xR and xA 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?

Question

Prepp Team · Accepted Answer

The question asks us to determine the relationship between the root mean square (RMS) and average values of two signals, x(t) and y(t), where y(t) is a scaled version of x(t), specifically y(t) = kx(t). The signal x(t) is periodic with period T, meaning x(t) = x(t - T). We need to find how the average value of y(t) (denoted as yA) relates to the average value of x(t) (denoted as xA), and how the RMS value of y(t) (denoted as yR) relates to the RMS value of x(t) (denoted as xR).

Understanding Average and RMS Values

Before deriving the relationships, let's recall the definitions:

Average Value (xA): The average value of a periodic signal x(t) over one period T is calculated by integrating the signal over the period and dividing by the period. The formula is: $${x_A} = \frac{1}{T} \int_{0}^{T} x(t) dt$$
RMS Value (xR): The RMS value of a periodic signal x(t) over one period T is the square root of the average of the square of the signal. The formula is: $${x_R} = \sqrt{\frac{1}{T} \int_{0}^{T} [x(t)]^2 dt}$$

We are given that k and T are constants independent of time t.

Calculating Average Value of y(t)

We need to find yA, the average value of y(t) = kx(t).

Using the definition of the average value:

$${y_A} = \frac{1}{T} \int_{0}^{T} y(t) dt$$

Substitute y(t) = kx(t) into the equation:

$${y_A} = \frac{1}{T} \int_{0}^{T} [kx(t)] dt$$

Since k is a constant, we can take it out of the integral:

$${y_A} = k \left( \frac{1}{T} \int_{0}^{T} x(t) dt \right)$$

Recognize that the term in the parenthesis is the definition of xA:

$${y_A} = k {x_A}$$

Therefore, the average value of y(t) is k times the average value of x(t).

Calculating RMS Value of y(t)

Next, we need to find yR, the RMS value of y(t) = kx(t).

Using the definition of the RMS value:

$${y_R} = \sqrt{\frac{1}{T} \int_{0}^{T} [y(t)]^2 dt}$$

Substitute y(t) = kx(t) into the equation:

$${y_R} = \sqrt{\frac{1}{T} \int_{0}^{T} [kx(t)]^2 dt}$$

Simplify the term inside the integral:

$${y_R} = \sqrt{\frac{1}{T} \int_{0}^{T} k^2 [x(t)]^2 dt}$$

Since k^2 is a constant, we can take it out of the integral:

$${y_R} = \sqrt{k^2 \left( \frac{1}{T} \int_{0}^{T} [x(t)]^2 dt \right)}$$

Recognize that the term inside the second square root is the definition of xR^2:

$${y_R} = \sqrt{k^2 \cdot {x_R}^2}$$

Simplify the square root:

$${y_R} = \sqrt{k^2} \sqrt{{x_R}^2}$$ $${y_R} = |k| {x_R}$$

The RMS value is scaled by the absolute value of the constant k. However, looking at the options provided, the relationship is expressed as yR = k * xR. This implies that either k is assumed to be positive, or the question implicitly considers the magnitude scaling effect represented by k.

Conclusion on Relationships

Based on the derivations:

The average value relationship is yA = k * xA.
The RMS value relationship is yR = |k| * xR, which is presented as yR = k * xR in the options.

Both conditions, yA = k * xA and yR = k * xR, must be true according to the options.

Comparing with Options

Option 1: ${y_A} = k{x_A}\;;{y_R} = k{x_R}$ - This matches our derived relationships.
Option 2: ${y_A} = k{x_A};{y_R} \ne k{x_R}$ - Incorrect, as yR is related to xR.
Option 3: ${y_A} \ne k{x_A};{y_R} = k{x_R}$ - Incorrect, as yA is equal to kxA.
Option 4: ${y_A} \ne k{x_A};{y_R} \ne k{x_R}$ - Incorrect, as both are related.