Show that the impulse response of an LTI system can be determined by observing the output and inputting white noise

Question

Problem

Show that the impulse response of an LTI system can be determined by observing the output and inputting white noise.

Attempt

Let $x(t)$ be the input signal (white noise) and $y(t)$ be the output signal. If the system is linear we have $$y(t) = x(t) \ast h(t) $$ Performing cross-correlation on the input signal and output signal we have

$$\begin{align*} R_{xy}(\tau) &= \mathbb{E}\bigg[ x(t)y(t+\tau) \bigg] \\\\ &= \mathbb{E}\bigg[ x(t) \Big(x(t+\tau) \ast h(t+\tau)\Big) \bigg] \\\\ &= \mathbb{E}\bigg[ x(t)x(t+\tau) \ast x(t)h(t+\tau) \bigg] \\\\ &= \mathbb{E}\bigg[ x(t)x(t+\tau) \bigg] \ast \mathbb{E}\bigg[ x(t)h(t+\tau) \bigg]\end{align*}$$

Because we are dealing with white noise, the input autocorrelation function is an impulse function.

$$\begin{align*}R_{xy}(\tau) &= \delta(t) \ast \mathbb{E}\bigg[ x(t)h(t+\tau) \bigg] \\\\ &= \mathbb{E}\bigg[ x(t)h(t+\tau) \bigg] \end{align*} $$

And here I don't see what else can be done to arrive at an expression for the impulse response. Is it possible to proceed from here?

Edit

As Dilip writes in his comment I've made an error in line 3. Correting it yields

$$\begin{align*} R_{xy}(\tau) &= \mathbb{E}\bigg[ x(t) \Big(x(t+\tau) \ast h(t+\tau) \Big) \bigg] \\\\ &= \mathbb{E}\bigg[ x(t) \int_{-\infty}^{\infty} x(t+\tau)h(t-\tau) \: \text{d}\tau\bigg] \\\\ &= \mathbb{E}\bigg[ \int_{-\infty}^{\infty} x(t)x(t+\tau)h(t-\tau) \: \text{d}\tau\bigg] \end{align*} $$

But I still don't know where to go from here.

Edit 2

Call $\beta = t+\tau$ and let the variable of integration in the convolution integral be $\alpha$, then

$$\begin{align*}R_{xy}(\tau) &= \mathbb{E}\bigg[ x(t) \Big(x(t+\tau) \ast h(t+\tau) \Big) \bigg] \\\\ &= \mathbb{E}\bigg[ x(t) \int_{-\infty}^{\infty} x(\alpha)h(\beta-\alpha) \: \text{d}\alpha\bigg] \\\\ &= \mathbb{E}\bigg[ \int_{-\infty}^{\infty} x(t)x(\alpha)h(\beta-\alpha) \: \text{d}\alpha\bigg] \end{align*}$$

And now I'm stuck again.

Answer 1

first let's find the cross-power spectral density between the input and output $$y(t)=x(t)*h(t) = \int{x(k)h(t-k)dk}$$

$$R_{xy}(\tau) = E[x(t)y(t+\tau)] = E[x(t)x(t+\tau)*h(t+\tau)]$$ from the commutativity property of convolution $x(t+\tau)*h(t+\tau) = h(t+\tau)*x(t+\tau)$ so we have: $$ = E[x(t)\int{h(k)x(t+\tau-k) \, dk}]$$

Expected value is linear so can go into the integral. We can insert $x(t)$ into the integral as well. h is deterministic so can be outside expected value. We rearrange the elements and get:

$$=\int{h(k)E[x(t)x(t+\tau-k) \, dk}] = \int{h(k)R_{xx}(\tau-k), dk} = h(\tau)*R_{xx}(\tau)$$

we can interprate this in the time domain, if x is white noise then $R_{xx}$ is $\sigma_x\delta(\tau)$, we plug that in and get that:

$$R_{xy}(\tau) = \sigma_xh(\tau)$$

we can also continue in the frequency domain. If we do the Fourier Transform we get that $P_{xy}(j\omega) = H(j\omega)P_{xx}(j\omega)$ (we could have started from here but I can't comment yet so I did the entire derivation. You got something mixed up in the convolution integral with the arguments).

So if our input is white noise then $P_{xx}(j\omega) = \sigma_x$ and then if we observe the output spectrum we get the system spectrum scaled by the variance of the white noise, $H(j\omega)=\frac{P_{xy}(j\omega)} {\sigma_x}$.