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{{quiz-Header|Buchseite=Modulationsverfahren/Qualitätskriterien
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{{quiz-Header|Buchseite=Modulation_Methods/Quality_Criteria
 
}}
 
}}
  
[[File:|right|]]
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[[File:P_ID953__Mod_Z_1_3.png|right|frame|Examplary signals for low-pass and band-pass noise]]
 +
A fundamental disturbance and one that occurs in any communication system is&nbsp; "thermal noise",&nbsp; since any resistance &nbsp;R&nbsp; with an absolute temperature of &nbsp;θ&nbsp; (in "degrees Kelvin")&nbsp; produces a noise signal &nbsp;n(t)&nbsp; with a&nbsp; (one-sided)&nbsp; noise power-spectral density
 +
:$${N_{\rm 0, \hspace{0.05cm}min}}= k_{\rm B} \cdot \theta
 +
\hspace{0.3cm}\left(k_{\rm B} = 1.38 \cdot 10^{-23}
 +
\hspace{0.05cm}{\rm Ws}/{\rm K}\right)$$
 +
where k_{\rm B}&nbsp; denotes the&nbsp; "Boltzmann constant" (from German&nbsp; "Konstante").
  
 +
However,&nbsp; this noise is limited to &nbsp;6\text{ THz}&nbsp; for physical reasons.&nbsp; Furthermore,&nbsp; it can be observed that this minimum value can only be achieved with exact impedance matching.
  
===Fragebogen===
+
In the realization of a circuit unit - for example, an amplifier - the effective noise power-spectral density is usually significantly greater,&nbsp; since several noise sources add up,&nbsp; and mismatches also play a role.&nbsp;  This effect is captured by the noise factor &nbsp;F \ge 1&nbsp; .&nbsp; It holds that:
 +
:$$N_0 = F \cdot {N_{\rm 0, \hspace{0.05cm}min}}= F \cdot k_{\rm B} \cdot \theta \hspace{0.05cm}.$$
 +
With a bandwidth &nbsp;B,&nbsp; the effective noise power is characterized by:
 +
:$$N = N_0 \cdot B \hspace{0.1cm} \hspace{0.01cm}.$$
 +
:N = N_0 \cdot B\cdot R = \sigma_n^2 \hspace{0.01cm}.
 +
*According to the first equation,&nbsp; the result is the actual,&nbsp; physical power in "watts"&nbsp; \rm (W).
 +
*According to the second equation,&nbsp; the result has the unit &nbsp; "\rm V^{ 2 }".
 +
*This means that the power is here converted to the reference resistance &nbsp;R = 1\ Ω&nbsp; – as is often the case in Communications Engineering.
 +
*This equation must also be used to calculate the standard deviation&nbsp; σ_n&nbsp; of the noise signal &nbsp;n(t)&nbsp;.
 +
 
 +
 
 +
All equations apply regardless of whether the noise is low-pass or band-pass.  The diagram shows two noise signals &nbsp;n_1(t)&nbsp; and &nbsp;n_2(t)&nbsp; of equal bandwidth.&nbsp; Question &nbsp; '''(4)'''&nbsp; asks which of these signals will appear at the output of a low-pass and a band-pass, respectively.
 +
 
 +
The two-sided noise power-spectral density of band-limited low-pass noise &nbsp;n_{\rm TP}(t)&nbsp; is:
 +
: {\it \Phi}_{n, {\hspace{0.05cm}\rm LP}}(f) = \left\{ \begin{array}{c} N_0/2 \\ 0 \\ \end{array} \right. \begin{array}{*{10}c} {\rm{for}} \\ \\ \end{array}\begin{array}{*{20}c} {\left| \hspace{0.005cm} f\hspace{0.05cm} \right| < B,} \\ {\rm otherwise.} \\ \end{array}
 +
In contrast,&nbsp; for band-pass noise &nbsp;n_{\rm BP}(t)&nbsp; with center frequency &nbsp;f_{\rm M}, it holds that:
 +
:{\it \Phi}_{n, {\hspace{0.05cm}\rm BP}}(f) = \left\{ \begin{array}{c} N_0/2 \\ 0 \\ \end{array} \right. \begin{array}{*{10}c} {\rm{for}} \\ \\ \end{array}\begin{array}{*{20}c} {\left| \hspace{0.005cm} f - f_{\rm M}\hspace{0.05cm} \right| < B/2,} \\ {\rm otherwise.} \\ \end{array}.
 +
For all subsequent numerical calculations it is assumed:
 +
: F = 10, \hspace{0.2cm}\theta = 290\,{\rm K},\hspace{0.2cm}R = 50\,{\rm \Omega},\hspace{0.2cm}B = 30\,{\rm kHz},\hspace{0.2cm}f_{\rm M} = 0 \hspace{0.5cm}{\rm or}\hspace{0.5cm}f_{\rm M} =100\,{\rm kHz}\hspace{0.05cm}.
 +
 
