# 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

- 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

**(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:

- $$E_{\rm B} = \frac {s_0^2 \cdot T_{\rm B} }{2 \cdot R} \hspace{0.15cm}\underline {= 0.04 \,\,{\rm µ Ws}}.$$