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

enter image description here

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.


2 Answers 2


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.


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.

  • $\begingroup$ Thanks Dear Marcus, Here is my autocorrelation and PSD ,but as you see we can't expect an impulse function obrazki.elektroda.pl/2572012200_1471945541.png , Yes RL is our system's load but why do we assume integral of psd as voltage not as output power? $\endgroup$ Aug 23, 2016 at 9:48
  • $\begingroup$ Simple: You observe ADC voltages, not power. $\endgroup$ Aug 23, 2016 at 10:58

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