Difference between revisions of "Aufgaben:Exercise 4.7Z: Signal Shapes for ASK, BPSK and DPSK"
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− | {{quiz-Header|Buchseite= | + | {{quiz-Header|Buchseite=Modulation_Methods/Linear_Digital_Modulation |
}} | }} | ||
− | [[File:P_ID1702__Mod_Z_4_6.png|right|]] | + | [[File:P_ID1702__Mod_Z_4_6.png|right|frame|Specified transmitted signals]] |
− | + | The figure shows, starting from the same source signal $q(t)$, the transmitted signals with | |
− | [ | + | * [[Modulation_Methods/Linear_Digital_Modulation#ASK_.E2.80.93_Amplitude_Shift_Keying| Amplitude Shift Keying]] $\rm (ASK)$, |
+ | * [[Modulation_Methods/Linear_Digital_Modulation#BPSK_.E2.80.93_Binary_Phase_Shift_Keying| Binary Phase Shift Keying ]] $\rm (BPSK)$, | ||
+ | * [[Modulation_Methods/Linear_Digital_Modulation#DPSK_.E2.80.93_Differential_Phase_Shift_Keying| Differential Phase Shift Keying ]] $\rm (DPSK)$. | ||
− | |||
− | + | The transmitted signals are generally designated here as $s_1(t)$, $s_2(t)$ and $s_3(t)$. The assignment to the given modulation methods is to be done by you. | |
− | + | In addition, the respective average energy per bit ⇒ $E_{\rm B}$ in "Ws" is to be specified for all signals, whereby the following assumptions can be made: | |
+ | * The (maximum) envelope of all carrier frequency modulated signals is $s_0 = 2\ \rm V$. | ||
+ | * The bit rate of the redundancy-free source signal is $R_{\rm B} = 1 \ \rm Mbit/s$. | ||
+ | * The modulators operate with a working resistance of $R = 50 \ \rm Ω$. | ||
− | |||
− | + | For example, for (bipolar) baseband transmission with symbol duration $T_{\rm } = 1/R_{\rm }$ would be: | |
− | : | + | :$$ E_{\rm B} = \frac {s_0^2 \cdot T_{\rm B} }{R} = \frac {(2\,{\rm V})^2 \cdot 10^{-6} \,{\rm s}}{50 \,{\rm V/A}}= 8 \cdot 10^{-8} \,{\rm Ws}= 0.08 \,\,{\rm µ Ws}.$$ |
− | |||
− | |||
− | |||
− | $ | + | |
− | + | ||
− | === | + | |
+ | Notes: | ||
+ | *The exercise belongs to the chapter [[Modulation_Methods/Linear_Digital_Modulation|Linear Digital Modulation]]. | ||
+ | *However, 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 given in $\rm V^2$; they thus refer to the reference resistor $R = 1 \ \rm \Omega$. | ||
+ | |||
+ | |||
+ | |||
+ | |||
+ | ===Questions=== | ||
<quiz display=simple> | <quiz display=simple> | ||
− | { | + | {Which signal describes ASK? |
− | |type=" | + | |type="()"} |
− | - $s_1(t)$ | + | - $s_1(t)$, |
− | - $s_2(t)$ | + | - $s_2(t)$, |
− | + $s_3(t)$ | + | + $s_3(t)$. |
− | { | + | {What is the average energy per bit ⇒ $E_{\rm B}$ for ASK? |
|type="{}"} | |type="{}"} | ||
− | $ | + | $E_{\rm B} \ = \ $ { 0.02 3% } $\ \rm µ Ws$ |
− | { | + | {Which signal describes BPSK? |
− | type=" | + | |type="()"} |
− | + $s_1(t)$ | + | + $s_1(t),$ |
− | - $s_2(t)$ | + | - $s_2(t),$ |
− | - $s_3(t)$ | + | - $s_3(t).$ |
− | { | + | {What is the average energy per bit ⇒ $E_{\rm B}$ for BPSK? |
|type="{}"} | |type="{}"} | ||
− | $ | + | $E_{\rm B} \ = \ $ { 0.04 3% } $\ \rm µ Ws$ |
− | { | + | {Which signal describes DPSK? |
− | type=" | + | |type="()"} |
− | - $s_1(t)$ | + | - $s_1(t),$ |
− | + $s_2(t)$ | + | + $s_2(t),$ |
− | - $s_3(t)$ | + | - $s_3(t)$. |
− | { | + | {What is the average energy per bit ⇒ $E_{\rm B}$ for DPSK? |
|type="{}"} | |type="{}"} | ||
