Coherent orthogonal binary FSK modulation is used to transmit two equiprobable symbol waveforms 𝑠1(𝑡) = 𝛼 cos 2𝜋𝑓1𝑡 and 𝑠2(𝑡) = 𝛼 cos 2𝜋𝑓2𝑡, where 𝛼 = 4 mV. Assume an AWGN channel with two-sided noise power spectral density N02=0.5×1012W/Hz. Using an optimal receiver and the relation Q(v)=12πveu2/2du, the bit error probability for a data rate of 500 kbps is

This question was previously asked in
GATE EC 2014 Official Paper: Shift 2
View all GATE EC Papers >
  1. Q(2)
  2. Q(22)
  3. Q(4)
  4. Q(42)

Answer (Detailed Solution Below)

Option 3 : Q(4)
Free
GATE EC 2023: Full Mock Test
3.4 K Users
65 Questions 100 Marks 180 Mins

Detailed Solution

Download Solution PDF

Concept:

FSK Modulation:

In FSK: transmission of 1 is represented as:

s1(t) = Ac cos 2π fHt

Transmission of 0 is represented as:

s2(t) = Ac cos 2π fLt

and the Bit error probability =Q[Ed2N0]

Where Ed is the energy of s1(t) – s2(t)

Ed=0Tb{s1(t)s2(t)}2dt 

Ed=0Tbs12(t)dt+0Tbs22(t)dt20Tbs1(t)s2(t)dt 

Since s1(t) & s2(t) are orthogonal, we cn write:

0Tbs1(t)s2(t)dt=0 

Ed=0Tbs12(t)dt+0Tbs22(t)dt 

Ed=Ac2Tb2+Ac2Tb2=Ac3Tb 

BER=Q(Ac2Tb2N0) 

Analysis:

Given:

Ac = α = 4 mV

N02=0.5×1012w/Hz 

N0 = 10-12 w/Hz

Tb=1Rb=1800×103 

Tb = 0.2 × 10-5

Tb = 2 × 10-6 sec.

BER=Q(Ac2Tb2N0) 

BER=Q((4×103)2×2×1062×1012) 

BER=Q(16×106×2×1062×1012) 

BER=Q(16) 

BER = Q(4)

Latest GATE EC Updates

Last updated on Jan 8, 2025

-> The GATE EC Call Letter has been released on 7th January 2025.

-> The GATE EC 2025 Exam will be held on 15th February 2025.

-> The mode of the GATE EC exam will be a Computer-Based test of 100 marks. 

-> Candidates preparing for the exam can refer to the GATE EC Previous Year Papers to improve their preparation and increase the chances of selection. 

-> Candidates must attempt the GATE EC Mock tests

More Frequency Shift Keying (FSK) Questions

Get Free Access Now
Hot Links: teen patti neta teen patti game paisa wala all teen patti game teen patti wealth