View source for Aufgaben:Exercise 4.7Z: Signal Shapes for ASK, BPSK and DPSK


{{quiz-Header|Buchseite=Modulation_Methods/Linear_Digital_Modulation
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[[File:P_ID1702__Mod_Z_4_6.png|right|frame|Specified transmitted signals]]
The figure shows,&nbsp; starting from the same source signal &nbsp;$q(t)$,&nbsp; the transmitted signals with

* [[Modulation_Methods/Linear_Digital_Modulation#ASK_.E2.80.93_Amplitude_Shift_Keying| Amplitude Shift Keying]]&nbsp; $\rm (ASK)$, 
* [[Modulation_Methods/Linear_Digital_Modulation#BPSK_.E2.80.93_Binary_Phase_Shift_Keying| Binary Phase Shift Keying ]]&nbsp; $\rm (BPSK)$,
* [[Modulation_Methods/Linear_Digital_Modulation#DPSK_.E2.80.93_Differential_Phase_Shift_Keying| Differential Phase Shift Keying ]]&nbsp; $\rm (DPSK)$.


The transmitted signals are generally designated here as &nbsp;$s_1(t)$, &nbsp;$s_2(t)$&nbsp; and &nbsp;$s_3(t)$.&nbsp; The assignment to the given modulation methods is to be done by you.

In addition,&nbsp; the respective average energy per bit  &nbsp; &rArr; &nbsp; $E_{\rm B}$&nbsp; in "Ws" is to be specified for all signals,&nbsp; whereby the following assumptions can be made:
* The&nbsp; (maximum)&nbsp; envelope of all carrier frequency modulated signals is &nbsp;$s_0 = 2\ \rm V$.
* The bit rate of the redundancy-free source signal is &nbsp;$R_{\rm B} = 1 \ \rm Mbit/s$.
* The modulators operate with a working resistance of &nbsp;$R = 50 \ \rm  Ω$.


For example,&nbsp; for (bipolar) baseband transmission with symbol duration &nbsp;$T_{\rm } = 1/R_{\rm }$&nbsp; would be:
:$$ E_{\rm B} = \frac {s_0^2 \cdot T_{\rm B} }{R} = \frac {(2\,{\rm V})^2 \cdot 10^{-6} \,{\rm s}}{50 \,{\rm V/A}}= 8 \cdot 10^{-8} \,{\rm Ws}= 0.08 \,\,{\rm &micro; Ws}.$$





Notes: 
*The exercise belongs to the chapter&nbsp; [[Modulation_Methods/Linear_Digital_Modulation|Linear Digital Modulation]].
*However, reference is also made to the chapter&nbsp;  [[Digital_Signal_Transmission/Grundlagen_der_codierten_Übertragung|Basics of Coded Transmission]]&nbsp;  in the book "Digital Signal Transmission".
*The powers are given in  &nbsp;$\rm V^2$;&nbsp; they thus refer to the reference resistor &nbsp;$R = 1 \ \rm \Omega$.
 



===Questions===

<quiz display=simple>
{Which signal describes ASK?
|type="()"}
- $s_1(t)$,
- $s_2(t)$,
+ $s_3(t)$.

{What is the average energy per bit  &nbsp; &rArr; &nbsp; $E_{\rm B}$&nbsp; for ASK?
|type="{}"}
 $E_{\rm B} \ = \ $ { 0.02 3% } $\ \rm &micro; Ws$

{Which signal describes BPSK?
|type="()"}
+ $s_1(t),$
- $s_2(t),$
- $s_3(t).$

{What is the average energy per bit  &nbsp; &rArr; &nbsp; $E_{\rm B}$&nbsp; for BPSK?
|type="{}"}
$E_{\rm B} \ = \ $ { 0.04 3% } $\ \rm &micro; Ws$

{Which signal describes DPSK?
|type="()"}
- $s_1(t),$
+ $s_2(t),$
- $s_3(t)$.

