endgroup$ Jan 8, 2020 at 22:03
  • $\begingroup$ That's right sir, thank you. The CTFT case seems to be closed. After the DFT case is closed I'll mark my question answered. What can we say about the DFT case ? $\endgroup$
    – user740171
    Jan 8, 2020 at 22:18
  • $\begingroup$ Thank you for your response. Isn't it z^(n-1) rather than x^(z-1) in the integral ?? $\endgroup$
    – user740171
    Jan 9, 2020 at 12:04
  • $\begingroup$ I did the calculations myself and in the last identity in your answer i did not obtain 1/2π multiplier but i obtained 1/N instead. Am I mistaken ? $\endgroup$
    – user740171
    Jan 9, 2020 at 15:02

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