Aufgaben:Exercise 4.11: Frequency Domain Consideration of the 4-QAM

• Return to book

Power-spectral densities of
BPSK and 4-QAM

Taking as our starting point  $\rm BPSK$  ("binary phase modulation")  with a rectangular basic pulse  $g_s(t)$  of width  $T_{\rm B} = 1 \ \rm µ s$  and amplitude  $s_0 = 2 \ \rm V$,  this exercise aims to determine the power-spectral density  $\rm (PSD)$  of the 4–QAM step by step.

In  Exercise 4.7,  the power-spectral density  ${\it Φ}_s(f)$  of the BPSK was determined for exactly these parameters.  Using

$$A = s_0^2 \cdot T_{\rm B} = 4 \cdot 10^{-6}\,{\rm V^2/Hz},$$

one obtains the actual power-spectral density  (in the bandpass range):

$${{\it \Phi}_s(f)} = {A}/{4} \cdot {\big [ {\rm sinc}^2( T_{\rm B}\cdot (f - f_{\rm T})) + {\rm sinc}^2( T_{\rm B}\cdot (f + f_{\rm T}))\big ]}\hspace{0.05cm}.$$

However,  the top graph shows the power-spectral density  ${{\it \Phi}_{s, \hspace{0.05cm}\rm TP}(f)}$  of the equivalent low-pass signal  German:  "äquivalentes Tiefpass–Signal"   ⇒   subscript:  "TP).  This is obtained from  ${\it Φ}_s(f)$  by

• truncating all components at negative frequencies,
• quadrupling the components at positive frequencies  (note: a spectrum must be doubled, a PSD quadrupled),
• shifting by  $f_{\rm T}$ to the left:

$${{\it \Phi}_{s, \hspace{0.05cm}\rm TP}(f)} = A \cdot {\rm sinc}^2(f \cdot T_{\rm B}). \hspace{0.2cm}$$

4–QAM differs from the BPSK regarding the following details:

• Splitting the binary source signal into two partial signals,  each with half the bit rate,  that is,  with symbol duration  $T = 2 · T_{\rm B}$.
• Multiplication of the partial signals with cosine and minus-sine,  whose amplitudes  $g_0$  are each smaller than  $s_0$ by a factor of  $\sqrt{2}$.
• Summation of the two partial signals denoted by  $s_{\cos}(t)$  nd  $s_{–\sin}(t)$ :

$$s(t) = s_{\rm cos}(t)+ s_{\rm -sin}(t) \hspace{0.05cm}.$$

Hints:

• This exercise belongs to the chapter  "Quadrature Amplitude Modulation"  $\rm (QAM)$.
• Reference is also made to the page  "Binary Phase Shift Keying"  $\rm (BPSK)$  in the previous chapter.
• The power-spectral density  $\rm (PSD)$  of a QAM component is identical to the comparable BPSK PSD:
• Energies are to be specified in  $\rm V^2s$;  they thus refer to the reference resistance $R = 1 \ \rm \Omega$.

Questions

1

What is the energy per bit ⇒ $E_{\rm B}$  for binary phase shift keying (BPSK)?

$E_{\rm B} \ = \ $ $\ \cdot 10^{-6}\ \rm V^2/Hz$

2

What is the power-spectral density  ${\it \Phi}_{s,\hspace{0.08cm} \cos, \hspace{0.08cm}{\rm TP}}(f )$ of the 4–QAM subsignal   $s_{\cos}(t)$  in the equivalent low-pass representation?
what value $B_0 = {\it \Phi}_{s, \hspace{0.08cm}\cos, \hspace{0.08cm}{\rm TP}}(f = 0) $  is obtained at frquency  $f = 0$?

$B_0 \ = \ $ $\ \cdot 10^{-6}\ \rm V^2/Hz$

3

What is the power-spectral density  ${\it \Phi}_{s,\hspace{0.08cm}{\rm TP}}(f )$  of the total 4–QAM signal $s(t)$?
What value  $Q_0 = {\it \Phi}_{s, \hspace{0.08cm}{\rm TP}}(f = 0) $  results here at frequency  $f = 0$?

$Q_0 \ = \ $ $\ \cdot 10^{-6}\ \rm V^2/Hz$

4

What is the energy per bit nbsp; ⇒   $E_{\rm B}$  for quadrature amplitude modulation (4-QAM)?