 +
 
 +
 
 +
 
 +
Hints:
 +
*This exercise belongs to the chapter&nbsp; [[Modulation_Methods/Quality_Criteria|Quality Criteria]].
 +
*Particular reference is made to the page&nbsp;  [[Modulation_Methods/Quality_Criteria#Some_remarks_on_the_AWGN_channel_model|Some remarks on the AWGN channel model]].
 +
*By specifying the powers in &nbsp;\rm Watts&nbsp;, they are independent of the reference resistance &nbsp;R,&nbsp; while power with the unit  &nbsp;\rm V^2&nbsp; can only be evaluated directly for &nbsp;R = 1\ \Omega.
 +
 +
 
 +
 
 +
 
 +
===Questions===
  
 
<quiz display=simple>
 
<quiz display=simple>
{Multiple-Choice Frage
+
{Calculate the noise power-spectral density &nbsp;N_0&nbsp; with a noise factor of  &nbsp;F = 10&nbsp; and &nbsp;θ = 290^\circ&nbsp; Kelvin.
|type="[]"}
+
|type="{}"}
- Falsch
+
N_0 \ = \ { 4 3% }  \ \cdot 10^{ -20 }\ \text{W/Hz}
+ Richtig
+
 
 +
{What is the maximum noise power&nbsp; (without bandwidth limits)?
 +
|type="{}"}
 +
N_{\rm max} \ = \ { 0.24 3% } $\ \cdot 10^{ -6 }\ \text{W/Hz}$
 +
 
  
 +
{What is the noise power &nbsp;N&nbsp;  with bandwidth &nbsp;B = 30\text{ kHz}?&nbsp; What is the standard deviation σ_n?
 +
|type="{}"}
 +
N \ = \ { 12 3% }  \ \cdot 10^{ -16 }\ \text{W/Hz}
 +
σ_n \ = \ { 0.245 3% } \ \cdot 10^{ -6 }\ \text{V}
 +
 +
{Which of the signals &ndash; &nbsp;n_1(t)&nbsp; or &nbsp;n_2(t)&nbsp; &ndash;  shows low-pass noise and which shows band-pass noise?
 +
|type="()"}
 +
+ The noise signal &nbsp;n_1(t)&nbsp; is characteristically low-pass.
 +
- The noise signal &nbsp;n_1(t)&nbsp; is characteristically band-pass.
  
{Input-Box Frage
+
{What is the value of the noise power-spectral density of the low-pass&nbsp; \rm (LP)&nbsp; noise at frequency &nbsp;f = 20\text{ kHz}?&nbsp; Let &nbsp;B = 30\text{ kHz}.
 
|type="{}"}
 
|type="{}"}
$\alpha$ = { 0.3 }
+
${\it Φ}_{n, \hspace{0.05cm}\rm LP}(f = 20 \ \rm kHz) \ = \ $ { 2 3% } \ \cdot 10^{ -12 }\ \text{W/Hz}
  
 +
{What is the value of the noise power-spectral density of the band-pass&nbsp; \rm (BP)&nbsp; noise at  &nbsp;f = 120\text{ kHz}?&nbsp; Let &nbsp;f_{\rm M} = 100\text{ kHz}&nbsp; and &nbsp;B = 30\text{ kHz}.
 +
|type="{}"}
 +
{\it Φ}_{n, \hspace{0.05cm}\rm BP}(f = 120 \ \rm kHz) \ = \ { 0. } \ \cdot 10^{ -12 }\ \text{W/Hz}
  
  
 
</quiz>
 
</quiz>
  
===Musterlösung===
+
===Solution===
 
{{ML-Kopf}}
 
{{ML-Kopf}}
'''1.'''
+
'''(1)'''&nbsp; Using the Boltzmann constant&nbsp; k_{\rm B}&nbsp; it holds that:
'''2.'''
+
:$$N_0 = F \cdot k_{\rm B} \cdot \theta = 10 \cdot
'''3.'''
+
1.38\hspace{0.05cm}\cdot 10^{-23} \hspace{0.05cm}\frac{\rm
'''4.'''
+
Ws}{\rm K}\cdot 290\,{\rm K} \hspace{0.15cm}\underline {\approx 4\hspace{0.05cm}\cdot
'''5.'''
+
10^{-20} \hspace{0.05cm}{\rm W}/{\rm Hz}}\hspace{0.05cm}.$$
'''6.'''
+
 
