Difference between revisions of "Aufgaben:Exercise 2.1: ACF and PSD with Coding"
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− | {{quiz-Header|Buchseite= | + | {{quiz-Header|Buchseite=Digital_Signal_Transmission/Basics_of_Coded_Transmission |
}} | }} | ||
− | [[File:P_ID1308__Dig_A_2_1.png|right|frame| | + | [[File:P_ID1308__Dig_A_2_1.png|right|frame|Power-spectral density with coding]] |
− | + | We consider the digital signal $s(t)$, using the following descriptive quantities: | |
− | + | *aν are the amplitude coefficients, | |
− | + | *gs(t) indicates the basic transmission pulse, | |
− | *aν | + | *T is the symbol duration (spacing of the pulses). |
− | *gs(t) | ||
− | *T | ||
− | + | Then holds: | |
− | * | + | :s(t)=+∞∑ν=−∞aν⋅gs(t−ν⋅T). |
+ | |||
+ | To characterize the spectral properties resulting from the coding and pulse shaping, one uses, among other things | ||
+ | *the auto-correlation function $\rm (ACF)$ | ||
:φs(τ)=+∞∑λ=−∞1/T⋅φa(λ)⋅φ∙gs(τ−λ⋅T), | :φs(τ)=+∞∑λ=−∞1/T⋅φa(λ)⋅φ∙gs(τ−λ⋅T), | ||
− | * | + | *the power-spectral density $\rm (PSD)$ |
:Φs(f)=1/T⋅Φa(f)⋅Φ∙gs(f). | :Φs(f)=1/T⋅Φa(f)⋅Φ∙gs(f). | ||
− | + | Here, φa(λ) denotes the discrete ACF of the amplitude coefficients related to the power-spectral density ${\it \Phi}_{a}(f)$ via the Fourier transform. Thus, for this holds: | |
:Φa(f)=+∞∑λ=−∞φa(λ)⋅e−j2πfλT. | :Φa(f)=+∞∑λ=−∞φa(λ)⋅e−j2πfλT. | ||
− | + | Furthermore, the energy ACF and energy spectrum are used in above equations: | |
:φ∙gs(τ)=∫+∞−∞gs(t)⋅gs(t+τ)dt∘−−−∙Φ∙gs(f)=|Gs(f)|2. | :φ∙gs(τ)=∫+∞−∞gs(t)⋅gs(t+τ)dt∘−−−∙Φ∙gs(f)=|Gs(f)|2. | ||
− | In | + | |
+ | |||
+ | In the present exercise, the following function is to be assumed for the power-spectral density of the amplitude coefficients (see graph): | ||
:Φa(f)=1/2−1/2⋅cos(4πfT). | :Φa(f)=1/2−1/2⋅cos(4πfT). | ||
− | + | The following assumptions are made for the basic transmission pulse: | |
− | * | + | *In question '''(2)''', let gs(t) be an NRZ rectangular pulse, so that there is a triangular energy ACF confined to the range |τ|≤T. The maximum value here is |
:φ∙gs(τ=0)=s20⋅T. | :φ∙gs(τ=0)=s20⋅T. | ||
− | * | + | *For question '''(3)''', assume a root-Nyquist characteristic with rolloff factor r=0. In this case holds: |
− | :$$|G_s(f)|^2 = \left\{ s20⋅T20 \right.\quad \begin{array}{*{1}c} {\rm{ | + | :$$|G_s(f)|^2 = \left\{ s20⋅T20 \right.\quad \begin{array}{*{1}c} {\rm{for}} \\ {\rm{for}} \\ \end{array} |f|<1/(2T),|f|>1/(2T).$$ |
− | * | + | *For numerical calculations, use always s20=10 mW. |
− | |||
− | |||
− | === | + | Notes: |
+ | *The exercise belongs to the chapter [[Digital_Signal_Transmission/Basics_of_Coded_Transmission|"Basics of Coded Transmission"]]. | ||
+ | |||
+ | *Consider that the transmit power PS is equal to the ACF φs(τ) at the point $\tau = 0, but can also be calculated as an integral over the PSD {\it \Phi}_{s}(f)$. | ||
+ | |||
+ | ===Questions=== | ||
<quiz display=simple> | <quiz display=simple> | ||
