Power spectral density of sinusoidal signal in noise

Question

In a literature I face with this input and power spectral density (PSD) $$x(t)=s(t)+n(t)=A\cos\left(\omega_c t +\phi\right) + n(t)$$

first I want to know

1. How can we find PSD of $\cos\left(\omega_c t +\phi\right)$?(I have found the autocorrelation function of $\cos\left(\omega_c t +\phi\right)$ but when I apply the PSD with Fourier transform it doesn't return the same result as the figure)
2. Why do we divided PSD of $\cos\left(\omega_c t +\phi\right)$ by $R_L$?
3. What is a bandpass filter's transfer function in time and frequency? And how can we find PSD of a bandpass?

Sorry for the simple question but I think I got confusing.

Answer 1

It looks like you are looking at the result of a combination of things.

First of all let us observe the image and try to understand it. The shredded horizontal line with value $\frac{N_{in}}{2B}$ represents $n(t)$ in the frequency domain and the signal $s(t)$ is represented by the two vertical arrows valued $\frac{A^2}{4R_L}$. Two arrows one in the sinus's frequency value and its mirror in the negative side is a representation of a sinuous wave in the frequency domain. The cosinuous wave will have the a similar representation as the sinuous wave in the frequency domain.I will reffer you to the table here

A BPF is a band pass filter and it is a combination of a low pass filter (will only pass frequencies up to a specified $f_0$ frequency) and a high pass filter (will only pass frequencies up from a specified $f_0$ frequency). In short, a BPF will only pass frequencies from a specific band bounded between some $f_1$ and $f_2$ as in $f_1\leq f\leq f_2$. The BPF is represented as a window in the frequency domain. You can see in your image the window with $B$ width around the sine arrows. This is also the reason the noise $n(t)$ is shredded, The filter is cutting it off outside the window's boundary. In the time domain a BPF will be represented by a $sinc()$ function, as can be seen in the table in this page.

I think your question needs additional data, since it lacks an explanation of what exactly $R_L$ is. I am assuming it is a resistor and this is some sort of an electronic circuit but if you want any further help you need to supply more detailed on the system.

Answer 2

How can we find PSD of cos(ωct+ϕ)?(I have been found autocorrelation function of cos(ωct+ϕ) but when I want to find PSD with Fourier transform it doesn't return the same result as the figure)

The Wiener-Chinchin Theorem states that for wide-sense stationary processes, the two should be equivalent. It's possible you have a normalization or scaling error, but definitely, the autocorrelation of a cosine should have a peak at its periods, shouldn't it?

Why do we divided PSD of cos(ωct+ϕ) by RL?

Completely different topic. "We" don't divide by anything. Obviously, something in your literature does that, and you forget to mention. Probably, this $R_L$ is the line/sink impedance of a hypothetical measurement device, and thus, gives the relation between Voltage squared asobserved (as per Ohm: $P=\frac{U^2}{R}$) and power.

What is bandpass filter transfer function in time and frequency? And how can we find PSD of bandpass?

You should definitely ask this in a different question, or just go and read up on what a filter is. A filter doesn't have a PSD – signals do, and a filter is a system, not a signal.