− | $ | + | $E_{\rm B} \ = \ $ { 0.04 3% } $\ \rm µ Ws$ |
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</quiz> | </quiz> | ||
− | === | + | ===Solution=== |
{{ML-Kopf}} | {{ML-Kopf}} | ||
− | '''1 | + | '''(1)''' <u>Solution 3</u> is correct: |
+ | *The ASK signal results from the multiplication of the here sinusoidal carrier signal $z(t)$ with the unipolar source signal $q(t)$. | ||
+ | *It is obvious that $s_3(t)$ describes such an ASK signal. | ||
+ | *The unipolar amplitude coefficients of the source signal are $1,\ 1,\ 0,\ 1,\ 0,\ 1,\ 1$. | ||
+ | |||
+ | |||
+ | |||
+ | '''(2)''' Compared to the bipolar baseband transmission, the following changes can be seen in ASK: | ||
+ | * The energy is halved because of multiplication by the sinusoidal signal. | ||
+ | * Since $q(t)$ is assumed to be redundancy-free, in half the time $s_3(t) = 0$, which halves the energy again. | ||
+ | |||
+ | |||
+ | This gives: | ||
+ | :$$E_{\rm B} = \frac {s_0^2 \cdot T_{\rm B} }{4 \cdot R} = \frac {(2\,{\rm V})^2 \cdot 10^{-6} \,{\rm s}}{4 \cdot 50 \,{\rm V/A}}= 2 \cdot 10^{-8} \,{\rm Ws}\hspace{0.15cm}\underline {= 0.02 \,\,{\rm µ Ws}}.$$ | ||
+ | |||
+ | |||
+ | |||
+ | '''(3)''' <u>Solution 1</u> is correct: | ||
+ | *Phase jumps are typical for BPSK. | ||
+ | *Since the same source signal was always assumed, these phase jumps occur exactly when there is a symbol change in the ASK signal $s_3(t)$. | ||
+ | |||
+ | |||
+ | |||
+ | '''(4)''' Of the changes mentioned under '''(2)''' compared to baseband transmission, only the first one is applicable for BPSK. Thus: | ||
+ | :$$E_{\rm B} = \frac {s_0^2 \cdot T_{\rm B} }{2 \cdot R} \hspace{0.15cm}\underline {= 0.04 \,\,{\rm µ Ws}}.$$ | ||
− | |||
− | |||
− | |||
− | |||
− | |||
− | ''' | + | '''(5)''' As can already be assumed, the correct answer is $s_2(t)$ ⇒ <u>solution 2</u>: |
+ | *The DPSK modulator operates as follows, assuming $m_0 = -1$: | ||
+ | :$$ m_0 = -1, \hspace{0.1cm}a_1 = +1 \hspace{0.3cm} \Rightarrow \hspace{0.3cm}m_1 = -1,$$ | ||
+ | :$$m_1 = -1, \hspace{0.1cm}a_2 = +1 \hspace{0.3cm} \Rightarrow \hspace{0.3cm}m_2 = -1,$$ | ||
+ | :$$m_2 = -1, \hspace{0.1cm}a_3 = -1 \hspace{0.3cm} \Rightarrow \hspace{0.3cm}m_3 = +1,$$ | ||
+ | :$$m_3 = +1, \hspace{0.1cm}a_4 = +1 \hspace{0.3cm} \Rightarrow \hspace{0.3cm}m_4 = +1,$$ | ||
+ | :$$m_4 = +1, \hspace{0.1cm}a_5 = -1 \hspace{0.3cm} \Rightarrow \hspace{0.3cm}m_5 = -1,$$ | ||
+ | :$$m_5 = -1, \hspace{0.1cm}a_6 = +1 \hspace{0.3cm} \Rightarrow \hspace{0.3cm}m_6 = -1, \,\,{\rm etc.}$$ | ||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | '''6 | + | '''(6)''' A comparison of the two signals $s_1(t)$ and $s_2(t)$ shows that nothing changes with respect to the signal energy. |
+ | *It follows: DPSK has exactly the same signal energy as BPSK: | ||
+ | :$$E_{\rm B} = \frac {s_0^2 \cdot T_{\rm B} }{2 \cdot R} \hspace{0.15cm}\underline {= 0.04 \,\,{\rm µ Ws}}.$$ | ||
{{ML-Fuß}} | {{ML-Fuß}} | ||
Line 95: | Line 122: | ||
− | [[Category: | + | [[Category:Modulation Methods: Exercises|^4.2 Linear Digital Modulation^]] |
Latest revision as of 15:06, 15 April 2022
The figure shows, starting from the same source signal $q(t)$, the transmitted signals with
- Amplitude Shift Keying $\rm (ASK)$,
- Binary Phase Shift Keying $\rm (BPSK)$,
- Differential Phase Shift Keying $\rm (DPSK)$.
