# Exercise 4.16: Comparison between Binary PSK and Binary FSK

Bit error probability curves:
Binary PSK and binary FSK

The graph shows the bit error probability for binary  FSK modulation  $\rm (BFSK)$  in

in comparison with  binary phase modulation  $\rm (BPSK)$.

Orthogonality is always assumed. For coherent demodulation, the modulation index can be a multiple of  $h = 0.5$ , so that the middle plot can also apply to   Minimum Shift Keying  $\rm (MSK)$ .  In contrast, for non-coherent demodulation of BFSK, the modulation index must be a multiple of  $h = 1$ .

This system comparison is once again based on the  AWGN channel , characterized by the relationship  $E_{\rm B}/N_0$. The equations for the bit error probabilities are as follows

• for Binary Phase Shift Keying  $\rm (BPSK)$:
$$p_{\rm B} = {\rm Q}\left ( \sqrt{{2 \cdot E_{\rm B}}/{N_0 }} \hspace{0.1cm}\right ) = {1}/{2}\cdot {\rm erfc}\left ( \sqrt{{E_{\rm B}}/{N_0 }} \hspace{0.1cm}\right ),$$
• for Binary Frequency Shift Keying  $\rm (BFSK)$  with coherent demodulation:
$$p_{\rm B} = {\rm Q}\left ( \sqrt{{E_{\rm B}}/{N_0 }} \hspace{0.1cm}\right ) = {1}/{2}\cdot {\rm erfc}\left ( \sqrt{{E_{\rm B}}/(2 N_0 )} \hspace{0.1cm}\right ),$$
• for Binary Frequency Shift Keying  $\rm (BFSK)$  with incoherent demodulation:
$$p_{\rm B} = {1}/{2} \cdot {\rm e}^{- E_{\rm B}/{(2N_0) }}\hspace{0.05cm}.$$

It was shown in  Exercise 4.8 , that for BPSK, the log ratio $10 · \lg \ E_{\rm B}/N_0$  must be at least  $9.6 \ \rm dB$  so that the bit error probability does not exceed  $p_{\rm B} = 10^{–5}$ .

Hints:

• Use the approximation  $\lg(2) ≈ 0.3$.

### Questions

1

What  $E_{\rm B}/N_0$  (in dB) is required in for MSK and coherent demodulation to satisfy  $p_{\rm B} \le 10^{–5}$ ?

 $10 · \lg \ E_{\rm B}/N_0 \ = \$ $\ \rm dB$

2

Which of the following statements is correct:   The same result is obtained for

 FSK with modulation index  $h = 0.7$, FSK with modulation index  $h = 1$?

3

What  $E_{\rm B}/N_0$  (in dB) is required for BFSK with  $h = 1$  and incoheren demodulation to satisfy  $p_{\rm B} \le 10^{–5}$ ?

 $10 · \lg \ E_{\rm B}/N_0 \ = \$ $\ \rm dB$

4

What is the bit error probability  $p_{\rm B}$  results from incoherent BFSK demodulation when  $10 · \lg \ E_{\rm B}/N_0 = 12.6 \ \rm dB$?

 $p_{\rm B} \ = \$ $\ \cdot 10^{-4}$

### Solution

#### Solution

(1)  A comparison of the first two equations on the exercise page makes it clear that for MSK with coherent demodulation, the AWGN ratio   $E_{\rm B}/N_0$  must be doubled to achieve the same error probability as for BPSK.

• In other words:  the coherent BFSK curve is  $10 · \lg (2) ≈ 3 \ \rm dB$  to the right of the the BPSK curve.
• Thus, to guarantee  $p_{\rm B} \le 10^{–5}$  it must hold that:
$$10 \cdot {\rm lg}\hspace{0.05cm}{E_{\rm B}} /{N_{\rm 0}}= 9.6\,\,{\rm dB} + 3\,\,{\rm dB} = \underline{12.6\,\,{\rm dB}}\hspace{0.05cm}.$$

• The equation given does not just hold for MSK  $($this is a FSK with  $h = 0.5)$, but also for every form of orthogonal FSK.
• This is the case whenever the modulation index  $h$  is an integer multiple of $0.5$ , such as when   $h = 1$.
• When   $h = 0.7$ , there is no orthogonal FSK.

(3)  From the inverse function of the equation given we get:

$$\frac{E_{\rm B}} {2 \cdot N_{\rm 0}}= {\rm ln}\hspace{0.05cm}\frac{1}{2 p_{\rm B}}= {\rm ln}(50000)\approx 10.82 \hspace{0.3cm} \Rightarrow \hspace{0.3cm}{E_{\rm B}} /{N_{\rm 0}}= 21.64 \hspace{0.3cm}\Rightarrow \hspace{0.3cm} 10 \cdot {\rm lg}\hspace{0.05cm}{E_{\rm B}}/ {N_{\rm 0}}\approx \underline{13.4\,\,{\rm dB}}\hspace{0.05cm}.$$

(4)  From $10 · \lg \ E_{\rm B}/N_0 = 12.6 \ \rm dB$ it follows that:

$${E_{\rm B}} /{N_{\rm 0}}= 10^{1.26} \approx 16.8 \hspace{0.25cm}\Rightarrow \hspace{0.25cm} ({E_{\rm B}} /{N_{\rm 0}})/2 \approx 8.4 \hspace{0.25cm} \Rightarrow \hspace{0.25cm} p_{\rm B} = {1}/{2} \cdot {\rm e}^{- 8.4} \approx \underline{1.12 \cdot 10^{-4}}\hspace{0.05cm}.$$

This means that for the same $E_{\rm B}/N_0$, the error probability of incoherent demodulation is increased by a factor of $11$ compared to that of coherent demodulation (see answer to question 1) .