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

From LNTwww
Line 1: Line 1:
  
{{quiz-Header|Buchseite=Stochastische Signaltheorie/Leistungsdichtespektrum (LDS)
+
{{quiz-Header|Buchseite=Theory_of_Stochastic_Signals/Power-Spectral_Density
 
}}
 
}}
[[File:P_ID409__Sto_Z_4_12.png|right|frame|Leistungsdichtespektren <br>von Weißem Rauschen]]
+
[[File:P_ID409__Sto_Z_4_12.png|right|frame|Power spectral densities <br> of white noise]]
Man bezeichnet ein Rauschsignal&nbsp; $n(t)$&nbsp; 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 <i>white;</i> if it contains all spectral components without preference of any frequencies.
* Das physikalische, nur f&uuml;r positive Frequenzen $f$&nbsp; definierte Leistungsdichtespektrum&nbsp; ${\it \Phi}_{n+}(f)$&nbsp; ist konstant&nbsp; $($gleich&nbsp; $N_0)$&nbsp; und reicht frequenzm&auml;&szlig;ig bis ins Unendliche.
+
* The physical Power spectral density defined only for positive frequencies $f$&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; ist in der oberen Grafik grün dargestellt.&nbsp; Das Pluszeichen im Index soll anzeigen, dass die Funktion nur f&uuml;r positive Werte von $f$&nbsp; g&uuml;ltig ist.
+
* ${\it \Phi}_{n+}(f)$&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 $f$&nbsp; .
* Zur mathematischen Beschreibung verwendet man meist das zweiseitige Leistungsdichtespektrum&nbsp; ${\it \Phi}_{n}(f)$.&nbsp; Hier gilt f&uuml;r alle Frequenzen von&nbsp; $-\infty$&nbsp; bis&nbsp; $+\infty$&nbsp; (blauer Kurvenzug im oberen  Bild):
+
* For mathematical description one usually uses the two-sided Power spectral density spectrum&nbsp; ${\it \Phi}_{n}(f)$.&nbsp; Here applies for;all frequencies from&nbsp; $-\infty$&nbsp; to&nbsp; $+\infty$&nbsp; (blue curve in the upper picture):
 
:$${\it \Phi}_n (f) ={N_0}/{2}.$$
 
:$${\it \Phi}_n (f) ={N_0}/{2}.$$
  
  
In der unteren Grafik sind die beiden Leistungsdichtespektren&nbsp; ${\it \Phi}_{b}(f)$&nbsp; und&nbsp; ${\it \Phi}_{b+}(f)$&nbsp; eines bandbegrenzten wei&szlig;en Rauschsignals&nbsp; $b(t)$&nbsp; dargestellt.&nbsp; Es gilt mit der einseitigen Bandbreite&nbsp; $B$:
+
The bottom graph shows the two Power spectral densities&nbsp; ${\it \Phi}_{b}(f)$&nbsp; and&nbsp; ${\it \Phi}_{b+}(f)$&nbsp; of a bandlimited white noise signal&nbsp; $b(t)$&nbsp; It holds with the one-sided bandwidth&nbsp; $B$:
:$${\it \Phi}_b(f)=\left\{ {N_0/2\atop 0}{\hspace{0.5cm} {\rm f\ddot{u}r}\quad |f|\le B \atop {\rm sonst}}\right.,$$
+
:$${\it \Phi}_b(f)=\left\{ {N_0/2\atop 0}{\hspace{0.5cm} {\rm f\ddot{u}r}\quad |f|\le B \atop {\rm else}}\right.,$$
:$${\it \Phi}_{b+}(f)=\left\{ {N_0\atop 0}{\hspace{0.5cm} {\rm f\ddot{u}r}\quad 0 \le f\le B \atop {\rm sonst}}\right.$$
+
:$${\it \Phi}_{b+}(f)=\left\{ {N_0\atop 0}{\hspace{0.5cm} {\rm f\ddot{u}r}\quad 0 \le f\le B \atop {\rm else}}\right.$$
  
