Difference between revisions of "Aufgaben:Exercise 4.2: Low-Pass for Signal Reconstruction"

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[[File:|right|]]
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[[File:EN_Mod_A_4_2.png|right|frame|Examples of continuous and discrete spectra]]
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We consider in this exercise two different source signals&nbsp; $q_{\rm con}(t)$&nbsp; and&nbsp; $q_{\rm dis}(t)$ whose magnitude spectra&nbsp; $|Q_{\rm con}(f)|$&nbsp; and&nbsp; $|Q_{\rm dis}(f)|$&nbsp; are plotted. &nbsp; The highest frequency occurring in the signals is in each case&nbsp; $4 \rm kHz$.
 +
* Nothing more is known of the spectral function&nbsp; $Q_{\rm con}(f)$&nbsp; than that it is a continuous spectrum,&nbsp; where:
 +
:$$Q_{\rm con}(|f| \le 4\,{\rm kHz}) \ne 0 \hspace{0.05cm}.$$
 +
* The spectrum&nbsp; $Q_{\rm dis}(f)$&nbsp; contains spectral lines at&nbsp; $±1 \ \rm kHz$,&nbsp; $±2 \ \rm kHz$,&nbsp; $±3 \ \rm kHz$&nbsp; and&nbsp; $±4 \ \rm kHz$.&nbsp; Thus:
 +
:$$q_{\rm dis}(t) = \sum_{i=1}^{4}C_i \cdot \cos (2 \pi \cdot f_i \cdot t - \varphi_i),$$
 +
:Amplitude values: &nbsp; $C_1 = 1.0 \ \rm V$, $C_2 = 1.8 \ \rm V$, $C_3 = 0.8 \ \rm V$, $C_4 = 0.4 \ \rm V.$
  
 +
:The phase values&nbsp; $φ_1$,&nbsp; $φ_2$&nbsp; and&nbsp; $φ_3$&nbsp; are respectively in the range&nbsp; $±180^\circ$&nbsp; and it holds&nbsp; $φ_4 = 90^\circ$.
  
===Fragebogen===
+
 
 +
The signals are each sampled at frequency&nbsp; $f_{\rm A}$&nbsp; and immediately fed to an ideal rectangular low-pass filter with cutoff frequency&nbsp; $f_{\rm G}$&nbsp; This scenario applies,&nbsp; for example,&nbsp; to
 +
* the interference-free pulse amplitude modulation&nbsp; $\rm (PAM)$&nbsp; and
 +
* the interference-free pulse code modulation&nbsp; $\rm (PCM)$&nbsp; at infinitely large quantization stage number&nbsp; $M$.
 +
 
 +
 
 +
The output signal of the&nbsp; (rectangular)&nbsp; low-pass filter is called the sink signal&nbsp; $v(t)$&nbsp; and for the error signal:&nbsp;
 +
:$$ε(t) = v(t) - q(t).$$
 +
This is different from zero only if the parameters of the sampling&nbsp; $($sampling frequency $f_{\rm A})$&nbsp; and/or the signal reconstruction&nbsp; $($cutoff frequency $f_{\rm G})$&nbsp; are not dimensioned in the best possible way.
 +
 
 +
 
 +
 
 +
 
 +
 
 +
Hints:
 +
*The exercise belongs to the chapter&nbsp; [[Modulation_Methods/Pulse_Code_Modulation|"Pulse Code Modulation"]].
 +
*Reference is made in particular to the page&nbsp; [[Modulation_Methods/Pulse_Code_Modulation#Sampling_and_signal_reconstruction|"Sampling and Signal Reconstruction"]].
 +
 +
 
 +
 
 +
===Questions===
  
 
<quiz display=simple>
 
<quiz display=simple>
{Multiple-Choice Frage
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{Which statements are true for &nbsp;$f_{\rm A} = 8\ \rm kHz$&nbsp; and &nbsp;$f_{\rm G} = 4\ \rm kHz$&nbsp;?
 
|type="[]"}
 
|type="[]"}
- Falsch
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+ The signal &nbsp;$q_{\rm con}(t)$&nbsp; can be completely reconstructed: &nbsp; $ε_{\rm con}(t) = 0$.
+ Richtig
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- The signal &nbsp;$q_{\rm dis}(t)$&nbsp; can be completely reconstructed: &nbsp; $ε_{\rm dis}(t) = 0$.
  
