This presentation covers noise performance of Continuous wave modulation systems; It explains modelling of white noise , noise figure of DSB-SC, SSB, AM, FM system
Study on Air-Water & Water-Water Heat Exchange in a Finned Tube Exchanger
Noise Performance of CW system
1. Noise performance of CW system
31-08-2016 IEC 503 ANALOG COMMUNICATION SYSTEM BY DR N R KIDWAI, INTEGRAL UNIVERSITY 1
2. Narrow band noise
31-08-2016 IEC 503 ANALOG COMMUNICATION SYSTEM BY DR N R KIDWAI, INTEGRAL UNIVERSITY 2
Narrow Band
BPF Hn(f)
Pass band 2B
Channel
white noise w(t)
Narrow band
Noise n(t)
Receiver
demod
ulator
LPF
cut off W
fc-B fc fc+B-fc-B -fc -fc+B
Sn(f)
fc-B fc fc+B-fc-B -fc -fc+B
Hn(f)
Sw(f)
• First stage at the receiver end is BPF of pass band 2B to limit the noise.
• spectral density of white Noise Sw(f) at input of BPF is constant (/2)
• spectral density of Noise Sn(f) at output of BPF is narrow band
3. Narrow band noise
31-08-2016 IEC 503 ANALOG COMMUNICATION SYSTEM BY DR N R KIDWAI, INTEGRAL UNIVERSITY 3
From the narrow band spectral density function of noise , it is evident that narrow band noise n(t) is
approximately a sinusoidal function of frequency fc with amplitude and phase varying randomly.
𝑛 𝑡 = 𝐴 𝑛 𝑡 cos 2𝜋𝑓𝑐 𝑡 + ∅ 𝑛 𝑡 = 𝐴 𝑛 𝑡 cos ∅ 𝑛 𝑡 cos 2𝜋𝑓𝑐 𝑡 − 𝐴 𝑛 𝑡 sin ∅ 𝑛 𝑡 sin 2𝜋𝑓𝑐 𝑡
= 𝑛𝐼(𝑡)cos 2𝜋𝑓𝑐 𝑡 − 𝑛 𝑄(𝑡)sin 2𝜋𝑓𝑐 𝑡
Where In phase component of narrow band noise, 𝑛𝐼 𝑡 = 𝐴 𝑛 𝑡 cos ∅ 𝑛 𝑡
And quadrature phase component of narrow band noise, 𝑛 𝑄 𝑡 = 𝐴 𝑛 𝑡 sin ∅ 𝑛 𝑡
Thus narrow band noise can be viewed as sum of In-
phase component of noise modulated on carrier
signal and quadrature phase component of noise
modulated with quadrature shifted version of carrier
signal
4. Narrow band noise
31-08-2016 IEC 503 ANALOG COMMUNICATION SYSTEM BY DR N R KIDWAI, INTEGRAL UNIVERSITY 4
Properties of narrow band noise
• If 𝑛 𝑡 has zero mean then 𝑛𝐼 𝑡 and 𝑛 𝑄 𝑡 has zero mean
• 𝑆 𝑛 𝑓 =
𝜂
2
𝑓𝑜𝑟 𝑓𝑐 − 𝐵 ≤ 𝑓 ≤ 𝑓𝑐 + 𝐵
