The Fourier transform of $f(ax)$ is $\frac{1}{|a|}F(\frac{u}{|a|})$. So the frequencies are scaled horizontally but the magnitudes are also scaled when the graph of $f$ is scaled horizontally.

On the other hand the fourier series of $f(\theta)$ on $[-\pi,\pi]$ is $\sum_{-\infty}^{\infty} c_n e^{in\theta}$ whereas the fourier series of $f(a\theta)$ on $[\frac{-\pi}{a},\frac{\pi}{a}]$ is $\sum_{-\infty}^{\infty} c_n e^{ina\theta}$ so that the frequencies are scaled horizontally but the magnitudes are left unchanged at $c_n$.

Now these are easy to show mathematically, but I'm left feeling uneasy conceptually. Is there same intuition geometrical view to leave my intuition satisfied that explains why there is this lack of symmetry?

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    $\begingroup$ I'm pretty sure that should be $F\left(\frac{u}{a}\right)$, not $F\left(\frac{u}{\left|a\right|}\right)$. $\endgroup$ – Henry Gomersall Jul 30 '12 at 9:33

Let $g(x) = f(ax)$. Then, using a change of variables $y = ax$, $$\begin{align*} G(u) &= \int_{-\infty}^{\infty} g(x)\exp(-j u x)\,\mathrm dx\\ &= \int_{-\infty}^{\infty} f(ax)\exp(-j u x)\,\mathrm dx\\ &= \frac{1}{|a|}\int_{-\infty}^{\infty} f(y)\exp(- j (u/a) y)\,\mathrm dy\\ &= \frac{1}{|a|}F\left(\frac{u}{a}\right) \end{align*}$$ not $\displaystyle \frac{1}{|a|}F\left(\frac{u}{|a|}\right)$ as you say (cf. Henry Gomersall's comment on your question). Now, if $f(x)$ is a periodic function with period $2\pi$, then its Fourier transform exists only in a generalized sense (involves impulses). The Fourier series representation $\displaystyle \sum_n c_n \exp(j nx)$ of $f(x)$ has Fourier coefficients $$\displaystyle c_n = \frac{1}{2\pi}\int_{-\pi}^{\pi} f(x)\exp(-j n x)\,\mathrm dx, ~ -\infty < n < \infty$$ and these also show up in its Fourier transform which is given by $\displaystyle F(u) = \sum_n c_n \delta(u -n)$. Note that the inverse Fourier transform is $$\int_{-\infty}^\infty F(u)\exp(jux)\,\mathrm du = \sum_n \int_{-\infty}^\infty c_n \delta(u-n)\exp(jux)\,\mathrm du = \sum_n c_n \exp(j nx) = f(x).$$ The function $g(x) = f(ax)$ of period $2\pi/a$ thus has Fourier transform $$G(u) = \frac{1}{|a|}\sum_{n=-\infty}^{\infty} c_n \delta(u/a -n) = \frac{1}{|a|}\sum_{n=-\infty}^{\infty} c_n \delta\left(\frac{u-an}{a}\right).$$ But, $\delta(\alpha x) = \frac{1}{|\alpha|}\delta(x)$, and so with $\alpha = a^{-1}$, we have $$G(u) = \sum_n c_n \delta(u - an)$$ whose inverse Fourier transform is $$\int_{-\infty}^\infty G(u)\exp(jux)\,\mathrm du = \sum_n \int_{-\infty}^\infty c_n \delta(u-an)\exp(jux)\,\mathrm du = \sum_n c_n \exp(j nax) = g(x) = f(ax).$$

So, you are not missing anything. Scaling an ordinary function scales both the magnitude and the frequency axis of the Fourier transform, but when impulses are involved, scaling the argument of the impulse leads to a magnitude scaling that exactly compensates for the magnitude scaling imposed on the Fourier transform, and leaves the Fourier coefficients unchanged. frequency scaling also scales the amplitude

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  • $\begingroup$ But intuitively why should this happen? $\endgroup$ – user782220 Jul 30 '12 at 21:19
  • $\begingroup$ Intuitively, continuous time (or continuous frequency) are irrelevant concepts when dealing with discrete time and discrete frequencies. Consider a lowpass digital filter with transfer function $H(z)$. Its cutoff frequency (or 3 dB frequency) is specified only in terms of the normalized frequency, not in terms of real frequencies. If pumping in data at 1000 samples/second gives a cutoff frequency of 10 Hz, the same filter with data coming in at 2000 samples/second will cut off at 20 Hz. No change in coefficients, right? Similarly, Fourier series represent $f(x)$ over a normalized interval .... $\endgroup$ – Dilip Sarwate Jul 31 '12 at 0:49
  • $\begingroup$ ...and changing the interval does not change the Fourier coefficients at all. The spacing of the Fourier coefficients on the frequency axis changes (compare 2000 samples/s vs 1000 samples/s) but not their values (filter coefficients are the same). $\endgroup$ – Dilip Sarwate Jul 31 '12 at 0:52

1) Fourier series is applicable to periodic signals

And the periodic signal means, it has constant average power. (Parseval’s theorem) That means whether you stretch or squeeze your original signal your average power is going to be the same, which is given as the following equation, and will be same in un-stretched and stretched state in the time domain.

\begin{equation} \frac{ 1 }{ T_0 } \int_{T_0} |x(t)|^2 dt = \sum_{k={-\infty}}^{\infty} |C_{k}|^2 \end{equation}

2) For Fourier transform, which is applicable to non-periodic signals,

those signals are with constant energy (not power). Which also means if you stretch or squeeze the signal in the time domain the energy of the signal will remain the same.

Stretching a signal in time domain means the signal will move slowly, implying its component frequencies will be shifted to a slow varying mode (scale down). The fact that the original signal component which is a fast varying one will have more energy and if you convert that fast pace signal component to a slowly varying component, the amplitude must be scaled up to make its energy even in both the cases. Similarly, if you squeeze a signal in time, slow frequency components have to be fast paced (scaled up) and to balance its energy, its amplitude must be scaled down.

Thus the frequency spectrum of a non-periodic signal will act like a material made out of jello or soft-rubber, if you stretch it, it will elongate but to maintain the same volume/area it will get compressed in its perpendicular direction. Vice-versa if you squeeze the spectrum it will bulge to maintain the equal volume/area.


$\textit{Stretching in the time domain} \rightarrow \textit{Squeezing in the frequency domain (with a bulge)}$

To get a bulge, you have to scale the amplitude, that’s why there is a scaling factor in the frequency domain.

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