  
{Input-Box Frage
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{Which statements are true for &nbsp;$f_{\rm A} = 10\ \rm kHz$&nbsp; and &nbsp;$f_{\rm G} = 5\ \rm kHz$&nbsp;?
|type="{}"}
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|type="()"}
$\alpha$ = { 0.3 }
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+ The signal &nbsp;$q_{\rm dis}(t)$&nbsp; can be completely reconstructed: &nbsp; $ε_{\rm dis}(t) = 0$.
 +
- $ε_{\rm dis}(t)$&nbsp; is a harmonic oscillation with &nbsp;$4 \rm kHz$.
 +
- $ε_{\rm dis}(t)$&nbsp; is a harmonic oscillation with &nbsp;$6 \rm kHz$.
  
 +
{Which statements are true for &nbsp;$f_{\rm A} = 10\ \rm kHz$&nbsp; and &nbsp;$f_{\rm G} = 3.5\ \rm kHz$&nbsp;?
 +
|type="()"}
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- The signal &nbsp;$q_{\rm dis}(t)$&nbsp; can be completely reconstructed: &nbsp; $ε_{\rm dis}(t) = 0$.
 +
+ $ε_{\rm dis}(t)$&nbsp; is a harmonic oscillation with &nbsp;$4 \rm kHz$.
 +
- $ε_{\rm dis}(t)$&nbsp; is a harmonic oscillation with &nbsp;$6 \rm kHz$.
  
 +
{Which statements are true for &nbsp;$f_{\rm A} = 10\ \rm kHz$&nbsp; and &nbsp;$f_{\rm G} = 6.5\ \rm kHz$?
 +
|type="()"}
 +
- The signal &nbsp;$q_{\rm dis}(t)$&nbsp; can be completely reconstructed: &nbsp; $ε_{\rm dis}(t) = 0$.
 +
- $ε_{\rm dis}(t)$&nbsp; is a harmonic oscillation with &nbsp;$4 \rm kHz$.
 +
+ $ε_{\rm dis}(t)$&nbsp; is a harmonic oscillation with &nbsp;$6 \ \rm kHz$.
  
 
</quiz>
 
</quiz>
  
===Musterlösung===
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===Solution===
 
{{ML-Kopf}}
 
{{ML-Kopf}}
'''1.'''
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'''(1)'''&nbsp; Only the&nbsp; <u>first statement</u>&nbsp; is correct:
'''2.'''
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*Sampling&nbsp; $q_{\rm dis}(t)$&nbsp; with sampling frequency&nbsp; $f_{\rm A} = 8 \ \rm kHz$&nbsp; leads to an irreversible error,&nbsp; since&nbsp; $Q_{\rm dis}(f)$&nbsp; involves a discrete spectral component&nbsp; (Dirac delta line)&nbsp; at&nbsp; $f_4 = 4\ \rm kHz$&nbsp; and the phase value is&nbsp; $φ_4 ≠ 0$.
'''3.'''
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*With the phase value&nbsp; $φ_4 = 90^\circ$&nbsp; $(4 \ \rm kHz$&nbsp; sinusoidal component$)$&nbsp;  given here holds&nbsp; $ε_{\rm dis}(t) = v_{\rm dis}(t) - q_{\rm dis}(t) = -0. 4 \ \rm V - \sin(2π \cdot f_4 \cdot t)$.&nbsp; See also solution to Exercise 4.2Z.
'''4.'''
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*On the other hand,&nbsp; the signal&nbsp; $q_{\rm con}(t)$&nbsp; with the continuous spectrum&nbsp; $Q_{\rm con}(f)$&nbsp; can also then be measured with a rectangular low-pass filter&nbsp; $($with cutoff frequency&nbsp; $f_{\rm G} = 4\ \rm kHz)$&nbsp; be completely reconstructed if sampling frequency&nbsp; $f_{\rm A} = 8\ \rm kHz$&nbsp; was used. &nbsp; For all frequencies not equal to&nbsp; $f_4$&nbsp; the sampling theorem is satisfied.
'''5.'''
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*The contribution of the&nbsp; $f_4$ component to the total spectrum&nbsp; $Q_{\rm con}(f)$&nbsp; is only vanishingly small &nbsp; ⇒ &nbsp; ${\rm Pr}(f_4) → 0$&nbsp; as long as the spectrum has no Dirac delta line  at&nbsp; $f_4$.
'''6.'''
+
 
