Question

Consider a real baseband signal \(x(t)=e^{-2t}\), for \(t\) (in seconds) \(\ge0\).
If \(99\%\) of the energy of \(x(t)\) lies within \(B\) Hz, then which of the following options is TRUE for the value of \(B\)?

• A. \(B>1\) kHz
• B. \(63/\pi\) Hz \(<B<\) \(64/\pi\) Hz
• C. \(126/\pi\) Hz \(<B<\) \(128/\pi\) Hz
• D. \(B<1\) Hz

Hint:

Use Parseval's theorem, integrate the energy spectral density from -B to B, and set the fraction equal to 0.99.

Answer 1

The Correct Option is B

Step 1: Find the total energy in the time domain.
\[ E_{total}=\int_0^{\infty}\left(e^{-2t}\right)^2dt=\int_0^{\infty}e^{-4t}dt=\frac{1}{4} \]

Step 2: Find the Fourier transform of x(t).
\[ X(f)=\int_0^{\infty}e^{-2t}e^{-j2\pi ft}dt=\frac{1}{2+j2\pi f} \] so the energy spectral density is \[ |X(f)|^2=\frac{1}{4+4\pi^2f^2} \]

Step 3: Write the energy contained in the band from -B to B.
\[ E(B)=\int_{-B}^{B}\frac{1}{4+4\pi^2f^2}\,df=2\int_0^{B}\frac{1}{4}\cdot\frac{1}{1+\pi^2f^2}\,df \]

Step 4: Carry out the integral.
Using $\int\frac{df}{1+\pi^2f^2}=\frac{1}{\pi}\arctan(\pi f)$, \[ E(B)=\frac{1}{2}\cdot\frac{1}{\pi}\arctan(\pi B)=\frac{1}{2\pi}\arctan(\pi B) \]

Step 5: Form the energy fraction.
\[ \frac{E(B)}{E_{total}}=\frac{\frac{1}{2\pi}\arctan(\pi B)}{\frac{1}{4}}=\frac{2}{\pi}\arctan(\pi B) \]

Step 6: Set this fraction equal to 0.99 and solve for B.
\[ \frac{2}{\pi}\arctan(\pi B)=0.99 \ \Rightarrow\ \arctan(\pi B)=\frac{0.99\pi}{2}\approx1.5551\text{ rad} \]

Step 7: Solve for \(\pi B\) using the fact that 1.5551 is close to \(\pi/2\).
Let $\varepsilon=\frac{\pi}{2}-1.5551\approx0.0157$. Since $\tan\left(\frac{\pi}{2}-\varepsilon\right)=\cot(\varepsilon)\approx\frac{1}{\varepsilon}$ for small $\varepsilon$, \[ \pi B\approx\frac{1}{0.0157}\approx63.7 \]

Step 8: Read off B.
\[ B\approx\frac{63.7}{\pi} \] which lies between $\dfrac{63}{\pi}$ Hz and $\dfrac{64}{\pi}$ Hz.

Step 9: Check the other options.
Option (A) and (D) are far off in scale, since the actual bandwidth is only about $20$ Hz. Option (C) is exactly double the correct range, the kind of error that appears if the factor of $2$ from integrating over $-B$ to $B$ is applied twice.

Step 10: Final conclusion.
\[ \boxed{\frac{63}{\pi}\text{ Hz}<B<\frac{64}{\pi}\text{ Hz}} \]