• Both 𝑛𝐼 𝑡 and 𝑛 𝑄 𝑡 has same power spectral density which is related to spectral
density of narrow band noise, 𝑆 𝑛 𝑓 as
𝑆 𝑛𝐼 𝑓 =𝑆 𝑛𝑄 𝑓 =𝑆 𝑛 𝑓 − 𝑓𝑐 +𝑆 𝑛 𝑓 + 𝑓𝑐 = 𝜂 𝑓𝑜𝑟 𝑓 ≤ 𝐵
Thus 𝑛2 𝑡 = 𝑛𝐼
2
𝑡 = 𝑛 𝑄
2
𝑡
f
𝑆 𝑛𝐼 𝑓 𝑜𝑟𝑆 𝑛𝑄 𝑓
B-B
fc-B fc+B
𝜂
2
𝑆 𝑛 𝑓
f-fc-B -fc+B
5. Noise performance of DSB-SC system (coherent detection)
31-08-2016 IEC 503 ANALOG COMMUNICATION SYSTEM BY DR N R KIDWAI, INTEGRAL UNIVERSITY 5
For DSB-SC, 𝑠 𝑡 = 𝐴 𝑐m(t)cos 2𝜋𝑓𝑐 𝑡
Input to demodulator, 𝑠𝑖 𝑡 = 𝐴 𝑐m t cos 2𝜋𝑓𝑐 𝑡 + 𝑛(𝑡)
Signal power at input, 𝑆𝑖 = 𝐴 𝑐m(t)cos 2𝜋𝑓𝑐 𝑡 2 =
𝐴 𝑐
2
2
𝑚2(𝑡)
Noise power at input, 𝑁𝑖 = 𝑛2(𝑡) = 2 𝑥 2𝑊 𝑥
𝜂
2
= 2𝑊𝜂
Input of multiplier =𝑠 𝑡 + 𝑛 𝑡 = 𝐴 𝑐m t cos 2𝜋𝑓𝑐 𝑡 + 𝑛𝐼(𝑡)cos 2𝜋𝑓𝑐 𝑡 − 𝑛 𝑄(𝑡)sin 2𝜋𝑓𝑐 𝑡
Output of the multiplier=𝐴 𝑐m t 𝑐𝑜𝑠2
2𝜋𝑓𝑐 𝑡 + 𝑛𝐼(𝑡)𝑐𝑜𝑠2
2𝜋𝑓𝑐 𝑡 − 𝑛 𝑄(𝑡)sin 2𝜋𝑓𝑐 𝑡 cos(2𝜋𝑓𝑐 𝑡)
Narrow Band
BPF Hn(f)
Pass band 2W
Channel
white noise w(t)
s(t)+n(t)
Receiver
LPF
cut off W
s(t) +nw(t)
demodulator
cos(2fct)
sd(t)+nd(t)
Si
Ni
So
No
s(t)
6. Noise performance of DSB-SC system (coherent detection)
31-08-2016 IEC 503 ANALOG COMMUNICATION SYSTEM BY DR N R KIDWAI, INTEGRAL UNIVERSITY 6
Output of the multiplier=
𝐴 𝑐
2
m t +
𝐴 𝑐
2
m t 𝑐𝑜𝑠 4𝜋𝑓𝑐 𝑡 +
𝑛 𝐼 𝑡
2
+
𝑛 𝐼 𝑡
2
𝑐𝑜𝑠 4𝜋𝑓𝑐 𝑡 −
𝑛 𝑄(𝑡)
2
sin 4𝜋𝑓𝑐 𝑡
Output of Low pass filter=
𝐴 𝑐
2
m t +
𝑛 𝐼 𝑡
2
• It can be observed that message component and in phase component of noise appear in the
output with half magnitude
• Quadrature component of noise is fully rejected by coherent receiver
at the output, Signal power , 𝑆 𝑜 =
𝐴 𝑐
2
m(t)
2
=
𝐴 𝑐
2
4
𝑚2(𝑡)
Noise power, 𝑁𝑜 =
𝑛 𝐼 𝑡
2
2
=
1
4
𝑛𝐼
2
(𝑡) =
2𝑊𝜂
4
=
𝑊𝜂
2
=
1
4
𝑛2(𝑡)
7. Noise performance of DSB-SC system (coherent detection)
31-08-2016 IEC 503 ANALOG COMMUNICATION SYSTEM BY DR N R KIDWAI, INTEGRAL UNIVERSITY 7
𝐹 =
𝑆𝑖
𝑆 𝑜
𝑥
𝑁𝑜
𝑁𝑖
=
𝐴 𝑐
2
2
𝑚2(𝑡)
𝐴 𝑐
2
4
𝑚2(𝑡)
𝑥
1
4
𝑛2(𝑡)
𝑛2(𝑡)2
noise figure of DSB-SC receiver, 𝑭 =
𝟏
𝟐
Figure of merit of DSB-SC receiver =1/ F =2
