Difference between revisions of "Aufgaben:Exercise 2.7Z: Power-Spectral Density of Pseudo-Ternary Codes"
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− | {{quiz-Header|Buchseite= | + | {{quiz-Header|Buchseite=Digital_Signal_Transmission/Symbolwise_Coding_with_Pseudo-Ternary_Codes |
}} | }} | ||
− | [[File:P_ID1354__Dig_Z_2_7.png|right|frame| | + | [[File:P_ID1354__Dig_Z_2_7.png|right|frame|Power-spectral densities of <br>three different pseudo-ternary codes]] |
− | In | + | In the graph you can see the power-spectral densities $\rm (PSD)$ of three different pseudo-ternary codes, which result from the general description according to [[Aufgaben:Exercise_2.7:_AMI_Code|"Exercise 2.7"]] by different values of the parameters $N_{\rm C}$ and $K_{\rm C}$. |
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
+ | In different colors the power-spectral densities | ||
+ | :$${\it \Phi}_s(f) = \ \frac{s_0^2 \cdot T}{2} \cdot {\rm sinc}^2 (f T) \cdot \big [1 - K_{\rm C} \cdot \cos (2\pi f N_{\rm C} T)\big ]$$ | ||
+ | are shown for the following variants: | ||
+ | *AMI code $(N_{\rm C} = 1,\ K_{\rm C} = +1)$, | ||
+ | *duobinary code $(N_{\rm C} = 1,\ K_{\rm C} = -1)$, | ||
+ | *second order bipolar code $ (N_{\rm C} = 2,\ K_{\rm C} = +1)$. | ||
− | |||
− | + | The above PSD equation assumes the use of rectangular NRZ basic transmission pulses. | |
− | + | ||
− | [[ | + | All pseudo-ternary codes considered here have the same probability distribution: |
− | === | + | :$${\rm Pr}\big[s(t) = 0\big]= {1}/{2},\hspace{0.2cm}{\rm Pr}\big[s(t) = +s_0\big]= {\rm Pr}\big[s(t) = -s_0\big]={1}/{4}\hspace{0.05cm}.$$ |
+ | |||
+ | |||
+ | |||
+ | |||
+ | Notes: | ||
+ | *The exercise belongs to the chapter [[Digital_Signal_Transmission/Symbolwise_Coding_with_Pseudo-Ternary_Codes|"Symbolwise Coding with Pseudo-Ternary Codes"]]. | ||
+ | |||
+ | *You can check the results with the (German language) SWF applet [[Applets:Pseudoternaercodierung|"Signals, ACF and PSD of pseudo-ternary codes"]]. | ||
+ | |||
+ | |||
+ | |||
+ | |||
+ | |||
+ | ===Questions=== | ||
<quiz display=simple> | <quiz display=simple> | ||
− | |||
− | |||
− | |||
− | |||
+ | {Which curve belongs to the '''AMI code'''? | ||
+ | |type="()"} | ||
+ | + red, | ||
+ | - blue, | ||
+ | - green. | ||
+ | |||
+ | {Which curve belongs to the '''duobinary code'''? | ||
+ | |type="()"} | ||
+ | - red, | ||
+ | + blue, | ||
+ | - green. | ||
− | { | + | {Which curve belongs to the '''second order bipolar code'''? |
− | |type=" | + | |type="()"} |
− | + | - red, | |
+ | - blue, | ||
+ | + green. | ||
+ | {Which code has the highest transmit power? | ||
+ | |type="()"} | ||
+ | - AMI code, | ||
+ | - duobinary code, | ||
+ | - 2nd order bipolar code. | ||
+ | + The transmit power is the same for all codes. | ||
+ | {Which of these codes has no DC component? | ||
+ | |type="[]"} | ||
+ | + AMI code, | ||
+ | - duobinary code, | ||
+ | + 2nd order bipolar code. | ||
+ | |||
+ | {Why do you need DC signal free codes for the "telephone channel"? | ||
+ | |type="[]"} | ||