'''7.'''
+
 
 +
'''(2)'''&nbsp; The specified noise power density&nbsp; N_0&nbsp; is physically limited to&nbsp; 6&nbsp; THz.&nbsp;  Thus the maximum noise power is:
 +
:$$N_{\rm max} =  4\hspace{0.05cm}\cdot 10^{-20}
 +
\hspace{0.08cm}\frac{\rm W}{\rm Hz}\cdot 6 \cdot10^{12}
 +
\hspace{0.08cm}{\rm Hz}\hspace{0.15cm}\underline {= 0.24\hspace{0.08cm}\cdot 10^{-6}\;{\rm
 +
W}}\hspace{0.05cm}.$$
 +
 
 +
 
 +
'''(3)'''&nbsp; Now the resulting noise power is:
 +
:$$N = N_0 \cdot B =  4\hspace{0.08cm}\cdot 10^{-20}
 +
\hspace{0.08cm}\frac{\rm W}{\rm Hz}\cdot 3 \cdot10^{4}
 +
\hspace{0.08cm}{\rm Hz}\hspace{0.15cm}\underline {= 12\hspace{0.05cm}\cdot 10^{-16}\;{\rm
 +
W}}\hspace{0.05cm}.$$
 +
 
 +
* Converting to the reference resistance&nbsp; R = 1 \ Ω:
 +
:$$N = N_0 \cdot B \cdot R = 12\hspace{0.05cm}\cdot 10^{-16}\;{\rm
 +
W}\hspace{0.05cm} \cdot 50 \; {\rm \Omega}= 6\hspace{0.05cm}\cdot
 +
10^{-14}\;{\rm V^2}\hspace{0.05cm}.$$
 +
[[File:EN_Mod_Z_1_3_e.png|rechts|frame|Power-spectral densities with band-limited noise]]
 +
*The noise standard deviation σ_n&nbsp; is the square root of this:
 +
:\sigma_n= \sqrt{6\hspace{0.05cm}\cdot 10^{-14}\;{\rm V^2}} \hspace{0.15cm}\underline {= 0.245 \hspace{0.05cm}\cdot 10^{-6}\;{\rm V}}\hspace{0.05cm}.
 +
 
 +
 
 +
'''(4)'''&nbsp; <u>Answer 1</u>&nbsp; is correct:
 +
*In the random signal&nbsp; n_2(t)&nbsp;one can recognize certain regularities similar to a harmonic oscillation &nbsp; &rArr; &nbsp;  it is bandpass noise.
 +
*In contrast, the signal&nbsp; n_1(t)&nbsp; is low-pass noise.
 +
 
 +
 
 +
'''(5)'''&nbsp; The noise power density of the random signal&nbsp; n_1(t)&nbsp; is constant in the frequency range&nbsp; |f| < 30&nbsp; kHz:
 +
:$${\it \Phi}_{n,\hspace{0.05cm}{  \rm LP} }(f) \hspace{-0.05cm}=\hspace{-0.05cm} \frac{N_0}{2}  \hspace{0.15cm}\underline {=2\hspace{0.05cm}\hspace{-0.05cm}\cdot \hspace{-0.05cm}
 +
10^{-12} \hspace{0.05cm}{\rm W}/{\rm Hz}}\hspace{0.05cm}.$$
 +
*Thus,&nbsp; this value is also valid for the frequency&nbsp; f = 20&nbsp; kHz.
 +
 
 +
 
 +
'''(6)'''&nbsp; As can be seen from the diagram, &nbsp; {\it Φ}_{n, \hspace{0.05cm}\rm BP}(f)&nbsp; is non-zero only in the range between&nbsp; 85&nbsp; kHz and&nbsp; 115&nbsp; kHz,&nbsp; when the bandwidth is&nbsp; B = 30&nbsp; kHz.
 +
*Thus,&nbsp; at the frequency &nbsp; f = 120&nbsp; kHz, the noise power density is zero:
 +
:$${\it Φ}_{n, \hspace{0.05cm}\rm BP}(f = 120 \ \rm kHz)\hspace{0.15cm}\underline{=0}.$$
 +
 
 
{{ML-Fuß}}
 
{{ML-Fuß}}
  
  
  
[[Category:Aufgaben zu Modulationsverfahren|^1.2 Qualitätskriterien^]]
+
[[Category:Modulation Methods: Exercises|^1.2 Quality Criteria^]]

Latest revision as of 18:27, 10 April 2022

Examplary signals for low-pass and band-pass noise

A fundamental disturbance and one that occurs in any communication system is  "thermal noise",  since any resistance  R  with an absolute temperature of  θ  (in "degrees Kelvin")  produces a noise signal  n(t)  with a  (one-sided)  noise power-spectral density

{N_{\rm 0, \hspace{0.05cm}min}}= k_{\rm B} \cdot \theta \hspace{0.3cm}\left(k_{\rm B} = 1.38 \cdot 10^{-23} \hspace{0.05cm}{\rm Ws}/{\rm K}\right)

where k_{\rm B}  denotes the  "Boltzmann constant" (from German  "Konstante").