− | |||
− | |||
− | |||
− | |||
+ | {What are the discrete ACF values φa(λ) of the amplitude coefficients? Enter the numerical values for λ=0, λ=1 and λ=2. | ||
+ | |type="{}"} | ||
+ | φa(λ=0) = { 0.5 3% } | ||
+ | φa(λ=1) = { 0. } | ||
+ | φa(λ=2) = { -0.2575--0.2425 } | ||
+ | |||
+ | {What is the transmit power with the <u>NRZ basic transmission pulse</u>? | ||
+ | |type="{}"} | ||
+ | PS = { 5 3% } mW | ||
− | { | + | {What is the transmit power with <u>root-Nyquist characteristic</u> (r=0)? |
|type="{}"} | |type="{}"} | ||
− | $\ | + | $P_{\rm S} \ = \ $ { 5 3% } mW |
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</quiz> | </quiz> | ||
− | === | + | ===Solution=== |
{{ML-Kopf}} | {{ML-Kopf}} | ||
− | '''(1)''' | + | '''(1)''' Since Φa(f) as a power-spectral density is always real (plus even and positive, but that does not matter here) and the ACF values φa(λ) are symmetric about λ=0, the given equation can be transformed as follows: |
− | '''(2)''' | + | :Φa(f)=+∞∑λ=−∞φa(λ)⋅e−j2πfλT=φa(0)+∞∑λ=12⋅φa(λ)⋅cos(2πfλT). |
− | '''(3)''' | + | *By comparison with the sketched function |
− | + | :Φa(f)=1/2−1/2⋅cos(4πfT). | |
− | + | :one obtains: | |
− | + | :φa(λ=0)=0.5_,φa(λ=2)=φa(λ=−2)=−0.25_. | |
+ | *All other ACF values result to zero, so also φa(λ=±1)=0_. | ||
+ | |||
+ | |||
+ | |||
+ | '''(2)''' For the rectangular NRZ basic pulse, due to the limitation of the energy ACF to the range |τ|≤T, we obtain: | ||
+ | :PS=φs(τ=0)=1T⋅φa(λ=0)⋅φ∙gs(τ=0)=1T⋅12⋅s20⋅T=s202=5mW_. | ||
+ | |||
+ | |||
+ | '''(3)''' For rectangular spectral function, it is more convenient to calculate the transmit power by integration over the power-spectral density: | ||
+ | :PS= ∫+1/(2T)−1/(2T)Φs(f)df=1T⋅∫+1/(2T)−1/(2T)Φa(f)⋅Φ∙gs(f)df | ||
+ | :$$\Rightarrow\hspace{0.3cm}P_{\rm S} = \ \frac{1}{T} \cdot \left [ s_0^2 \cdot T^2 \right ] \cdot \int_{-1/(2T)}^{+1/(2T)} \left( {1}/{2} - {1}/{2} \cdot \cos (4 \pi f \hspace{0.02cm} T)\right ) \,{\rm d} f\hspace{0.05cm} = {s_0^2}/{2}\hspace{0.15cm}\underline { = 5\,\,{\rm mW}} .$$ | ||
+ | *Here it is considered that the energy PSD |Gs(f)|2 is given as constant (within the integration interval) and thus can be drawn in front of the integral. | ||
+ | *In spite of a completely different signal form $s(t)$, the same transmit power results here, since the integral yields the value 1/(2T). | ||
+ | *It should be noted that this simple calculation is only possible for the rolloff factor r=0. | ||
+ | |||
{{ML-Fuß}} | {{ML-Fuß}} | ||
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− | [[Category: | + | [[Category:Digital Signal Transmission: Exercises|^2.1 Basics of Coded Transmission^]] |
Latest revision as of 16:23, 23 May 2022
We consider the digital signal s(t), using the following descriptive quantities:
- aν are the amplitude coefficients,
- gs(t) indicates the basic transmission pulse,
- T is the symbol duration (spacing of the pulses).
