Aufgaben:Exercise 4.16: Binary Frequency Shift Keying

• Return to book

Band-pass signals of the FSK

In binary FSK, the two messages  $m_0$  and  $m_1$  are represented by two different frequencies. For the two possible band–pass signals then applies in each case in the range  $0 ≤ t ≤ T$  with  $f_0 = f_{\rm T} + \Delta f_{\rm A}$  as well as  $f_1 = f_{\rm T} \, – \Delta f_{\rm A}$:

$$s_{\rm BP0}(t) \hspace{-0.1cm} \ = \ \hspace{-0.1cm} \sqrt{2E/T} \cdot \cos( 2\pi f_0 t)\hspace{0.05cm},$$

$$ s_{\rm BP1}(t) \hspace{-0.1cm} \ = \ \hspace{-0.1cm} \sqrt{2E/T} \cdot \cos( 2\pi f_1 t)\hspace{0.05cm}.$$

The graph shows example signals $($in each case only one symbol duration  $T)$.

• In the above equation,  $f_{\rm T}$  indicates the carrier frequency  and  $\Delta f_{\rm A}$  indicates the frequency deviation  as the maximum deviation of the  "instantaneous frequency"  from the carrier frequency.
• $E$  is the signal energy. Here, it is equally true for the mean symbol energy and the mean bit energy:

$$E_{\rm S} = E_{\rm B} = E\hspace{0.05cm}.$$

Mostly, however, one works with the modulation index, which is defined as the ratio of total frequency deviation and symbol rate:

$$h = \frac{2 \cdot \Delta f_{\rm A}}{1/T} = 2 \cdot \Delta f_{\rm A} \cdot T \hspace{0.05cm}.$$

Using  $h$,  the equivalent low-pass representation leads to the two complex signals

$$ s_{\rm TP0}(t) \hspace{-0.1cm} \ = \ \hspace{-0.1cm} \sqrt{E/T} \cdot {\rm e}^{\hspace{0.05cm}+{\rm j} \hspace{0.03cm}\cdot \hspace{0.03cm} \pi \hspace{0.03cm}\cdot \hspace{0.03cm} h \hspace{0.03cm}\cdot \hspace{0.03cm}t/T}\hspace{0.05cm},\hspace{0.2cm} 0 \le t \le T\hspace{0.05cm},$$

$$ s_{\rm TP1}(t) \hspace{-0.1cm} \ = \ \hspace{-0.1cm} \sqrt{E/T} \cdot {\rm e}^{\hspace{0.05cm}-{\rm j} \hspace{0.03cm}\cdot \hspace{0.03cm} \pi \hspace{0.03cm}\cdot \hspace{0.03cm} h \hspace{0.03cm}\cdot \hspace{0.03cm}t/T}\hspace{0.05cm},\hspace{0.2cm} 0 \le t \le T\hspace{0.05cm}.$$

An orthogonal FSK exists when the inner product gives the value  $0$: 

$$< \hspace{-0.05cm}s_{\rm TP0}(t) \cdot s_{\rm TP1}(t) \hspace{-0.05cm}> \hspace{0.2cm}=

\int_{0}^{T} s_{\rm TP0}(t) \cdot s_{\rm TP1}^{\star}(t) \,{\rm d} t =0 \hspace{0.05cm}.$$

In this case, non-coherent demodulation as described in the chapter  "Carrier Frequency Systems with Non-Coherent Demodulation"  is also possible.

The inner product of the band–pass signals can be determined from the inner product of the low–pass signals by real partitioning:

$$< \hspace{-0.05cm}s_{\rm BP0}(t) \cdot s_{\rm BP1}(t) \hspace{-0.05cm}> \hspace{0.2cm}=

{\rm Re}\left [ \hspace{0.1cm}<  \hspace{-0.05cm}s_{\rm TP0}(t) \cdot s_{\rm TP1}(t) \hspace{-0.05cm}> \hspace{0.15cm} \right ]\hspace{0.05cm}.$$

If  $〈 s_{\rm BP0}(t) \cdot s_{\rm BP1}(t)〉 = 0$, but at the same time  $〈 s_{\rm TP0}(t) \cdot s_{\rm TP1}(t)〉 ≠ 0$,

• so coherent demodulation is possible,
• but no non-coherent demodulation.

Notes:

• The exercise belongs to the chapter  "Carrier Frequency Systems with Coherent Demodulation".
• Reference is made in particular to the sections  "Binary Frequency Shift Keying" (BFSK) and  "Minimum Shift Keying"  (MSK).

Questions

1 Which carrier frequency  $f_{\rm T}$  and which frequency deviation  $\Delta f_{\rm A}$  are the basis of the graph in the information section?

$f_{\rm T}\hspace{0.48cm} = \ $ $\ \cdot 1/T$

$\Delta f_{\rm A}\ = \ $ $\ \cdot 1/T$

2 To what modulation index  $h$  does this frequency deviation correspond?

$h\ = \ $

3 For which values of  $h$  is the orthogonality of the low–pass signals given?

	$h = 0.5$,
	$h = \pi/4$,
	$h = 1$,
	$h = 2$.

