PSD and average power of A cos(2πfct + θ)

Energy spectral density

For an energy signal $g(t)$ (finite energy, zero average power), the ESD is the squared magnitude of its Fourier transform:

$$\Psi_g(f)=\lvert G(f)\rvert^2\ \ (\text{J/Hz}),\qquad E=\int_{-\infty}^{\infty}\Psi_g(f)df$$

It is also the Fourier transform of the energy autocorrelation $R_g(\tau)=\int g(t)g^*(t-\tau)dt$.

Power spectral density

For a power signal (finite non-zero average power, infinite energy), the PSD is the power per unit bandwidth, defined with the truncated signal $g_T(t)$:

$$S_g(f)=\lim_{T\to\infty}\frac{\lvert G_T(f)\rvert^2}{T}\ \ (\text{W/Hz}),\qquad P=\int_{-\infty}^{\infty}S_g(f)df$$

By the Wiener–Khinchin theorem, $S_g(f)$ is the Fourier transform of the time-averaged autocorrelation $R_g(\tau)=\lim_{T\to\infty}\frac{1}{T}\int_{-T/2}^{T/2}g(t)g^*(t-\tau)dt$.

PSD and average power of $g(t)=A\cos(2\pi f_ct+\theta)$

The signal is periodic with period $T_0=1/f_c$, so the time average over one period is enough.

Step 1 – autocorrelation:

$$\begin{aligned} R_g(\tau)&=\frac{1}{T_0}\int_{-T_0/2}^{T_0/2}A\cos(2\pi f_ct+\theta)\,A\cos(2\pi f_c(t-\tau)+\theta)\,dt\\ &=\frac{A^2}{2T_0}\int_{-T_0/2}^{T_0/2}\left[\cos(2\pi f_c\tau)+\cos(4\pi f_ct-2\pi f_c\tau+2\theta)\right]dt\\ &=\frac{A^2}{2}\cos(2\pi f_c\tau) \end{aligned}$$

The second cosine integrates to zero over a full period. Note that $R_g(\tau)$ does not depend on $\theta$.

Step 2 – PSD (Fourier transform of $R_g(\tau)$, using $\cos 2\pi f_c\tau\leftrightarrow\frac{1}{2}[\delta(f-f_c)+\delta(f+f_c)]$):

$$S_g(f)=\frac{A^2}{4}\left[\delta(f-f_c)+\delta(f+f_c)\right]$$

So the PSD consists of two impulses, each of weight $A^2/4$, at $f=\pm f_c$.

          S(f)
  A^2/4 ^   |   ^ A^2/4
        |   |   |
 -------+---+---+-------> f
      -fc   0   fc

Step 3 – average power:

$$\begin{aligned} P&=\int_{-\infty}^{\infty}S_g(f)\,df=\frac{A^2}{4}+\frac{A^2}{4}=\frac{A^2}{2}\\ P&=R_g(0)=\frac{A^2}{2}\ \ (\text{check}) \end{aligned}$$

Answer: $S_g(f)=\frac{A^2}{4}[\delta(f-f_c)+\delta(f+f_c)]$ and average power $P=\frac{A^2}{2}$ (watts into a 1 $\Omega$ load). The phase $\theta$ has no effect on PSD or power.