Essential Bandwidth of rect(t/T)

Question

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.

Answer 1

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.

Answer 2

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.