+ | + Transformers are needed to connect lines of different impedance. These have high-pass character. | ||
+ | + Since power is often supplied via the signal line, the message signal must not contain any DC signal components. | ||
</quiz> | </quiz> | ||
− | === | + | ===Solution=== |
{{ML-Kopf}} | {{ML-Kopf}} | ||
− | '''(1)''' | + | '''(1)''' With the AMI code, the PSD can be transformed as follows: |
− | '''(2)''' | + | :$${\it \Phi}_s(f) = {s_0^2 \cdot T} \cdot \sin^2 (\pi f T) \cdot {\rm sinc}^2 (f T) \hspace{0.05cm}.$$ |
− | '''(3)''' | + | *This curve shape is shown in <u>red</u>. |
− | '''(4)''' | + | |
− | '''(5)''' | + | *The PSD of the amplitude coefficients is ${\it \Phi}_{a}(f) = \sin^2(\pi fT)$. |
− | '''(6)''' | + | |
+ | |||
+ | '''(2)''' For the duobinary code we obtain : | ||
+ | :$${\it \Phi}_s(f) = {s_0^2 \cdot T} \cdot \cos^2 (\pi f T) \cdot {\rm sinc}^2 (f T) \hspace{0.05cm}.$$ | ||
+ | *In the graph, the duobinary code is drawn in <u>blue</u>. | ||
+ | |||
+ | *Furthermore, for the duobinary code ${\it \Phi}_{a}(f) = \cos^2(\pi fT)$. | ||
+ | |||
+ | |||
+ | '''(3)''' The second order bipolar code differs from the AMI code only by the factor $2$ in the argument of the $\sin^{2}$–function: | ||
+ | :$${\it \Phi}_s(f) = {s_0^2 \cdot T} \cdot \sin^2 (2\pi f T) \cdot {\rm sinc}^2 (f T) \hspace{0.05cm}.$$ | ||
+ | *The <u>green</u> curve represents this function. | ||
+ | |||
+ | *Compared to AMI code, ${\it \Phi}_{a}(f)$ is exactly half as wide. | ||
+ | |||
+ | |||
+ | '''(4)''' The transmit power $P_{\rm S}$ is equal to the integral over the power-spectral density ${\it \Phi}_{s}(f)$ and is the same for all codes considered here ⇒ <u>solution 4</u>. | ||
+ | *This also follows from the power calculation by coulter averaging: | ||
+ | :$$P_{\rm S} = \ {\rm Pr}[s(t) = +s_0] \cdot (+s_0)^2 + {\rm Pr}[s(t) = -s_0] \cdot (-s_0)^2= {1}/{4}\cdot s_0^2 + {1}/{4}\cdot s_0^2 = {1}/{2}\cdot s_0^2\hspace{0.05cm}.$$ | ||
+ | |||
+ | |||
+ | '''(5)''' <u>Solutions 1 and 3</u> are correct: | ||
+ | *DC signal freedom exists if the power-spectral density has no component at frequency $f = 0$. | ||
+ | |||
+ | *This is true for the AMI code and the second order bipolar code. | ||
+ | |||
+ | *This statement does not only mean that $s(t)$ has no DC component, i.e. that ${\it \Phi}_{s}(f)$ has no Dirac delta function at $f = 0$. | ||
+ | |||
+ | *Moreover, it also means that the continuous PSD component vanishes at $f = 0$. | ||
+ | |||
+ | *This is achieved exactly when both the long "$+1$" and the long "$–1$" sequences are excluded by the coding rule. | ||
+ | |||
+ | |||
+ | '''(6)''' <u>Both solutions</u> apply in practice. | ||
{{ML-Fuß}} | {{ML-Fuß}} | ||
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− | [[Category: | + | [[Category:Digital Signal Transmission: Exercises|^2.4 Pseudo-Ternary Codes^]] |
Latest revision as of 17:18, 25 May 2022
In the graph you can see the power-spectral densities $\rm (PSD)$ of three different pseudo-ternary codes, which result from the general description according to "Exercise 2.7" by different values of the parameters $N_{\rm C}$ and $K_{\rm C}$.