4 For which values of  $h$  is the orthogonality of the band–pass signals given?

	$h = 0.5$,
	$h = \pi/4$,
	$h = 1$,
	$h = 2$.

5 Which statements are true regarding coherent or non-coherent demodulation?

	Coherent demodulation is always possible.
	If non-coherent demodulation is possible, coherent demodulation is also possible.
	If coherent demodulation is possible, non-coherent demodulation is also possible.

Solution

Solution

(1)  By counting the oscillations within a symbol duration $T$, the two frequencies $f_0 = 4.5/T$ and $f_1 = 3.5/T$ are obtained.

• From this carrier frequencies and frequency deviation are calculated to

$$f_{\rm T} \hspace{-0.1cm} \ = \ \hspace{-0.1cm} {1}/{2}\cdot (f_0 + f_1) = \underline{4 \cdot 1/T}\hspace{0.05cm},$$

$$ \Delta f_{\rm A} \hspace{-0.1cm} \ = \ \hspace{-0.1cm} {1}/{2}\cdot (f_0 - f_1)= \underline{0.5 \cdot 1/T }\hspace{0.05cm}.$$

(2)  With the given equation, the modulation index is:

$$h = 2 \cdot \Delta f_{\rm A} \cdot T = 2 \cdot 0.5 \cdot 1/T \cdot T \hspace{0.3cm} \Rightarrow \hspace{0.3cm} \underline{h= 1}\hspace{0.05cm}. $$

(3)  The inner product of the low–pass signals is:

$$< \hspace{-0.05cm} s_{\rm TP0}(t) \hspace{0.01cm} \ \cdot \ \hspace{0.01cm} s_{\rm TP1}(t) \hspace{-0.05cm} > \hspace{0.2cm} =

\int_{0}^{T} s_{\rm TP0}(t) \cdot s_{\rm TP1}^{\star}(t) \,{\rm d} t = \frac{E}{T}  \cdot \int_{0}^{T} {\rm e}^{\hspace{0.05cm}{\rm j}  2\pi  h \hspace{0.03cm}\cdot \hspace{0.03cm}t/T} \,{\rm d} t = \frac{E}{{\rm j}2\pi   h} \cdot \left [ {\rm e}^{\hspace{0.05cm}{\rm j}  2\pi  h} - 1 \right ] 
 \hspace{0.05cm}.$$

Orthogonality means that this inner product must be $0$:

$$< \hspace{-0.05cm} s_{\rm TP0}(t) \cdot s_{\rm TP1}(t) \hspace{-0.05cm} > \hspace{0.2cm} =

 \frac{E}{{\rm j}2\pi   h} \cdot \left [ {\rm e}^{\hspace{0.05cm}{\rm j}  2\pi  h} - 1 \right ] = 0

\hspace{0.3cm}\Rightarrow \hspace{0.3cm} h = 1, 2, 3, ...$$

Consequently, solutions 3 and 4 are correct:

• If the modulation index $h$ is integer, non-coherent demodulation can be performed without violating orthogonality.

(4)  Solutions 1, 3 and 4 are correct:

• For the inner product of the band–pass signals, according to the explanations in the information section, it can be written:

$$< \hspace{-0.05cm}s_{\rm BP0}(t) \hspace{0.01cm} \ \cdot \ \hspace{0.01cm} s_{\rm BP1}(t) \hspace{-0.05cm}> \hspace{0.2cm}=

{\rm Re}\left [ \hspace{0.1cm}<  \hspace{-0.05cm}s_{\rm TP0}(t) \cdot s_{\rm TP1}(t) \hspace{-0.05cm}> \hspace{0.2cm} \right ] = {\rm Re}\left [ \frac{E}{{\rm j}2\pi   h} \cdot \left ( {\rm e}^{\hspace{0.05cm}{\rm j}  2\pi  h} - 1 \right ) \right ]$$

$$ \Rightarrow\hspace{0.3cm}< \hspace{-0.05cm}s_{\rm BP0}(t) \hspace{0.01cm} \ \cdot \ \hspace{0.01cm} s_{\rm BP1}(t) \hspace{-0.05cm}> \hspace{0.2cm}= {\rm Re}\left [ \frac{E}{2\pi h} \cdot \left ( \sin( 2\pi h) - {\rm j} \cdot [\cos( 2\pi h) - 1 ]\right ) \right ] = \frac{E \cdot \sin( 2\pi h)}{2\pi h} \hspace{0.05cm}.$$

• This result is $0$ whenever the modulation index $h$ is an integer multiple of $0.5$.

(5)  Solution 2 is correct.

• For coherent demodulation, $h$ must be a multiple of $0.5$.
• If non-coherent demodulation is possible, as for example in the case considered here ($h = 1$), coherent demodulation is also applicable.
• In contrast, for $h = 0.5$, coherent demodulation can be applied, but non-coherent demodulation (which relies on the envelope) fails.