• Noise figure of DSB-SC receiver implies that noise power is reduced by half. IT eliminates
quadrature component of noise power
8. Noise performance of SSB-SC system (coherent detection)
31-08-2016 IEC 503 ANALOG COMMUNICATION SYSTEM BY DR N R KIDWAI, INTEGRAL UNIVERSITY 8
For SSB-SC, s t =
𝐴 𝑐
2
[m t cos 2𝜋𝑓𝑐 𝑡 ± 𝑚 𝑡 sin 2𝜋𝑓𝑐 𝑡 ], +ve sign for LSB and –ve sign for USB
Input to demodulator, 𝑠𝑖 𝑡 =
𝐴 𝑐
2
m t cos 2𝜋𝑓𝑐 𝑡 ±
𝐴 𝑐
2
𝑚 𝑡 sin 2𝜋𝑓𝑐 𝑡 + 𝑛(𝑡)
Signal power at input, 𝑆𝑖 =
𝐴 𝑐
2
m(t)cos 2𝜋𝑓𝑐 𝑡
2
+
𝐴 𝑐
2
𝑚 𝑡 sin 2𝜋𝑓𝑐 𝑡
2
=
𝐴 𝑐
2
8
𝑚2(𝑡) +
𝐴 𝑐
2
8
𝑚
2
(𝑡)
or, 𝑆𝑖 =
𝐴 𝑐
2
4
𝑚2(𝑡) as 𝑚2(𝑡) = 𝑚
2
(𝑡) [ 𝑚 𝑡 is Hilbert transform of m(t)]
Thus signal power is half of that in DSB-SC
Noise power at input, 𝑁𝑖 = 𝑛2(𝑡) = 2 𝑥 𝑊 𝑥
𝜂
2
= 𝑊𝜂
Narrow Band
BPF Hn(f)
Pass band W
Channel
white noise w(t)
s(t)+n(t)
Receiver
LPF
cut off W
s(t) +nw(t)
demodulator
cos(2fct)
sd(t)+nd(t)
Si
Ni
So
No
s(t)
9. Noise performance of SSB-SC system (coherent detection)
31-08-2016 IEC 503 ANALOG COMMUNICATION SYSTEM BY DR N R KIDWAI, INTEGRAL UNIVERSITY 9
Input of multiplier
= 𝑠 𝑡 + 𝑛 𝑡 =
𝐴 𝑐
2
m t cos 2𝜋𝑓𝑐 𝑡 ±
𝐴 𝑐
2
𝑚 𝑡 sin 2𝜋𝑓𝑐 𝑡 + 𝑛𝐼(𝑡)cos 2𝜋𝑓𝑐 𝑡 − 𝑛 𝑄(𝑡)sin 2𝜋𝑓𝑐 𝑡
Output of the multiplier
=
𝐴 𝑐
2
m t 𝑐𝑜𝑠2
2𝜋𝑓𝑐 𝑡 ±
𝐴 𝑐
2
𝑚 𝑡 sin 2𝜋𝑓𝑐 𝑡 cos 2𝜋𝑓𝑐 𝑡 + 𝑛𝐼(𝑡)𝑐𝑜𝑠2
2𝜋𝑓𝑐 𝑡 − 𝑛 𝑄(𝑡)sin 2𝜋𝑓𝑐 𝑡 cos 2𝜋𝑓𝑐 𝑡
=
𝐴 𝑐
4
m t +
𝐴 𝑐
4
m t 𝑐𝑜𝑠 4𝜋𝑓𝑐 𝑡 ±
𝐴 𝑐
4
𝑚 𝑡 sin 4𝜋𝑓𝑐 𝑡 +
𝑛𝐼 𝑡
2
+
𝑛𝐼 𝑡
2
𝑐𝑜𝑠 4𝜋𝑓𝑐 𝑡 −
𝑛 𝑄(𝑡)
2
sin 4𝜋𝑓𝑐 𝑡
Output of Low pass filter=
𝐴 𝑐
4
m t +
𝑛 𝐼 𝑡
2
at the output, Signal power , 𝑆 𝑜 =
𝐴 𝑐
4
m(t)
2
=
𝐴 𝑐
2
16
𝑚2(𝑡) ,
Noise power, 𝑁𝑜 =
𝑛 𝐼 𝑡
2
2
=
1
4
𝑛𝐼
2
(𝑡) =
1
4
𝑥 2𝑥
𝑊
2
𝑥 𝜂 =
𝑊𝜂
4
=
1
4
𝑛2(𝑡)
10. Noise performance of SSB-SC system (coherent detection)
31-08-2016 IEC 503 ANALOG COMMUNICATION SYSTEM BY DR N R KIDWAI, INTEGRAL UNIVERSITY 10
𝐹 =
𝑆𝑖
𝑆 𝑜
𝑥
𝑁𝑜
𝑁𝑖
=
𝐴 𝑐
2
4
𝑚2(𝑡)
𝐴 𝑐
2
16
𝑚2(𝑡)
𝑥
1
4
𝑛𝐼
2
(𝑡)
𝑛2(𝑡)2
= 1
noise figure of SSB-SC receiver, 𝐹 = 1
Figure of merit of DSB-SC receiver =1/ F =1
• It implies that noise performance of SSB-SC is inferior to DSB-SC.
