Sum of Sine and Cosine with Random Phase as LTI System

Question

I have the following system:
Where $ {H}_{1} \left( f \right) = {H}_{2} \left( f \right) $ and $ \theta \sim U[0, 2\pi]$ independent of any other factor in the system.

Given the input is identical, Is an LTI system?
Could you prove it?

Otherwise I don't have explanation why if the input is White Gaussian Noise with the same Variance to each of the inputs (Same statistics, not the same signal) $ {Y}_{1} $ and $ {Y}_{2} $ aren't Gaussian Process yet $ r(t) $ is a Gaussian Process.

I have the feeling the random phase is offset by the sum of the Sine and Cosine.

Thank You.

Answer 1

So let's just look at it from a linear systems perspective: $$ \begin{array}{ccl} r(t) &=& Y_1(t) + Y_2(t)\\ &=& Z_1(t) \cos(2\pi f_0 t + \theta) + Z_2(t) \sin(2\pi f_0 t + \theta)\\ &=& [ X_1(t) * H_1(t) ] \cos(2\pi f_0 t + \theta) + [ X_2(t) * H_1(t)] \sin(2\pi f_0 t + \theta) \end{array} $$ where $X_1$ and $X_2$ are the respective (unlabeled) inputs, $H_1$ and $H_2$ are the impulse responses of the filters, and $*$ is convolution.

You appear to be asking whether the system is LTI when $H_1 = H_2 = H$ and $X_1 = X_2 = X$. In that case, we get $$ \begin{array}{ccl} r(t) &=& [ X(t) * H(t) ] [\cos(2\pi f_0 t + \theta) + \sin(2\pi f_0 t + \theta)]\\ &=& \sqrt{2} [ X(t) * H(t) ] \cos(2\pi f_0 t + \theta - \pi/4) \end{array} $$

So it's linear: $$ \begin{array}{ccl} r_a(t) &=& \sqrt{2} [ a(t) * H(t) ] \cos(2\pi f_0 t + \theta - \pi/4)\\ r_b(t) &=& \sqrt{2} [ b(t) * H(t) ] \cos(2\pi f_0 t + \theta - \pi/4)\\ r_{a+b}(t) &=& \sqrt{2} [ (a(t) + b(t)) * H(t) ] \cos(2\pi f_0 t + \theta - \pi/4) = r_a(t) + r_b(t)\\ \end{array} $$ But it's not time-invariant: $$ \begin{array}{ccl} r(t) &=& \sqrt{2} [ X(t) * H(t) ] \cos(2\pi f_0 t + \theta - \pi/4)\\ r_\delta(t) &=& \sqrt{2} [ X(t-\delta) * H(t) ] \cos(2\pi f_0 t + \theta - \pi/4)\\ r(t-\delta) &=& \sqrt{2} [ X(t-\delta) * H(t) ] \cos(2\pi f_0 (t-\delta) + \theta - \pi/4)\\ \end{array} $$ as $r_\delta(t) \not = r(t-\delta)$.

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EDIT

So, what happens if $X_1 = X_2 = X$ and $X\ \tilde{\ }\ N(0,\sigma^2)$ and $\theta\ \tilde{\ }\ U[0,2\pi)$?

I can see why $Y_1(t) = [ X(t) * H(t) ] \cos(2\pi f_0 t + \theta)$ might be non-Gaussian.

However, I cannot see why $Y_1 + Y_2 = \sqrt{2} [ X(t) * H(t) ] \cos(2\pi f_0 t + \theta - \pi/4)$ will be Gaussian?

Can you please update your question to explain?

Answer 2

Just a different way of proofing time variance. Not nearly as neat and thorough as Dilip's but a lot quicker :-)

Assume the input is x(t) = 1 (a constant) and H1(f) = H2(f) = 1 (flat, basically a wire. Output is then r(t) = sin(om+phi)+cos(om+phi) which is sqrt(2)*sin(om +phi + pi/4). So we have DC in and a different frequency out. The output is a function of time when the input isn't, so it's necessarily time variant.

Answer 3

Solution:

1. The processes $ {Y}_{1} $ and $ {Y}_{2} $ are dependent.
   $ {Y}_{1} $ and $ {Y}_{2} $ are dependent intuitively by looking at the problem from a different angle.
   Given a uniformly distributed random variable $ \theta $ the random variables $ sin(\theta) $ and $ cos(\theta) $ are clearly dependent. Multiplying them by a Gaussian Random Variable doesn't decorrelate them.
2. The processes $ {Y}_{1} $ and $ {Y}_{2} $ aren't Gaussian (And not Ergodic).
   Clearly over time (The Phase is constant) the process $ {Y}_{i} $ is Gaussian. Yet the ensemble isn't Gaussian since each of its realization is retrieved from Gaussian Distribution with different parameters.
3. The process $ r(t) $ is indeed Gaussian.
   This could be proved by using the Characteristic Function of the random process.
   $ X \sim N(0, 1) $, $ Y \sim N(0, 1) $, $ \theta \sim U(0, 2\pi) $ all are independent.
   Let $ Z = X sin(\theta) + Y cos(\theta) $. Looking at its Characteristic Function and applying the Smoothing Theorem yields: \begin{align} { \varphi }_{Z}(t) & = & \mathbb{E}[{e}^{itZ}] \\ & = & \mathbb{E} [ \mathbb{E}[{e}^{itZ} | \theta] ] \\ & = & \mathbb{E} [ \mathbb{E}[{e}^{it(X cos(\theta) + Y sin(\theta))} | \theta] ] \end{align}
   Looking at the last equation per realization of $ \theta $:
   $$ \mathbb{E}[{e}^{it(X cos(\theta) + Y sin(\theta))} | \Theta = \theta] = \mathbb{E}[{e}^{itX sin(\theta)}] \mathbb{E}[{e}^{itY cos(\theta)}] $$ Each of the item is the Characteristic Function of a scaled Normally Distributed Random Process: $$ \mathbb{E}[{e}^{itX sin(\theta)}] \mathbb{E}[{e}^{itY cos(\theta)}] = {e}^{-\frac{{t}^{2}}{2} ({sin}^{2}(\theta) + {cos}^{2}(\theta))} = {e}^{-\frac{{t}^{2}}{2}} $$ Hence we get: $$ { \varphi }_{Z}(t) = \mathbb{E} [ \mathbb{E}[{e}^{itZ} | \Theta = \theta] ] = \mathbb{E} [{e}^{-\frac{{t}^{2}}{2}}] = {e}^{-\frac{{t}^{2}}{2}} $$ Namely, it has the Characteristic Function of a Normalized Gaussian Variable -> $ Z \sim N(0, 1) $.