Exercise 4.13: Gaussian ACF and PSD

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Two Gaussian ACFs

Let the random process considered here  $\{x_i(t)\}$  be characterized by the auto-correlation function  $\rm (ACF)$  outlined above  This random process is mean-free and the equivalent ACF duration is  ${ {\rm \nabla} }\tau_x = 5 \hspace{0.08cm} \rm µ s$:

$$\varphi_x(\tau)=\rm 0.25 V^2\cdot \rm e^{-\pi \hspace{0.03cm}\cdot \hspace{0.03cm} ({\tau}{/ 5 \hspace{0.08cm}{\rm µ}s })^2} .$$

The bottom graph shows the ACF of the process  $\{y_i(t)\}$.  This reads with the equivalent ACF duration  ${ {\rm \nabla} }\tau_y = 10 \hspace{0.08cm} \rm µ s$:

$$ \varphi_y(\tau)=\rm 0.16 V^2 + \rm 0.09 V^2\cdot\rm e^{-\pi \hspace{0.03cm}\cdot \hspace{0.03cm} ({\tau}/{\nabla \it \tau_y})^2} .$$

In this exercise,  the power-spectral densities of the two processes are sought.

Hints:

• This exercise belongs to the chapter  Power-spectral density  $\rm (PSD)$.
• Reference is also made to the chapter  Auto-correlation function  $\rm (ACF)$.
• To solve this exercise you can use the following Fourier correspondence:

$$\rm e^{-\pi \hspace{0.03cm}\cdot \hspace{0.03cm} ({\it f}/{\rm \Delta\it f})^2}\ \bullet\!\!-\!\!\!-\!\!\!-\!\!\circ\,\ {\rm \Delta \it f} \cdot \rm e^{-\pi \hspace{0.05cm}\cdot \hspace{0.05cm} ({\rm \Delta\it f} \hspace{0.05cm}\cdot \hspace{0.05cm}\it t {\rm )}^{\rm 2}}.$$

Questions

1

What is the equivalent PSD bandwidth of the process  $\{x_i(t)\}$?

$ {\rm \nabla} \hspace{-0.05cm} f_x \ = \ $ $\ \rm kHz$

2

What is the power-spectral density  ${\it \Phi}_x(f)$?  Give the PSD values for  $f= 0$  and  $f = 200 \hspace{0.08cm} \rm kHz$.

${\it \Phi}_x(f = 0)\ = \ $ $\ \cdot 10^{-6} \ \rm V^2\hspace{-0.1cm}/Hz$

${\it \Phi}_x(f = 200 \hspace{0.08cm} \rm kHz)\ = \ $ $\ \cdot 10^{-6} \ \rm V^2\hspace{-0.1cm}/Hz$

3

Which statements are valid,  if the random process has no periodic parts?  Furthermore,  a constant power is assumed.

	The process power is the integral over the PSD.
	If the process is zero mean,  the PSD is always continuous.
	The wider the ACF,  the wider the PSD.
	A wider ACF results in higher PSD values.

4

Calculate the power-spectral density  ${\it \Phi}_y(f)$.  What are the values for the continuous PSD component at  $f= 0$  and  $f = 200 \hspace{0.08cm} \rm kHz$?

${\it \Phi}_y(f = 0)\ = \ $ $\ \cdot 10^{-6} \ \rm V^2\hspace{-0.1cm}/Hz$

${\it \Phi}_y(f = 200 \hspace{0.08cm} \rm kHz)\ = \ $ $\ \cdot 10^{-6} \hspace{0.05cm} \rm V^2\hspace{-0.1cm}/Hz$

5

Which of the following statements are true regarding the process  $\{y_i(t)\}$?

	The PSD involves a Dirac delta function at frequency $ f = {\rm \nabla} \hspace{-0.05cm} f_y$.
	The PSD involves a Dirac delta function at frequency $f= 0$.
	The Dirac weight and the continuous PSD have the same unit.

Solution

Solution

(1)  The equivalent PSD bandwidth is the reciprocal of the equivalent ACF duration:

$$\nabla f_x = 1 / \nabla \tau_x \hspace{0.15cm}\underline{= {\rm 200\hspace{0.1cm}kHz}}.$$

(2)  One can adapt the given Fourier correspondence to the task as follows:

$$K\cdot{\rm e}^{-\pi({\tau}/{\nabla\tau_x})^2}\ \circ\!\!-\!\!\!-\!\!\!-\!\!\bullet\,\ \frac{\it K}{\nabla \it f_x}\cdot{\rm e}^{-\pi({f}/{\nabla f_x})^2}.$$

• With  $K = 0.25 \hspace{0.05cm}\rm V^2$  and  $ {\rm \nabla} \hspace{-0.05cm} f_x = 200\hspace{0.05cm} \rm kHz$  one obtains:

$${\it \Phi_x}(f)=1.25\cdot\rm 10^{-\rm 6}\hspace{0.1cm}\frac{V^2}{Hz}\cdot\rm e^{-\pi({\it f}/{\nabla\it f_x})^2}$$

$$\Rightarrow \hspace{0.3cm}{\it \Phi_x}(f = 0)=\hspace{0.15cm}\underline{\rm 1.25 \cdot 10^{-6} \hspace{0.1cm} V^2\hspace{-0.1cm}/Hz}, \hspace{0.5cm}{\it \Phi_x}(f = 200 \hspace{0.05cm} \rm kHz)=\hspace{0.15cm}\underline{\rm 0.054 \cdot 10^{-6} \hspace{0.1cm} V^2\hspace{-0.1cm}/Hz}.$$

(3)  Correct are the  solutions 1, 2, and 4:

• A mean-free process always results in a continuous PSD.  This is narrower the wider the ACF is  ("Reciprocity Law").
• The process power is equal to the integral of the PSD.
• Therefore,  at constant power,  a wider ACF  (narrower PSD)  must be compensated by higher PSD values.
• A DC component or periodic components always result in Dirac delta functions in the PSD;  otherwise, the PSD is always continuous in value.

(4)  Analogous to subtask  (2)  holds with  $ {\rm \nabla} \hspace{-0.05cm} f_y = 100\hspace{0.05cm} \rm kHz$:

$${\it \Phi_y}(f)=\frac{\rm 0.09 V^2}{\nabla\it f_y}\cdot\rm e^{-\pi({\it f}/{\nabla\it f_y})^2}+\it m_y^{\rm 2}\cdot\delta(f).$$

• Because of the DC component,  there is a Dirac delta function at frequency  $f = 0$  in addition to the continuous PSD component.
• The continuous PSD–part at $f= 0$ is  ${\it \Phi_y}(f = 0)=\hspace{0.15cm}\underline{\rm 0.9 \cdot 10^{-6} \hspace{0.1cm} V^2\hspace{-0.1cm}/Hz}.$
• The continuous PSD–part at  $f = 2 \cdot {\rm \nabla} \hspace{-0.05cm} f_y = 200 \hspace{0.05cm}\rm kHz$  is lower by a factor  ${\rm e}^{-4} \approx 0.0183$   ⇒   ${\it \Phi_y}(f )=\hspace{0.15cm}\underline{\rm 0.0165 \cdot 10^{-6} \hspace{0.1cm} V^2\hspace{-0.1cm}/Hz}.$

(5)  Correct is  only the second proposed solution:

• The PSD of a mean-valued process generally involves a Dirac delta function at $f=0$  with weight  $m_y^2$.
• In the present case,  this value is equal to  $0.16 \ \rm V^2$.
• Since  $\delta(f)$  has unit  $\rm 1/Hz = s$,  the units of the continuous and discrete PSD components differ.