Consider a real baseband signal x(t) = e ^−2t , for t (in seconds) ≥ 0. If 99% of energy of x(t) lies within B Hz, then which of the following options is TRUE for the value of B?

The correct option is 63/π Hz < B < 64/π Hz.

To understand why this is correct, we will find the total energy of the signal and then determine the bandwidth $B$ (in Hz) that contains 99% of this total energy.

Step 1: Find the total energy of the signal
The given signal is:
$x\left(t\right)=e^{-2t}\text{for}t\ge 0$
The total energy $E$ of a continuous-time signal $x\left(t\right)$ is defined as:
$E={\int }_{-\infty }^{\infty }{\left|x\right(t\left)\right|}^{2}dt$
For our signal, this becomes:
$E={\int }_0^{\infty }{\left(e^{-2t}\right)}^{2}dt={\int }_0^{\infty }{e}^{-4t}dt$
Evaluating the integral:
$E={\left[\frac{e^{-4t}}{-4}\right]}_0^{\infty }=0-(-\frac{1}{4})=\frac{1}{4}=0.25$

Step 2: Find the Fourier Transform of the signal
The Fourier transform of the causal exponential signal $x\left(t\right)=e^{-at}u\left(t\right)$ is given by:
$X\left(f\right)=\frac{1}{a+j2\pi f}$
For $a=2$:
$X\left(f\right)=\frac{1}{2+j2\pi f}$
The energy spectral density is the square of the magnitude of the Fourier transform:
${\left|X\right(f\left)\right|}^2=\frac{1}{2^2+{\left(2\pi f\right)}^{}}=\frac{1}{4+4{\pi }^2{f}^2}$

Step 3: Calculate the energy in the bandwidth B Hz
Since $x\left(t\right)$ is a real baseband signal, its energy is distributed symmetrically around 0 Hz. The energy $E_B$ within the frequency range $-B\le f\le B$ is:
$E_B={\int }_{-B}^B{\left|X\right(f\left)\right|}^{2}df=2{\int }_0^B\frac{1}{4+4{\pi }^2{f}^2}df=\frac{1}{2}{\int }_0^B\frac{1}{1+{\pi }^2{f}^2}df$
Using the integration formula $\int \frac{1}{1+u^2}du={tan}^{-1}\left(u\right)$ with substitution $u=\pi f$ and $df=\frac{du}{\pi }$:
$E_B=\frac{1}{2\pi }{\left[{tan}^{-1}\right(\pi f\left)\right]}_0^B=\frac{1}{2\pi }{tan}^{-1}\left(\pi B\right)$

Step 4: Solve for B when energy within B is 99% of the total energy
We are given that $E_B=0.99\times E$:
$\frac{1}{2\pi }{tan}^{-1}\left(\pi B\right)=0.99\times \frac{1}{4}$
Multiplying both sides by $2\pi$:
${tan}^{-1}\left(\pi B\right)=\frac{0.99\pi }{2}$
Taking the tangent of both sides:
$\pi B=tan\left(0.495\pi \right)$
We can express the angle in terms of its complement to evaluate it:
$tan\left(0.495\pi \right)=tan(\frac{\pi }{2}-0.005\pi )=cot\left(0.005\pi \right)=\frac{1}{tan\left(0.005\pi \right)}$
Since $0.005\pi$ is a very small angle, we can use the small-angle approximation $tan\left(\theta \right)\approx \theta$:
$tan\left(0.005\pi \right)\approx 0.005\pi$
Therefore:
$\pi B\approx \frac{1}{0.005\pi }=\frac{200}{\pi }$
Solving for $B$:
$B\approx \frac{200}{{\pi }^2}$
Using ${\pi }^2\approx 9.8696$:
$B\approx \frac{200}{9.8696}\approx 20.26\text{Hz}$

Step 5: Check the options in terms of $\frac{1}{\pi }$
Let us evaluate the limits given in the correct option $\frac{63}{\pi }<B<\frac{64}{\pi }$ numerically using $\pi \approx 3.14159$:
$\frac{63}{\pi }\approx \frac{63}{3.14159}\approx 20.05\text{Hz}$
$\frac{64}{\pi }\approx \frac{64}{3.14159}\approx 20.37\text{Hz}$
Since $B\approx 20.26\text{Hz}$ falls directly inside this range ($20.05<20.26<20.37$), the inequality is satisfied:
$\frac{63}{\pi }<B<\frac{64}{\pi }$