x_R and x_A are, respectively, the rms and average values of x(t) = x(t - T), and similarly, yR and yA are, respectively, the rms and average values of y(t) = kx(t), k, T are independent of t. Which of the following is true?

The correct option is y_A=kx_A; y_R≠kx_R.

Let us analyze the given mathematical definitions for the average value and root-mean-square (rms) value of a periodic signal x(t) with period T, and determine how scaling by a constant k affects these values.

1. Average Value Derivation:
The average value x_A of a periodic signal x(t) over a period T is defined as:

$x_A=\frac{1}{T}{\int }_0^{T}x\left(t\right)dt$

For the signal y(t) = kx(t), where k is a constant independent of t, its average value y_A is calculated as:

$y_A=\frac{1}{T}{\int }_0^{T}y\left(t\right)dt=\frac{1}{T}{\int }_0^{T}kx\left(t\right)dt$

Since k is constant, it can be factored out of the integral:

$y_A=k·\left(\frac{1}{T}{\int }_0^{T}x\left(t\right)dt\right)=k{x}_A$

Thus, y_A = kx_A is always true.

2. RMS Value Derivation:
The rms value x_R of a periodic signal x(t) over a period T is defined as:

$x_R=\sqrt{\frac{1}{T}{\int }_0^T{\left[x\left(t\right)\right]}^{2}dt}$

For the signal y(t) = kx(t), its rms value y_R is:

$y_R=\sqrt{\frac{1}{T}{\int }_0^T{\left[kx\left(t\right)\right]}^{2}dt}=\sqrt{k^2·\frac{1}{T}{\int }_0^T{\left[x\left(t\right)\right]}^{2}dt}$

Taking k² out of the square root gives:

$y_R=\left|k\right|·x_R$

Since the absolute value function |k| is involved, if k is negative (i.e., k < 0), then |k| = -k ≠ k. Therefore, in general for any arbitrary real constant k, y_R = |k|x_R, which means y_R ≠ kx_R (unless k ≥ 0).

Hence, the relation y_A = kx_A holds strictly, while y_R = kx_R does not hold generally for all real numbers k.