Difference between revisions of "Aufgaben:Exercise 4.7: Spectra of ASK and BPSK"
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− | {{quiz-Header|Buchseite= | + | {{quiz-Header|Buchseite=Modulation_Methods/Linear_Digital_Modulation |
}} | }} | ||
− | [[File:P_ID1701__Mod_A_4_6.png|right|]] | + | [[File:P_ID1701__Mod_A_4_6.png|right|frame|Power-spectral densities of q(t) and s(t) – valid for ASK and BPSK]] |
− | + | The transmitted signals of $\rm ASK$ ("Amplitude Shift Keying") and $\rm BPSK$ ("Binary Phase Shift Keying") can both be expressed in the form | |
+ | :$$s(t) = q(t) · z(t),$$ | ||
+ | where the carrier z(t) represents a harmonic oscillation with frequency fT and amplitude $1$. The carrier phase $ϕ_{\rm T}$ is not important for the power-spectral densities considered here. | ||
− | + | *In each case, the source is redundancy-free, which means that the two possible symbols are equally probable and the symbols are statistically independent of each other. | |
− | q(t)=+∞∑ν=−∞aν⋅gq(t−ν⋅T) | + | *For ASK, the unipolar amplitude coefficients – that is: $a_ν ∈ \{0, 1\}$ – of the source signal are |
− | + | :q(t)=+∞∑ν=−∞aν⋅gq(t−ν⋅T) | |
+ | :while in the case of BPSK $a_ν ∈ \{-1, +1\}$ has to be considered. | ||
− | |||
− | |||
− | |||
− | + | In the diagram, the power-spectral densities Φq(f) and Φs(f) of the source signal and the transmitted signal are given, respectively, for an NRZ rectangular pulse gq(t) with amplitude s0=2 V and duration T = 1 \ \rm µ s. Thus the spectral function is: | |
− | === | + | :$$G_q(f) = s_0 \cdot T \cdot {\rm si}(\pi f T)\hspace{0.05cm}.$$ |
+ | The constants A, B, C and D for the ASK and BPSK modulation methods are to be determined. | ||
+ | |||
+ | |||
+ | |||
+ | |||
+ | |||
+ | |||
+ | Notes: | ||
+ | *The exercise belongs to the chapter [[Modulation_Methods/Linear_Digital_Modulation|Linear Digital Modulation]]. | ||
+ | *Reference is also made to the chapter [[Digital_Signal_Transmission/Grundlagen_der_codierten_Übertragung|Basics of Coded Transmission]] in the book "Digital Signal Transmission". | ||
+ | *The powers are to be specified in $\rm V^2; they thus refer to the reference resistance R = 1 \ \rm \Omega$. | ||
+ | |||
+ | |||
+ | |||
+ | |||
+ | ===Questions=== | ||
<quiz display=simple> | <quiz display=simple> | ||
− | { | + | {Which values result with ASK for the parameters $A = {\it Φ}_q(f = 0)$ and $B (Dirac weight at f = 0)$? |
|type="{}"} | |type="{}"} | ||
− | $ | + | $A \ = \ { 1 3% }\ \cdot 10^{-6} \ \rm V^2/Hz$ |
− | B | + | $B \ = \ $ { 1 3% } $\ \rm V^2$ |
− | { | + | {Determine the parameters $C = {\it Φ}_s(f = f_{\rm T})$ and $D (Diracweightatf = f_{\rm T})$ for the ASK transmitted signal. |
|type="{}"} | |type="{}"} | ||
− | $ | + | $C \ = \ { 0.25 3% }\ \cdot 10^{-6} \ \rm V^2/Hz$ |
− | D | + | $D \ = \ { 0.25 3% }\ \rm V^2$ |
− | { | + | {What are the values for parameters A and $B$ for BPSK? |
|type="{}"} | |type="{}"} | ||
− | $ | + | $A \ = \ { 4 3% }\ \cdot 10^{-6} \ \rm V^2/Hz$ |
− | B | + | $B \ = \ $ { 0. } $\ \rm V^2$ |
− | { | + | {What are the values of parameters C and $D$ for BPSK? |
|type="{}"} | |type="{}"} | ||
− | $ | + | $C \ = \ { 1 3% }\ \cdot 10^{-6} \ \rm V^2/Hz$ |
− | D | + | $D \ = \ $ { 0. } $\ \rm V^2$ |
− | { | + | {Which statements are always true for BPSK even if gq(t) is not an NRZ rectangular pulse? |
− | |type=" | + | |type="[]"} |
− | + | + | + The continuous part of $ {\it Φ}_q(f)$ is equal in form to $|G_q(f)|^2$. |
− | - $ | + | - ${\it Φ}_q(f)$ contains a single Dirac delta line at ASK for BPSK (at $f = 0)$. |
− | - $ | + | - ${\it Φ}_q(f)$ contains a single Dirac delta line at BPSK for BPSK (at $f = 0)$. |
