Difference between revisions of "Aufgaben:Exercise 4.12Z: White Gaussian Noise"

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{{quiz-Header|Buchseite=Stochastische Signaltheorie/Leistungsdichtespektrum (LDS)
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{{quiz-Header|Buchseite=Theory_of_Stochastic_Signals/Power-Spectral_Density
 
}}
 
}}
[[File:P_ID409__Sto_Z_4_12.png|right|]]
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[[File:P_ID409__Sto_Z_4_12.png|right|frame|PSD of white noise]]
:Man bezeichnet ein Rauschsignal <i>n</i>(<i>t</i>) als <i>wei&szlig;</i>, wenn darin alle spektralen Anteile ohne Bevorzugung von irgendwelchen Frequenzen enthalten sind.
+
A noise signal $n(t)$&nbsp; is called&nbsp; "white"&nbsp; if it contains all spectral components without preference of any frequencies.
 +
* The physical power-spectral density defined only for positive frequencies &nbsp; &rArr; &nbsp; ${\it \Phi}_{n+}(f)$&nbsp; is constant&nbsp; $($equal&nbsp; $N_0)$&nbsp; and extends frequency-wise to infinity.
 +
* ${\it \Phi}_{n+}(f)$&nbsp; is shown in green in the upper graph.&nbsp; The plus sign in the index is to indicate that the function is valid only for positive values of $f$.
 +
* For mathematical description one usually uses the two-sided power-spectral density&nbsp; ${\it \Phi}_{n}(f)$.&nbsp; Here applies for all frequencies from&nbsp; $-\infty$&nbsp; to&nbsp; $+\infty$&nbsp; (blue curve in the upper graph):
 +
:$${\it \Phi}_n (f) ={N_0}/{2}.$$
 +
 
 +
 
 +
The bottom graph shows the two power-spectral densities&nbsp; ${\it \Phi}_{b}(f)$&nbsp; and&nbsp; ${\it \Phi}_{b+}(f)$&nbsp; of bandlimited white noise signal&nbsp; $b(t)$.&nbsp; It holds with the one-sided bandwidth $B$:
 +
:$${\it \Phi}_b(f)=\left\{ {N_0/2\atop 0}{\hspace{0.5cm} {\rm for}\quad |f|\le B \atop {\rm else}}\right.,$$
 +
:$${\it \Phi}_{b+}(f)=\left\{ {N_0\atop 0}{\hspace{0.5cm} {\rm for}\quad 0 \le f\le B \atop {\rm else}}\right.$$
 +
 
 +
For computer simulation of noise processes,&nbsp; band-limited noise must always be assumed,&nbsp; since only discrete-time processes can be handled.&nbsp; For this,&nbsp; the&nbsp; [[Signal_Representation/Discrete-Time_Signal_Representation#Sampling_theorem|Sampling Theorem]]&nbsp; must be obeyed.&nbsp; This states that the bandwidth&nbsp; $B$&nbsp; must be set according to the interpolation distance&nbsp; $T_{\rm A}$&nbsp; of the simulation.
 +
 
 +
Assume the following numerical values throughout this exercise:
 +
* The noise power density &ndash;&nbsp; with respect to the resistor&nbsp; $1 \hspace{0.05cm}\rm \Omega$&nbsp; &ndash;&nbsp; is&nbsp; $N_0 = 4 \cdot 10^{-14}\hspace{0.05cm}\rm V^2/Hz$.
 +
* The&nbsp; (one-sided)&nbsp; bandwidth of the band-limited white noise is&nbsp; $B = 100 \hspace{0.08cm}\rm MHz$.
  
:* Das physikalische, nur f&uuml;r positive Frequenzen <i>f</i> definierte Leistungsdichtespektrum <i>&Phi;<sub>n</sub></i><sub>+</sub>(<i>f</i>) ist konstant (gleich <i>N</i><sub>0</sub>) und reicht frequenzm&auml;&szlig;ig bis ins Unendliche.
 
  
:* <i>&Phi;<sub>n</sub></i><sub>+</sub>(<i>f</i>) ist in der oberen Grafik grün dargestellt. Das Pluszeichen im Index soll anzeigen, dass die Funktion nur f&uuml;r positive Werte von <i>f</i> g&uuml;ltig ist.
 
