auto-correlation of differentiator

Question

How to do this entirely in the time-domain, without using frequency domain?

Let's say I have a continuous-time system that is a differentiator, and X(t) is WSS:

$$x(t) \rightarrow \boxed{\frac{d}{dt}} \rightarrow y(t)$$

I can change the differentiator operator into convolution as follows:

$$X(t) \rightarrow \boxed{\frac{d}{dt} \delta(t)} \rightarrow Y(t)$$

$$h(t) = \frac{d}{dt}\delta(t)$$

$$Y(t) = h(t) * X(t)=\frac{d}{dt}x(t)$$

$$\begin{matrix}R_{XX}(\tau)\end{matrix}\longrightarrow \boxed{\begin{matrix}h(\tau)\end{matrix}}\longrightarrow \begin{matrix}R_{XY}(\tau)\end{matrix}\longrightarrow \boxed{\begin{matrix}h(-\tau)\end{matrix}}\longrightarrow \begin{matrix}R_{YY}(\tau)\end{matrix}$$

---

cross-correlation of X(t) and Y(t):

$$R_{XY}(\tau) = h(\tau)*R_{XX}(\tau)$$

$$R_{XY}(\tau) = \Big(\frac{d}{d\tau}\delta(\tau)\Big) * R_{XX}(\tau)$$

$$R_{XY}(\tau) = \frac{d}{d\tau} R_{XX}(\tau)$$

---

autocorrelation of Y(t):

$$R_{YY}(\tau) = h(-\tau)*R_{XY}(\tau)$$

$$R_{YY}(\tau) = h(-\tau)* \frac{d}{d\tau} R_{XX}(\tau)$$

$$h(\tau) = \frac{d}{dt}\delta(\tau)$$

$$h(-\tau) = \frac{d}{dt}\delta(-\tau)$$

$$\delta(\tau) = \delta(-\tau)$$

$$h(-\tau) = \frac{d}{dt}\delta(\tau)$$

$$R_{YY}(\tau) = \Big(\frac{d}{d\tau}\delta(\tau)\Big) * \Big(\frac{d}{d\tau} R_{XX}(\tau)\Big)$$

$$R_{YY}(\tau) = \frac{d^2}{d\tau^2} R_{XX}(\tau)$$

However, the book says the answer should be:

$$R_{YY}(\tau) = -\frac{d^2}{d\tau^2} R_{XX}(\tau)$$

Somewhere I lost a sign...

and the reason why this sign is important because if I take the Fourier of $R_{YY}(\tau)$, it needs this sign to make $S_{YY}(\omega)$ positive as is required for PSD.

Answer 1

The derivative of the Dirac delta impulse is written as $\delta'(\tau)$. This helps with notation because the mistake you made is to write $h(-\tau)=\frac{d}{d\tau}\delta(-\tau)$, which is not the case because $\delta(\tau)$ is an even (generalized) function, whereas the derivative operator $\delta'(\tau)$ is an odd (generalized) function:

$$\delta'(\tau)=-\delta'(-\tau)\tag{1}$$

Consequently, the auto-correlation function of the output of the differentiator is

$$\begin{align}R_{YY}(\tau)&=R_{XX}(\tau)\star\delta'(\tau)\star\delta'(-\tau)\\&=-R_{XX}(\tau)\star\delta'(\tau)\star\delta'(\tau)\\&=-\frac{d^2}{d\tau^2}R_{XX}(\tau)\tag{2}\end{align}$$