Bei der Rechnersimulation von Rauschvorg&auml;ngen muss stets von bandbegrenztem Rauschen ausgegangen werden, da nur zeitdiskrete Vorg&auml;nge behandelt werden k&ouml;nnen. Dazu muss das&nbsp; [[Signal_Representation/Time_Discrete_Signal_Representation#Das_Abtasttheorem|Abtasttheorem]]&nbsp; eingehalten werden.&nbsp; Dieses sagt aus, dass die Bandbreite&nbsp; $B$&nbsp; gemäß dem St&uuml;tzstellenabstand&nbsp; $T_{\rm A}$&nbsp; der Simulation eingestellt werden muss.
+
For computer simulation of noise processes, band-limited noise must always be assumed, since only discrete-time processes can be handled. For this, the&nbsp; [[Signal_Representation/Discrete-Time_Signal_Representation#Sampling_theorem|SamplingTheorem]]&nbsp; must be obeyed.&nbsp; This states that the bandwidth&nbsp; $B$&nbsp; must be set according to the pitch&nbsp; $T_{\rm A}$&nbsp; of the simulation.
  
Gehen Sie in der gesamten Aufgabe von folgenden Zahlenwerten aus:
+
Assume the following numerical values throughout the exercise:
* Die Rauschleistungsdichte &ndash;&nbsp; bezogen auf den Widerstand&nbsp; $1 \hspace{0.05cm}\rm \Omega$&nbsp; &ndash;&nbsp; betr&auml;gt&nbsp; $N_0 = 4 \cdot 10^{-14}\hspace{0.05cm}\rm V^2/Hz$.
+
* The noise power density &ndash;&nbsp; with respect to the resistor&nbsp; $1 \hspace{0.05cm}\rm \Omega$&nbsp; &ndash;&nbsp; betr&auml;gt&nbsp; $N_0 = 4 \cdot 10^{-14}\hspace{0.05cm}\rm V^2/Hz$.
* Die (einseitige) Bandbreite des bandbegrenzten wei&szlig;en Rauschens betr&auml;gt&nbsp; $B = 100 \hspace{0.08cm}\rm MHz$.
+
* The (one-sided) bandwidth of the band-limited white noise is $B = 100 \hspace{0.08cm}\rm MHz$.
  
  
Line 27: Line 27:
  
  
''Hinweise:''
+
Hints:  
*Die Aufgabe gehört zum  Kapitel&nbsp; [[Theory_of_Stochastic_Signals/Leistungsdichtespektrum_(LDS)|Leistungsdichtespektrum]].
+
*This exercise belongs to the chapter&nbsp; [[Theory_of_Stochastic_Signals/Power-Spectral_Density|Power-Spectral Density]].
*Bezug genommen wird auch auf das  Kapitel&nbsp; [[Theory_of_Stochastic_Signals/Autokorrelationsfunktion_(AKF)|Autokorrelationsfunktion]].
+
*Reference is also made to the chapter&nbsp; [[Theory_of_Stochastic_Signals/Auto-Correlation_Function|Auto-correlation function]].
*Die Eigenschaften von weißem Rauschen werden im zweiten Teil des Lernvideos&nbsp; [[Der_AWGN-Kanal_(Lernvideo)|Der AWGN-Kanal]]&nbsp; zusammengefasst.
+
*The properties of white noise are summarized in the second part of the tutorial video&nbsp; [[Der_AWGN-Kanal_(Lernvideo)|The AWGN channel (German)]]&nbsp;.
 
   
 
   
  
  
  
===Fragebogen===
+
===Questions===
  
 
<quiz display=simple>
 
<quiz display=simple>
{Welche Aussagen treffen bei einem wei&szlig;en Rauschsignal&nbsp; $n(t)$&nbsp; immer zu?&nbsp; 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&nbsp; $\varphi_n(\tau)$&nbsp; hat einen si-f&ouml;rmigen Verlauf.
+
- The ACF&nbsp; $\varphi_n(t)$&nbsp; has a si-shaped progression.
+ Die AKF&nbsp; $\varphi_n(\tau)$&nbsp; ist ein Dirac bei&nbsp; $\tau = 0$&nbsp; mit Gewicht&nbsp; $N_0/2$.
+
+ The ACF&nbsp; $\varphi_n(\tau)$&nbsp; is a Dirac at&nbsp; $\tau = 0$&nbsp; with weight&nbsp; $N_0/2$.
+ In der Praxis gibt es kein (exakt) wei&szlig;es Rauschen.
+
+ In practice, there is no (exact) 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&nbsp; $\varphi_b(\tau)$&nbsp; des auf&nbsp; $B = 100 \hspace{0.08cm}\rm MHz$&nbsp; bandbegrenzten Zufallssignals $b(t)$.&nbsp; Welcher Wert ergibt sich f&uuml;r&nbsp; $\tau = 0$?
+
{Calculate the ACF&nbsp; $\varphi_b(\tau)$&nbsp; of the random signal $b(t)$ bandlimited to&nbsp; $B = 100 \hspace{0.08cm}\rm MHz$&nbsp; What value results for&nbsp; $\tau = 0$?
 
