# How can we prove that Spectral density of In-Phase component and Quad-Phase component of a sample function of narrow-band noise have same definition?

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?

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

• I don't know that book, but it's probably just defined to be that way, not derived, hence, no proof can be given. – 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 $$S_{N_I}(f) = S_{N_Q}(f) = \begin{cases} S_N(f-f_c) + S_N(f+f_c) & -B\le f \le B \\ 0 & \text{otherwise} \end{cases}$$