Question: State and explain the condition for orthogonality of BFSK signal determine its spectrum and hence bandwidth requirement for transmission of signal.
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Mumbai University > Electronics and Telecommunication Engineering > Sem 6 > Data Communication

Marks: 10 Marks

Year: Dec 2016

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modified 22 months ago  • written 22 months ago by gravatar for Sayali Bagwe Sayali Bagwe1.9k
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Binary Frequency Shift Keying (BFSK)

  • FSK is treated as non-linear modulation since information is put into the frequency.

Time domain representation:

The transmitted signal can be represented as:

$$s_i(t)=\bigg\{^{\sqrt{\dfrac{2T_b}{T_b}}\cos(2 \pi f_il), 0\lt1\le T_b} _{0,\hspace{2cm} elsewhere}$$

  • S1(t) represents symbol “1”.
  • S2(t) represents symbol “0”.

$$f_1=f_c+\dfrac{1}{2T_b} \ \ \ \ f_2=f_c-\dfrac{1}{2T_b}$$

The minimum frequency spacing between two adjacent frequency slots to ensure orthogonality for the continuous phase BFSK is

$\dfrac{1}{2T_b}$

Phase Discontinuous Frequency Shift Keying

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Generation of BFSK

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Coherent Detection of BFSK

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Bandwidth of FSK:

The null to null bandwidth of FSK is

$B.W \gt 2R_b$

Power Spectral density of BFSK

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We need two phasor

Two carrier are used .The frequency are selected in such a way that two carrier will be orthogonal to each other.

$∫_0^{Tb}\sqrt{2p} \cos(2πf_Ht) \sqrt{2p} \cos(2πf_Lt) = 0$

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written 22 months ago by gravatar for Sayali Bagwe Sayali Bagwe1.9k
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