Principles of Communication Systems – Part II
Prof. Aditya K. Jagannatham
Department of Electrical Engineering
Indian Institute of Technology, Kanpur
Lecture - 16
Optimal Decision Rule for FSK, Bit-Error Rate (BER) and Comparison with
BPSK, ASK
Hello. Welcome to another module in this massive open online course. So, we are looking
at frequency shift keying all right. And in frequency shift keying well we are considering
2 different wave forms, to represent the information bit 0 and one which is shifted in
frequency correct.
(Refer Slide Time: 00:45)
That is, we have,
𝐴𝑝1 (𝑡) + 𝑛(𝑡)𝑓𝑜𝑟 𝑏𝑖𝑡 0
𝑦(𝑡) = {
𝐴𝑝2 (𝑡) + 𝑛(𝑡)𝑓𝑜𝑟 𝑏𝑖𝑡 1
(Refer Slide Time: 01:23).
The SNR optimizing receive filter is given by,
ℎ(𝑡) = 𝑝1 (𝑇 − 𝑡) − 𝑝2 (𝑇 − 𝑡)
(Refer Slide Time: 02:27)
(Refer Slide Time: 03:02)
Sampler output for information bit 0 will be given by,
∞
𝑟(𝑇) = ∫ 𝑦(𝜏)ℎ(𝑇 − 𝜏)𝑑𝜏
−∞
(Refer Slide Time: 04:05)
∞
= ∫ (𝐴𝑝1 (𝜏) + 𝑛(𝜏)) × (𝑝1 (𝜏) − 𝑝2 (𝜏))𝑑𝜏
−∞
∞ ∞
= ∫ 𝐴𝑝1 (𝜏)(𝑝1 (𝜏) − 𝑝1 (𝜏))𝑑𝜏 + ∫ 𝑛(𝜏)(𝑝1 (𝜏) − 𝑝2 (𝜏))𝑑𝜏
−∞ −∞
The first term in the above expression is the signal component and the second term are the
noise component.
(Refer Slide Time: 05:26)
(Refer Slide Time: 06:15)
Let us first simplify the signal component.
∞
𝑠𝑖𝑔𝑛𝑎𝑙 = ∫ 𝐴𝑝1 (𝜏)(𝑝1 (𝜏) − 𝑝2 (𝜏))𝑑𝜏
−∞
∞ ∞
=∫ 𝐴𝑝12 (𝜏)𝑑𝜏 − ∫ 𝐴𝑝1 (𝜏)𝑝2 (𝜏)𝑑𝜏
−∞ −∞
= 𝐴𝐸𝑃
(Refer Slide Time: 07:32)
Substituting 𝐸𝑃 = 1 the signal power would simply be 𝐴.
The noise part is given by
∞
𝑛̃ = ∫ 𝑛(𝜏)(𝑝1 (𝜏) − 𝑝2 (𝜏))𝑑𝜏
−∞
(Refer Slide Time: 08:15)
We know from our previous discussions the above expression will be a gaussian with mean
zero (since n(t) is a gaussian with mean zero).
Variance of the above expression will be given by,
𝜂 ∞
𝑣𝑎𝑟 = ∫ |ℎ(𝜏)|2 𝑑𝜏
2 −∞
(Refer Slide Time: 09:11)
𝜂 ∞ 2
= ∫ (𝑝1 (𝜏) − 𝑝2 (𝜏)) 𝑑𝜏
2 −∞
𝜂 ∞ 2
= ∫ (𝑝1 (𝜏) + 𝑝22 (𝜏))𝑑𝜏
2 −∞
Since the signals are orthonormal the last term 2𝑝1 (𝑡)𝑝2 (𝑡) will be zero.
=𝜂
(Refer Slide Time: 10:13)
(Refer Slide Time: 11:54)
This is the noise power after sampling
(Refer Slide Time: 12:37)
Corresponding to bit 1,
𝑦(𝑡) = 𝐴𝑝2 (𝑡) + 𝑛(𝑡)
(Refer Slide Time: 13:45)
Signal after sampling will be given by,
𝑟(𝑇) = −𝐴𝐸𝑃 + 𝑛̃
= −𝐴 + 𝑛̃
(Refer Slide Time: 14:58)
Thus, overall, we will have,
𝐴 + 𝑛̃ 𝑓𝑜𝑟 𝑏𝑖𝑡 0
𝑟(𝑇) = {
−𝐴 + 𝑛̃ 𝑓𝑜𝑟 𝑏𝑖𝑡 1
The above expression is similar to BPSK except 𝜎 2 = 𝜂 the noise term is 𝜂/2 for BPSK.
(Refer Slide Time: 16:26)
By symmetry we can say that the optimal decision rule will be,
𝑟(𝑇) ≥ 0 ⇒ 𝑑𝑒𝑐𝑖𝑑𝑒 𝑏𝑖𝑡 0
{
𝑟(𝑇) < 0 ⇒ 𝑑𝑒𝑐𝑖𝑑𝑒 𝑏𝑖𝑡 1
(Refer Slide Time: 17:53)
This is the optimal threshold-based decision rule for FSK.
(Refer Slide Time: 18:48)
The probability of bit error or BER is given by,
𝐴
𝐵𝐸𝑅 = 𝑄 ( )
𝜎
𝐴2
= 𝑄 (√ )
𝜂
(Refer Slide Time: 19:54)
Or,
𝐸𝑏
𝑃𝑒 = 𝑄 (√ )
𝜂
(Refer Slide Time: 21:13)
Hence BER for FSK is equal to BER for ASK (for equal average energy per bit). And both
of these are greater than BER of BPSK.
We need twice the average bit energy for ASK and FSK to achieve the same BER as
BPSK.
(Refer Slide Time: 23:55)
Or we can say ASK and FSK need 3dB more power to achieve the same BER as BPSK
So, BPSK is the amongst these 3 BPSK binary phase shift keying is the most efficient
modulations scheme and the reason for that if you explore it will be because binary phase
shift keying for the same average bit error rate uses antipodal signalling that is it uses
signals plus A and minus A all right which maximises the distance between the
constipation points.
You can note this in formally all though we have not shown this rigorously it maximises
the distance between the conciliation points for the same average bit error rate. While both
amplitude shift keying and frequency shift keying do not do this
So, we will stop here.