• But closer examination reveals that signal power and bandwidth at the input of SSB-
SC is half of that in DSB-SC (due to one sideband removed in filtering method).
• coherent receiver rejects quadrature component of noise. Also message and in
phase component of noise appear in the output with half magnitude .
• Thus for same average input power and same transmission bandwidth performance
of both DSB-SC and SSB-SC receivers will be same
11. Noise performance of AM system
(Envelope detector)
31-08-2016 IEC 503 ANALOG COMMUNICATION SYSTEM BY DR N R KIDWAI, INTEGRAL UNIVERSITY 11
For AM, s t = 𝐴 𝑐 + 𝑘 𝑎m t 𝑐𝑜𝑠 2𝜋𝑓𝑐 𝑡
Input to demodulator, 𝑠𝑖 𝑡 = 𝐴 𝑐 + 𝑘 𝑎m t 𝑐𝑜𝑠 2𝜋𝑓𝑐 𝑡 + 𝑛(𝑡)
Signal power at input, 𝑆𝑖 = 𝐴 𝑐 + 𝑘 𝑎m t 𝑐𝑜𝑠 2𝜋𝑓𝑐 𝑡
2
= 𝐴 𝑐 𝑐𝑜𝑠 2𝜋𝑓𝑐 𝑡 2 + 𝑘 𝑎m t 𝑐𝑜𝑠 2𝜋𝑓𝑐 𝑡 2
or, 𝑆𝑖 =
𝐴 𝑐
2
2
+
𝑘 𝑎
2
2
𝑚2
(𝑡) =
𝐴 𝑐
2
+𝑘 𝑎
2
𝑚2(𝑡)
2
Noise power at input, 𝑁𝑖 = 𝑛2(𝑡) = 2 𝑥 2𝑊 𝑥
𝜂
2
= 2𝑊𝜂
Narrow Band
BPF Hn(f)
Pass band 2W
Channel
white noise w(t)
s(t)+n(t)
Receiver
s(t) +nw(t)
demodulator
Si
Ni
So
No
s(t) Envelope
detector
12. Noise performance of AM system
(Envelope detector)
Input of multiplier
𝑆𝑖 𝑡 = 𝑠 𝑡 + 𝑛 𝑡 = 𝐴 𝑐 + 𝑘 𝑎m t 𝑐𝑜𝑠 2𝜋𝑓𝑐 𝑡 + 𝑛𝐼(𝑡)cos 2𝜋𝑓𝑐 𝑡 − 𝑛 𝑄(𝑡)sin 2𝜋𝑓𝑐 𝑡
= 𝐴 𝑐 + 𝑘 𝑎m t + 𝑛𝐼(𝑡) 𝑐𝑜𝑠 2𝜋𝑓𝑐 𝑡 − 𝑛 𝑄(𝑡)sin 2𝜋𝑓𝑐 𝑡
𝑆𝑖 𝑡 can be represented in polar form as 𝑆𝑖 𝑡 = 𝑒(𝑡)𝑐𝑜𝑠 2𝜋𝑓𝑐 𝑡 + 𝜓(𝑡)
where 𝑒 𝑡 = 𝐴 𝑐 + 𝑘 𝑎m t + 𝑛𝐼(𝑡) 2 + 𝑛 𝑄
2
(𝑡) is envelope of AM wave
And 𝜓 𝑡 = 𝑡𝑎𝑛−1 𝑛 𝑄(𝑡)
𝐴 𝑐+𝑘 𝑎m t +𝑛 𝐼(𝑡)
is phase angle
31-08-2016 IEC 503 ANALOG COMMUNICATION SYSTEM BY DR N R KIDWAI, INTEGRAL UNIVERSITY 12
