Question:hard
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 \(B\)?
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 \(B\)?
If \(99\%\) of the energy of \(x(t)\) lies within \(B\) Hz, then which of the following options is TRUE for the value of \(B\)?
Show Hint
Use Parseval's theorem, integrate the energy spectral density from -B to B, and set the fraction equal to 0.99.
Updated On: Jul 20, 2026
- \(B>1\) kHz
- \(63/\pi\) Hz \(<B<\) \(64/\pi\) Hz
- \(126/\pi\) Hz \(<B<\) \(128/\pi\) Hz
- \(B<1\) Hz
Show Solution
The Correct Option is B
Solution and Explanation
Step 1: Recall the standard result for a one sided exponential pulse.
For $x(t)=e^{-at}u(t)$, the fraction of energy contained within $B$ Hz is a known result: \[ \frac{E(B)}{E_{total}}=\frac{2}{\pi}\arctan\left(\frac{2\pi B}{a}\right) \]
Step 2: Plug in the decay constant.
Here $a=2$, so \[ \frac{E(B)}{E_{total}}=\frac{2}{\pi}\arctan(\pi B) \] which matches the expression derived directly from the Fourier transform.
Step 3: Use the complement, since 99% inside means 1% outside.
\[ \frac{2}{\pi}\arctan(\pi B)=0.99\ \Rightarrow\ \frac{\pi}{2}-\arctan(\pi B)=\frac{\pi}{2}(1-0.99)=0.005\pi\approx0.0157 \]
Step 4: Use the identity \(\frac{\pi}{2}-\arctan(x)=\arctan(1/x)\) for \(x>0\).
\[ \arctan\left(\frac{1}{\pi B}\right)\approx0.0157 \] For a small angle, $\arctan(y)\approx y$, so \[ \frac{1}{\pi B}\approx0.0157\ \Rightarrow\ \pi B\approx63.7 \]
Step 5: Conclude.
\[ B\approx\frac{63.7}{\pi} \] \[ \boxed{\frac{63}{\pi}\text{ Hz}<B<\frac{64}{\pi}\text{ Hz}} \] This matches the value found by direct integration.
For $x(t)=e^{-at}u(t)$, the fraction of energy contained within $B$ Hz is a known result: \[ \frac{E(B)}{E_{total}}=\frac{2}{\pi}\arctan\left(\frac{2\pi B}{a}\right) \]
Step 2: Plug in the decay constant.
Here $a=2$, so \[ \frac{E(B)}{E_{total}}=\frac{2}{\pi}\arctan(\pi B) \] which matches the expression derived directly from the Fourier transform.
Step 3: Use the complement, since 99% inside means 1% outside.
\[ \frac{2}{\pi}\arctan(\pi B)=0.99\ \Rightarrow\ \frac{\pi}{2}-\arctan(\pi B)=\frac{\pi}{2}(1-0.99)=0.005\pi\approx0.0157 \]
Step 4: Use the identity \(\frac{\pi}{2}-\arctan(x)=\arctan(1/x)\) for \(x>0\).
\[ \arctan\left(\frac{1}{\pi B}\right)\approx0.0157 \] For a small angle, $\arctan(y)\approx y$, so \[ \frac{1}{\pi B}\approx0.0157\ \Rightarrow\ \pi B\approx63.7 \]
Step 5: Conclude.
\[ B\approx\frac{63.7}{\pi} \] \[ \boxed{\frac{63}{\pi}\text{ Hz}<B<\frac{64}{\pi}\text{ Hz}} \] This matches the value found by direct integration.
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