1
$\begingroup$

I need help with this question. I am sure this is the right StackExchange forum for this type of question.

Consider a nonlinear device such that the output is $Y(t) = aX^2(t)$, where the input X(t) consists of a signal plus a noise component, $X(t) = S(t) + N(t)$.
Determine the mean and autocorrelation function for $Y(t)$ when the signal $S(t)$ and the noise $N(t)$ are both Gaussian random processes and wide sense stationary (WSS) with zero mean, and $S(t)$ is independent of $N(t)$.

I know that for a WSS signal, the mean and autocorrelation depends on the time difference, $\tau$ only. Meaning that the mean of $S(t)$ and $N(t)$ is a constant and their individual autocorrelation is not dependent on $t$.
I've been working only with linear systems. I don't know how to solve this problem since $Y(t)$ a nonlinear system. I will be happy if you can help me with the solution

UPDATE 1

I followed the suggestion from @Hilmar and came up with the following:
1. The mean of $Y(t) = 0$ since $S(t)$ and $N(t)$ have 0 mean.
2. For the autocorrelation, I used this formula $R_{yy}(\tau)=R_{xx}(\tau)*h(\tau)*h(-\tau)$.
looking for $R_{xx}$, I used $R_{xx}=E[X(t)X(t+\tau)]$.
I ended up having $R_{xx} = E[S(t)S(t+\tau) +S(t)N(t+\tau)+N(t)S(t+\tau) + N(t)N(t+\tau)]$.
Since $S(t)$ and $N(t)$ are independent, $R_{SN} = R_{NS} = 0$
$=> R_{XX}= E[S(t)S(t+\tau) + N(t)N(t+\tau)]$
$=> R_{XX} = R_{SS} + R_{NN}$

$R_{YY} = (R_{SS} + R_{NN})*h(\tau)*h(-\tau)$
Now I am stuck. From the definition of WSS signal, $R_{SS}$ and $R_{NN}$ are constants. i.e. they are dependent only on $\tau$. I can assume their sum to a some constant value, $C$
I don't know how to move from here. I will need to find $h(\tau)$ I don't know how to go about it.
$\endgroup$
2
$\begingroup$

The OP's updated working is incorrect. Following up what Hilmar suggested gives \begin{align} Y(t) &= a\left(X(t)\right)^2\\ &= a\left(S(t) + N(t)\right)^2\\ &= a\left(S(t)\right)^2 + 2aS(t)N(t) + a\left(N(t)\right)^2\\ &{\large\Downarrow}\\ E[Y(t)]&= aE\left[\left(S(t)\right)^2 \right] + 2aE\left[S(t)N(t)\right] + aE\left[\left(S(t)\right)^2 \right]. \tag{1} \end{align} The OP claims that $E[Y(t)] = 0$ on the grounds that $S(t)$ and $N(t)$ have mean $0$. However, while $S(t)$ and $N(t)$ are indeed zero-mean random variables, and independence of $S(t)$ and $N(t)$ gives $$E[\left[S(t)N(t)\right]= E\left[S(t)\right]E\left[N(t)\right] = 0,$$ it is not true that the other terms in $(1)$ have value $0$ also. In fact, since $S(t)$ is a zero-mean random variable, $E\left[\left(S(t)\right)^2 \right]$ equals the variance of $S(t)$ which is $R_{SS}(0)$ in this case. Similarly, $E\left[\left(N(t)\right)^2 \right]$ equals the variance of $N(t)$ which is $R_{NN}(0)$; in short, $$E[Y(t)] = R_{SS}(0) + R_{NN}(0) > 0.\tag{2}$$ Well, at least the mean is constant (does not depend on the value of $t$) but it is nonzero, contrary to the OP's calculations.