13. Noise performance of AM system
(Envelope detector)
31-08-2016 IEC 503 ANALOG COMMUNICATION SYSTEM BY DR N R KIDWAI, INTEGRAL UNIVERSITY 13
𝐴 𝑐 + 𝑘 𝑎m(t) 𝑛𝐼(𝑡)
𝑛 𝑄(𝑡)𝑅(𝑡)𝑒(𝑡)
𝜓(𝑡) 𝜃(𝑡)
small noise
Small noise case
In this case, 𝐴 𝑐 + 𝑘 𝑎m t ≫ 𝑛𝐼 𝑡 𝑜𝑟 𝑛 𝑄 𝑡
Under this condition e(t) can be approximated as
‘𝑒 𝑡 = 𝐴 𝑐 + 𝑘 𝑎m t 2 + 2 𝐴 𝑐 + 𝑘 𝑎m t 𝑛𝐼 𝑡 + 𝑛𝐼
2
(𝑡) + 𝑛 𝑄
2
(𝑡)
Ignoring higher order terms being very small
𝑒 𝑡 ≅ 𝐴 𝑐 + 𝑘 𝑎m t 1 +
2𝑛𝐼 𝑡
𝐴 𝑐 + 𝑘 𝑎m t
1
2
= 𝐴 𝑐 + 𝑘 𝑎m t 1 +
𝑛𝐼 𝑡
𝐴 𝑐 + 𝑘 𝑎m t
Or output of envelope detector. 𝑒 𝑡 ≅ 𝐴 𝑐 + 𝑘 𝑎m t + 𝑛𝐼 𝑡
and 𝜓 𝑡 ≅ 0
14. Noise performance of AM system
(Envelope detector)
31-08-2016 IEC 503 ANALOG COMMUNICATION SYSTEM BY DR N R KIDWAI, INTEGRAL UNIVERSITY 14
Signal power at the output 𝑆 𝑜 = 𝑘 𝑎
2
𝑚2(𝑡) = 𝑘 𝑎
2 𝑚2(𝑡)
Noise power at the output 𝑁𝑜 = 𝑛𝐼
2
(𝑡) = 𝑛2 (𝑡)
Noise figure 𝐹 =
𝐴 𝑐
2+𝑘 𝑎
2 𝑚2(𝑡)
2
𝑘 𝑎
2 𝑚2(𝑡)
𝑥
𝑛2 (𝑡)
𝑛2(𝑡)
=
1+
𝐴 𝑐
2
𝑘 𝑎
2 𝑚2(𝑡)
2
As
𝐴 𝑐
𝑘 𝑎 𝑚(𝑡)
> 1 (modulation index is less than 100 %), noise figure of AM receiver is always greater than 1
For single tone message 𝑚 𝑡 = 𝐴 𝑚 cos 2𝜋𝑓𝑚 𝑡 , 𝑚2(𝑡) = 𝐴 𝑚
2
2
Noise figure 𝐹 =
1+
2
𝑚2
2
where m is modulation index
For maximum modulation (m=1), noise figure will be minimum i.e. Minimum 𝐹 =
3
2
• Noise figure of AM with envelope detector (small noise) is same as that of Noise
figure of AM with coherent detector
15. Noise performance of AM system
(Envelope detector)
31-08-2016 IEC 503 ANALOG COMMUNICATION SYSTEM BY DR N R KIDWAI, INTEGRAL UNIVERSITY 15
𝐴 𝑐 + 𝑘 𝑎m(t) 𝑛𝐼(𝑡)
𝑛 𝑄(𝑡)
𝑅(𝑡)
𝑒(𝑡)
𝜓(𝑡) 𝜃(𝑡)
Large noise
