I am currently analysing my signal by looking at its spectrogram to determine the sinusoidal frequency content at the local sections of my signal overtime. I can do my spectrogram in 2 ways:

  1. Directly get the FFT of each local sections. Same as I multiply each section by a rectangular window with a height 1 and get the FFT.
  2. Multiply each section by a Gaussian window and get the FFT.

Apparently, I get 2 different pictures of my spectrogram. Number 2 is known as the Gabor Transform.

My question here is, what is the motivation for doing the Gabor Transform vs directly doing the FFT at each section of a signal?

  • $\begingroup$ Open up any standard DSP book (e.g. Oppenheim & Schafer's Discrete-time signal processing) and read the section on windowing. $\endgroup$
    – Batman
    Commented Jun 17, 2015 at 14:04

3 Answers 3


Using a Gaussian window has one advantage over all other windows-- it yields the most compact time-frequency view of the spectrum. In other words, the product of the uncertainty in the time and frequency domains is minimized. This is due to the uncertainty principle of the Fourier transform.

Incidentally, in physics this property of the Fourier transform is the mathematical underpinning behind the Heisenberg uncertainty principle, which states that the position and momentum cannot simultaneously be known to infinite precision. Replace "position" with "time" and "momentum" with "frequency" in that statement and you can see how this applies to spectral analysis.

(There are other reasons not to use a Gaussian window as well, depending on what your application requires.)

Also, incidentally, Gabor was a physicist who first applied many mathematical tools from physics to signal processing problems, such as the analytic signal representation. It would not surprise me if his use of the Gaussian window in signal processing was carried over from quantum mechanics.


from the POV of sinusoidal modeling or identifying sinusoids, two very good basic reasons why a Gaussian window is good are:

  1. The Fourier Transform of a Gaussian is a Gaussian. (and Gaussians have essentially no side lobes.)

$$ \mathscr{F} \{ e^{-\pi t^2} \} = e^{-\pi f^2} $$

  1. The Gaussian function is just like the linearly-swept chirp, except for an imaginary unit, so they share lotsa stuff in common and can be modeled together elegantly in the math.

just FYI, the exact definition of (unitary) Fourier Transform used is:

$$ X(f) \triangleq \mathscr{F} \{ x(t) \} = \int\limits_{-\infty}^{\infty} x(t) \, e^{-j 2 \pi f t} \ dt $$

and inverse Fourier Transform:

$$ x(t) \triangleq \mathscr{F}^{-1} \{ X(f) \} = \int\limits_{-\infty}^{\infty} X(f) \, e^{+j 2 \pi f t} \ df $$

i am taking advantage of the symmetry of the forward and inverse Fourier Transforms. they are identical except on replaces $-j$ for $j$, but $-j$ and $j$ are qualitatively identical. they both have equal claim to squaring to be $-1$ and to call themselves "the imaginary unit".

this old paper of mine spells out some of this but i might add to this answer a little mathematical expression to the reasons 1 and 2.

Gaussian window of width $\sqrt{\frac{1}{\alpha}}$ :

$$ \mathscr{F} \{ e^{-\pi \alpha t^2} \} = \sqrt{\frac{1}{\alpha}} e^{-(\pi/\alpha) f^2} $$

we gotta restrict $\alpha > 0$.

Linear-swept chirp with sweep rate of $\beta$ :

$$ \mathscr{F} \{ e^{j \pi \beta t^2} \} = \sqrt{\frac{j}{\beta}} e^{-j (\pi/\beta) f^2} $$

so here's a linearly-swept chirp windowed with a Gaussian window:

$$\begin{align} \mathscr{F} \{ e^{-\pi \alpha t^2} e^{j \pi \beta t^2} \} &= \mathscr{F} \{ e^{-\pi (\alpha - j \beta) t^2} \} \\ \\ &= \sqrt{\tfrac{1}{\alpha - j \beta}} e^{-\pi \frac{1}{\alpha - j \beta} f^2} \\ \\ &= \sqrt{\tfrac{\alpha + j \beta}{\alpha^2 + \beta^2}} e^{-\pi \frac{\alpha + j \beta}{\alpha^2 + \beta^2} f^2} \\ \end{align}$$

