In [1, p. 62 exercise 3.16] we are asked to show that the random process $x[n]= A \cos(2\pi f_{0}n+\phi)$ (1)

, where $\phi$ is uniformly distributed on $(0,2\pi)$, is WSS by finding its mean and ACF. Using the same assumptions we are asked to repeat the exercise for a single complex sinusoid $x[n]=A\exp[j(2\pi f_{0}n+\phi)].$ (2)

Solution: The mean of the random process is given by: $\mu_{x}[n]$ $=$ $\int_{0}^{2\pi}A\cos(2\pi f_{0}n+\phi)\frac{1}{2\pi} d\phi$ $=$ $\frac{A}{2\pi}\Big[\sin(2\pi f_{0}n+\phi)\Big]_{\phi=0}^{\phi=2\pi}$ $=$ $0$

and is thus independent of the sampling variable. Let the p.d.f of $\phi$ be $f_{pdf}(\phi)=\frac{1}{2\pi}$ then the ACF of the random process is given by $r_{xx}[n,k]$ $=$ $E\{x^{\star}[n]x[n+k]\}$ $=$ $\int_{0}^{2\pi}x^{\star}[n]x[n+k]f_{pdf}d\phi$ $=$ $\int_{0}^{2\pi}A \cos(2\pi f_{0}n+\phi)A \cos(2\pi f_{0}(n+k)+\phi)\frac{1}{2\pi}d\phi$ $=$ $\frac{A^{2}}{2\pi} \int_{0}^{2\pi}\cos(2\pi f_{0}n+\phi) \cos(2\pi f_{0}(n+k)+\phi)d\phi$ (3)

From the trigonometric formula [2, p. 810 rel 21.2-12]: $\cos\mathfrak{A} \cos\mathfrak{B}=\frac{cos(\mathfrak{A-B})+cos(\mathfrak{A+B})}{2}$ (4)

by setting $\mathfrak{A}=2\pi f_{0}(n+k)+\phi$ and $\mathfrak{B}=2\pi f_{0}n+\phi$ we obtain because $\int_{0}^{2\pi}cos(4\pi f_{0}n+ 2\pi f_{0}k+2 \phi)d\phi =0$: $r_{xx}[n,k]$ $=$ $\frac{A^{2}}{2} \cos(2\pi f_{0}k)+ \frac{A^{2}}{4\pi}\int_{0}^{2\pi}cos(4\pi f_{0}n+ 2\pi f_{0}k+2 \phi)d\phi$ $=$ $\frac{A^{2}}{2}\cos(2\pi f_{0}k)$ $=$ $r_{xx}[k].$

Thus it is shown that the random process $x[n]$ is WSS. Repeating the same for a single complex sinusoid $x[n]=Ae^{j2\pi f_{0}n+\phi}$ we obtain for the mean that it is equal to zero and independent of $n$: $\mu_{x}[n]$ $=$ $E\left\{x[n]\right\}$ $=$ $\int_{0}^{2\pi}\frac{A}{2\pi}e^{j(2\pi f_{0}n+\phi)}d\phi$ $=$ $\frac{A}{2\pi j}e^{j2\pi f_{0}n}\left[e^{j\phi}\right]_{0}^{2\pi}$ $=$ $0.$ (5)

The autocorrelation of the complex sinusoid is obtained by: $r_{xx}[n,k]$ $=$ $E\left\{x[n+k]x^{\star}[n]\right\}$ $=$ $\frac{A^{2}}{2\pi}\int_{0}^{2\pi}e^{j(2\pi f_{0}(n+k)+\phi)}e^{-j(2\pi f_{0}n+\phi)}d\phi$ $=$ $A^{2}e^{j2\pi f_{0}k}$ $=$ $r_{xx}[k],$

which again is independent of $n$ and thus the complex sinusoid is also wide sense stationary. We see that in this case, the real part of the complex sinusoid equals the double of the autocorrelation of the real sinusoid.

 Steven M. Kay: “Modern Spectral Estimation – Theory and Applications”, Prentice Hall, ISBN: 0-13-598582-X.
 Granino A. Korn and Theresa M. Korn: “Mathematical Handbook for Scientists and Engineers”, Dover, ISBN: 978-0-486-41147-7.