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).

The correct answer is 2.0.

To find the value of the constant c, we need to analyze how scaling and shifting affect the energy of a continuous-time signal.

The energy E of a continuous-time signal x(t) is defined by the following integral:

$$E={\int }_{-\infty }^{\infty }{\left|x\right(t\left)\right|}^{2}dt$$

Let y(t) represent the modified signal:

$$y\left(t\right)=2x(2t-1)$$

The energy of the modified signal, E_y, is given by:

$$E_y={\int }_{-\infty }^{\infty }{\left|y\right(t\left)\right|}^{2}dt$$

Substitute the expression for y(t) into the energy equation:

$$E_y={\int }_{-\infty }^{\infty }{\left|2x\right(2t-1\left)\right|}^{2}dt$$

$$E_y={\int }_{-\infty }^{\infty }4{\left|x\right(2t-1\left)\right|}^{2}dt$$

To evaluate this integral, we apply a change of variables (substitution). Let:

$$u=2t-1$$

Differentiating both sides with respect to t gives:

$$du=2dt\Rightarrow dt=\frac{du}{2}$$

Since the limits of integration are from -∞ to ∞, scaling and shifting do not change the span of the limits. The new limits for u will also be from -∞ to ∞.

Now, substitute u and dt back into the integral:

$$E_y={\int }_{-\infty }^{\infty }4{\left|x\right(u\left)\right|}^2\frac{du}{2}$$

$$E_y=2{\int }_{-\infty }^{\infty }{\left|x\right(u\left)\right|}^{2}du$$

Since u is a dummy variable of integration, the integral represents the original energy E of the signal x(t):

$${\int }_{-\infty }^{\infty }{\left|x\right(u\left)\right|}^{2}du=E$$

Thus, we can express the energy of the modified signal as:

$$E_y=2E$$

We are given that the energy of the modified signal is cE. Comparing the equations:

$$cE=2E$$

$$c=2$$

Rounding to one decimal place as requested, we get:

$$c=2.0$$