Question

If the energy of a continuous-time signal \( x(t) \) is \( E \) and the energy of the signal \( 2x(2t - 1) \) is \( cE \), then \( c \) is (rounded off to 1 decimal place).

Hint:

For scaled signals \( kx(at - b) \), the energy scaling factor is \( c = k^2 / |a| \). Always account for both amplitude and time scaling effects.

Answer 1

Step 1: Energy scaling properties of signals. The energy of a continuous-time signal \( x(t) \) is given by: \[ E = \int_{-\infty}^\infty |x(t)|^2 \, dt. \] If the signal is scaled as \( x(at - b) \), the energy scales as: \[ E' = \frac{1}{|a|} \int_{-\infty}^\infty |x(t)|^2 \, dt = \frac{E}{|a|}. \] Step 2: Analyze the given signal \( 2x(2t - 1) \). For the signal \( 2x(2t - 1) \): - The amplitude scaling factor is \( 2 \), so the energy scales by \( 2^2 = 4 \). - The time scaling factor is \( a = 2 \), so the energy scales by \( \frac{1}{|2|} = 0.5 \). Step 3: Calculate the total scaling factor \( c \). The total scaling factor is: \[ c = 4 \cdot 0.5 = 2.0. \]