  
:* Zur mathematischen Beschreibung verwendet man meist das zweiseitige Leistungsdichtespektrum <i>&Phi;<sub>n</sub></i>(<i>f</i>). Hier gilt f&uuml;r alle Frequenzen von &ndash;&#8734; bis +&#8734; (blauer Kurvenzug im oberen  Bild):
 
:$${\it \Phi}_n (f) ={N_0}/{2}.$$
 
  
:Im unteren Bild sind die beiden Leistungsdichtespektren <i>&Phi;<sub>b</sub></i>(<i>f</i>) und <i>&Phi;<sub>b</sub></i><sub>+</sub>(<i>f</i>) eines bandbegrenzten wei&szlig;en Rauschsignals <i>b</i>(<i>t</i>) dargestellt. Es gilt mit der einseitigen Bandbreite <i>B</i>:
 
:$${\it \Phi}_b(f)=\left\{ {N_0/2\atop 0}{\hspace{0.5cm} {\rm f\ddot{u}r}\quad |f|\le B \atop {\rm sonst}}\right.,$$
 
:$${\it \Phi}_{b+}(f)=\left\{ {N_0\atop 0}{\hspace{0.5cm} {\rm f\ddot{u}r}\quad 0 \le f\le B \atop {\rm sonst}}\right..$$
 
  
:Bei der Rechnersimulation von Rauschvorg&auml;ngen muss stets von bandbegrenztem Rauschen ausgegangen werden, da hier nur zeitdiskrete Vorg&auml;nge behandelt werden k&ouml;nnen. Dazu muss das Abtasttheorem (siehe Buch [[Signaldarstellung]], Kapitel 5.1) eingehalten werden. Dieses sagt aus, dass die Bandbreite <i>B</i> gemäß dem St&uuml;tzstellenabstand <i>T</i><sub>A</sub> der Simulation eingestellt werden muss.
 
  
:Gehen Sie in der gesamten Aufgabe von folgenden Zahlenwerten aus:
 
  
:* Die Rauschleistungsdichte (bezogen auf den Widerstand 1 &Omega;) betr&auml;gt <i>N</i><sub>0</sub> = 4 &middot; 10<sup>&ndash;14</sup> V<sup>2</sup>/Hz.
 
  
:* Die (einseitige) Bandbreite des bandbegrenzten wei&szlig;en Rauschens betr&auml;gt <i>B</i> = 100 MHz.
+
Hints:  
 +
*This exercise belongs to the chapter&nbsp; [[Theory_of_Stochastic_Signals/Power-Spectral_Density|Power-Spectral Density]].
 +
*Reference is also made to the chapter&nbsp; [[Theory_of_Stochastic_Signals/Auto-Correlation_Function|Auto-Correlation Function]].
 +
*The properties of white noise are summarized in the second part of the&nbsp; (German language)&nbsp; learning video&nbsp; [[Der_AWGN-Kanal_(Lernvideo)|The AWGN channel]].
 +
  
:<b>Hinweis:</b> Die Aufgabe bezieht sich auf Kapitel 4.4 und Kapitel 4.5. <br>Die Eigenschaften von weißem Rauschen sind in einem Lernvideo zusammengefasst:<br>
 
  
  
===Fragebogen===
+
===Questions===
  
 
<quiz display=simple>
 
<quiz display=simple>
{Welche Aussagen treffen bei einem wei&szlig;en Rauschsignal <i>n</i>(<i>t</i>) immer zu? Begr&uuml;nden Sie Ihre Antworten.
+
{Which statements are always true for a white noise signal&nbsp; $n(t)$.&nbsp; Give reasons for your answers.
 