For the autocorrelation calculation, the OP insists on fitting the problem posed into the Procrustean bed of linear system theory and drags in impulse responses which are inappropriate in this context. Just using the definitions gives \begin{align} E\left[Y(t)Y(t+\tau)\right] &= a^2E\left[\big(X(t)\big)^2 \big(X(t+\tau)\big)^2\right]\\ &= a^2E\left[\big(S(t)+N(t)\big)^2 \big(S(t+\tau)+N(t+\tau)\big)^2\right]\\ &= a^2E\left[\big(S(t)S(t+\tau)+ N(t)N(t+\tau) + S(t)N(t+\tau)+S(t+\tau)N(t)\big)^2\right]\\ &= a^2E\big[16~\text{terms involving products of} ~S(t),S(t+\tau), N(t), N(t+\tau)\\ & \qquad\quad\text{and their squares.}\quad \big]\tag{3} \end{align} Since $S(t),S(t+\tau), N(t), N(t+\tau)$ are jointly Gaussian random variables, the expectations of the 16 terms in $(3)$ can be computed. Some of the expectations are $0$, for example, one of the 16 terms is $S(t)\left(S(t+\tau)\right)^2N(t)$ and since the processes $\{S(t)\}$ and $\{N(t)\}$ are independent, we have that $$E\big[S(t)\left(S(t+\tau)\right)^2N(t)\big] = E\big[S(t)\left(S(t+\tau)\right)^2\big]E\big[N(t)\big] = 0.$$ Similarly, $$E\big[\left(S(t)\right)^2S(t+\tau)N(t+\tau)\big] = E\big[\left(S(t)\right)^2S(t+\tau)\big]E\big[N(t+\tau)\big] = 0.$$ However, other terms are not $0$. For example, $$E\big[\left(S(t)S(t+\tau)\right)^2\big] = E\big[\left(S(t)\right)^2\left(S(t+\tau)\right)^2\big]$$ is the expected value of the product of the squares of two jointly Gaussian and correlated random variables, and so is guaranteed to be positive, and similarly for $E\big[\left(N(t)N(t+\tau)\right)^2\big]$. I will leave it to the OP tp work out the details and determine whether $\Y(t)\}$ is also a WSS process.

$\endgroup$
3
  • $\begingroup$ Thanks for the correction. I did not come to my mind that I could use this other method. I will redo again following your solution. @DilipSarwate $\endgroup$ Feb 10 at 5:49
  • $\begingroup$ Is $E[(S(t)N(t+\tau))^2] == 0$? since $S(t)$ and $N(t)$ are independent. $\endgroup$ Feb 10 at 6:15
  • $\begingroup$ @KofiMokome Absolutely NOT! It doesn't matter in the least whether $S(t)$ and $N(t+\tau)$ are independent or not. The random variable $\big(S(t)N(t+\tau)\big)^2$ cannot have expected value $0$ except in the trivial case when at least one of $S(t)$ and $N(t+\tau)$ is itself a degenerate random variable that takes on value $0$ with probability $1$. Statistically illiterate people call such random variables (that take on a fixed value $c$ with probability $1$) as constants and they denote such a constant by $c$. Once again, NO, your thought is completely incorrect. $\endgroup$ Feb 10 at 23:58
2
$\begingroup$
  1. Put your second equations into your first equation, express $Y(t)$ as a function of $S(t)$ and $N(t)$
  2. Apply the definition for mean and autocorrelation.
  3. Simplify and solve
$\endgroup$
13
  • $\begingroup$ is $X^2(t)$ same as $S^2(t) + N^2(t)$? $\endgroup$ Feb 8 at 16:46
  • 1
    $\begingroup$ @KofiMokome um, high school math says: No. $\endgroup$ Feb 8 at 18:59
  • $\begingroup$ so, how will I calculate $X^2(t)$? I am new to digital signal processing $\endgroup$ Feb 8 at 19:46
  • 1
    $\begingroup$ i will use $X^2(t) == [X(t)]^2$ to solve it, and update my question with the solution I will have $\endgroup$ Feb 9 at 9:08
  • 1
    $\begingroup$ @MarcusMüller $(x+y)^2 = x^2+y^2$ is called the freshman's dream by teachers of freshman calculus (in the US). Perhaps things are different in Europe.... $\endgroup$ Feb 10 at 3:33

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.