Large noise case
In this case, 𝐴 𝑐 + 𝑘 𝑎m t ≪ 𝑛𝐼 𝑡 𝑜𝑟 𝑛 𝑄 𝑡
Under this condition e(t) can be approximated as
‘ 𝑒 𝑡 = 𝐴 𝑐 + 𝑘 𝑎m t 2 + 2 𝐴 𝑐 + 𝑘 𝑎m t 𝑛𝐼 𝑡 + 𝑛𝐼
2
(𝑡) + 𝑛 𝑄
2
(𝑡)
Ignoring higher order terms being very small and using 𝑅 𝑡 = 𝑛𝐼
2
(𝑡) + 𝑛 𝑄
2
(𝑡)
𝑒 𝑡 ≅ 𝑅(𝑡) 1 +
2 𝐴 𝑐 + 𝑘 𝑎m t 𝑅 𝑡 𝑐𝑜𝑠 𝜃 𝑡
𝑅2 𝑡
1
2
= 𝑅(𝑡) 1 +
𝐴 𝑐 + 𝑘 𝑎m t 𝑐𝑜𝑠 𝜃 𝑡
𝑅 𝑡
Or output of envelope detector. 𝑒 𝑡 ≅ 𝑅 𝑡 + 𝐴 𝑐 𝑐𝑜𝑠 𝜃 𝑡 + 𝑘 𝑎m t 𝑐𝑜𝑠 𝜃 𝑡
As the term containing message 𝑘 𝑎m t 𝑐𝑜𝑠 𝜃 𝑡 is multiplied by noise term, message
signal can not be recovered.
16. Noise performance of FM system
31-08-2016 IEC 503 ANALOG COMMUNICATION SYSTEM BY DR N R KIDWAI, INTEGRAL UNIVERSITY 16
For FM, s t = 𝐴 𝑐 𝑐𝑜𝑠 2𝜋𝑓𝑐 𝑡 + ϕ(𝑡) , where ϕ 𝑡 = 2𝜋𝑘 𝑓 0
𝑡
𝑚 𝑡 𝑑𝑡
Also narrow band noise, 𝑛 𝑡 = 𝑛𝐼 𝑡 cos 2𝜋𝑓𝑐 𝑡 − 𝑛 𝑄 𝑡 sin 2𝜋𝑓𝑐 𝑡 = 𝑅 𝑡 𝑐𝑜𝑠 2𝜋𝑓𝑐 𝑡 + 𝜃(𝑡)
Where 𝑅 𝑡 = 𝑛𝐼
2
(𝑡) + 𝑛 𝑄
2
(𝑡) and 𝜃 𝑡 = 𝑡𝑎𝑛−1 𝑛 𝑄(𝑡)
𝑛 𝐼(𝑡)
Signal power at input, 𝑆𝑖 = 𝐴 𝑐 𝑐𝑜𝑠 2𝜋𝑓𝑐 𝑡 + ϕ(𝑡) 2 = 𝐴 𝑐
2
2
Noise power at input, 𝑁𝑖 = 𝑛2(𝑡) = 2 𝑥 𝐵 𝑇 𝑥
𝜂
2
= 𝐵 𝑇 𝜂
Input to demodulator, 𝑆𝑖 𝑡 = 𝑠 𝑡 + 𝑛(𝑡) = 𝐴 𝑐 𝑐𝑜𝑠 2𝜋𝑓𝑐 𝑡 + ϕ(𝑡) + 𝑅 𝑡 𝑐𝑜𝑠 2𝜋𝑓𝑐 𝑡 + 𝜃(𝑡)
Envelope of𝑆𝑖 𝑡 is of no interest and any variation in envelope will be removed by hard limiter,
while relative phasor is of the interest.
Narrow Band
BPF Hn(f)
Pass band BT
Channel
white noise w(t)
s(t)+n(t)
Receiver
s(t) +nw(t)
demodulator
Si
Ni
So
No
s(t) Hard
limiter
1
2𝜋
𝑑
𝑑𝑡
LPF
Cutoff W
sd(t) +nd(t)
17. Noise performance of FM system
Small noise case
In this case, 𝐴 𝑐 ≫ 𝑛 𝑡 𝑜𝑟 𝑛𝐼 𝑡 𝑜𝑟 𝑛 𝑄 𝑡
Choosing 2𝜋𝑓𝑐 𝑡 + ϕ(𝑡) as reference for phasor diagram.