However,  this noise is limited to  6\text{ THz}  for physical reasons.  Furthermore,  it can be observed that this minimum value can only be achieved with exact impedance matching.

In the realization of a circuit unit - for example, an amplifier - the effective noise power-spectral density is usually significantly greater,  since several noise sources add up,  and mismatches also play a role.  This effect is captured by the noise factor  F \ge 1  .  It holds that:

N_0 = F \cdot {N_{\rm 0, \hspace{0.05cm}min}}= F \cdot k_{\rm B} \cdot \theta \hspace{0.05cm}.

With a bandwidth  B,  the effective noise power is characterized by:

N = N_0 \cdot B \hspace{0.1cm} \hspace{0.01cm}.
N = N_0 \cdot B\cdot R = \sigma_n^2 \hspace{0.01cm}.
  • According to the first equation,  the result is the actual,  physical power in "watts"  \rm (W).
  • According to the second equation,  the result has the unit   "\rm V^{ 2 }".
  • This means that the power is here converted to the reference resistance  R = 1\ Ω  – as is often the case in Communications Engineering.
  • This equation must also be used to calculate the standard deviation  σ_n  of the noise signal  n(t) .


All equations apply regardless of whether the noise is low-pass or band-pass. The diagram shows two noise signals  n_1(t)  and  n_2(t)  of equal bandwidth.  Question   (4)  asks which of these signals will appear at the output of a low-pass and a band-pass, respectively.

The two-sided noise power-spectral density of band-limited low-pass noise  n_{\rm TP}(t)  is:

{\it \Phi}_{n, {\hspace{0.05cm}\rm LP}}(f) = \left\{ \begin{array}{c} N_0/2 \\ 0 \\ \end{array} \right. \begin{array}{*{10}c} {\rm{for}} \\ \\ \end{array}\begin{array}{*{20}c} {\left| \hspace{0.005cm} f\hspace{0.05cm} \right| < B,} \\ {\rm otherwise.} \\ \end{array}

In contrast,  for band-pass noise  n_{\rm BP}(t)  with center frequency  f_{\rm M}, it holds that:

{\it \Phi}_{n, {\hspace{0.05cm}\rm BP}}(f) = \left\{ \begin{array}{c} N_0/2 \\ 0 \\ \end{array} \right. \begin{array}{*{10}c} {\rm{for}} \\ \\ \end{array}\begin{array}{*{20}c} {\left| \hspace{0.005cm} f - f_{\rm M}\hspace{0.05cm} \right| < B/2,} \\ {\rm otherwise.} \\ \end{array}.

For all subsequent numerical calculations it is assumed:

F = 10, \hspace{0.2cm}\theta = 290\,{\rm K},\hspace{0.2cm}R = 50\,{\rm \Omega},\hspace{0.2cm}B = 30\,{\rm kHz},\hspace{0.2cm}f_{\rm M} = 0 \hspace{0.5cm}{\rm or}\hspace{0.5cm}f_{\rm M} =100\,{\rm kHz}\hspace{0.05cm}.



Hints:

  • This exercise belongs to the chapter  Quality Criteria.
  • Particular reference is made to the page  Some remarks on the AWGN channel model.
  • By specifying the powers in  \rm Watts , they are independent of the reference resistance  R,  while power with the unit  \rm V^2  can only be evaluated directly for  R = 1\ \Omega.



Questions

1

Calculate the noise power-spectral density  N_0  with a noise factor of  F = 10  and  θ = 290^\circ  Kelvin.

N_0 \ = \

\ \cdot 10^{ -20 }\ \text{W/Hz}

2

What is the maximum noise power  (without bandwidth limits)?

N_{\rm max} \ = \

\ \cdot 10^{ -6 }\ \text{W/Hz}

3

What is the noise power  N  with bandwidth  B = 30\text{ kHz}?  What is the standard deviation σ_n?

N \ = \

\ \cdot 10^{ -16 }\ \text{W/Hz}
σ_n \ = \

\ \cdot 10^{ -6 }\ \text{V}

4

Which of the signals –  n_1(t)  or  n_2(t)  – shows low-pass noise and which shows band-pass noise?

