Signal Energy Bandwidth for $x(t) = e^{-2t}$ Signal

The problem requires finding the bandwidth $B$ (in Hz) where 99% of the total energy of the signal $x(t) = e^{-2t}$ (for $t \ge 0$) resides. We assume bandwidth $B$ refers to the frequency range $[-B, B]$.

Calculate Total Signal Energy

The total energy $E_{total}$ of a signal $x(t)$ is calculated as:

$ E_{total} = \int_{-\infty}^{\infty} |x(t)|^2 dt $

For the given signal $x(t) = e^{-2t}$ where $t \ge 0$, this becomes:

$ E_{total} = \int_{0}^{\infty} (e^{-2t})^2 dt = \int_{0}^{\infty} e^{-4t} dt $

Evaluating the integral yields:

$ E_{total} = \left[ -\frac{1}{4} e^{-4t} \right]_{0}^{\infty} = 0 - \left( -\frac{1}{4} \right) = \frac{1}{4} $

Determine Power Spectral Density

The Fourier Transform of $x(t) = e^{-at}u(t)$ is $X(f) = \frac{1}{a+j2\pi f}$. With $a=2$, we have:

$ X(f) = \frac{1}{2+j2\pi f} $

The Power Spectral Density (PSD) is the magnitude squared of the Fourier Transform:

$ |X(f)|^2 = \left| \frac{1}{2+j2\pi f} \right|^2 = \frac{1}{2^2 + (2\pi f)^2} = \frac{1}{4 + 4\pi^2 f^2} $

Calculate Energy within Bandwidth B

The energy $E_B$ within the frequency band $[-B, B]$ is found by integrating the PSD:

$ E_B = \int_{-B}^{B} |X(f)|^2 df = \int_{-B}^{B} \frac{1}{4 + 4\pi^2 f^2} df $

Using the substitution $u = 2\pi f$, $du = 2\pi df$, the integral transforms to:

$ E_B = \frac{1}{2\pi} \int_{-2\pi B}^{2\pi B} \frac{1}{4 + u^2} du = \frac{1}{2\pi} \left[ \frac{1}{2} \arctan\left(\frac{u}{2}\right) \right]_{-2\pi B}^{2\pi B} $

This simplifies to:

$ E_B = \frac{1}{4\pi} \left( \arctan\left(\frac{2\pi B}{2}\right) - \arctan\left(\frac{-2\pi B}{2}\right) \right) = \frac{1}{2\pi} \arctan(\pi B) $

Apply 99% Energy Condition

The condition is that $E_B$ must contain 99% of the total energy:

$ E_B = 0.99 \times E_{total} $

Substituting the expressions for $E_B$ and $E_{total}$:

$ \frac{1}{2\pi} \arctan(\pi B) = 0.99 \times \frac{1}{4} $

Rearranging to solve for $\arctan(\pi B)$:

$ \arctan(\pi B) = \frac{0.99 \pi}{2} $

Solve for Bandwidth B

Taking the tangent of both sides gives:

$ \pi B = \tan\left(\frac{0.99 \pi}{2}\right) $

Since $\frac{0.99 \pi}{2}$ is very close to $\frac{\pi}{2}$, we can approximate $\tan\left(\frac{0.99 \pi}{2}\right)$. Let $\alpha = \frac{0.99 \pi}{2} = \frac{\pi}{2} - \frac{0.01 \pi}{2}$.

Using $\tan(\frac{\pi}{2} - z) = \cot(z)$ and the approximation $\cot(z) \approx 1/z$ for small $z$ (where $z = \frac{0.01 \pi}{2}$):

$ \tan\left(\frac{0.99 \pi}{2}\right) = \cot\left(\frac{0.01 \pi}{2}\right) \approx \frac{1}{\frac{0.01 \pi}{2}} = \frac{2}{0.01 \pi} = \frac{200}{\pi} $

Therefore, $\pi B \approx \frac{200}{\pi}$, which leads to:

$ B \approx \frac{200}{\pi^2} $

Calculating the approximate value:

$ B \approx \frac{200}{(3.14159)^2} \approx 20.26 \text{ Hz} $

Compare with Options

Evaluate the bounds of the relevant options:

• Option 2: Lower bound $63/\pi \approx 20.06$ Hz. Upper bound $64/\pi \approx 20.37$ Hz.
• Option 3: Lower bound $126/\pi \approx 40.11$ Hz. Upper bound $128/\pi \approx 40.74$ Hz.

The calculated value $B \approx 20.26$ Hz falls within the range $(20.06 \text{ Hz}, 20.37 \text{ Hz})$.