|type="{}"}
 
|type="{}"}
$\varphi_b(\tau = 0) \ = \ $ { 4 3% } $\ \cdot 10^{-6} \ \rm V^2$
+
$\varphi_b(\tau = 0) \ = \ $ { 4 3% } $\ \cdot 10^{-6} \ \rm V^2$
  
  
{Wie gro&szlig; ist der Effektivwert dieses bandbegrenzten Zufallssignals&nbsp; $b(t)$?
+
{What is the rms value of this bandlimited random signal&nbsp; $b(t)$?
 
|type="{}"}
 
|type="{}"}
$\sigma_b \ = \ $ { 2 3% }   $\ \rm mV$
+
$\sigma_b \ = \ $ { 2 3% } $\ \rm mV$
  
  
{Welcher Abtastabstand&nbsp; $T_{\rm A}$&nbsp; ist (höchstens) zu w&auml;hlen, wenn das bandbegrenzte Signal&nbsp; $b(t)$&nbsp; zur zeitdiskreten Simulation von wei&szlig;em Rauschen eingesetzt wird?
+
{What sampling distance&nbsp; $T_{\rm A}$&nbsp; should be chosen (at most) if the band-limited signal&nbsp; $b(t)$&nbsp; is used for discrete-time simulation of white noise?
 
|type="{}"}
 
|type="{}"}
$T_{\rm A} \ = \ $ { 5 3% } $\ \rm ns$
+
$T_{\rm A} \ = \ $ { 5 3% } $\ \rm ns$
  
  
{Gehen Sie vom Abtastabstand&nbsp; $T_{\rm A} = 1 \hspace{0.05cm}\rm   ns$&nbsp; aus.&nbsp; Welche der Aussagen treffen dann f&uuml;r zwei aufeinanderfolgende Abtastwerte des Signals&nbsp; $b(t)$&nbsp; zu?
+
{Assume sampling distance&nbsp; $T_{\rm A} = 1 \hspace{0.05cm}\rm ns$&nbsp; Then, 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.
  
  
Line 72: Line 72:
 
