View source for Aufgaben:Exercise 4.4Z: Signal-to-Noise Ratio with PCM


{{quiz-Header|Buchseite=Modulation_Methods/Pulse_Code_Modulation
}}

[[File:EN_Mod_Z_4_4.png|right|frame|Signal-to-noise ratio of PCM 30/32 compared to ZSB amplitude modulation]]
The graph shows the signal-to-noise ratio &nbsp;$10 ⋅ \lg \ ρ_v$&nbsp; for pulse code modulation&nbsp; $\rm (PCM)$&nbsp; compared to analog double-sideband amplitude modulation, abbreviated as&nbsp; "$\rm DSB-AM$".&nbsp; 

For the latter,&nbsp; &nbsp;$ρ_v = ξ$,&nbsp; where the persormanc parameter
:$$\xi = \frac{\alpha^2 \cdot P_{\rm S}}{N_0 \cdot f_{\rm N}} $$
summarizes the following system parameters:
:* the frequency-independent attenuation factor &nbsp;$α$&nbsp; of the transmission channel,
:* the power &nbsp;$P_{\rm S}$&nbsp; of the transmitted signal &nbsp;$s(t)$,&nbsp, also called&nbsp; "transmit power"&nbsp; for short,
:* the message frequency &nbsp;$f_{\rm N}$&nbsp; (bandwidth)&nbsp; of the cosine source signal &nbsp;$q(t)$,
:* the noise power density &nbsp;$N_0$&nbsp; of the AWGN noise.


For the PCM system,&nbsp; the following approximation for the sink SNR was given in the section&nbsp; [[Modulation_Methods/Pulse_Code_Modulation#Estimation_of_SNR_degradation_due_to_transmission_errors|"Estimation of SNR degradation due to bit errors"]],&nbsp; which also takes into account transmission errors due to AWGN noise:
:$$ \rho_{\upsilon}= \frac{1}{ 2^{-2N } + 4 \cdot p_{\rm B}} \hspace{0.05cm}.$$
*Here &nbsp;$N$&nbsp; denotes the number of bits per sample and &nbsp;$p_{\rm B}$&nbsp; the bit error probability.
* Since &nbsp;$ξ$&nbsp; can in digital modulation also be interpreted as the&nbsp; "signal energy per bit"&nbsp; related to the&nbsp; "noise power density"&nbsp; $(E_{\rm B}/N_0)$,&nbsp; with the complementary Gaussian error signal &nbsp;${\rm Q}(x)$&nbsp; approximately:
:$$ p_{\rm B}= {\rm Q} \left ( \sqrt{2 \xi }\right ) \hspace{0.05cm}.$$





Hints:
*The exercise belongs to the chapter&nbsp; [[Modulation_Methods/Pulse_Code_Modulation|"Pulse Code Modulation"]].
*Reference is made in particular to the sections&nbsp; [[Modulation_Methods/Pulse_Code_Modulation#Influence_of_transmission_errors|"Influence of transmission errors"]] &nbsp; and &nbsp; [[Modulation_Methods/Pulse_Code_Modulation#Estimation_of_SNR_degradation_due_to_transmission_errors|"Estimation of SNR degradation due to transmission errors"]].
*The PCM considered here is the&nbsp; '''PCM 30/32''',&nbsp; whose system parameters are given,&nbsp; e.g. in&nbsp; [[Aufgaben:Exercise_4.1:_PCM_System_30/32 |Exercise 4.1]]&nbsp;.
 


===Questions===

<quiz display=simple>
{How many bits per sample &nbsp; &rArr; &nbsp; $N = N_1$&nbsp; does the PCM system under consideration use?
|type="{}"}
$N_1 \ = \ $ { 8 3% } 

{How many bits per sample &nbsp; &rArr; &nbsp; $N = N_2$&nbsp; would have to be used to achieve &nbsp;$10 ⋅ \lg \ ρ_v > 64 \ \rm dB$&nbsp; ("music quality")?
|type="{}"}
$N_2 \ = \ $ { 11 3% } 

{What&nbsp; (logarithmized)&nbsp; performance parameter &nbsp;$ξ_{40\ \rm dB}$&nbsp; is required for 8-bit PCM to have a signal-to-noise ratio of &nbsp;$40\ \rm dB$&nbsp;?
|type="{}"}
$10 ⋅ \lg \ ξ_{40\ \rm dB} \ = \ $ { 10 3% } $\ \rm dB$ 

{By what factor could the transmit power be reduced for PCM compared to&nbsp; "DSB-AM"&nbsp; to still achieve &nbsp;$10 ⋅ \lg \ ρ_v = 40\ \rm dB$&nbsp;?
|type="{}"}
$K_\text{AM → PCM} \ = \ $ { 1000 3% } 

{What is the bit error probability &nbsp;$p_{\rm B}$&nbsp; for &nbsp;$10 ⋅ \lg \ ξ = 6\ \rm dB$&nbsp; and &nbsp;$N = N_1$ &nbsp; &rArr; &nbsp; result of the subtask&nbsp; '''(1)'''?
|type="{}"}
$p_{\rm B} \ = \ $ { 2.5 3% } $\ \%$