now here's a linearly-swept chirp windowed with a Gaussian window that has, in the center of the window a specific frequency $f_0$ for the sinusoid:

$$\begin{align} \mathscr{F} \{ e^{-\pi \alpha t^2} e^{j \pi \beta t^2} e^{j 2 \pi f_0 t} \} &= \mathscr{F} \{ e^{-\pi \alpha t^2} e^{j \pi \beta t^2} \} \Bigg|_{f \leftarrow f-f_0} \\ \\ &= \sqrt{\tfrac{\alpha + j \beta}{\alpha^2 + \beta^2}} e^{-\pi \frac{\alpha + j \beta}{\alpha^2 + \beta^2} (f-f_0)^2} \\ \end{align}$$

finally we can generalize it a little more by adding a ramp in the amplitude in addition to linearly swept frequency. we can think of it as sorta a linear ramp, but we're really gonna use an exponential ramp because it makes the math do much easier.

$$ 1 + 2 \pi \lambda t \ \approx \ e^{2 \pi \lambda t} \qquad \text{for } |\lambda t| \ll 1 $$

$$\begin{align} \mathscr{F} \{ e^{-\pi \alpha t^2} e^{j \pi \beta t^2} e^{j 2 \pi f_0 t} e^{2 \pi \lambda t}\} &= \mathscr{F} \{ e^{-\pi \alpha t^2} e^{j \pi \beta t^2} e^{j 2 \pi (f_0 - j \lambda) t} \} \\ \\ &= \mathscr{F} \{ e^{-\pi \alpha t^2} e^{j \pi \beta t^2} \} \Bigg|_{f \leftarrow f-(f_0-j\lambda)} \\ \\ &= \sqrt{\tfrac{\alpha + j \beta}{\alpha^2 + \beta^2}} e^{-\pi \frac{\alpha + j \beta}{\alpha^2 + \beta^2} (f-f_0 + j\lambda)^2} \\ \\ &= \sqrt{\tfrac{\alpha + j \beta}{\alpha^2 + \beta^2}} e^{-\pi \frac{\alpha + j \beta}{\alpha^2 + \beta^2} (f-f_0)^2} e^{\pi \frac{\alpha + j \beta}{\alpha^2 + \beta^2} \lambda (\lambda - j 2 (f-f_0) )} \\ \end{align}$$

so, if you use a Gaussian window, you can model each sinusoidal component with frequency $f_0$ and sweep rate of $\beta$ and ramp rate of $2 \pi \lambda$. and you have a function of the very same form in the frequency domain. the paper pointed to above says how you can extract $f_0$ and $\beta$ and $\lambda$ out of the $\log(\cdot)$ of each Gaussian lobe in the frequency domain data.

this is why you might consider using the Gaussian window with the Short Time Fourier Transform.

what's really tits is that this is basically true for any exponential raised to a quadratic power:

$$ \mathscr{F} \{ e^{a t^2 + b t + c} \} = e^{A f^2 + B f + C} $$

where $A, B, C$ are constants that are some deterministic functions of $a, b, c$.


Here is the answer to my question. The explanations in this page is quite clear: http://www.ni.com/white-paper/4844/en/#toc2

  • 1
    $\begingroup$ The material at that 'www.ni.com' link tells a plausible story, but beware. That material contains several wildly incorrect descriptions of the relationships between a sequence of discrete time-domain amplitude samples and the fast Fourier transform (FFT) of that sequence. For one example, that material states that an FFT's output samples can contain spectral energy that did not exist in the FFT's input time-domain sequence. That is simply not true. $\endgroup$ Commented Sep 16, 2015 at 12:14

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