$E_{\rm B} \ = \ $ $\ \cdot 10^{-6}\ \rm V^2/Hz$

Solution

Solution

(1)  The power of the BPSK transmit signal is equal to the intergral over the power-spectral density.

• If one integrates over the equivalent low-pass PSD, the factor  $1/2$ must still be taken into account:

$$P_{\rm BPSK} = \int_{ - \infty }^{+\infty} {{\it \Phi}_{s}(f)}\hspace{0.1cm} {\rm d}f = \frac{1}{2} \cdot \int_{ - \infty }^{+\infty} {{\it \Phi}_{s, \hspace{0.05cm}\rm TP}(f)}\hspace{0.1cm} {\rm d}f = \frac{A}{2} \cdot \int_{ - \infty }^{+\infty} {\rm si}^2(\pi f T_{\rm B})\hspace{0.1cm} {\rm d}f = \frac{A}{2T_{\rm B}} \cdot \int_{ - \infty }^{+\infty} {\rm si}^2(\pi x)\hspace{0.1cm} {\rm d}x =\frac{A}{2T_{\rm B}}$$

$$\text{Mit} \ \ A = 4 \cdot 10^{-6}\,{\rm V^2/Hz}\hspace{0.05cm}, \hspace{0.2cm} T_{\rm B}= 10^{-6}\,{\rm s} \hspace{0.3cm}\Rightarrow \hspace{0.3cm} P_{\rm BPSK} = 2\,{\rm V^2} ( = {s_0^2 }/{2})\hspace{0.05cm}.$$

• Accordingly, the energy per bit is for BPSK:

$$E_{\rm B} = {P_{\rm BPSK} \cdot T_{\rm B}}\hspace{0.15cm}\underline {= 2 \cdot 10^{-6}\,{\rm V^2/Hz}}\hspace{0.05cm}.$$

• Here again, the reference resistance is$1\ \rm Ω$.

(2)  Due to the double symbol duration of 4-QAM   $(T = 2 · T_{\rm B})$ , the spectral function is only half as wide as compared to BPSK, but twice as high, and instead of   $s_0$ , the smaller value   $g_0$  must now be considered.

• The PSD value at frequency   $f = 0$  is thus:

$${\it \Phi}_{s, \hspace{0.05cm}\rm cos,\hspace{0.05cm}\rm TP}(f = 0 ) = \left ({s_0}/{\sqrt{2}} \right )^2 \cdot 2 \cdot T_{\rm B} ={s_0^2 \cdot T_{\rm B}} = B_0 \hspace{0.05cm}.$$

• Therefore, the result is exactly the same value as for BPSK:

$$B_0 = {\it \Phi}_{s, \hspace{0.05cm}\rm cos,\hspace{0.05cm}\rm TP}(f = 0 ) \hspace{0.15cm}\underline {= 4 \cdot 10^{-6}\,{\rm V^2/Hz}}$$

(3)  The second partial signal  $s_{–\sin}(t)$  yields exactly the same contribution as the signal   $s_{\cos}(t)$ just considered.

• Due to the orthogonality between the cosine and the minus-sine functions, the powers can be added and we get:

$$Q_0 = {\it \Phi}_{s, \hspace{0.05cm}\rm TP}(f = 0 ) = 2 \cdot B_0 \hspace{0.15cm}\underline {= 8 \cdot 10^{-6}\,{\rm V^2/Hz}}\hspace{0.05cm}.$$

(4)  Analogously to in question  (1)  we get an energy per bit of:

$$E_{\rm B} = \frac{1}{2} \cdot T_{\rm B} \cdot \int_{ - \infty }^{+\infty} {{\it \Phi}_{s, \hspace{0.05cm}\rm TP}(f)}\hspace{0.1cm} {\rm d}f = \frac{Q_0 \cdot T_{\rm B}}{2T} \cdot \int_{ - \infty }^{+\infty} {\rm si}^2(\pi f T_{\rm B})\hspace{0.1cm} {\rm d}f = \frac{Q_0 \cdot T_{\rm B}}{2T} = \frac{8 \cdot 10^{-6}\,{\rm V^2/Hz} \cdot 1\,{\rm \mu s}}{ 2 \cdot 2\,{\rm \mu s}}\hspace{0.15cm}\underline {= 2 \cdot 10^{-6}\,{\rm V^2/Hz}}\hspace{0.05cm}.$$

• It can be seen that with the assumptions made here, the "energy per bit" of BPSK and 4-QAM coincide.