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|frame| | + | [[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 | |
− | :$$s(t) = q(t) · z(t)$$ | + | :$$s(t) = q(t) · z(t),$$ |
− | + | where the carrier $z(t)$ represents a harmonic oscillation with frequency $f_{\rm T}$ 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. |
− | * | + | *For ASK, the unipolar amplitude coefficients – that is: $a_ν ∈ \{0, 1\}$ – of the source signal are |
:$$ q(t) = \sum_{\nu = - \infty}^{+\infty}a_\nu \cdot g_q (t - \nu \cdot T)$$ | :$$ q(t) = \sum_{\nu = - \infty}^{+\infty}a_\nu \cdot g_q (t - \nu \cdot T)$$ | ||
− | : | + | :while in the case of BPSK $a_ν ∈ \{-1, +1\}$ has to be considered. |
− | In | + | In the diagram, the power-spectral densities ${\it Φ}_q(f)$ and ${\it Φ}_s(f)$ of the source signal and the transmitted signal are given, respectively, for an NRZ rectangular pulse $g_q(t)$ with amplitude $s_0 = 2 \ \rm V$ and duration $T = 1 \ \rm µ s$. Thus the spectral function is: |
:$$G_q(f) = s_0 \cdot T \cdot {\rm si}(\pi f T)\hspace{0.05cm}.$$ | :$$G_q(f) = s_0 \cdot T \cdot {\rm si}(\pi f T)\hspace{0.05cm}.$$ | ||
− | + | The constants $A$, $B$, $C$ and $D$ for the $\rm ASK$ and $\rm BPSK$ modulation methods are to be determined. | |
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− | + | 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% } $\ \cdot 10^{-6} \ \rm V^2/Hz$ | $A \ = \ $ { 1 3% } $\ \cdot 10^{-6} \ \rm V^2/Hz$ | ||
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− | { | + | {Determine the parameters $C = {\it Φ}_s(f = f_{\rm T})$ and $D$ $($Dirac weight at $f = f_{\rm T})$ for the ASK transmitted signal. |
|type="{}"} | |type="{}"} | ||
$C \ = \ $ { 0.25 3% } $\ \cdot 10^{-6} \ \rm V^2/Hz$ | $C \ = \ $ { 0.25 3% } $\ \cdot 10^{-6} \ \rm V^2/Hz$ | ||
$D \ = \ $ { 0.25 3% } $\ \rm V^2$ | $D \ = \ $ { 0.25 3% } $\ \rm V^2$ | ||
− | { | + | {What are the values for parameters $A$ and $B$ for BPSK? |
|type="{}"} | |type="{}"} | ||
$A \ = \ $ { 4 3% } $\ \cdot 10^{-6} \ \rm V^2/Hz$ | $A \ = \ $ { 4 3% } $\ \cdot 10^{-6} \ \rm V^2/Hz$ | ||
$B \ = \ $ { 0. } $\ \rm V^2$ | $B \ = \ $ { 0. } $\ \rm V^2$ | ||
− | { | + | {What are the values of parameters $C$ and $D$ for BPSK? |
|type="{}"} | |type="{}"} | ||
$C \ = \ $ { 1 3% } $\ \cdot 10^{-6} \ \rm V^2/Hz$ | $C \ = \ $ { 1 3% } $\ \cdot 10^{-6} \ \rm V^2/Hz$ | ||
$D \ = \ $ { 0. } $\ \rm V^2$ | $D \ = \ $ { 0. } $\ \rm V^2$ | ||
− | { | + | {Which statements are always true for BPSK even if $g_q(t)$ is not an NRZ rectangular pulse? |
|type="[]"} | |type="[]"} | ||
− | + | + | + The continuous part of $ {\it Φ}_q(f)$ is equal in form to $|G_q(f)|^2$. |
− | - ${\it Φ}_q(f)$ | + | - ${\it Φ}_q(f)$ contains a single Dirac delta line at ASK for BPSK $($at $f = 0)$. |
− | - ${\it Φ}_q(f)$ | + | - ${\it Φ}_q(f)$ 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 $m_q = s_0/2$. Thus, the Dirac weight is $B = m_q^2 = s_0^2/4\hspace{0.15cm}\underline{ = 1 \ \rm V^2}$. |
− | * | + | *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: |
:$$A = \frac {s_0^2 \cdot T }{4} = \frac {(2\,{\rm V})^2 \cdot | :$$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}}.$$ | 10^{-6} \,{\rm s}}{4}\hspace{0.15cm}\underline {= 10^{-6} \,{\rm V^{2}/Hz}}.$$ | ||
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− | '''(2)''' | + | '''(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: |
:$$C = {A}/{4} \hspace{0.15cm}\underline { = 0.25 \cdot 10^{-6} \,{\rm | :$$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}}}.$$ | V^{2}/Hz}},\hspace{0.2cm}D = {B}/{4}\hspace{0.15cm}\underline { = 0.25 \,{\rm V^{2}}}.$$ | ||
− | + | Note: The power per bit is obtained as the integral over ${\it Φ}_s(f)$: | |
:$$P_{\rm S} = \int_{ - \infty }^\infty \hspace{-0.3cm} {{\it \Phi}_s(f)}\hspace{0.1cm} {\rm d}f | :$$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 | + | = 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 ] = | d}f= 2 \cdot \left [ \frac{C}{T} + D \right ] = | ||
2 \cdot \left [ \frac{0.25 \cdot 10^{-6} \,{\rm | 2 \cdot \left [ \frac{0.25 \cdot 10^{-6} \,{\rm | ||
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− | '''(3)''' | + | '''(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: |
:$$A = {s_0^2 \cdot T }\hspace{0.15cm}\underline { = 4 \cdot 10^{-6} \,{\rm V^{2}/Hz}}.$$ | :$$A = {s_0^2 \cdot T }\hspace{0.15cm}\underline { = 4 \cdot 10^{-6} \,{\rm V^{2}/Hz}}.$$ | ||
− | '''(4)''' | + | '''(4)''' For the PSD parameters of the BPSK transmitted signal, the following applies analogously to the ASK: |
:$$C = \frac {A}{4}\hspace{0.15cm}\underline { = 10^{-6} \,{\rm V^{2}/Hz}},\hspace{0.2cm}D = | :$$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}.$$ | \frac {B}{4} \hspace{0.15cm}\underline {= 0}.$$ | ||
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− | '''(5)''' | + | '''(5)''' Only the <u>first statement</u> 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". | |
− | |||
− | |||
{{ML-Fuß}} | {{ML-Fuß}} | ||
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− | [[Category:Modulation Methods: Exercises|^4.2 | + | [[Category:Modulation Methods: Exercises|^4.2 Linear Digital Modulation^]] |
Latest revision as of 13:31, 18 January 2023
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 $f_{\rm T}$ 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.
- For ASK, the unipolar amplitude coefficients – that is: $a_ν ∈ \{0, 1\}$ – of the source signal are
- $$ q(t) = \sum_{\nu = - \infty}^{+\infty}a_\nu \cdot g_q (t - \nu \cdot T)$$
- while in the case of BPSK $a_ν ∈ \{-1, +1\}$ has to be considered.
In the diagram, the power-spectral densities ${\it Φ}_q(f)$ and ${\it Φ}_s(f)$ of the source signal and the transmitted signal are given, respectively, for an NRZ rectangular pulse $g_q(t)$ with amplitude $s_0 = 2 \ \rm V$ 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".