Exercise 4.14Z: Echo Detection

Echo measuring arrangement

To measure acoustic echoes in rooms – for example caused by reflections at a wall – the adjacent setup can be used:

The noise generator produces a "white noise in the relevant frequency range" $x(t)$ with power density $N_0 = 10^{-6} \hspace{0.08cm} \rm W/Hz$.
This is bandlimited to $B_x = 20 \hspace{0.08cm} \rm kHz$ and is given to a loudspeaker.
The entire measurement setup is designed for the resistance value $R = 50 \hspace{0.08cm} \rm \Omega$.

In the most general case, the signal recorded by the microphone can be described as follows:

$$y(t) = \sum_{\mu = 1}^M \alpha_\mu \cdot x ( t - t_\mu ) .$$

Here the $\alpha_\mu$ denote damping factors and $t_\mu$ delay times.

Hints:

The exercise belongs to the chapter Cross-Correlation Function and Cross Power-Spectral Density.

Use for numerical calculations the parameter values.

$$\alpha_1 = 0.5, \hspace{0.2cm}t_1 = 200 \,{\rm ms}, \hspace{0.2cm} \alpha_2 = 0.1, \hspace{0.2cm}t_2 = 250 \,{\rm ms}.$$

Questions

$\sigma_x \ = \ $

$\ \rm V$

$\varphi_{xy}(\tau= 0) \ = \ $

$\ \rm V^2$

$\varphi_{xy}(\tau= t_1) \ = \ $

$\ \rm V^2$

$\varphi_{xy}(\tau= t_2) \ = \ $

$\ \rm V^2$

${\it \Phi}_{xy}(f =0)\ = \ $

$\ \cdot 10^{-6}\ \rm V^2\hspace{-0.1cm}/Hz$

	The noise is now "true white" – so it is not bandlimited.
	The noise power is reduced compared to subtask (1).
	The cross-correlation function is the sum of weighted and shifted Dirac delta functions.
	The cross power-spectral density is calculated as in subtask (3).

$\varphi_{y}(\tau= 0) \ = \ $

$\ \cdot 10^{-6}\ \rm W\hspace{-0.1cm}/Hz$

$\varphi_{y}(\tau= \Delta t) \ = \ $

$\ \cdot 10^{-6}\ \rm W\hspace{-0.1cm}/Hz$

Solution

(1) The two-sided power spectral density spectrum ${\it \Phi}_{x}(f)$ is constantly equal $N_0/2$ in the range $\pm B_x$ .

Whose Fourier transform is the ACF:

$$\varphi_x (\tau) = {N_0}/{2} \cdot 2 B_x \cdot {\rm si} (2 \pi B_x \tau) = 0.02 \hspace {0.08cm}{\rm W} \cdot {\rm si} (2 \pi B_x \tau).$$

Converted from $R = 50 \hspace{0.08cm} \rm \Omega$ to $R = 1 \hspace{0.08cm} \rm \Omega$ thus one obtains $($multiplication by $R = 50 \hspace{0.08cm} \rm \Omega)$:

$$\varphi_x (\tau) = 0.02 \hspace {0.05cm}{\rm VA} \cdot 50 \hspace {0.05cm}{\rm V/A}\cdot {\rm si} (2 \pi B_x \tau)= 1 \hspace {0.05cm}{\rm V}^2 \cdot {\rm si} (2 \pi B_x \tau).$$

The rms value is the square root of the ACF value at $\tau = 0$:

$$\sigma_x \hspace{0.15cm}\underline{= 1 \hspace {0.08cm}{\rm V}}.$$

(2) For the cross correlation function (CCF), in the present case:

$$\varphi_{xy} (\tau) = \overline {x(t) \hspace{0.05cm}\cdot \hspace{0.05cm}y(t+\tau)} = \overline {x(t) \hspace{0.05cm}\cdot \hspace{0.05cm}\big [ \alpha_1 \cdot x(t- t_1+ \tau)\hspace{0.1cm}+\hspace{0.1cm} \alpha_2 \cdot x(t- t_2+ \tau)\big] } . $$