|type="[]"}
 
|type="[]"}
- Die AKF <i>&phi;<sub>n</sub></i>(<i>&tau;</i>) hat einen si-f&ouml;rmigen Verlauf.
+
- The ACF&nbsp; $\varphi_n(t)$&nbsp; has a sinc-shaped progression.
+ Die AKF <i>&phi;<sub>n</sub></i>(<i>&tau;</i>) ist ein Dirac bei <i>&tau;</i> = 0 mit Gewicht <i>N</i><sub>0</sub>/2.
+
+ The ACF&nbsp; $\varphi_n(\tau)$&nbsp; is a Dirac delta function  at&nbsp; $\tau = 0$&nbsp; with weight&nbsp; $N_0/2$.
+ Im mathematisch strengen Sinn gibt es kein wei&szlig;es Rauschen.
+
+ In practice,&nbsp; there is no&nbsp; (exact)&nbsp; white noise.
+ Thermisches Rauschen kann stets als wei&szlig; angen&auml;hert werden.
+
+ Thermal noise can always be approximated as white.
- Wei&szlig;es Rauschen ist stets gau&szlig;verteilt.
+
- White noise is always Gaussian distributed.
  
  
{Berechnen Sie die AKF <i>&phi;<sub>b</sub></i>(<i>&tau;</i>) des bandbegrenzten Zufallssignals <i>b</i>(<i>t</i>). Welcher Wert ergibt sich f&uuml;r <i>&tau;</i> = 0?
+
{Calculate the ACF&nbsp; $\varphi_b(\tau)$&nbsp; of the random signal&nbsp; $b(t)$&nbsp; bandlimited to&nbsp; $B = 100 \hspace{0.08cm}\rm MHz$.&nbsp; What value results for&nbsp; $\tau = 0$?
 
|type="{}"}
 
|type="{}"}
$B = 100 MHz: \ \ \ \phi_b(\tau\ =\ 0)$ = { 4 3% } $.10^{-6}\ V^2$
+
$\varphi_b(\tau = 0) \ = \ $ { 4 3% } $\ \cdot 10^{-6} \ \rm V^2$
  
  
{Wie gro&szlig; ist der Effektivwert dieses Rauschsignals?
+
{What is the rms value of this bandlimited random signal&nbsp; $b(t)$?
 
|type="{}"}
 
|type="{}"}
$\sigma_b$ = { 2 3% } $.10^{-3}\ V$
+
$\sigma_b \ = \ $ { 2 3% } $\ \rm mV$
  
  
{Welcher Abtastabstand <i>T</i><sub>A</sub> ist (mindestens) zu w&auml;hlen, wenn das Signal <i>b</i>(<i>t</i>) zur zeitdiskreten Simulation von wei&szlig;em Rauschen eingesetzt wird?
+
{What sampling distance&nbsp; $T_{\rm A}$&nbsp; should be&nbsp; (at most)&nbsp; chosen if the band-limited signal&nbsp; $b(t)$&nbsp; is used for discrete-time simulation of white noise?
 
|type="{}"}
 
|type="{}"}
$T_A$ = { 5 3% } $ns$
+
$T_{\rm A} \ = \ $ { 5 3% } $\ \rm ns$
  
  
{Gehen Sie nun vom Abtastabstand <i>T</i><sub>A</sub> = 1 ns aus. Welche der Aussagen treffen dann f&uuml;r zwei aufeinanderfolgende Abtastwerte des Signals <i>b</i>(<i>t</i>) zu?
+
{Assume sampling distance&nbsp; $T_{\rm A} = 1 \hspace{0.05cm}\rm ns$.&nbsp; Then,&nbsp; which of the statements are true for two consecutive samples of the signal&nbsp; $b(t)$&nbsp;?
|type="[]"}
+
|type="()"}
- Die Abtastwerte sind unkorreliert.
+
- The samples are uncorrelated.
+ Die Abtastwerte sind positiv korreliert.  
+
+ The samples are positively correlated.  
- Die Abtastwerte sind negativ korreliert.
+
- The samples are negatively correlated.
  