Relative phase angle 𝜓 𝑡 = 𝜙 𝑡 + 𝑡𝑎𝑛−1 𝑅 𝑡 𝑠𝑖𝑛 𝜃 𝑡 −𝜙(𝑡)
𝐴 𝑐+𝑅 𝑡 𝑠𝑖𝑛 𝜃 𝑡 −𝜙(𝑡)
≅ 𝜙 𝑡 + 𝑡𝑎𝑛−1 𝑅 𝑡 𝑠𝑖𝑛 𝜃 𝑡 −𝜙(𝑡)
𝐴 𝑐
≅ 𝜙 𝑡 +
𝑅 𝑡
𝐴 𝑐
𝑠𝑖𝑛 𝜃 𝑡 − 𝜙(𝑡) [for small . tan = ]
The differentiator output = 𝑆 𝑑 𝑡 + 𝑛 𝑑 𝑡 = 1
2𝜋
𝑑𝜓 𝑡
𝑑𝑡
= 𝑘 𝑓 𝑚 𝑡 +
1
2𝜋𝐴 𝑐
𝑑
𝑑𝑡
𝑅 𝑡 𝑠𝑖𝑛 𝜃 𝑡 − 𝜙(𝑡)
The signal component 𝑘 𝑓 𝑚 𝑡 will pass through LPF, so signal power in the output
𝑆 𝑜 = 𝑘 𝑓
2
𝑚2(𝑡)
31-08-2016 IEC 503 ANALOG COMMUNICATION SYSTEM BY DR N R KIDWAI, INTEGRAL UNIVERSITY 17
𝐴 𝑐
𝑅(𝑡)
𝑒(𝑡)
𝜓 𝑡 − 𝜙(𝑡) 𝜃 𝑡 − 𝜙(𝑡)
18. Noise performance of FM system
Noise component 𝑛 𝑑 𝑡 =
1
2𝜋𝐴 𝑐
𝑑
𝑑𝑡
𝑅 𝑡 𝑠𝑖𝑛 𝜃 𝑡 − 𝜙(𝑡)
As 𝜃 𝑡 is random and uniformly distributed, so it can be fairly assumed that 𝜃 𝑡 − 𝜙(𝑡) is
uniformly distributed over 2. So noise component will be independent of message signal.
So noise component 𝑛 𝑑 𝑡 =
1
2𝜋𝐴 𝑐
𝑑
𝑑𝑡
𝑅 𝑡 𝑠𝑖𝑛 𝜃 𝑡 =
1
2𝜋𝐴 𝑐
𝑑
𝑑𝑡
𝑛 𝑄(𝑡)
Thus noise component at differentiator output. 𝑛 𝑑 𝑡 can be obtained passing 𝑛 𝑄(𝑡) through a
system whose frequency response is given as 𝐻 𝑓 =
𝑗2𝜋𝑓
2𝜋𝐴 𝑐
=
𝑗𝑓
𝐴 𝑐
Power spectral density of noise component 𝑛 𝑑 𝑡 is given as 𝑆 𝑑 𝑓 = 𝐻(𝑓) 2
𝑆 𝑛𝑄 𝑓
Where 𝑆 𝑛𝑄 𝑓 =𝑆 𝑛 𝑓 − 𝑓𝑐 +𝑆 𝑛 𝑓 + 𝑓𝑐 = 𝜂 𝑓𝑜𝑟 𝑓 ≤ 𝐵 𝑇
2
31-08-2016 IEC 503 ANALOG COMMUNICATION SYSTEM BY DR N R KIDWAI, INTEGRAL UNIVERSITY 18
19. Noise performance of FM system
31-08-2016 IEC 503 ANALOG COMMUNICATION SYSTEM BY DR N R KIDWAI, INTEGRAL UNIVERSITY 19
−𝑓𝑐 − 𝐵 𝑇
2
−𝑓𝑐 + 𝐵 𝑇
2
𝜂
2
𝑆 𝑛 𝑓
f
−𝑓𝑐 − 𝐵 𝑇
2
−𝑓𝑐 + 𝐵 𝑇
2
f
𝑆 𝑛𝑄 𝑓
− 𝐵 𝑇
2
𝐵 𝑇
2
f
𝐵 𝑇
2𝐴 𝑐
2
𝑆 𝑛𝑑 𝑓
− 𝐵 𝑇
2
𝐵 𝑇
2
f
𝑊
𝐴 𝑐
2
𝑆 𝑛𝑜 𝑓
− 𝑊 𝑊
𝑆 𝑛𝑑 𝑓 =
𝑓
𝐴 𝑐
2
𝜂 𝑓𝑜𝑟 𝑓 ≤ 𝐵 𝑇
2
Again 𝑆 𝑛𝑜 𝑓 =
𝑓
𝐴 𝑐
2
𝜂 𝑓𝑜𝑟 𝑓 ≤ 𝑊
Output noise power, 𝑁𝑜 = −𝑊
𝑊
𝑆 𝑛𝑜 𝑓 𝑑𝑓
Or 𝑁𝑜 = 𝜂
𝐴 𝑐
2 −𝑊
𝑊
𝑓2
𝑑𝑓 =
2𝜂𝑊3
3𝐴 𝑐
2
20. Noise performance of FM system
31-08-2016 IEC 503 ANALOG COMMUNICATION SYSTEM BY DR N R KIDWAI, INTEGRAL UNIVERSITY 20
Noise figure 𝐹 =
𝐴 𝑐
2
2
𝑘 𝑓
2 𝑚2(𝑡)
𝑥
2𝜂𝑊3
3𝐴 𝑐
2
𝐵 𝑇 𝜂
=
𝑊3
3𝐵 𝑇 𝑘 𝑓
2 𝑚2(𝑡)
For single tone message 𝑚 𝑡 = 𝐴 𝑚 cos 2𝜋𝑊𝑡
𝑚2(𝑡) = 𝐴 𝑚
2
2
. modulation index 𝛽 =