'''7.'''
+
 
 +
 
 +
'''(2)'''&nbsp; Only the&nbsp; <u>proposed solution 1</u>&nbsp; is correct:
 +
*With&nbsp; $f_{\rm A} = 10\ \rm kHz$&nbsp; the sampling theorem is satisfied in both cases.
 +
*With&nbsp; $f_{\rm G} = f_{\rm A} /2$&nbsp; both error signals&nbsp; $ε_{\rm con}(t)$&nbsp; and&nbsp; $ε_{\rm dis}(t)$&nbsp; are identically zero.
 +
*In addition,&nbsp; the signal reconstruction also works as long as&nbsp; $f_{\rm G} > 4 \ \rm kHz$&nbsp; and&nbsp; $f_{\rm G} < 6 \ \rm kHz$&nbsp; holds.
 +
 
 +
 
 +
 
 +
'''(3)'''&nbsp; The correct solution here is&nbsp; <u>suggested solution 2</u>:
 +
*With&nbsp; $f_{\rm G} = 3.5 \ \rm kHz$&nbsp; the low-pass incorrectly removes the&nbsp; $4\ \rm kHz$ component,&nbsp; that is,&nbsp; then holds:
 +
:$$ v_{\rm dis}(t) = q_{\rm dis}(t) - 0.4\,{\rm V} \cdot \sin (2 \pi \cdot f_{\rm 4} \cdot t)\hspace{0.3cm}\Rightarrow \hspace{0.3cm} \varepsilon_{\rm dis}(t) = - 0.4\,{\rm V} \cdot \sin (2 \pi \cdot f_{\rm 4} \cdot t)\hspace{0.05cm}.$$
 +
 
 +
 
 +
 
 +
[[File:EN_Mod_A_4_2d_neu.png|P_ID1609__Mod_A_4_2d.png|right|frame|Signal reconstruction with too large cutoff frequency]]
 +
'''(4)'''&nbsp; The correct solution here is&nbsp; <u>suggested solution 3</u>:
 +
*Sampling with&nbsp; $f_{\rm A} = 10\ \rm kHz$&nbsp; yields the periodic spectrum sketched on the right.
 +
*The low-pass with&nbsp; $f_{\rm G} = 6.5 \ \rm kHz$&nbsp; removes all discrete frequency components with&nbsp; $|f| ≥ 7\ \rm kHz$,&nbsp; but not the&nbsp; $6\ \rm kHz$ component.  
 +
 
 +
 
 +
The error signal&nbsp; $ε_{\rm dis}(t) = v_{\rm dis}(t) - q_{\rm dis}(t)$&nbsp; is then a harmonic oscillation with
 +
* the frequency&nbsp; $f_6 = f_{\rm A} - f_4 = 6\ \rm kHz$,
 +
* the amplitude&nbsp; $A_4$ of the&nbsp; $f_4$ component,
 +
* the phase&nbsp; $φ_{-4} = -φ_4$&nbsp; of the&nbsp; $Q(f)$&nbsp; component at&nbsp; $f = -f_4$.
 +
 
