# Why does multiplying a real signal by a random complex phase term result in "spreading" in the Fourier domain?

Suppose I have some real-valued signal $$x\mapsto f(x)$$. The amplitude of its Fourier transform $$\mathcal{F}[f]$$ then looks like a peak around the DC-term, decaying as we move towards higher frequencies. If however we pre-multiply with $$g(x)=f(x)e^{i N(x)}$$ where $$N(x)$$ is random noise, then the Fourier domain has information about $$f$$ "spread" across it, or seen differently, its amplitude distribution is a lot less biased. Why does this happen?

• Are you familiar with the convolution property of the Fourier transform? Commented Nov 6, 2023 at 19:12
• @MarcusMüller Yes, so we are equivalently convolving the Fourier domain with the Fourier transform of this amplitude-1 noise, which very intuitively I can see why that would 'spread', but I don't think I have a good mental picture of what the Fourier domain of this noise is... Commented Nov 6, 2023 at 19:36
• Start with a a simple case, like $N(x)$ being a pulse train with amplitudes $0$ and $\pi$, which results in sign changes in $g(x)$. Can you see how those sudden changes will spread the spectrum of $g(x)$?
– MBaz
Commented Nov 6, 2023 at 22:16
• @user2983473: that will actually depend a lot on the probability density function of $N(x)$. Are you assuming that it's uniformly distributed on $[0,2\pi]$, Gaussian with a given mean and variance or something else ? Commented Nov 7, 2023 at 1:30
• It depends on the probability density and dynamic behavior of $N(x)$ Note that because $e^{i \theta}$ is nonlinear in $\theta$, you can't go straight from the probability density and frequency spectrum of $N(x)$ to the spectrum of $e^{i N(x)}$. Commented Nov 7, 2023 at 3:37

I'm not sure I correctly understand what you mean by DC term, but will assume you mean fundamental frequency $$f_0$$ of some periodic signal, i.e. an unmodulated carrier.

So your real-valued signal $$f(x)$$ is periodic, this is a consequence of my assumption in my first point. This means it must have some kind of sinusoidal approximation of $$f(x)$$. This could be a basic sine wave, or even a Fourier series of something more complicated.

I will take a simplified approach for illustrative purposes and say $$f(x) = \cos (2 \pi f_0 x)$$. So then we can see that: $$f(x) e^{i N(x)} = \cos (2 \pi f_0 x) \cdot e^{i N(x)} = \frac{1}{2} \left( e^{i(2 \pi f_0 x)} + e^{-i(2 \pi f_0 x)} \right)\cdot e^{i N(x)}$$

and if we simplify some more $$f(x) e^{i N(x)} = \frac{1}{2} \left( e^{i(2 \pi f_0 x + N(x))} + e^{-i(2 \pi f_0 x + N(x))} \right) = \cos (2 \pi f_0 x + N(x)).$$

So $$N(x)$$ being noise and being situated where inside the cosine function is now acting as a random phase. We call this phase noise. In the time domain, it's referred to as jitter.

It's good to remember that a randomised phase will generally look like a frequency change in the frequency domain. Because of this, you get randomised "spreading" as you mentioned, because it is essentially looks like the frequency is randomly changing.

What you're asking about (phase noise) is a very important phenomenon and is often the main specification when building things like a Phase Locked Loop (PLL). It can be a limiting factor in the quality of a modulated signal (i.e. its Error Vector Magnitude (EVM)).