Then holds:
- s(t)=+∞∑ν=−∞aν⋅gs(t−ν⋅T).
To characterize the spectral properties resulting from the coding and pulse shaping, one uses, among other things
- the auto-correlation function (ACF)
- φs(τ)=+∞∑λ=−∞1/T⋅φa(λ)⋅φ∙gs(τ−λ⋅T),
- the power-spectral density (PSD)
- Φs(f)=1/T⋅Φa(f)⋅Φ∙gs(f).
Here, φa(λ) denotes the discrete ACF of the amplitude coefficients related to the power-spectral density Φa(f) via the Fourier transform. Thus, for this holds:
- Φa(f)=+∞∑λ=−∞φa(λ)⋅e−j2πfλT.
Furthermore, the energy ACF and energy spectrum are used in above equations:
- φ∙gs(τ)=∫+∞−∞gs(t)⋅gs(t+τ)dt∘−−−∙Φ∙gs(f)=|Gs(f)|2.
In the present exercise, the following function is to be assumed for the power-spectral density of the amplitude coefficients (see graph):
- Φa(f)=1/2−1/2⋅cos(4πfT).
The following assumptions are made for the basic transmission pulse:
- In question (2), let gs(t) be an NRZ rectangular pulse, so that there is a triangular energy ACF confined to the range |τ|≤T. The maximum value here is
- φ∙gs(τ=0)=s20⋅T.
- For question (3), assume a root-Nyquist characteristic with rolloff factor r=0. In this case holds:
- |Gs(f)|2={s20⋅T20forfor|f|<1/(2T),|f|>1/(2T).
- For numerical calculations, use always s20=10 mW.
Notes:
- The exercise belongs to the chapter "Basics of Coded Transmission".
- Consider that the transmit power PS is equal to the ACF φs(τ) at the point τ=0, but can also be calculated as an integral over the PSD Φs(f).
Questions
Solution
- Φa(f)=+∞∑λ=−∞φa(λ)⋅e−j2πfλT=φa(0)+∞∑λ=12⋅φa(λ)⋅cos(2πfλT).
- By comparison with the sketched function
- Φa(f)=1/2−1/2⋅cos(4πfT).
- one obtains:
- φa(λ=0)=0.5_,φa(λ=2)=φa(λ=−2)=−0.25_.
- All other ACF values result to zero, so also φa(λ=±1)=0_.
(2) For the rectangular NRZ basic pulse, due to the limitation of the energy ACF to the range |τ|≤T, we obtain:
- PS=φs(τ=0)=1T⋅φa(λ=0)⋅φ∙gs(τ=0)=1T⋅12⋅s20⋅T=s202=5mW_.
(3) For rectangular spectral function, it is more convenient to calculate the transmit power by integration over the power-spectral density:
- PS= ∫+1/(2T)−1/(2T)Φs(f)df=1T⋅∫+1/(2T)−1/(2T)Φa(f)⋅Φ∙gs(f)df
- ⇒PS= 1T⋅[s20⋅T2]⋅∫+1/(2T)−1/(2T)(1/2−1/2⋅cos(4πfT))df=s20/2=5mW_.
- Here it is considered that the energy PSD |Gs(f)|2 is given as constant (within the integration interval) and thus can be drawn in front of the integral.
- In spite of a completely different signal form s(t), the same transmit power results here, since the integral yields the value 1/(2T).
- It should be noted that this simple calculation is only possible for the rolloff factor r=0.