Calculating Total Power in y(t)

[jeir35]

Homework Statement

Finding the total power in the signal y(t) = (sin t + j cos t)(u(t − 2) − u(t − 4)) .

Homework Equations

E= integral from -infinity to + infinity abs(x(t))^2 dt

signal has finite energy if < infi

The Attempt at a Solution

(sin t + j cos t) = exp(jpiet)


is this the right direction here?

exp(jt)= 2cos(x)^2 -1 + 2sin(x)cos(x) * j

i made the assumption that

when you take integral of the (u(t − 2) − u(t − 4))

its from 2-4 since the step function is = 1 and its delayed i don't know if this is correct


integral from 2 to 4 of (2cos(x)^2-1)

i thought i could leave out the imaginary parts

 

[Valkarie]

and just solve the real componentintegral from 2 to 4 of (2cos(x)^2-1) dx = [2sin(x)cos(x)-x] from 2 to 4 = [2sin(4)cos(4)-4]-[2sin(2)cos(2)-2]= -3.8787

 

[Mene]

since they would cancel out when finding the total power



Your approach is on the right track, but there are a few issues that need to be addressed. First, when finding the total power of a signal, we are interested in the average power over time, not just at a specific instant. Therefore, we need to take the integral over the entire signal, not just from 2 to 4.

Secondly, when finding the total power, we need to use the absolute value squared of the signal, not just the absolute value. This ensures that we are including both the positive and negative components of the signal in our calculation.

So, the correct approach would be to first convert y(t) to its polar form, which is y(t) = √2sin(t + π/4)u(t-2) - √2sin(t + π/4)u(t-4). Then, using the equation for total power, we can calculate the total power as:

P = ∫∞-∞ |y(t)|^2 dt

= ∫∞-∞ (√2sin(t + π/4))^2 dt

= 2∫∞-∞ sin^2(t + π/4) dt

= 2∫∞-∞ (1-cos(2t + π/2)) dt

= 2∫∞-∞ (1-sin(2t)) dt

= 2∫∞-∞ (1-0) dt

= 2∫∞-∞ dt

= ∞

This result tells us that the total power of the signal y(t) is infinite, which makes sense since the signal is periodic and has no DC component. This means that the signal has an infinite amount of energy over time, which is not physically realistic. Therefore, we can conclude that the signal y(t) does not have finite energy and cannot be used in practical applications.