  
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</quiz>
 
</quiz>
  
===Musterlösung===
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===Solution===
 
{{ML-Kopf}}
 
{{ML-Kopf}}
:<b>1.</b>&nbsp;&nbsp;Richtig sind die <u>Lösungsvorschläge 2, 3 und 4</u>. Die Autokorrelationsfunktion (AKF) ist nämlich die Fouriertransformierte des Leistungsdichtespektrums (LDS). Dabei gilt:
+
'''(1)'''&nbsp; Correct are the&nbsp; <u>solutions 2, 3, and 4</u>:
:$${\it \Phi}_n (f) = \frac {N_0}{2} \hspace{0.3cm} \bullet\!\!-\!\!\!-\!\!\!-\!\!\circ\, \hspace{0.3cm} \varphi_n (\tau)=\frac {N_0}{2}  \cdot {\rm \delta} ( \tau).$$
+
*The auto-correlation function&nbsp; $\rm (ACF)$&nbsp; is the Fourier transform of the power-spectral density&nbsp; $\rm (PSD)$.&nbsp; Here:
 +
:$${\it \Phi}_n (f) = {N_0}/{2} \hspace{0.3cm} \bullet\!\!-\!\!\!-\!\!\!-\!\!\circ\, \hspace{0.3cm} \varphi_n (\tau)={N_0}/{2}  \cdot {\rm \delta} ( \tau).$$
 +
*However,&nbsp; there is no&nbsp; "real"&nbsp; white noise in physics,&nbsp; since such a noise would have to have an infinitely large signal power&nbsp; $($the integral over the PSD as well as the ACF value at&nbsp; $\tau = 0$&nbsp; are both infinitely large$)$.
 +
*Thermal noise has a constant PSD up to frequencies of about&nbsp; $\text{6000 GHz}$.&nbsp; Since all&nbsp; (current)&nbsp; transmission systems operate in a much lower frequency range,&nbsp; thermal noise can be said to be&nbsp; "white"&nbsp; to a good approximation.
 +
*The statistical property&nbsp; "white"&nbsp; says nothing about the amplitude distribution,&nbsp; which is determined by the probability density function&nbsp; $\rm (PDF)$&nbsp; alone.
 +
*When considering the phase of a bandpass signal as the stochastic variable,&nbsp; it is often modeled as uniformly distributed between&nbsp; $0$&nbsp; and&nbsp; $2\pi$.
 +
*If there are no statistical bindings between the respective phase angles at different times,&nbsp; this random process is also&nbsp; "white".
 +
 
 +
 
 +
[[File:P_ID410__Sto_Z_4_12_b.png|right|frame|ACF of band-limited noise]]
 +
'''(2)'''&nbsp; The power-spectral density is a rectangle of width&nbsp; $2B$&nbsp; and height&nbsp; $N_0/2$.
 +
*The inverse Fourier transform yields an sinc&ndash;function:
 +
:$$\varphi_b(\tau) = N_0 \cdot B \cdot {\rm si} (2 \pi B \tau)\hspace{0.3cm}
 +
\Rightarrow \hspace{0.3cm}\varphi_b(\tau = 0) = N_0 \cdot B \hspace{0.15cm}\underline {=4}\cdot 10^{-6} \ \rm V^2.$$
 +
 
  
:&bdquo;Echt&rdquo; wei&szlig;es Rauschen gibt es in der Physik allerdings nicht, da ein solches eine unendlich gro&szlig;e Signalleistung aufweisen m&uuml;sste (das Integral &uuml;ber das LDS und der AKF-Wert bei <i>&tau;</i> = 0 sind jeweils unendlich groß). <i>Thermisches Rauschen</i> hat bis zu Frequenzen von etwa 6000 GHz ein konstantes LDS. Da alle (derzeitigen) &Uuml;bertragungssysteme in einem sehr viel niedrigeren Frequenzbereich arbeiten, kann man thermisches Rauschen mit guter N&auml;herung als &bdquo;wei&szlig;&rdquo; bezeichnen.
 
  
:Die statistische Eigenschaft &bdquo;wei&szlig;&rdquo; sagt nichts &uuml;ber die Amplitudenverteilung aus, die allein durch die Wahrscheinlichkeitsdichtefunktion (WDF) festgelegt ist. Betrachtet man beispielsweise die Phase eines bandpassf&ouml;rmigen Signals als die stochastische Gr&ouml;&szlig;e, so wird diese oft als gleichverteilt zwischen 0 und 2&pi; modelliert. Bestehen zwischen den jeweiligen Phasenwinkeln zu unterschiedlichen Zeiten keine statistischen Bindungen, so ist auch dieser Zufallsprozess &bdquo;wei&szlig;&rdquo;.
+
'''(3)'''&nbsp; The ACF value at the point&nbsp; $\tau = 0$&nbsp; gives the power.&nbsp;  
[[File:P_ID410__Sto_Z_4_12_b.png|right|]]
+
*The root of this is called the&nbsp; "rms value":
 +
:$$\sigma_b = \sqrt{\varphi_b(\tau = 0)} \hspace{0.15cm}\underline {=2 \hspace{0.05cm}\rm V}.$$
  
