Suppose we have a narrow-band noise $N(t)$. We analyse this to In-Phase and Quad phase components by the equation:

$$N(t) = N_I(t) \cos(2\pi f_c t) - N_Q(t) \sin(2\pi f_c t)$$

Now a book (by Simon Haykin) states that:

$$ S_{N_I}(f) = S_{N_Q}(f) = \begin{cases} S_N(f-f_c) + S_N(f+f_c) & -B\le f \le B \\ 0 & \text{otherwise} \end{cases} $$

How do we prove this?

Image of Power Spectral Density of Narrowband Noise

This is the Image of Power Spectral Density of Narrowband Noise.

  • 1
    $\begingroup$ I don't know that book, but it's probably just defined to be that way, not derived, hence, no proof can be given. $\endgroup$ – Marcus Müller Jul 13 '19 at 19:56

Synthetic Method:

If $\{\hat X(t)\}$ and $\{\hat Y(t)\}$ are zero-mean uncorrelated low-pass WSS processes with identical autocorrelation function $R(\tau)$ and identical power spectral density $S(f)$ enjoying the property that $S(f) = 0$ for $|f|>B$, then $$\hat{N}(t) = \hat X(t)\cos(2\pi f_ct) - \hat Y(t)\sin(2\pi f)ct$$ is a band-pass process whose autocorrelation function is $R(\tau)\cos(2\pi f_c\tau)$ and power spectral density is $${S}_{\hat{N}}(f) = S(f-f_c)+S(f+f_c) = S_N(f).$$ There is no pretense that $\hat{N}(t)$ is the same as $N(t)$ for any $t$ or that $\hat X(t)$ and $\hat Y(t)$ are the complex baseband representation of $N(t)$ but the point is that by looking only at the power spectral densities, one cannot tell whether the synthesized process $\{\hat N(t)\}$ is the same as, or different from, the given bandpass process $\{N(t)\}$.

Analytic Method: (Suspend disbelief and take with a large grain of NaCl)

Suppose that $\{N(t)\}$ is a zero-mean bandpass WSS process whose power spectral density $S_N(f)$ has value $0$ except when $|f| \in [f_c-B, f_c+B]$ and consider the random variables $N(t)\cos(2\pi f_ct)$ and $-N(t)\sin(2\pi f_ct)$. They obviously have zero mean, and if $N(t)$ has variance $\sigma^2 = R_N(0)$, then these variables have variances \begin{align}\sigma^2\cos^2(2\pi f_ct) &= \frac 12\big(\sigma^2+\cos(2\pi (2f_c)t)\big)\tag{1}\\ \sigma^2\sin^2(2\pi f_ct) &= \frac 12\big(\sigma^2-\cos(2\pi (2f_c)t)\big)\tag{2}. \end{align} Their covariance is $$-\sigma^2\cos(2\pi f_c t)\sin(2\pi f_c t) = -\frac 12\sigma^2\sin(2\pi (2f_c)t).\tag{3}$$

We take the viewpoint that each of the random processes $\{N(t)\cos(2\pi f_ct)\}$ and $\{-N(t)\sin(2\pi f_ct)\}$, neither of which is a WSS process, is the sum of a low-pass WSS process $\{X(t)\}$ (respectively $\{Y(t)\}$) and a "double-frequency" process which not WSS, and which we can eliminate by (ideal) low-pass filtering these two processes with bandwidth $B$. Put another way, the result of (separately) low-pass filtering the processes $\{N(t)\cos(2\pi f_ct)\}$ and $\{-N(t)\sin(2\pi f_ct)\}$ leaves us with WSS processes with variance $\frac 12\sigma^2$ each (the double-frequency stuff in $(1)$ and $(2)$ has been eliminated by the low-pass filtering), and these processes are uncorrelated (the nonzero covariance in $(3)$ is double-frequency stuff that has been eliminated by the low-pass filtering).

Multiplying by $\cos(2\pi f_ct)$ or $\sin(2\pi f_ct)$ is modulation (frequency content at frequency $f$ is moved to frequencies $f+f_c$ and $f-f_c$). Applying this notion to the frequency band $[f_c-B, f_c+B]$, we see that any content there is moved down to be centered at $0$ and also moved to the band $[2f_c-B, 2f_c+B]$. Similarly, any content in the band $[-f_c-B, -f_c+B]$ is moved down to be centered at $0$ and also moved to the frequency band $[-f_c-B, -f_c+B]$. The latter is the stuff that gets eliminated by the low-pass filtering. What is left is the stuff moved down to DC and this "explains" why Haykin claims that $$ S_{N_I}(f) = S_{N_Q}(f) = \begin{cases} S_N(f-f_c) + S_N(f+f_c) & -B\le f \le B \\ 0 & \text{otherwise} \end{cases} $$

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