 
{{ML-Fuß}}
 
{{ML-Fuß}}
  
  
  
[[Category:Aufgaben zu Modulationsverfahren|^4.1 Pulscodemodulation^]]
+
[[Category:Modulation Methods: Exercises|^4.1 Pulse Code Modulation^]]

Latest revision as of 16:31, 18 January 2023

Examples of continuous and discrete spectra

We consider in this exercise two different source signals  $q_{\rm con}(t)$  and  $q_{\rm dis}(t)$ whose magnitude spectra  $|Q_{\rm con}(f)|$  and  $|Q_{\rm dis}(f)|$  are plotted.   The highest frequency occurring in the signals is in each case  $4 \rm kHz$.

  • Nothing more is known of the spectral function  $Q_{\rm con}(f)$  than that it is a continuous spectrum,  where:
$$Q_{\rm con}(|f| \le 4\,{\rm kHz}) \ne 0 \hspace{0.05cm}.$$
  • The spectrum  $Q_{\rm dis}(f)$  contains spectral lines at  $±1 \ \rm kHz$,  $±2 \ \rm kHz$,  $±3 \ \rm kHz$  and  $±4 \ \rm kHz$.  Thus:
$$q_{\rm dis}(t) = \sum_{i=1}^{4}C_i \cdot \cos (2 \pi \cdot f_i \cdot t - \varphi_i),$$
Amplitude values:   $C_1 = 1.0 \ \rm V$, $C_2 = 1.8 \ \rm V$, $C_3 = 0.8 \ \rm V$, $C_4 = 0.4 \ \rm V.$
The phase values  $φ_1$,  $φ_2$  and  $φ_3$  are respectively in the range  $±180^\circ$  and it holds  $φ_4 = 90^\circ$.


The signals are each sampled at frequency  $f_{\rm A}$  and immediately fed to an ideal rectangular low-pass filter with cutoff frequency  $f_{\rm G}$  This scenario applies,  for example,  to

  • the interference-free pulse amplitude modulation  $\rm (PAM)$  and
  • the interference-free pulse code modulation  $\rm (PCM)$  at infinitely large quantization stage number  $M$.


The output signal of the  (rectangular)  low-pass filter is called the sink signal  $v(t)$  and for the error signal: 

$$ε(t) = v(t) - q(t).$$

This is different from zero only if the parameters of the sampling  $($sampling frequency $f_{\rm A})$  and/or the signal reconstruction  $($cutoff frequency $f_{\rm G})$  are not dimensioned in the best possible way.



Hints:


Questions

1

Which statements are true for  $f_{\rm A} = 8\ \rm kHz$  and  $f_{\rm G} = 4\ \rm kHz$ ?

The signal  $q_{\rm con}(t)$  can be completely reconstructed:   $ε_{\rm con}(t) = 0$.
The signal  $q_{\rm dis}(t)$  can be completely reconstructed:   $ε_{\rm dis}(t) = 0$.

2

Which statements are true for  $f_{\rm A} = 10\ \rm kHz$  and  $f_{\rm G} = 5\ \rm kHz$ ?

The signal  $q_{\rm dis}(t)$  can be completely reconstructed:   $ε_{\rm dis}(t) = 0$.
$ε_{\rm dis}(t)$  is a harmonic oscillation with  $4 \rm kHz$.
$ε_{\rm dis}(t)$  is a harmonic oscillation with  $6 \rm kHz$.

3

Which statements are true for  $f_{\rm A} = 10\ \rm kHz$  and  $f_{\rm G} = 3.5\ \rm kHz$ ?

The signal  $q_{\rm dis}(t)$  can be completely reconstructed:   $ε_{\rm dis}(t) = 0$.
$ε_{\rm dis}(t)$  is a harmonic oscillation with  $4 \rm kHz$.
$ε_{\rm dis}(t)$  is a harmonic oscillation with  $6 \rm kHz$.

4

Which statements are true for  $f_{\rm A} = 10\ \rm kHz$  and  $f_{\rm G} = 6.5\ \rm kHz$?