</quiz> | </quiz> | ||
− | === | + | ===Solution=== |
{{ML-Kopf}} | {{ML-Kopf}} | ||
− | '''1 | + | '''(1)''' The DC component of the unipolar redundancy-free source signal is mq=s0/2. Thus, the Dirac weight is $B = m_q^2 = s_0^2/4\hspace{0.15cm}\underline{ = 1 \ \rm V^2}$. |
− | $$A = \frac{s_0^2 \cdot T}{4} = \frac{( | + | |
+ | *Without this DC component, the stochastic signal $q(t) - m_q ∈ \{+s_0/2, -s_0/2\}$ would be obtained. | ||
+ | *This DC-free signal has the continuous PSD component $(s_0/2)^2 · T · {\rm sinc}^2(fT)$. | ||
+ | *From this, the value we are looking for at frequency $f = 0$ can be determined: | ||
+ | :$$A = \frac {s_0^2 \cdot T }{4} = \frac {(2\,{\rm V})^2 \cdot | ||
+ | 10^{-6} \,{\rm s}}{4}\hspace{0.15cm}\underline {= 10^{-6} \,{\rm V^{2}/Hz}}.$$ | ||
+ | |||
+ | |||
+ | |||
+ | '''(2)''' The spectrum Z(f) of a cosine signal z(t) consists of two Dirac delta functions at ±fT, each with weight 1/2. | ||
+ | *The power-spectral density Φz(f) also consists of the two Dirac delta functions, but now with respective weights 1/4. | ||
+ | *The convolution Φq(f)∗Φz(f) gives the power-spectral density Φs(f) of the transmitted signal. It follows that: | ||
+ | :$$C = {A}/{4} \hspace{0.15cm}\underline { = 0.25 \cdot 10^{-6} \,{\rm | ||
+ | V^{2}/Hz}},\hspace{0.2cm}D = {B}/{4}\hspace{0.15cm}\underline { = 0.25 \,{\rm V^{2}}}.$$ | ||
+ | |||
+ | Note: The power per bit is obtained as the integral over Φs(f): | ||
+ | :$$P_{\rm S} = \int_{ - \infty }^\infty \hspace{-0.3cm} {{\it \Phi}_s(f)}\hspace{0.1cm} {\rm d}f | ||
+ | = 2 \cdot \int_{ 0 }^\infty \hspace{-0.3cm} {\left [ C \cdot {\rm sinc}^2(f T) + D \cdot \delta (f - f_{\rm T}]\right ]}\hspace{0.1cm} {\rm | ||
+ | d}f= 2 \cdot \left [ \frac{C}{T} + D \right ] = | ||
+ | 2 \cdot \left [ \frac{0.25 \cdot 10^{-6} \,{\rm | ||
+ | V^{2}/Hz}}{10^{-6} \,{\rm s}} + 0.25 \,{\rm V^{2}} \right ] \hspace{0.15cm}\underline {= 1 | ||
+ | \,{\rm V^{2}}}.$$ | ||
+ | |||
+ | |||
+ | |||
+ | '''(3)''' For BPSK, the source signal q(t) is to be bipolar. | ||
+ | *Therefore, the Dirac delta line is missing in the power-spectral density ⇒ B=0_. | ||
+ | * The continuous PSD component is four times larger than with ASK: | ||
+ | :$$A = {s_0^2 \cdot T }\hspace{0.15cm}\underline { = 4 \cdot 10^{-6} \,{\rm V^{2}/Hz}}.$$ | ||
+ | |||
+ | |||
− | ''' | + | '''(4)''' For the PSD parameters of the BPSK transmitted signal, the following applies analogously to the ASK: |
− | $$C = \frac{A}{4} | + | :$$C = \frac {A}{4}\hspace{0.15cm}\underline { = 10^{-6} \,{\rm V^{2}/Hz}},\hspace{0.2cm}D = |
+ | \frac {B}{4} \hspace{0.15cm}\underline {= 0}.$$ | ||
− | |||
− | ''' | + | '''(5)''' Only the <u>first statement</u> is correct: |
+ | * For BPSK, Φq(f) does not contain a single Dirac delta line even if gq(t) deviates from the rectangular form (assuming equally probable symbols). | ||
+ | *In contrast, the unipolar ASK source signal contains infinitely many Dirac delta lines at all multiples of 1/T. | ||
+ | *For more information on this topic, see the page "ACF and PSD for unipolar binary signals" in the book "Digital Signal Transmission". | ||
− | |||
{{ML-Fuß}} | {{ML-Fuß}} | ||
− | [[Category: | + | [[Category:Modulation Methods: Exercises|^4.2 Linear Digital Modulation^]] |
Latest revision as of 14:31, 18 January 2023
The transmitted signals of ASK ("Amplitude Shift Keying") and BPSK ("Binary Phase Shift Keying") can both be expressed in the form
- s(t)=q(t)·z(t),
where the carrier z(t) represents a harmonic oscillation with frequency fT and amplitude 1. The carrier phase ϕT is not important for the power-spectral densities considered here.
