# Time-Bandwidth Product

The following text is cited from a textbook, "Spotlight Mode Synthetic Aperture Radar: A Signal Processing Approach", I would like to ask if anyone knows the proof to the following statements, as the proof is not outlined in the book, and I am unable to prove it myself.

The function $f(t)$ is represents a pulse envelope waveform. The effective duration of a pulse envelope waveform is given as:

$$T_{e}=\frac{\int_{-\infty}^{\infty}f(t)dt}{f(0)}$$

A corresponding measure of effective bandwidth of the pulse envelope waveform is given by:

$$B_{e}=\frac{1}{2\pi}\cdot \frac{\int_{-\infty}^{\infty}F(\omega )d\omega }{F(0)}$$

From the defining equation for the Fourier transform, the above measures can be shown to have a product which is constant.

$$B_{e}T_{e}=1$$

I have read elsewhere that the above is the time-bandwidth product. I have no idea what that means or what the physical significance of it is. Can anyone shed light on how the last statement can actually be proven? Thank you!

• Thank you Stanley. I am familiar with the Fourier Uncertainty Principle, but doesn't that give me an inequality where the product of dispersions of f(t) and F(w) are greater than 1/4? Also, the integrand is squared inside, whereas in the above case for the time-bandwidth product it is simply an... integral, in the L1 sense perhaps. Thanks for pointing it out to me regardless! – Sam Low Dec 28 '17 at 15:42
• As for its significance in SAR, it will tell you the SNR improvement that results from range compression. – AnonSubmitter85 Dec 28 '17 at 21:05

Note that these definitions of effective duration and effective bandwidth are only useful for very specific signals, namely for (real-valued) low-pass signals that are even and centered around $t=0$. This implies that their Fourier transform is also real-valued, even and centered around $\omega=0$, and, consequently, the same definition of width can be used.

Now for the proof, which is straightforward. Take the definition of the Fourier transform and its inverse:

$$F(\omega)=\int_{-\infty}^{\infty}f(t)e^{-j\omega t}dt\tag{1}$$

$$f(t)=\frac{1}{2\pi}\int_{-\infty}^{\infty}F(\omega)e^{j\omega t}d\omega\tag{2}$$

From $(1)$ and $(2)$ we get

$$F(0)=\int_{-\infty}^{\infty}f(t)dt\tag{3}$$

and

$$f(0)=\frac{1}{2\pi}\int_{-\infty}^{\infty}F(\omega)d\omega\tag{4}$$

from which we get

$$T_e=\frac{F(0)}{f(0)}\tag{5}$$

$$B_e=\frac{f(0)}{F(0)}\tag{6}$$

The result follows immediately.

For information on more general definitions of duration/bandwidth and on the uncertainty principle take a look at this answer.

• This is brilliant!!! Thank you, what a clear proof to an otherwise confusing statement in the question!!! – Sam Low Dec 28 '17 at 20:02