𝑘 𝑓 𝐴 𝑚
𝑊
and and 𝐵 𝑇 = 2W(β + 1)
So, 𝐹 =
1
3𝛽2(β+1)
For NBFM, β < 0.3, 𝐹 ≅
1
3𝛽2
For WBFM, β > 1, 𝐹 ≅
1
3𝛽3
Thus FM improves noise performance with increase in modulation index
21. Noise performance of FM system
31-08-2016 IEC 503 ANALOG COMMUNICATION SYSTEM BY DR N R KIDWAI, INTEGRAL UNIVERSITY 21
Best noise figure obtained in AM is 1.5 (for m=1), for better noise performance of FM than AM,
NBFM noise figure must be less than AM noise figure i.e.
1
3𝛽2 <1.5 or β ≻ 0.4702
Thus for β ≻ 0.5 noise performance of FM is similar to AM while for higher β noise performance of
FM is superior to AM and In FM SNR improves with increase in modulation index.
FM Threshold Effect: Above results were based on assumptions that carrier component is much
lather than noise component.
If input signal is decreased input SNR will also decrease, resulting in decrease in SNR improvement.
There will be a point below which if input SNR decreases, output SNR will decrease more rapidly.
That value of input SNR is referred as FM threshold.
FM threshold is 13 dB or 20
i.e . Input SNR 𝐴 𝑐
2 2
2𝜂𝐵 𝑇
> 20 implies that carrier power at the input 𝐴 𝑐
2
2
>40𝜂𝐵 𝑇
22. Threshold improvement in FM: Pre-emphasis de-emphasis
31-08-2016 IEC 503 ANALOG COMMUNICATION SYSTEM BY DR N R KIDWAI, INTEGRAL UNIVERSITY 22
• In practical signals amplitude decreases with increase in
frequency. This implies lower SNR (more effect of noise) for high
frequency components of message.
• High frequency component may suffer from FM threshold effect.
• High frequency components have lower amplitude but are most
important for quality of message.
This difficulty can be solved by use of Pre-emphasis de-emphasis.
Pre emphasis: High frequency components are boosted (amplified)
before modulation.
De emphasis: After demodulation High frequency components are
de-boosted to restore original signal.
dB
f
Pre-emphasis gain
dB
f
De-emphasis gain