{What is the average energy per bit  &nbsp; &rArr; &nbsp; $E_{\rm B}$&nbsp; for DPSK?
|type="{}"}
$E_{\rm B} \ = \ $ { 0.04 3% } $\ \rm &micro; Ws$




</quiz>

===Solution===
{{ML-Kopf}}
'''(1)'''&nbsp;  <u>Solution 3</u>&nbsp; is correct:
*The ASK signal results from the multiplication of the here sinusoidal carrier signal&nbsp; $z(t)$&nbsp; with the unipolar source signal&nbsp; $q(t)$. 
*It is obvious that&nbsp; $s_3(t)$&nbsp; describes such an ASK signal.
*The unipolar amplitude coefficients of the source signal are&nbsp; $1,\ 1,\ 0,\ 1,\ 0,\ 1,\ 1$.



'''(2)'''&nbsp;  Compared to the bipolar baseband transmission,&nbsp; the following changes can be seen in ASK:
* The energy is halved because of multiplication by the sinusoidal signal.
* Since&nbsp; $q(t)$&nbsp; is assumed to be redundancy-free, in half the time&nbsp; $s_3(t) = 0$,&nbsp; which halves the energy again.


This gives:
:$$E_{\rm B} = \frac {s_0^2 \cdot T_{\rm B} }{4 \cdot R} = \frac {(2\,{\rm V})^2 \cdot 10^{-6} \,{\rm s}}{4 \cdot 50 \,{\rm V/A}}= 2 \cdot 10^{-8} \,{\rm Ws}\hspace{0.15cm}\underline {= 0.02 \,\,{\rm &micro; Ws}}.$$



'''(3)'''&nbsp;  <u>Solution 1</u>&nbsp; is correct:
*Phase jumps are typical for BPSK.
*Since the same source signal was always assumed,&nbsp; these phase jumps occur exactly when there is a symbol change in the ASK signal&nbsp; $s_3(t)$.&nbsp; 



'''(4)'''&nbsp;  Of the changes mentioned under&nbsp; '''(2)'''&nbsp; compared to baseband transmission,&nbsp; only the first one is applicable for BPSK.&nbsp; Thus:
:$$E_{\rm B} = \frac {s_0^2 \cdot T_{\rm B} }{2 \cdot R} \hspace{0.15cm}\underline {= 0.04 \,\,{\rm &micro; Ws}}.$$



'''(5)'''&nbsp;  As can already be assumed,&nbsp; the correct answer is&nbsp; $s_2(t)$ &nbsp; ⇒ &nbsp; <u>solution 2</u>: 

*The DPSK modulator operates as follows,&nbsp; assuming&nbsp; $m_0 = -1$:&nbsp; 
:$$ m_0 = -1, \hspace{0.1cm}a_1 = +1 \hspace{0.3cm} \Rightarrow  \hspace{0.3cm}m_1 = -1,$$
:$$m_1 = -1, \hspace{0.1cm}a_2 = +1 \hspace{0.3cm} \Rightarrow  \hspace{0.3cm}m_2 = -1,$$ 
:$$m_2 = -1, \hspace{0.1cm}a_3 = -1 \hspace{0.3cm} \Rightarrow  \hspace{0.3cm}m_3 = +1,$$ 
:$$m_3 = +1, \hspace{0.1cm}a_4 = +1 \hspace{0.3cm} \Rightarrow  \hspace{0.3cm}m_4 = +1,$$
:$$m_4 = +1, \hspace{0.1cm}a_5 = -1 \hspace{0.3cm} \Rightarrow  \hspace{0.3cm}m_5 = -1,$$ 
:$$m_5 = -1, \hspace{0.1cm}a_6 = +1 \hspace{0.3cm} \Rightarrow  \hspace{0.3cm}m_6 = -1, \,\,{\rm etc.}$$



'''(6)'''&nbsp;  A comparison of the two signals&nbsp; $s_1(t)$&nbsp; and&nbsp; $s_2(t)$&nbsp; shows that nothing changes with respect to the signal energy.
*It follows: &nbsp; DPSK has exactly the same signal energy as BPSK:
:$$E_{\rm B} = \frac {s_0^2 \cdot T_{\rm B} }{2 \cdot R} \hspace{0.15cm}\underline {= 0.04 \,\,{\rm &micro; Ws}}.$$

{{ML-Fuß}}



[[Category:Modulation Methods: Exercises|^4.2 Linear Digital Modulation^]]