# Exercise 4.7Z: Signal Shapes for ASK, BPSK and DPSK

The figure shows,  starting from the same source signal  $q(t)$,  the transmitted signals with

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

1

 $s_1(t)$, $s_2(t)$, $s_3(t)$.

2

What is the average energy per bit   ⇒   $E_{\rm B}$  for ASK?

 $E_{\rm B} \ = \$ $\ \rm µ Ws$

3

Which signal describes BPSK?

 $s_1(t),$ $s_2(t),$ $s_3(t).$

4

What is the average energy per bit   ⇒   $E_{\rm B}$  for BPSK?

 $E_{\rm B} \ = \$ $\ \rm µ Ws$

5

Which signal describes DPSK?

 $s_1(t),$ $s_2(t),$ $s_3(t)$.

6

What is the average energy per bit   ⇒   $E_{\rm B}$  for DPSK?

 $E_{\rm B} \ = \$ $\ \rm µ Ws$

### Solution

#### Solution

(1)  Solution 3  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)  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:
$$E_{\rm B} = \frac {s_0^2 \cdot T_{\rm B} }{2 \cdot R} \hspace{0.15cm}\underline {= 0.04 \,\,{\rm µ Ws}}.$$