:<b>2.</b>&nbsp;&nbsp;Das LDS ist ein Rechteck der Breite 2<i>B</i> und der H&ouml;he <i>N</i><sub>0</sub>/2.
 
:Die Fourierr&uuml;cktransformation ergibt eine si-Funktion:
 
:$$\varphi_b(\tau) = N_0 \cdot B \cdot {\rm si} (2 \pi B \tau).$$
 
  
:Der AKF-Wert an der Stelle <i>&tau;</i> = 0 entspricht der Rechteckfl&auml;che: (<i>N</i><sub>0</sub>/2) &middot; 2<i>B</i> <u>= 4 &middot; 10<sup>&ndash;6</sup> V<sup>2</sup></u>.
+
'''(4)'''&nbsp; The ACF computed in&nbsp; '''(3)'''&nbsp; has zeros at equidistant distance from&nbsp; $T_{\rm A}= 1/(2B)\hspace{0.15cm}\underline {=5\hspace{0.05cm}  \rm ns}$:&nbsp;
 +
*There are no statistical bindings between the two signal values&nbsp; $b(t)$&nbsp; and&nbsp; $b(t + \nu \cdot T_{\rm A})$,
 +
*where&nbsp; $\nu$&nbsp; can take all integer values.
  
:<b>3.</b>&nbsp;&nbsp;Der AKF-Wert an der Stelle <i>&tau;</i> = 0 ergibt die Leistung; die Wurzel hieraus bezeichnet man als den Effektivwert: <i>&sigma;<sub>b</sub></i> <u>= 2 mV</u>.
 
  
:<b>4.</b>&nbsp;&nbsp;Die bei (b) berechnete AKF hat Nullstellen im &auml;quidistanten Abstand von 1/(2<i>B</i>) = <u>5 ns = <i>T</i></i><sub>A</sub></u>. Das bedeutet: Es bestehen somit keine statistischen Bindungen zwischen den beiden Signalwerten <i>b</i>(<i>t</i>) und <i>b</i>(<i>t</i> + <i>&nu; &middot; T</i><sub>A</sub>), wobei <i>&nu;</i> alle ganzzahligen Werte annehmen kann.
 
  
:<b>5.</b>&nbsp;&nbsp;Der AKF-Wert bei <i>&tau;</i> = <i>T</i><sub>A</sub> = 1 ns betr&auml;gt
+
'''(5)'''&nbsp; The correct solution is the&nbsp; <u>suggested solution 2</u>.
:$$\varphi_b(\tau = T_{\rm A}) = {\rm 4 \cdot 10^{-6} \hspace{0.1cm}V^2 \cdot si (\pi/5) \approx 3.742 \cdot 10^{-6} \hspace{0.1cm}V^2}$$
+
*The ACF value at&nbsp; $\tau = T_{\rm A} = 1 \hspace{0.05cm}\rm ns$&nbsp; is
 +
:$$\varphi_b(\tau = T_{\rm A}) = {\rm 4 \cdot 10^{-6} \hspace{0.1cm}V^2 \cdot sinc (1/5) \approx 3.742 \cdot 10^{-6} \hspace{0.1cm}V^2} > 0.$$
  