The transmitted signals are generally designated here as $s_1(t)$, $s_2(t)$ and $s_3(t)$. The assignment to the given modulation methods is to be done by you.
In addition, the respective average energy per bit ⇒ $E_{\rm B}$ in "Ws" is to be specified for all signals, whereby the following assumptions can be made:
- The (maximum) envelope of all carrier frequency modulated signals is $s_0 = 2\ \rm V$.
- The bit rate of the redundancy-free source signal is $R_{\rm B} = 1 \ \rm Mbit/s$.
- The modulators operate with a working resistance of $R = 50 \ \rm Ω$.
For example, for (bipolar) baseband transmission with symbol duration $T_{\rm } = 1/R_{\rm }$ would be:
- $$ E_{\rm B} = \frac {s_0^2 \cdot T_{\rm B} }{R} = \frac {(2\,{\rm V})^2 \cdot 10^{-6} \,{\rm s}}{50 \,{\rm V/A}}= 8 \cdot 10^{-8} \,{\rm Ws}= 0.08 \,\,{\rm µ Ws}.$$
Notes:
- The exercise belongs to the chapter Linear Digital Modulation.
- However, reference is also made to the chapter Basics of Coded Transmission in the book "Digital Signal Transmission".
- The powers are given in $\rm V^2$; they thus refer to the reference resistor $R = 1 \ \rm \Omega$.
Questions
Solution
- The ASK signal results from the multiplication of the here sinusoidal carrier signal $z(t)$ with the unipolar source signal $q(t)$.
- It is obvious that $s_3(t)$ describes such an ASK signal.
- The unipolar amplitude coefficients of the source signal are $1,\ 1,\ 0,\ 1,\ 0,\ 1,\ 1$.
(2) Compared to the bipolar baseband transmission, the following changes can be seen in ASK:
- The energy is halved because of multiplication by the sinusoidal signal.
- Since $q(t)$ is assumed to be redundancy-free, in half the time $s_3(t) = 0$, which halves the energy again.
This gives:
- $$E_{\rm B} = \frac {s_0^2 \cdot T_{\rm B} }{4 \cdot R} = \frac {(2\,{\rm V})^2 \cdot 10^{-6} \,{\rm s}}{4 \cdot 50 \,{\rm V/A}}= 2 \cdot 10^{-8} \,{\rm Ws}\hspace{0.15cm}\underline {= 0.02 \,\,{\rm µ Ws}}.$$
(3) Solution 1 is correct:
- Phase jumps are typical for BPSK.
- Since the same source signal was always assumed, these phase jumps occur exactly when there is a symbol change in the ASK signal $s_3(t)$.
(4) Of the changes mentioned under (2) compared to baseband transmission, only the first one is applicable for BPSK. Thus:
- $$E_{\rm B} = \frac {s_0^2 \cdot T_{\rm B} }{2 \cdot R} \hspace{0.15cm}\underline {= 0.04 \,\,{\rm µ Ws}}.$$
(5) As can already be assumed, the correct answer is $s_2(t)$ ⇒ solution 2:
- The DPSK modulator operates as follows, assuming $m_0 = -1$:
- $$ m_0 = -1, \hspace{0.1cm}a_1 = +1 \hspace{0.3cm} \Rightarrow \hspace{0.3cm}m_1 = -1,$$
- $$m_1 = -1, \hspace{0.1cm}a_2 = +1 \hspace{0.3cm} \Rightarrow \hspace{0.3cm}m_2 = -1,$$
- $$m_2 = -1, \hspace{0.1cm}a_3 = -1 \hspace{0.3cm} \Rightarrow \hspace{0.3cm}m_3 = +1,$$
- $$m_3 = +1, \hspace{0.1cm}a_4 = +1 \hspace{0.3cm} \Rightarrow \hspace{0.3cm}m_4 = +1,$$
- $$m_4 = +1, \hspace{0.1cm}a_5 = -1 \hspace{0.3cm} \Rightarrow \hspace{0.3cm}m_5 = -1,$$
- $$m_5 = -1, \hspace{0.1cm}a_6 = +1 \hspace{0.3cm} \Rightarrow \hspace{0.3cm}m_6 = -1, \,\,{\rm etc.}$$
(6) A comparison of the two signals $s_1(t)$ and $s_2(t)$ shows that nothing changes with respect to the signal energy.
- It follows: DPSK has exactly the same signal energy as BPSK:
- $$E_{\rm B} = \frac {s_0^2 \cdot T_{\rm B} }{2 \cdot R} \hspace{0.15cm}\underline {= 0.04 \,\,{\rm µ Ws}}.$$