The signal  $q_{\rm dis}(t)$  can be completely reconstructed:   $ε_{\rm dis}(t) = 0$.
$ε_{\rm dis}(t)$  is a harmonic oscillation with  $4 \rm kHz$.
$ε_{\rm dis}(t)$  is a harmonic oscillation with  $6 \ \rm kHz$.


Solution

(1)  Only the  first statement  is correct:

  • Sampling  $q_{\rm dis}(t)$  with sampling frequency  $f_{\rm A} = 8 \ \rm kHz$  leads to an irreversible error,  since  $Q_{\rm dis}(f)$  involves a discrete spectral component  (Dirac delta line)  at  $f_4 = 4\ \rm kHz$  and the phase value is  $φ_4 ≠ 0$.
  • With the phase value  $φ_4 = 90^\circ$  $(4 \ \rm kHz$  sinusoidal component$)$  given here holds  $ε_{\rm dis}(t) = v_{\rm dis}(t) - q_{\rm dis}(t) = -0. 4 \ \rm V - \sin(2π \cdot f_4 \cdot t)$.  See also solution to Exercise 4.2Z.
  • On the other hand,  the signal  $q_{\rm con}(t)$  with the continuous spectrum  $Q_{\rm con}(f)$  can also then be measured with a rectangular low-pass filter  $($with cutoff frequency  $f_{\rm G} = 4\ \rm kHz)$  be completely reconstructed if sampling frequency  $f_{\rm A} = 8\ \rm kHz$  was used.   For all frequencies not equal to  $f_4$  the sampling theorem is satisfied.
  • The contribution of the  $f_4$ component to the total spectrum  $Q_{\rm con}(f)$  is only vanishingly small   ⇒   ${\rm Pr}(f_4) → 0$  as long as the spectrum has no Dirac delta line at  $f_4$.


(2)  Only the  proposed solution 1  is correct:

  • With  $f_{\rm A} = 10\ \rm kHz$  the sampling theorem is satisfied in both cases.
  • With  $f_{\rm G} = f_{\rm A} /2$  both error signals  $ε_{\rm con}(t)$  and  $ε_{\rm dis}(t)$  are identically zero.
  • In addition,  the signal reconstruction also works as long as  $f_{\rm G} > 4 \ \rm kHz$  and  $f_{\rm G} < 6 \ \rm kHz$  holds.


(3)  The correct solution here is  suggested solution 2:

  • With  $f_{\rm G} = 3.5 \ \rm kHz$  the low-pass incorrectly removes the  $4\ \rm kHz$ component,  that is,  then holds:
$$ v_{\rm dis}(t) = q_{\rm dis}(t) - 0.4\,{\rm V} \cdot \sin (2 \pi \cdot f_{\rm 4} \cdot t)\hspace{0.3cm}\Rightarrow \hspace{0.3cm} \varepsilon_{\rm dis}(t) = - 0.4\,{\rm V} \cdot \sin (2 \pi \cdot f_{\rm 4} \cdot t)\hspace{0.05cm}.$$


Signal reconstruction with too large cutoff frequency

(4)  The correct solution here is  suggested solution 3:

  • Sampling with  $f_{\rm A} = 10\ \rm kHz$  yields the periodic spectrum sketched on the right.
  • The low-pass with  $f_{\rm G} = 6.5 \ \rm kHz$  removes all discrete frequency components with  $|f| ≥ 7\ \rm kHz$,  but not the  $6\ \rm kHz$ component.


The error signal  $ε_{\rm dis}(t) = v_{\rm dis}(t) - q_{\rm dis}(t)$  is then a harmonic oscillation with

  • the frequency  $f_6 = f_{\rm A} - f_4 = 6\ \rm kHz$,
  • the amplitude  $A_4$ of the  $f_4$ component,
  • the phase  $φ_{-4} = -φ_4$  of the  $Q(f)$  component at  $f = -f_4$.