Here is a question I have been trying to solve:

Estimate the "essential bandwidth" of a rectangular pulse

$$ g(t) = \operatorname{rect}\left(\frac{t}{T}\right), $$ with $T>0$, where this "essential" bandwidth contains 90% of the rectangular pulse energy.

What I have so far is that the Fourier Transform of $\operatorname{rect}\left(\frac{t}{T}\right)$ is

$$ G(f) = \mathcal{F}\{g(t)\} = \mathcal{F}\left\{ \operatorname{rect}\left(\frac{t}{T}\right) \right\} = T \operatorname{sinc}(fT) $$

where $$\operatorname{rect}(u) \triangleq \begin{cases} 0 & \text{if } |u| > \frac{1}{2} \\ \frac{1}{2} & \text{if } |u| = \frac{1}{2} \\ 1 & \text{if } |u| < \frac{1}{2} \\ \end{cases}$$

$$\operatorname{sinc}(u) \triangleq \begin{cases} \frac{\sin(\pi u)}{\pi u} & \text{if } u \ne 0 \\ 1 & \text{if } u = 0 \\ \end{cases}$$

$$ X(f) = \mathcal{F}\{x(t)\} \triangleq \int\limits_{-\infty}^{+\infty} x(t) \ e^{-i 2 \pi f t} \ dt $$ and $$ x(t) = \int\limits_{-\infty}^{+\infty} X(f) \ e^{+i 2 \pi f t} \ df. $$

Integrating $G(f)$ over $\pm \infty$ results in $1$. Also, integrating $|g(t)|^2$ over $\pm \infty$ results in $T$. This is about where I am lost.

Any help is appreciated.

  • $\begingroup$ Hint: can you apply Parseval's formula here? $\endgroup$
    – Jeff
    Nov 16, 2014 at 18:51
  • $\begingroup$ I tried setting it Eg=0.90Etot but I am lost in the Math part. Can you give more clues? $\endgroup$
    – Ary
    Nov 16, 2014 at 18:54
  • $\begingroup$ Since the essential bandwidth must contain 90% of the pulse energy, then that's a clue that you need to involve the energy in the time domain. But it's asking for the bandwidth, which is a frequency domain quantity, so that's another clue that you need to mix the two domains. Use Parseval's formula to figure out how much energy you need in the frequency domain to get 90% in the time domain. (You limit the energy by truncating the integral: instead of integrating over infinity, you integrate over [-B,+B], where B is the essential bandwidth and the result gives 90% energy in the time domain) $\endgroup$
    – Jeff
    Nov 16, 2014 at 19:03

2 Answers 2


I'll add another answer, even though MBaz's answer is correct, because I think that it doesn't actually address the problem you have with arriving at the final solution. Summarizing, you have to solve

$$\frac{1}{2\pi}\int_{-W}^W|P(\omega)|^2d\omega=0.9\int_{-\infty}^{\infty}\text{rect}^2(t/T)dt=0.9\cdot T\tag{1}$$

for $W$ (which is the bandwidth in radians). If you like, rewrite (1) using $f$ instead of $\omega$. $P(\omega)$ is the Fourier transform of the rectangular pulse:

$$P(\omega)=2\frac{\sin(\omega T/2)}{\omega}\tag{2}$$

If I understood your question and your comments correctly, then you understand all of this, but your problem is solving (1) for $W$. Note that the integral has no closed-form solution, unless you consider an expression including the sine integral $\text{Si}(x)=\int_0^x\sin(t)/t\;dt$ closed form. With a little help from WolframAlpha, Equation (1) can be written as

$$\frac{2}{\pi W}\left[TW\cdot \text{Si}(TW)+\cos(TW)-1\right]=0.9\cdot T\tag{3}$$

Equation (3) has no closed-form solution for $W$, so it is not surprising that you got confused. You could of course try to solve (3) numerically but I do not think that this is the idea of the exercise. What you need to know is that obviously the largest portion of the energy is contained in the main lobe of the spectrum, i.e. between $-W_0$ and $W_0$, where $W_0$ is the first zero crossing of $P(\omega)$. From (2) it is clear that

$$W_0T/2=\pi\quad\Longrightarrow W_0=\frac{2\pi}{T}\tag{4}$$

So this is the essential bandwidth in radians. Checking the exact percentage using the left-hand side of (3) we get

$$\frac{2}{\pi W_0}\left[TW_0\cdot \text{Si}(TW_0)+\cos(TW_0)-1\right]= \frac{2\;\text{Si}(2\pi)}{\pi}\cdot T=0.9028\cdot T$$

which seems close enough.

In sum, the essential bandwidth of a rectangular pulse is given by the width of the mainlobe of its spectrum, so you only need to be able to calculate the first zero of the spectrum and you're done.


You need to follow these steps:

  1. Read up on Parseval's Theorem.
  2. Find the correct Fourier Transform of $\operatorname{rect}(t)$ -- the one you give above doesn't seem right to me. (note from r b-j, it's because of $\omega$ vs. $f$ in the Fourier Transform. it wasn't "wrong", but messy, so i changed it.) Also, I'd recommend working on "ordinary frequency" $f$ instead of angular frequency $\omega$.
  3. Find $E_g$, the energy of $g(t)$.
  4. Find the frequency $f_0$ such that the integral of the square of the magnitude of $G(f)$ on the interval from $-f_0$ to $f_0$ is equal to $0.9E_g$. Then, relying on Parserval, $f_0$ is your answer.

Intuitively, what is happening is this. Say you filter $g(t)$ with an ideal low-pass filter. Since the filter removes some frequency content from $g(t)$, then the energy of the filter's output is less than that of $g(t)$. You're trying to adjust the filter's cutoff frequency so that the output's energy is exactly 90% of the input's.

  • $\begingroup$ i hope you guys don't mind that i tinkered in your posts. $\endgroup$ Nov 16, 2014 at 20:32
  • $\begingroup$ @robertbristow-johnson, I don't mind, quite the contrary! I tweaked your own edits to the question a little bit. I thought the Fourier Transform of the rect pulse in the original question was wrong because it was missing a $1/\sqrt{2\pi}$ factor that is needed when working with angular frequency. $\endgroup$
    – MBaz
    Nov 16, 2014 at 20:37
  • $\begingroup$ it depends on your definition of the F.T. with angular frequency. the unitary F.T. is the one where the forward F.T. and inverse F.T. are identical, except for the sign on $i$ (or $j$, whatever your religion is). if you have a unitary F.T. with angular frequency, you have $\frac{1}{\sqrt{2\pi}}$ in front of the integral in both the forward and inverse F.T. the nice thing about the $f$ version we like to use in the electrical engineering religion is that those factors are gone and our life is much easier scaling things. (but we have to remember the $2\pi$ in the exponent.) $\endgroup$ Nov 16, 2014 at 20:45
  • 1
    $\begingroup$ Guys, I don't think @Ary's problem has anything to do with the use of $f$ or $\omega$, but with the fact that the resulting integral has no closed form solution. $\endgroup$
    – Matt L.
    Nov 17, 2014 at 10:28
  • 1
    $\begingroup$ parseval's theorem is both intuitively and practically easier to deal with using "ordinary frequency" (as opposed to "cyclical frequency"). otherwise you have to worry about where to put the $2 pi$ factor. you can always look it up, but why bother when the unitary Fourier Transform loses the scaling factor (actually puts it in the exponent). $\endgroup$ Nov 17, 2014 at 15:17

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