</quiz>
 
</quiz>
  
===Musterlösung===
+
===Solution===
 
{{ML-Kopf}}
 
{{ML-Kopf}}
'''(1)'''&nbsp; Richtig sind die <u>Lösungsvorschläge 2, 3 und 4</u>:  
+
'''(1)'''&nbsp; Correct are <u>solutions 2, 3, and 4</u>:  
*Die Autokorrelationsfunktion (AKF) ist die Fouriertransformierte des Leistungsdichtespektrums (LDS). Dabei gilt:
+
*The auto-correlation function (ACF) is the Fourier transform of the Power spectral density (PSD). Here:
:$${\it \Phi}_n (f) = {N_0}/{2} \hspace{0.3cm} \bullet\!\!-\!\!\!-\!\!\!-\!\!\circ\, \hspace{0.3cm} \varphi_n (\tau)={N_0}/{2}  \cdot {\rm \delta} ( \tau).$$
+
:$${\it \Phi}_n (f) = {N_0}/{2} \hspace{0.3cm} \bullet\!\!-\!\!\!-\!\!\!-\!\!\circ\, \hspace{0.3cm} \varphi_n (\tau)={N_0}/{2}  \cdot {\rm \delta} ( \tau).$$
*Allerdings gibt es in der Physik kein "echt" wei&szlig;es Rauschen, da ein solches eine unendlich gro&szlig;e Signalleistung aufweisen m&uuml;sste&nbsp; $($das Integral &uuml;ber das LDS sowie der AKF-Wert bei&nbsp; $\tau = 0$&nbsp; sind jeweils unendlich groß$)$.  
+
*However, there is no "real" white noise in physics, 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$)$.  
*Thermisches Rauschen hat bis zu Frequenzen von etwa&nbsp; $\text{6000 GHz}$&nbsp; 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 "wei&szlig;" bezeichnen.
+
*Thermal noise has a constant PSD up to frequencies of about&nbsp; $\text{6000 GHz}$&nbsp;. Since all (current) &uuml;b transmission systems operate in a much lower frequency range, thermal noise can be said to be "white" to a good approximation.
*Die statistische Eigenschaft "wei&szlig;" sagt nichts &uuml;ber die Amplitudenverteilung aus, die allein durch die Wahrscheinlichkeitsdichtefunktion (WDF) festgelegt ist.  
+
*The statistical property "white" says nothing about the amplitude distribution, which is determined by the probability density function (PDF) alone.  
*Betrachtet man die Phase eines bandpassf&ouml;rmigen Signals als die stochastische Gr&ouml;&szlig;e, so wird diese oft als gleichverteilt zwischen&nbsp; $0$&nbsp; und&nbsp; $2\pi$&nbsp; modelliert.  
+
*When considering the phase of a bandpass signal as the stochastic variable, it is often modeled as uniformly distributed between&nbsp; $0$&nbsp; and&nbsp; $2\pi$&nbsp; .  
*Bestehen zwischen den jeweiligen Phasenwinkeln zu unterschiedlichen Zeiten keine statistischen Bindungen, so ist auch dieser Zufallsprozess "wei&szlig;".
+
*If there are no statistical bindings between the respective phase angles at different times, this random process is also "white".
  
  
  
[[File:P_ID410__Sto_Z_4_12_b.png|right|frame|AKF von bandbegrenztem Rauschen]]
+
 
'''(2)'''&nbsp; Das Leistungsdichtespektrum ist ein Rechteck der Breite&nbsp; $2B$&nbsp; und der H&ouml;he&nbsp; $N_0/2$.  
+
 
*Die Fourierr&uuml;cktransformation ergibt eine si-Funktion:
+
 
 +
[[File:P_ID410__Sto_Z_4_12_b.png|right|frame|ACF of band-limited noise]]
 +
'''(2)'''&nbsp; The Power spectral density spectrum is a rectangle of width&nbsp; $2B$&nbsp; and height&nbsp; $N_0/2$.  
 +
*The inverse Fourier transformation yields an si function:
 
:$$\varphi_b(\tau) = N_0 \cdot B \cdot {\rm si} (2 \pi B \tau)\hspace{0.3cm}
 
:$$\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.$$
 
\Rightarrow \hspace{0.3cm}\varphi_b(\tau = 0) = N_0 \cdot B \hspace{0.15cm}\underline {=4}\cdot 10^{-6} \ \rm V^2.$$
Line 93: Line 96:
  
  
'''(3)'''&nbsp; Der AKF-Wert an der Stelle&nbsp; $\tau = 0$&nbsp; ergibt die Leistung.&nbsp;  
+
'''(3)'''&nbsp; The ACF value at the point&nbsp; $\tau = 0$&nbsp; gives the power.&nbsp;  
*Die Wurzel hieraus bezeichnet man als den Effektivwert:  
+
*The root of this is called the rms value:  
 
:$$\sigma_b = \sqrt{\varphi_b(\tau = 0)} \hspace{0.15cm}\underline {=2 \hspace{0.05cm}\rm V}.$$
 
:$$\sigma_b = \sqrt{\varphi_b(\tau = 0)} \hspace{0.15cm}\underline {=2 \hspace{0.05cm}\rm V}.$$
+
'''(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; $b(t + \nu \cdot T_{\rm A})$,  
'''(4)'''&nbsp; Die in&nbsp; '''(3)'''&nbsp; berechnete AKF hat Nullstellen im &auml;quidistanten Abstand von&nbsp; $T_{\rm A}= 1/(2B)\hspace{0.15cm}\underline {=5\hspace{0.05cm} \rm ns}$:&nbsp;  
+
*where&nbsp; $\nu$&nbsp; can take all integer values.
*Es bestehen keine statistischen Bindungen zwischen den beiden Signalwerten&nbsp; $b(t)$&nbsp; und&nbsp; $b(t + \nu \cdot T_{\rm A})$,  
 
*wobei&nbsp; $\nu$&nbsp; alle ganzzahligen Werte annehmen kann.
 