After splitting the averaging on the two terms, we get:

$$\varphi_{xy} (\tau) = \alpha_1 \cdot \overline {x(t) \hspace{0.05cm}\cdot \hspace{0.05cm} x(t- t_1+ \tau)} \hspace{0.1cm}+\hspace{0.1cm} \alpha_2 \cdot \overline {x(t) \hspace{0.05cm}\cdot \hspace{0.05cm} x(t- t_2+ \tau)} .$$

Using the ACF $\varphi_x(\tau)$ can also be written herefor:

$$\varphi_{xy} (\tau) = \alpha_1 \cdot {\varphi_{x}(\tau- t_1)} \hspace{0.1cm}+\hspace{0.1cm} \alpha_2\cdot {\varphi_{x}(\tau- t_2)} = 1 \hspace {0.08cm}{\rm V}^2 \cdot \big[ \alpha_1 \cdot {\rm si} (2 \pi B_x (\tau - t_1)) + \alpha_2 \cdot {\rm si} (2 \pi B_x (\tau - t_2)) \big].$$

The si function exhibits equidistant zero crossings at multiples of $1/(2B_x) = 25 \hspace{0.08cm} µ \rm s$ respectively, related to the centers at $t_1 = 200 \hspace{0.08cm} {\rm ms}$ and $t_2 = 250 \hspace{0.08cm} {\rm ms}$.
This results in the CCF values to:

$$\varphi_{xy} (\tau = 0) \hspace{0.15cm}\underline{= 0},\hspace{0.5cm}\varphi_{xy} (\tau = t_1)= \alpha_1 \cdot \varphi_{x} (\tau = 0) \hspace{0.15cm}\underline{= 0.5\,{\rm V}^2} ,\hspace{0.5cm} \varphi_{xy} (\tau = t_2)= \alpha_2 \cdot \varphi_{x} (\tau = 0) \hspace{0.15cm}\underline{= 0.1\,{\rm V}^2} .$$

(3) The cross power spectral density is the Fourier transform of the CCF, just as the power spectral density spectrum (PSD) gives the Fourier transform of the ACF. For this holds:

$${\it \Phi}_{xy} (f) = \alpha_1 \cdot {\it \Phi}_{x} (f) \cdot {\rm e}^{-{\rm j}2 \pi f t_1} \hspace{0.15cm}+ \hspace{0.15cm}\alpha_2 \cdot {\it \Phi}_{x} (f) \cdot {\rm e}^{-{\rm j}2 \pi f t_2}. $$

Outside the range $|f| \le B_x$ the PSD ${\it \Phi}_{x}(f)$ – and correspondingly the KPSD ${\it \Phi}_{xy}(f)$ – is identically zero.
In contrast, inside this interval holds ${\it \Phi}_{x}(f) = N_0/2$. It follows in this range:

ACF and CCF in white noise

$${\it \Phi}_{xy} (f) = {N_0}/{2} \left( \alpha_1 \cdot {\rm e}^{-{\rm j}2 \pi f t_1} \hspace{0.15cm}+ \hspace{0.15cm}\alpha_2 \cdot {\rm e}^{-{\rm j}2 \pi f t_2} \right). $$

It is evident that ${\it \Phi}_{xy}(f)$ unlike ${\it \Phi}_{x}(f)$ is a complex function. For $f = 0$ holds:

$${\it \Phi}_{xy} (f = 0) = {N_0}/{2} \left( \alpha_1 \hspace{0.15cm}+ \hspace{0.15cm}\alpha_2 \right) = 0.3 \cdot 10^{-6}\hspace{0.05cm}{\rm W/Hz} \hspace{0.15cm}\underline{= 15 \cdot 10^{-6}\hspace{0.07cm}{\rm V^2/Hz}} . $$

(4) Correct are the proposed solutions 1 and 3:

The Fourier transform of a dirac-shaped ACF leads to a constant PSD for all frequencies $f$ that is, in fact, to "true white noise".
This has an infinitely large power, and for the CCF can then be written according to the above graph:

$$\varphi_{xy} (\tau) = \alpha_1 \cdot { N_0}/{2} \cdot {\rm \delta}( \tau - t_1) \hspace {0.1cm}+ \hspace {0.1cm} \alpha_2 \cdot { N_0}/{2} \cdot { \rm \delta}( \tau - t_2) .$$

In the frequency domain, indeed, no difference is detectable for $|f| \le B_x$ compared to the subtask (3) .
But since true white noise is now present, but here the cross power spectral density is not limited to this range.

(5) For the ACF of the echoed signal, $\varphi_{y} (\tau) = \overline {y(t) \hspace{0.05cm}\cdot \hspace{0.05cm}y(t+\tau)}$.

This ACF $\varphi_{y} (\tau)$ can consequently be represented as the following sum:

$$\alpha_1^2 \cdot \overline {x(t - t_1) \cdot x(t - t_1+ \tau)} \hspace{0.03cm} + \hspace{0.03cm} \alpha_1\hspace{0.02cm}\alpha_2 \cdot \overline {x(t - t_1) \cdot x(t - t_2+ \tau)} + \hspace{0.05cm} \alpha_2\hspace{0.02cm}\alpha_1 \cdot \overline {x(t - t_2) \cdot x(t - t_1+ \tau)}\hspace{0.03cm} + \hspace{0.03cm} \alpha_2^2 \cdot \overline {x(t - t_2) \cdot x(t - t_2+ \tau)}. $$

For the first and the last mean:

$$\overline {x(t - t_1) \hspace{0.05cm}\cdot \hspace{0.05cm}x(t - t_1+ \tau)} = \overline {x(t - t_2) \hspace{0.05cm}\cdot \hspace{0.05cm}x(t - t_2+ \tau)} = \overline {x(t ) \hspace{0.05cm}\cdot \hspace{0.05cm}x(t + \tau)} =\varphi_x(\tau).$$

In contrast, for the second and third mean values, we obtain with $\delta t = t_2 - t_1= 50 \ \rm ms$:

$$\overline {x(t - t_1) \hspace{0.05cm}\cdot \hspace{0.05cm}x(t - t_2+ \tau)} = \overline {x(t ) \hspace{0.05cm}\cdot \hspace{0.05cm}x(t + t_1- t_2+ \tau)} =\varphi_x(\tau - \Delta t),$$

$$\overline {x(t - t_2) \hspace{0.05cm}\cdot \hspace{0.05cm}x(t - t_1+ \tau)} = \overline {x(t ) \hspace{0.05cm}\cdot \hspace{0.05cm}x(t + t_2- t_1+ \tau)} =\varphi_x(\tau + \Delta t).$$

Total, this again results in a symmetric ACF, as shown in the graph below:

$$\varphi_{y} (\tau) = {N_0}/{2} \cdot \left[ ( \alpha_1^2 \hspace{0.1cm} + \hspace{0.1cm} \alpha_2^2 ) \cdot {\rm \delta} (\tau) \hspace{0.1cm} + \hspace{0.1cm} \alpha_1 \cdot \alpha_2 \cdot {\rm \delta}(\tau - \delta t) \hspace{0.1cm} + \hspace{0.1cm} \alpha_1 \cdot \alpha_2 \cdot {\rm \delta}(\tau + \Delta t) \right].$$

$$\Rightarrow \hspace{0.3cm}\varphi_{y} (\tau = 0 ) \hspace{0.15cm}\underline{= 0.13 \cdot 10^{-6}\, {\rm W/Hz}}, \hspace{0.3cm}\varphi_{y} (\tau = \Delta t )\hspace{0.15cm}\underline{ = 0.025 \cdot 10^{-6}\, {\rm W/Hz}}.$$