Question

Find the power and the rms value of the following signal square: x(t) = 10 sin(10t) sin(15t)
1. Px =
2. xrmx =

Answer 1

Answer:

$\mathbf{P_x =25 \ watts}$

$\mathbf{x_{rmx} = 5 \ unit}$

Explanation:

Given that:

x(t) = 10 sin(10t) . sin (15t)

the objective is to find the power and the rms value of the following signal square.

Recall that:

sin (A + B) + sin(A - B) = 2 sin A.cos B

x(t) = 10 sin(15t) . cos (10t)

x(t) = 5(2 sin (15t). cos (10t))

x(t) = 5 × ( sin (15t + 10t) +  sin (15t-10t)

x(t) = 5sin(25 t) + 5 sin (5t)

From the knowledge of sinusoidial signal  Asin (ωt), Power can be expressed as:

$P= \dfrac{A^2}{2}$

For the number of sinosoidial signals;

Power can be expressed as:

$P = \dfrac{A_1^2}{2}+ \dfrac{A_2^2}{2}+ \dfrac{A_3^2}{2}+ ...$

As such,

For x(t), Power  $P_x = \dfrac{5^2}{2}+ \dfrac{5^2}{2}$

$P_x = \dfrac{25}{2}+ \dfrac{25}{2}$

$P_x = \dfrac{50}{2}$

$\mathbf{P_x =25 \ watts}$

For the number of sinosoidial signals;

$RMS = \sqrt{(\dfrac{A_1}{\sqrt{2}})^2+(\dfrac{A_2}{\sqrt{2}})^2+(\dfrac{A_3}{\sqrt{2}})^2+...$

For x(t), the RMS value is as follows:

$x_{rmx} =\sqrt{(\dfrac{5}{\sqrt{2}} )^2 +(\dfrac{5}{\sqrt{2}} )^2 }$

$x_{rmx }=\sqrt{(\dfrac{25}{2} ) +(\dfrac{25}{2} ) }$

$x_{rmx }=\sqrt{(\dfrac{50}{2} )}$

$x_{rmx} =\sqrt{25}$

$\mathbf{x_{rmx} = 5 \ unit}$