  
  
  
'''(5)'''&nbsp; Richtig ist der <u>Lösungsvorschlag 2</u>.  
+
'''(5)'''&nbsp; The correct solution is <u>suggested solution 2</u>.  
*Der AKF-Wert bei&nbsp; $\tau = T_{\rm A} = 1 \hspace{0.05cm}\rm ns$&nbsp; betr&auml;gt
+
*The ACF value at&nbsp; $\tau = T_{\rm A} = 1 \hspace{0.05cm}\rm ns$&nbsp; amounts to.
:$$\varphi_b(\tau = T_{\rm A}) = {\rm 4 \cdot 10^{-6} \hspace{0.1cm}V^2 \cdot si (\pi/5) \approx 3.742 \cdot 10^{-6} \hspace{0.1cm}V^2} > 0.$$
+
:$$\varphi_b(\tau = T_{\rm A}) = {\rm 4 \cdot 10^{-6} \hspace{0.1cm}V^2 \cdot si (\pi/5) \approx 3.742 \cdot 10^{-6} \hspace{0.1cm}V^2} > 0.$$
  
*Dieses Ergebnis besagt: &nbsp; Zwei um&nbsp; $T_{\rm A} = 1 \hspace{0.05cm}\rm ns$&nbsp; auseinander liegende Signalwerte sind positiv korreliert:  
+
*This result says: &nbsp; Two signal values separated by&nbsp; $T_{\rm A} = 1 \hspace{0.05cm}\rm ns$&nbsp; are positively correlated:  
*Ist&nbsp; $b(t)$&nbsp; positiv und gro&szlig;, dann ist mit gro&szlig;er Wahrscheinlichkeit auch&nbsp; $b(t+1 \hspace{0.05cm}\rm ns)$&nbsp; positiv und gro&szlig;.  
+
*If&nbsp; $b(t)$&nbsp; is positive and large;, then with high probability&nbsp; $b(t+1 \hspace{0.05cm}\rm ns)$&nbsp; is also positive and large;.  
*Dagegen besteht zwischen&nbsp; $b(t)$&nbsp; und&nbsp; $b(t+7 \hspace{0.05cm}\rm ns)$&nbsp; eine negative Korrelation: &nbsp; Ist&nbsp; $b(t)$&nbsp; positiv, so ist&nbsp; $b(t+7 \hspace{0.05cm}\rm ns)$&nbsp; wahrscheinlich negativ.
+
*In contrast, 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, then&nbsp; $b(t+7 \hspace{0.05cm}\rm ns)$&nbsp; is probably negative.
 
{{ML-Fuß}}
 
{{ML-Fuß}}
  

Revision as of 19:59, 7 March 2022

Power spectral densities
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 $f$  ${\it \Phi}_{n+}(f)$  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 spectrum  ${\it \Phi}_{n}(f)$.  Here applies for;all frequencies from  $-\infty$  to  $+\infty$  (blue curve in the upper picture):
$${\it \Phi}_n (f) ={N_0}/{2}.$$


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

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

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

Assume the following numerical values throughout the exercise:

  • The noise power density –  with respect to the resistor  $1 \hspace{0.05cm}\rm \Omega$  –  beträgt  $N_0 = 4 \cdot 10^{-14}\hspace{0.05cm}\rm V^2/Hz$.
  • 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 si-shaped progression.
The ACF  $\varphi_n(\tau)$  is a Dirac 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 chosen (at most) 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 solutions 2, 3, and 4:

  • The auto-correlation function (ACF) is the Fourier transform of the Power spectral density (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) üb 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 (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 spectrum is a rectangle of width  $2B$  and height  $N_0/2$.

  • The inverse Fourier transformation yields an si 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 suggested solution 2.

  • The ACF value at  $\tau = T_{\rm A} = 1 \hspace{0.05cm}\rm ns$  amounts to.
$$\varphi_b(\tau = T_{\rm A}) = {\rm 4 \cdot 10^{-6} \hspace{0.1cm}V^2 \cdot si (\pi/5) \approx 3.742 \cdot 10^{-6} \hspace{0.1cm}V^2} > 0.$$
  • 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.