Answer :

Solution :

Given that,

x_R and x_A are, respectively, the RMS and average values of x(t) = x(t – T) y_R and

y_A are, respectively, the RMS and average values of y(t) = kx(t) k, T are independent of t.

The average value of x(t) is,

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

The average value of y(t) is,

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

The RMS value of x(t) is,

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

The RMS value of y(t) is,

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

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