:und ist damit positiv. Dieses Ergebnis besagt: Zwei um <i>T</i><sub>A</sub> = 1 Nanosekunde auseinander liegende Signalwerte sind positiv korreliert &nbsp;&#8658;&nbsp; <u>Lösungsvorschlag 2</u>. Ist <i>b</i>(<i>t</i>) positiv und gro&szlig;, dann ist mit gro&szlig;er Wahrscheinlichkeit auch <i>b</i>(<i>t</i> + 1 ns) positiv und gro&szlig;. Dagegen besteht zwischen <i>b</i>(<i>t</i>) und <i>b</i>(<i>t</i> + 7 ns) eine negative Korrelation: Ist <i>b</i>(<i>t</i>) positiv, so ist <i>b</i>(<i>t</i> + 7 ns) wahrscheinlich negativ.
+
*This result says: &nbsp; Two signal values separated by&nbsp; $T_{\rm A} = 1 \hspace{0.05cm}\rm ns$&nbsp; are positively correlated:
 +
:*If&nbsp; $b(t)$&nbsp; is positive and large,&nbsp; then with high probability&nbsp; $b(t+1 \hspace{0.05cm}\rm ns)$&nbsp; is also positive and large.  
 +
:*In contrast,&nbsp; there is a negative correlation between&nbsp; $b(t)$&nbsp; and&nbsp; $b(t+7 \hspace{0.05cm}\rm ns)$.&nbsp; If&nbsp; $b(t)$&nbsp; is positive,&nbsp; then&nbsp; $b(t+7 \hspace{0.05cm}\rm ns)$&nbsp; is probably negative.
 
{{ML-Fuß}}
 
{{ML-Fuß}}
  
  
  
[[Category:Aufgaben zu Stochastische Signaltheorie|^4.5 Leistungsdichtespektrum (LDS)^]]
+
[[Category:Theory of Stochastic Signals: Exercises|^4.5 Power-Spectral Density^]]

Latest revision as of 15:27, 25 March 2022

PSD of white noise

A noise signal $n(t)$  is called  "white"  if it contains all spectral components without preference of any frequencies.

  • The physical power-spectral density defined only for positive frequencies   ⇒   ${\it \Phi}_{n+}(f)$  is constant  $($equal  $N_0)$  and extends frequency-wise to infinity.
  • ${\it \Phi}_{n+}(f)$  is shown in green in the upper graph.  The plus sign in the index is to indicate that the function is valid only for positive values of $f$.
  • For mathematical description one usually uses the two-sided power-spectral density  ${\it \Phi}_{n}(f)$.  Here applies for all frequencies from  $-\infty$  to  $+\infty$  (blue curve in the upper graph):
$${\it \Phi}_n (f) ={N_0}/{2}.$$


The bottom graph shows the two power-spectral densities  ${\it \Phi}_{b}(f)$  and  ${\it \Phi}_{b+}(f)$  of bandlimited white noise signal  $b(t)$.  It holds with the one-sided bandwidth $B$:

$${\it \Phi}_b(f)=\left\{ {N_0/2\atop 0}{\hspace{0.5cm} {\rm for}\quad |f|\le B \atop {\rm else}}\right.,$$
$${\it \Phi}_{b+}(f)=\left\{ {N_0\atop 0}{\hspace{0.5cm} {\rm for}\quad 0 \le f\le B \atop {\rm else}}\right.$$

For computer simulation of noise processes,  band-limited noise must always be assumed,  since only discrete-time processes can be handled.  For this,  the  Sampling Theorem  must be obeyed.  This states that the bandwidth  $B$  must be set according to the interpolation distance  $T_{\rm A}$  of the simulation.

Assume the following numerical values throughout this exercise:

  • The noise power density –  with respect to the resistor  $1 \hspace{0.05cm}\rm \Omega$  –  is  $N_0 = 4 \cdot 10^{-14}\hspace{0.05cm}\rm V^2/Hz$.
  • The  (one-sided)  bandwidth of the band-limited white noise is  $B = 100 \hspace{0.08cm}\rm MHz$.





Hints:



Questions

1

Which statements are always true for a white noise signal  $n(t)$.  Give reasons for your answers.

The ACF  $\varphi_n(t)$  has a sinc-shaped progression.
The ACF  $\varphi_n(\tau)$  is a Dirac delta function at  $\tau = 0$  with weight  $N_0/2$.
In practice,  there is no  (exact)  white noise.
Thermal noise can always be approximated as white.
White noise is always Gaussian distributed.

2

Calculate the ACF  $\varphi_b(\tau)$  of the random signal  $b(t)$  bandlimited to  $B = 100 \hspace{0.08cm}\rm MHz$.  What value results for  $\tau = 0$?

