Fractional Bandwidth of a Gaussian Amplitude Modulated Signal

Question

a gaussian modulated sinusoidal signal may be expressed as $$x(t)=A\cdot e^{(j2\pi ft)}\cdot e^{\left[-\frac{1}{2\sigma^{2}}\cdot(t-t_{0})^{2}\right]}$$

Let's consider the case in which the gaussian peak is at 0 and the total amplitude is 1:

$$x(t)=e^{(j2\pi ft)}\cdot e^{\left[-\frac{1}{2\sigma^{2}}\cdot t^{2}\right]}$$

What is the fractional bandwidth of this function?

I'm asking this question since I think it's a quite used parameter. In fact, there are a proper function of MATLAB and a proper function of Python) which can generate this function, and both ask me to specify:

• the sine wave frequency (which is $f$);

• the signal fractional bandwidth

I think the last one is necessarily linked to the standard deviation $\sigma$. But by which mathematical relationship? It's a bandwidth in time domain (so the FWHM of the gaussian pulse: $2.35\cdot \sigma$) or the bandwidth in frequency domain (so ???)?

Answer 1

I'll give this a shot.

The Fourier Transform of a Gaussian is also a Gaussian. The standard deviations in each domain are related as $\sigma_t \cdot \sigma_F = \frac{1}{2\pi}$ The time standard deviation, $\sigma_t$ has units of time and the frequency domain standard deviation $\sigma_F$ has units of Hz.

We can define the "bandwidth" of a gaussion as the -3dB point, i.e the point where the energy has fallen to 50%. We solve for $f(x) = \sqrt2$ and we get

$$x_{-3dB} = \sigma \sqrt{ln(2)}$$

This holds in both domains. We can then find the -3dB frequency $f_{-3dB}$ as follows

$$f_{-3dB} = \sigma_F \sqrt{ln(2)} = \frac{\sqrt{ln(2)}}{2\pi\sigma_T}$$

The bandwidth is twice the -3dB frequency since, the energy extends above and below the modulation frequency. Fractional, I assume, means relative to modulation frequency, $f_{mod}$ and so we can define the fractional bandwidth as

$$B_{frac} = \frac{2f_{-3dB}}{f_{mod}} = \frac{\sqrt{ln(2)}}{\pi \cdot \sigma_T \cdot f_{mod}}$$

The scale factor depends on your exact definition of "bandwidth". For -3dB it comes out to be $\sqrt{ln(2)}$

Answer 2

The fact that it's modulated with a sinusoid doesn't change the FWHM bandwidth of your pulse – the $e^{jx}$ function has $\left\lvert e^{jx}\right\rvert\equiv 1$ at every point. That doesn't change the amplitude, so the FWHM of a sinusoid-modulated gaussian is just the same as of the unmodulated gaussian.