A signal $f(t)$ has energy $E$, the energy of the signal $f(2t)$ is equal to

Question

A periodic signal $x(t)$ of period $T_0$ is given by $ x(t) = \begin{cases} 1, & |t|<T_1 \\ 0, & T_1<|t|<\frac{T_0}{2} \end{cases} $. The dc component of $x(t)$ is

Answer 1

Step 1: Take a simple test signal, a rectangular pulse $ f(t) = 1 $ for $ 0 \leq t \leq 1 $ and zero elsewhere, so its energy is $ E = \displaystyle\int_0^1 1^2\,dt = 1 $.
Step 2: The compressed signal $ f(2t) $ equals $ 1 $ only when $ 0 \leq 2t \leq 1 $, i.e. for $ 0 \leq t \leq 0.5 $, so its energy is $ \displaystyle\int_0^{0.5} 1^2\,dt = 0.5 $.
Step 3: Comparing, the compressed signal's energy $ 0.5 $ is exactly half of the original energy $ E = 1 $, matching the general rule that compressing time by a factor $ a $ scales energy by $ 1/a $. \[ \boxed{E_{f(2t)} = \dfrac{E}{2}} \]

Answer 2

Here is a purely dimensional/scaling argument rather than an integral substitution, checked against each option.

Option E: For the energy to stay at $ E $ after compression, the time axis squeezing would have to be exactly compensated by an amplitude change, but $ f(2t) $ only re-indexes time, it does not rescale amplitude, so the energy cannot remain the same; this is inconsistent with how the area under the squared signal behaves.
Option E/2: Squeezing the time axis by a factor of $ 2 $ means every value of $ f $ is reached in half the time span; since energy is an integral of $ |f|^2 $ over time, shrinking the time span by a factor of $ 2 $ while the amplitude profile stays the same shrinks the total area, hence the energy, by that same factor of $ 2 $, giving $ E/2 $.
Option 2E: This would require the compressed signal to somehow spend more time accumulating energy than the original, which contradicts the fact that compression strictly shortens the interval of integration.
Option 4E: A factor of $ 4 $ would arise only if both the time axis were compressed and the amplitude were doubled simultaneously, neither of which happens in the transformation $ f(t) \to f(2t) $; amplitude is left untouched here.

Since only the time axis is rescaled and the area under the squared curve shrinks proportionally to that time-axis compression factor, the energy must reduce by exactly that factor.

So the correct answer is $ E/2 $.

Answer 3

Here is a purely dimensional/scaling argument rather than an integral substitution, checked against each option.

Option E: For the energy to stay at $ E $ after compression, the time axis squeezing would have to be exactly compensated by an amplitude change, but $ f(2t) $ only re-indexes time, it does not rescale amplitude, so the energy cannot remain the same; this is inconsistent with how the area under the squared signal behaves.
Option E/2: Squeezing the time axis by a factor of $ 2 $ means every value of $ f $ is reached in half the time span; since energy is an integral of $ |f|^2 $ over time, shrinking the time span by a factor of $ 2 $ while the amplitude profile stays the same shrinks the total area, hence the energy, by that same factor of $ 2 $, giving $ E/2 $.
Option 2E: This would require the compressed signal to somehow spend more time accumulating energy than the original, which contradicts the fact that compression strictly shortens the interval of integration.
Option 4E: A factor of $ 4 $ would arise only if both the time axis were compressed and the amplitude were doubled simultaneously, neither of which happens in the transformation $ f(t) \to f(2t) $; amplitude is left untouched here.

Since only the time axis is rescaled and the area under the squared curve shrinks proportionally to that time-axis compression factor, the energy must reduce by exactly that factor.

So the correct answer is $ E/2 $.

LIST-I	LIST-II
A. \( e^{-at}u(t), a>0 \)	I. \( \pi[\delta(\omega - \omega_0) + \delta(\omega + \omega_0)] \)
B. \( \cos \omega_0 t \)	II. \( \frac{1}{j\omega + a} \)
C. \( \sin \omega_0 t \)	III. \( \frac{1}{(j\omega + a)^2} \)
D. \( te^{-at}u(t), a>0 \)	IV. \( -j\pi[\delta(\omega - \omega_0) - \delta(\omega + \omega_0)] \)

A signal $f(t)$ has energy $E$, the energy of the signal $f(2t)$ is equal to

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The Correct Option is B

Approach Solution - 1

Approach Solution -2

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