- In each case, the source is redundancy-free, which means that the two possible symbols are equally probable and the symbols are statistically independent of each other.
- For ASK, the unipolar amplitude coefficients – that is: aν∈{0,1} – of the source signal are
- q(t)=+∞∑ν=−∞aν⋅gq(t−ν⋅T)
- while in the case of BPSK aν∈{−1,+1} has to be considered.
In the diagram, the power-spectral densities Φq(f) and Φs(f) of the source signal and the transmitted signal are given, respectively, for an NRZ rectangular pulse gq(t) with amplitude s0=2 V and duration T = 1 \ \rm µ s. Thus the spectral function is:
- G_q(f) = s_0 \cdot T \cdot {\rm si}(\pi f T)\hspace{0.05cm}.
The constants A, B, C and D for the \rm ASK and \rm BPSK modulation methods are to be determined.
Notes:
- The exercise belongs to the chapter Linear Digital Modulation.
- Reference is also made to the chapter Basics of Coded Transmission in the book "Digital Signal Transmission".
- The powers are to be specified in \rm V^2; they thus refer to the reference resistance R = 1 \ \rm \Omega.
Questions
Solution
- Without this DC component, the stochastic signal q(t) - m_q ∈ \{+s_0/2, -s_0/2\} would be obtained.
- This DC-free signal has the continuous PSD component (s_0/2)^2 · T · {\rm sinc}^2(fT).
- From this, the value we are looking for at frequency f = 0 can be determined:
- A = \frac {s_0^2 \cdot T }{4} = \frac {(2\,{\rm V})^2 \cdot 10^{-6} \,{\rm s}}{4}\hspace{0.15cm}\underline {= 10^{-6} \,{\rm V^{2}/Hz}}.
(2) The spectrum Z(f) of a cosine signal z(t) consists of two Dirac delta functions at \pm f_{\rm T}, each with weight 1/2.
- The power-spectral density {\it Φ}_z(f) also consists of the two Dirac delta functions, but now with respective weights 1/4.
- The convolution {\it Φ}_q(f) ∗ {\it Φ}_z(f) gives the power-spectral density {\it Φ}_s(f) of the transmitted signal. It follows that:
- C = {A}/{4} \hspace{0.15cm}\underline { = 0.25 \cdot 10^{-6} \,{\rm V^{2}/Hz}},\hspace{0.2cm}D = {B}/{4}\hspace{0.15cm}\underline { = 0.25 \,{\rm V^{2}}}.
Note: The power per bit is obtained as the integral over {\it Φ}_s(f):
- P_{\rm S} = \int_{ - \infty }^\infty \hspace{-0.3cm} {{\it \Phi}_s(f)}\hspace{0.1cm} {\rm d}f = 2 \cdot \int_{ 0 }^\infty \hspace{-0.3cm} {\left [ C \cdot {\rm sinc}^2(f T) + D \cdot \delta (f - f_{\rm T}]\right ]}\hspace{0.1cm} {\rm d}f= 2 \cdot \left [ \frac{C}{T} + D \right ] = 2 \cdot \left [ \frac{0.25 \cdot 10^{-6} \,{\rm V^{2}/Hz}}{10^{-6} \,{\rm s}} + 0.25 \,{\rm V^{2}} \right ] \hspace{0.15cm}\underline {= 1 \,{\rm V^{2}}}.
(3) For BPSK, the source signal q(t) is to be bipolar.
- Therefore, the Dirac delta line is missing in the power-spectral density ⇒ \underline{B = 0}.
- The continuous PSD component is four times larger than with ASK:
- A = {s_0^2 \cdot T }\hspace{0.15cm}\underline { = 4 \cdot 10^{-6} \,{\rm V^{2}/Hz}}.
(4) For the PSD parameters of the BPSK transmitted signal, the following applies analogously to the ASK:
- C = \frac {A}{4}\hspace{0.15cm}\underline { = 10^{-6} \,{\rm V^{2}/Hz}},\hspace{0.2cm}D = \frac {B}{4} \hspace{0.15cm}\underline {= 0}.
(5) Only the first statement is correct:
- For BPSK, {\it Φ}_q(f) does not contain a single Dirac delta line even if g_q(t) deviates from the rectangular form (assuming equally probable symbols).
- In contrast, the unipolar ASK source signal contains infinitely many Dirac delta lines at all multiples of 1/T.
- For more information on this topic, see the page "ACF and PSD for unipolar binary signals" in the book "Digital Signal Transmission".