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I have the following system: enter image description here
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.

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  • $\begingroup$ @DilipSarwate, How come it is not Time Invariant? $\endgroup$ – Royi Sep 19 '13 at 19:50
  • $\begingroup$ OK, It is not Time Invariant since the input frequency isn't identical to the output frequency. $\endgroup$ – Royi Sep 19 '13 at 19:57
  • $\begingroup$ @DilipSarwate, Have a look at my edit, I came to the question by not knowing what happens if the input is White Gaussian Noise with the same Variance to each of the inputs. $\endgroup$ – Royi Sep 19 '13 at 20:03
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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 $r_\delta(t) \not = r(t-\delta)$.


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?

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    $\begingroup$ Unless it's $\theta(t)$, that is $\theta$ changes with each time instant, the analysis doesn't change. For any particular $\theta$, the analysis still holds: the system is linear, but not time-invariant (except for degenerate cases like $f_0 = 0$). $\endgroup$ – Peter K. Sep 21 '13 at 16:38
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    $\begingroup$ Something is wrong. Per $ \theta $, It is linear. Yet the ensemble doesn't act as linear. Since Y1 and Y2 aren't Gaussian. $\endgroup$ – Royi Sep 21 '13 at 17:02
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    $\begingroup$ Apart from Something is wrong. None of the rest of your comment makes sense to me... Per theta, It is linear/ : The system is linear, regardless of whether $\theta$ is random or not (unless, as I said, $\theta$ changes with time). Yet the ensemble doesn't act as linear.: See my answer. It is a proof of linearity. Why do you think it is not linear? Since Y1 and Y2 aren't Gaussian. : I have assumed nothing about the form of $Y_1$ and $Y_2$. They can be Gaussian or not. Why are you concerned about their Gaussianity? If the input is Gaussian, they are (see Dilip's answer). $\endgroup$ – Peter K. Sep 21 '13 at 22:03
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    $\begingroup$ Because I can prove they aren't. And it is easy to prove it. Both by math and by simulation. Have a look here, Assuming the input to the upper branch is X~N(0, 1) and theta~U[0, 2*pi]. Hence Y won't comply a Gaussian Property: imgur.com/l017RrR Namely, it doesn't act as Linear System. $\endgroup$ – Royi Sep 22 '13 at 1:10
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    $\begingroup$ @PeterK. I think I have a solution. I will post it later. Maybe others will have a better one. $\endgroup$ – Royi Sep 23 '13 at 19:09
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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.

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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) $.
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  • $\begingroup$ -1 While obviously eminently acceptable to the OP, this is not an answer to the question posed: is this an LTI system? On the other hand, Moderator @PeterK.'s answer (and Hilmar's too for that matter) are dead on target. $\endgroup$ – Dilip Sarwate Sep 29 '13 at 15:17
  • $\begingroup$ @DilipSarwate, Why would -1 it? It does answer it since if it was an LTI system the output would be Gaussian. That's the context it was asked. By the answer it could be easily understood this is not an LTI system or even Linear system (The "Conditional System is Linear though it is Time Variant). This is not a nice play. $\endgroup$ – Royi Sep 29 '13 at 21:52
  • $\begingroup$ I repeat: your answer is not an answer. The output of a linear system (note the missing adjective time-invariant ) with a Gaussian noise process as input is a Gaussian noise process. PeterK and Hilmar have already proved to you that the system is linear but not time-invariant which answers your question "is it LTI? Can you prove it?" In contrast, your own answer nowhere mentions LTI property, and leaves it to the reader to infer that since you have proved the Gaussian property, the system must be linear, and you do not address the TI part at all. You may be perfectly... $\endgroup$ – Dilip Sarwate Sep 29 '13 at 22:35
  • $\begingroup$ satisfied and happy with your answer, but I continue to contend that it is not an answer to the question that you have asked. I don't downvote very frequently, but in this case, I see no reason to change my opinion or my vote. If you edit your answer suitably, I will be glad to reconsider the matter. $\endgroup$ – Dilip Sarwate Sep 29 '13 at 22:37
  • $\begingroup$ Again, you disregard the following: "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) Y1 and Y2 aren't Gaussian Process yet r(t) is a Gaussian Process.". This is what behind the question and the answer is targeted there. This isn't an LTI system and though Y1, Y2 aren't Gaussian the sum is and each of the branches of the system isn't Linear and not Time Invariant. $\endgroup$ – Royi Sep 30 '13 at 7:01

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