The noise signal  n_1(t)  is characteristically low-pass.
The noise signal  n_1(t)  is characteristically band-pass.

5

What is the value of the noise power-spectral density of the low-pass  \rm (LP)  noise at frequency  f = 20\text{ kHz}?  Let  B = 30\text{ kHz}.

{\it Φ}_{n, \hspace{0.05cm}\rm LP}(f = 20 \ \rm kHz) \ = \

\ \cdot 10^{ -12 }\ \text{W/Hz}

6

What is the value of the noise power-spectral density of the band-pass  \rm (BP)  noise at  f = 120\text{ kHz}?  Let  f_{\rm M} = 100\text{ kHz}  and  B = 30\text{ kHz}.

{\it Φ}_{n, \hspace{0.05cm}\rm BP}(f = 120 \ \rm kHz) \ = \

\ \cdot 10^{ -12 }\ \text{W/Hz}


Solution

(1)  Using the Boltzmann constant  k_{\rm B}  it holds that:

N_0 = F \cdot k_{\rm B} \cdot \theta = 10 \cdot 1.38\hspace{0.05cm}\cdot 10^{-23} \hspace{0.05cm}\frac{\rm Ws}{\rm K}\cdot 290\,{\rm K} \hspace{0.15cm}\underline {\approx 4\hspace{0.05cm}\cdot 10^{-20} \hspace{0.05cm}{\rm W}/{\rm Hz}}\hspace{0.05cm}.


(2)  The specified noise power density  N_0  is physically limited to  6  THz.  Thus the maximum noise power is:

N_{\rm max} = 4\hspace{0.05cm}\cdot 10^{-20} \hspace{0.08cm}\frac{\rm W}{\rm Hz}\cdot 6 \cdot10^{12} \hspace{0.08cm}{\rm Hz}\hspace{0.15cm}\underline {= 0.24\hspace{0.08cm}\cdot 10^{-6}\;{\rm W}}\hspace{0.05cm}.


(3)  Now the resulting noise power is:

N = N_0 \cdot B = 4\hspace{0.08cm}\cdot 10^{-20} \hspace{0.08cm}\frac{\rm W}{\rm Hz}\cdot 3 \cdot10^{4} \hspace{0.08cm}{\rm Hz}\hspace{0.15cm}\underline {= 12\hspace{0.05cm}\cdot 10^{-16}\;{\rm W}}\hspace{0.05cm}.
  • Converting to the reference resistance  R = 1 \ Ω:
N = N_0 \cdot B \cdot R = 12\hspace{0.05cm}\cdot 10^{-16}\;{\rm W}\hspace{0.05cm} \cdot 50 \; {\rm \Omega}= 6\hspace{0.05cm}\cdot 10^{-14}\;{\rm V^2}\hspace{0.05cm}.
Power-spectral densities with band-limited noise
  • The noise standard deviation σ_n  is the square root of this:
\sigma_n= \sqrt{6\hspace{0.05cm}\cdot 10^{-14}\;{\rm V^2}} \hspace{0.15cm}\underline {= 0.245 \hspace{0.05cm}\cdot 10^{-6}\;{\rm V}}\hspace{0.05cm}.


(4)  Answer 1  is correct:

  • In the random signal  n_2(t) one can recognize certain regularities similar to a harmonic oscillation   ⇒   it is bandpass noise.
  • In contrast, the signal  n_1(t)  is low-pass noise.


(5)  The noise power density of the random signal  n_1(t)  is constant in the frequency range  |f| < 30  kHz:

{\it \Phi}_{n,\hspace{0.05cm}{ \rm LP} }(f) \hspace{-0.05cm}=\hspace{-0.05cm} \frac{N_0}{2} \hspace{0.15cm}\underline {=2\hspace{0.05cm}\hspace{-0.05cm}\cdot \hspace{-0.05cm} 10^{-12} \hspace{0.05cm}{\rm W}/{\rm Hz}}\hspace{0.05cm}.
  • Thus,  this value is also valid for the frequency  f = 20  kHz.


(6)  As can be seen from the diagram,   {\it Φ}_{n, \hspace{0.05cm}\rm BP}(f)  is non-zero only in the range between  85  kHz and  115  kHz,  when the bandwidth is  B = 30  kHz.

  • Thus,  at the frequency   f = 120  kHz, the noise power density is zero:
{\it Φ}_{n, \hspace{0.05cm}\rm BP}(f = 120 \ \rm kHz)\hspace{0.15cm}\underline{=0}.