Finding power spectral density (ESD first)

[thomas49th]

Homework Statement

Find the Power Spectral Density Sg(w) of the power signal g(t) = cos wt.
(Hint: Compute the autocorrelation function first, and then use the prop-
erty Rg(T ) <=> Sg(w))

Homework Equations

Well, if I take the hint's path I need to calculate the ESD first

ESD(T) = $$\int_{-\infty}^{\infty}g(t)g(t+T)dt$$

The Attempt at a Solution

Firstly ESD(T) - why does the energy spectral density change as T changes. Isn't the energy spectral density the amount of energy that each frequency "requires" out of the signal? Do I simply integrate

$$\int_{-\infty}^{\infty}cos(w_{0}t) \cdot cos(w_{0}t+T)dt$$

Spent ages reading over this and don't understand it very well!

Thanks
Thomas

 

[mighty2000]

Hello Thomas,

Thank you for your question. The energy spectral density (ESD) is a measure of the energy contained in a signal at different frequencies. As the signal g(t) = cos wt is a power signal, the ESD will give us the power spectral density (PSD) instead. The PSD is a measure of the power contained in a signal at different frequencies.

To calculate the PSD, we first need to calculate the autocorrelation function Rg(T) of the signal g(t). This can be done by using the definition of the autocorrelation function:

Rg(T) = \int_{-\infty}^{\infty}g(t)g(t+T)dt

Substituting g(t) = cos wt into the integral, we get:

Rg(T) = \int_{-\infty}^{\infty}cos wt \cdot cos(wt+T)dt

Using the trigonometric identity cos a cos b = 1/2(cos(a-b) + cos(a+b)), we can simplify the integral to:

Rg(T) = 1/2\int_{-\infty}^{\infty}cos wt \cdot cos(wt+T)dt + 1/2\int_{-\infty}^{\infty}cos wt \cdot cos(wt-T)dt

Since the cosine function is an even function, the second integral will be equal to the first integral with T replaced by -T. Therefore, we can rewrite the integral as:

Rg(T) = \int_{-\infty}^{\infty}cos wt \cdot cos(wt+T)dt

Now, using the property Rg(T) <=> Sg(w), we can say that the PSD Sg(w) of the signal g(t) is equal to the autocorrelation function Rg(T) evaluated at T = 0. Therefore, we have:

Sg(w) = Rg(0) = \int_{-\infty}^{\infty}cos wt \cdot cos(wt+0)dt = \int_{-\infty}^{\infty}cos^{2}wt dt

Using the trigonometric identity cos^{2} a = 1/2(1 + cos2a), we can simplify the integral to:

Sg(w) = 1/2\int_{-\infty