In different colors the power-spectral densities
- $${\it \Phi}_s(f) = \ \frac{s_0^2 \cdot T}{2} \cdot {\rm sinc}^2 (f T) \cdot \big [1 - K_{\rm C} \cdot \cos (2\pi f N_{\rm C} T)\big ]$$
are shown for the following variants:
- AMI code $(N_{\rm C} = 1,\ K_{\rm C} = +1)$,
- duobinary code $(N_{\rm C} = 1,\ K_{\rm C} = -1)$,
- second order bipolar code $ (N_{\rm C} = 2,\ K_{\rm C} = +1)$.
The above PSD equation assumes the use of rectangular NRZ basic transmission pulses.
All pseudo-ternary codes considered here have the same probability distribution:
- $${\rm Pr}\big[s(t) = 0\big]= {1}/{2},\hspace{0.2cm}{\rm Pr}\big[s(t) = +s_0\big]= {\rm Pr}\big[s(t) = -s_0\big]={1}/{4}\hspace{0.05cm}.$$
Notes:
- The exercise belongs to the chapter "Symbolwise Coding with Pseudo-Ternary Codes".
- You can check the results with the (German language) SWF applet "Signals, ACF and PSD of pseudo-ternary codes".
Questions
Solution
- $${\it \Phi}_s(f) = {s_0^2 \cdot T} \cdot \sin^2 (\pi f T) \cdot {\rm sinc}^2 (f T) \hspace{0.05cm}.$$
- This curve shape is shown in red.
- The PSD of the amplitude coefficients is ${\it \Phi}_{a}(f) = \sin^2(\pi fT)$.
(2) For the duobinary code we obtain :
- $${\it \Phi}_s(f) = {s_0^2 \cdot T} \cdot \cos^2 (\pi f T) \cdot {\rm sinc}^2 (f T) \hspace{0.05cm}.$$
- In the graph, the duobinary code is drawn in blue.
- Furthermore, for the duobinary code ${\it \Phi}_{a}(f) = \cos^2(\pi fT)$.
(3) The second order bipolar code differs from the AMI code only by the factor $2$ in the argument of the $\sin^{2}$–function:
- $${\it \Phi}_s(f) = {s_0^2 \cdot T} \cdot \sin^2 (2\pi f T) \cdot {\rm sinc}^2 (f T) \hspace{0.05cm}.$$
- The green curve represents this function.
- Compared to AMI code, ${\it \Phi}_{a}(f)$ is exactly half as wide.
(4) The transmit power $P_{\rm S}$ is equal to the integral over the power-spectral density ${\it \Phi}_{s}(f)$ and is the same for all codes considered here ⇒ solution 4.
- This also follows from the power calculation by coulter averaging:
- $$P_{\rm S} = \ {\rm Pr}[s(t) = +s_0] \cdot (+s_0)^2 + {\rm Pr}[s(t) = -s_0] \cdot (-s_0)^2= {1}/{4}\cdot s_0^2 + {1}/{4}\cdot s_0^2 = {1}/{2}\cdot s_0^2\hspace{0.05cm}.$$
(5) Solutions 1 and 3 are correct:
- DC signal freedom exists if the power-spectral density has no component at frequency $f = 0$.
- This is true for the AMI code and the second order bipolar code.
- This statement does not only mean that $s(t)$ has no DC component, i.e. that ${\it \Phi}_{s}(f)$ has no Dirac delta function at $f = 0$.
- Moreover, it also means that the continuous PSD component vanishes at $f = 0$.
- This is achieved exactly when both the long "$+1$" and the long "$–1$" sequences are excluded by the coding rule.
(6) Both solutions apply in practice.