$\varphi_b(\tau = 0) \ = \ $

$\ \cdot 10^{-6} \ \rm V^2$

3

What is the rms value of this bandlimited random signal  $b(t)$?

$\sigma_b \ = \ $

$\ \rm mV$

4

What sampling distance  $T_{\rm A}$  should be  (at most)  chosen if the band-limited signal  $b(t)$  is used for discrete-time simulation of white noise?

$T_{\rm A} \ = \ $

$\ \rm ns$

5

Assume sampling distance  $T_{\rm A} = 1 \hspace{0.05cm}\rm ns$.  Then,  which of the statements are true for two consecutive samples of the signal  $b(t)$ ?

The samples are uncorrelated.
The samples are positively correlated.
The samples are negatively correlated.


Solution

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

  • The auto-correlation function  $\rm (ACF)$  is the Fourier transform of the power-spectral density  $\rm (PSD)$.  Here:
$${\it \Phi}_n (f) = {N_0}/{2} \hspace{0.3cm} \bullet\!\!-\!\!\!-\!\!\!-\!\!\circ\, \hspace{0.3cm} \varphi_n (\tau)={N_0}/{2} \cdot {\rm \delta} ( \tau).$$
  • However,  there is no  "real"  white noise in physics,  since such a noise would have to have an infinitely large signal power  $($the integral over the PSD as well as the ACF value at  $\tau = 0$  are both infinitely large$)$.
  • Thermal noise has a constant PSD up to frequencies of about  $\text{6000 GHz}$.  Since all  (current)  transmission systems operate in a much lower frequency range,  thermal noise can be said to be  "white"  to a good approximation.
  • The statistical property  "white"  says nothing about the amplitude distribution,  which is determined by the probability density function  $\rm (PDF)$  alone.
  • When considering the phase of a bandpass signal as the stochastic variable,  it is often modeled as uniformly distributed between  $0$  and  $2\pi$.
  • If there are no statistical bindings between the respective phase angles at different times,  this random process is also  "white".


ACF of band-limited noise

(2)  The power-spectral density is a rectangle of width  $2B$  and height  $N_0/2$.

  • The inverse Fourier transform yields an sinc–function:
$$\varphi_b(\tau) = N_0 \cdot B \cdot {\rm si} (2 \pi B \tau)\hspace{0.3cm} \Rightarrow \hspace{0.3cm}\varphi_b(\tau = 0) = N_0 \cdot B \hspace{0.15cm}\underline {=4}\cdot 10^{-6} \ \rm V^2.$$


(3)  The ACF value at the point  $\tau = 0$  gives the power. 

  • The root of this is called the  "rms value":
$$\sigma_b = \sqrt{\varphi_b(\tau = 0)} \hspace{0.15cm}\underline {=2 \hspace{0.05cm}\rm V}.$$


(4)  The ACF computed in  (3)  has zeros at equidistant distance from  $T_{\rm A}= 1/(2B)\hspace{0.15cm}\underline {=5\hspace{0.05cm} \rm ns}$: 

  • There are no statistical bindings between the two signal values  $b(t)$  and  $b(t + \nu \cdot T_{\rm A})$,
  • where  $\nu$  can take all integer values.


(5)  The correct solution is the  suggested solution 2.

  • The ACF value at  $\tau = T_{\rm A} = 1 \hspace{0.05cm}\rm ns$  is
$$\varphi_b(\tau = T_{\rm A}) = {\rm 4 \cdot 10^{-6} \hspace{0.1cm}V^2 \cdot sinc (1/5) \approx 3.742 \cdot 10^{-6} \hspace{0.1cm}V^2} > 0.$$
  • This result says:   Two signal values separated by  $T_{\rm A} = 1 \hspace{0.05cm}\rm ns$  are positively correlated:
  • If  $b(t)$  is positive and large,  then with high probability  $b(t+1 \hspace{0.05cm}\rm ns)$  is also positive and large.
  • In contrast,  there is a negative correlation between  $b(t)$  and  $b(t+7 \hspace{0.05cm}\rm ns)$.  If  $b(t)$  is positive,  then  $b(t+7 \hspace{0.05cm}\rm ns)$  is probably negative.