{What would be the SNR for the same &nbsp;$ξ$&nbsp; with a 3-bit PCM &nbsp; &rArr; &nbsp; $N = 3$&nbsp;?
|type="{}"}
$10 ⋅ \lg \ ρ_v \ = \ $ { 15.9 3% } $\ \rm dB$ 

</quiz>

===Solution===
{{ML-Kopf}}
'''(1)'''&nbsp; The horizontal section of the PCM curve is determined by the quantization noise alone.&nbsp; 
*Here,&nbsp; with the quantization step number&nbsp; $M = 2^N$:
:$$ \rho_{v} (\xi \rightarrow \infty) = \rho_{\rm Q} = M^2 = 2^{2N} \hspace{0.3cm}\rightarrow \hspace{0.3cm} 10 \cdot {\rm lg}\hspace{0.1cm}\rho_{v} \approx 6\,{\rm dB} \cdot N\hspace{0.05cm}.$$
*From the readable signal-to-noise ratio&nbsp; $10 ⋅ \lg \ ρ_v ≈ 48 \ \rm dB$&nbsp; it follows:<br>&nbsp; $N_1\hspace{0.15cm}\underline { = 8}$&nbsp; bits per sample and for the quantization level number&nbsp; $M = 256$.



'''(2)'''&nbsp; From the above approximation,&nbsp; we obtain for&nbsp; $N_2\hspace{0.15cm}\underline { = 11}$&nbsp; bits per sample &nbsp; ⇒ &nbsp; $M = 2048$&nbsp; the signal-to-noise ratio&nbsp; $66 \ \rm dB$. 
*With&nbsp; $N = 10$ &nbsp; ⇒ &nbsp; $M = 1024$&nbsp; one reaches only approx.&nbsp; $60 \ \rm dB$. 
*For the compact disc&nbsp; $\rm (CD)$,&nbsp; the PCM parameters&nbsp; $N = 16$ &nbsp; ⇒ &nbsp; $M = 65536$ &nbsp; ⇒ &nbsp; $10 ⋅ \lg \ ρ_v > 96 \ \rm dB$&nbsp; are used.



'''(3)'''&nbsp; For double-sideband amplitude modulation&nbsp; $\rm (DSB-AM)$,&nbsp; this would require&nbsp; $10 ⋅ \lg \ ξ = 40\ \rm dB$&nbsp;. 
*As can be seen from the graph in the data section,&nbsp; this abscissa value for the given PCM is lower by&nbsp; $30 \ \rm dB$&nbsp;⇒&nbsp; $10 ⋅ \lg \ ξ_{40\ \rm dB}\hspace{0.15cm}\underline { = 10 \ \rm dB}$. 



'''(4)'''&nbsp; The logarithmic value&nbsp; $30 \ \rm dB$&nbsp; corresponds to a power reduced by a factor&nbsp; $10^3\hspace{0.15cm}\underline { = 1000}$&nbsp; .



'''(5)'''&nbsp; From the graph in the information section,&nbsp; it can be seen that the abscissa value&nbsp; $10 ⋅ \lg \ ξ= 6 \ \rm dB$&nbsp; results in the signal-to-noise ratio&nbsp; $20 \ \rm dB$. 
*From&nbsp; $10 ⋅ \lg \ ρ_v = 20 \ \rm dB$&nbsp; follows&nbsp; $ρ_v = 100$&nbsp; and thus further&nbsp; $($with&nbsp; $N = N_1 = 8)$:
:$$\rho_{\upsilon}= \frac{1}{ 2^{-2N } + 4 \cdot p_{\rm B}} \approx \frac{1}{ 1.5 \cdot 10^{-5} + 4 \cdot p_{\rm B}} = 100 \hspace{0.3cm}
\Rightarrow \hspace{0.3cm} p_{\rm B} = \frac{0.01 - 1.5 \cdot 10^{-5}}{ 4} \hspace{0.15cm}\underline {\approx 2.5\%} \hspace{0.05cm}.$$


'''(6)'''&nbsp; With the same performance parameter&nbsp; $ξ$,&nbsp; the bit error probability is still&nbsp; $p_{\rm B} = 0.025$.&nbsp; Thus, with&nbsp; $N = 3$&nbsp; (bits per sample):
:$$\rho_{\upsilon}= \frac{1}{ 2^{-6 } + 4 \cdot p_{\rm B}} = \frac{1}{ 0.015625 + 0.01} \approx 39 \hspace{0.3cm} \Rightarrow \hspace{0.3cm}10 \cdot {\rm lg} \hspace{0.15cm}\rho_{\upsilon}\hspace{0.15cm}\underline {\approx 15.9\,{\rm dB}} \hspace{0.05cm}.$$
Further,&nbsp; it should be noted:
*With only three bits per sample,&nbsp; the quantization noise power&nbsp; $(P_{\rm Q} = 0.015625)$&nbsp; is already larger than the error noise power&nbsp; $(P_{\rm E} = 0.01)$. 
*By increasing the transmit power,&nbsp; the signal-to-noise ratio could be maximum&nbsp; $10 ⋅ \lg \ ρ_v =18 \ \rm dB$&nbsp; because of quantization, if no bit errors occur&nbsp; $(P_{\rm E} = 0)$.

{{ML-Fuß}}



[[Category:Modulation Methods: Exercises|^4.1 Pulse Code Modulation^]]