This document discusses synchronization and error performance of bandpass signals. It begins by defining synchronous and non-synchronous signals based on carrier frequency and phase between transmitted and received signals. It then examines the impact of non-synchronization in three cases: when phase is non-synchronous but frequency is synchronous, when frequency is non-synchronous but phase is synchronous, and when both phase and frequency are non-synchronous. Causes of non-synchronization are also discussed. The document derives mathematical relationships for probability of error for amplitude-shift keying, phase-shift keying, and frequency-shift keying signals. Graphs of probability of error versus signal-to-noise ratio are also presented.

1. Synchronization and Error Performance of Bandpass
System
Dr. Varun Kumar
Dr. Varun Kumar (IIIT Surat) Bandpass Signal Reception 1 / 15

2. Outlines
1 Introduction to Signal Synchronization:
2 Impact of Non-synchronous Signal During Reception:
3 Cause of Signal Non-synchronization
4 Error Performance of Bandpass Signal
5 References
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3. Introduction to Signal Synchronization:
Synchronous vs Non-synchronous Signal:
1. Synchronous Signal: If the transmitted signal and received should
have same frequency and phase it is called as synchronous signal
otherwise non-synchronous. Ex- carrier frequency (fc)T = (fc)R and
phase φT = φR
2. Non-synchronous Signal: Let (fc)T , (fc)R be the carrier frequency
of transmitted and received signal and φT , φR be the phase of
transmitted and received signal. Based on the given classification,
there are three cases that cause non-synchronization.
When (fc )T = (fc )R and φT = φR Synchronous freq and
Non-synchronous in phase
When (fc )T = (fc )R and φT = φR Non-synchronous freq and
Synchronous in phase
When (fc )T = (fc )R and φT = φR Both phase and freq are
Non-synchronous
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4. Impact of Non-synchronous Signal During Reception:[1],[2]
Case 1: When (fc)T = (fc)R = fc and φT = φR
Let baseband signal is md (t) and a bandpass amplitude shift keying (ASK)
signal is ΦASK (t). Let φT = φ1 and φR = φ2
ΦASK (t) = md (t)cos(2πfct + φ1) Transmitted Signal (1)
Block Diagram of ASK Receiver
y1(t) = md (t)cos(φ1−φ2)
2 + md (t)cos(4πfc t+φ1+φ2)
2
where, yR(t) = md (t)cos(φ1−φ2)
2
If φ1 − φ2 = π/2 then yR(t) = 0 {Drawback of Non-synchronization}
Dr. Varun Kumar (IIIT Surat) Bandpass Signal Reception 4 / 15

5. Continued:
Case 2: When (fc)T = (fc)R and φT = φR
Let baseband signal is md (t) and a bandpass amplitude shift keying (ASK)
signal is ΦASK (t). Let φT = φR = φ and (fc)T = fc and (fc)R = fc + ∆fc.
In this case transmitted ASK signal can be expressed as
ΦASK (t) = md (t)cos(2πfct + φ) Transmitted Signal (2)
Block Diagram of ASK Receiver
y1(t) = md cos(2π∆fc t)
2 + md cos(2π(2fc +∆fc )t+2φ)
2
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6. Continued–
yR(t) = md cos(2π∆fc t)
2 ⇒ ∆fc < fcutoff
yR(t) = 0 ⇒ ∆fc > fcutoff {This condition merely exist}
yR(t) = 0, At t = 2k+1
4∆fc
∀ k = 0, 1, 2, ...
Case 3: When (fc)T = (fc)R and φT = φR
Let baseband signal is md (t) and a bandpass amplitude shift keying (ASK)
signal is ΦASK (t). Let φT = φ1 and φR = φ2. Carrier frequency
(fc)T = fc and (fc)R = fc + ∆fc. In this case transmitted ASK signal can
be expressed as
ΦASK (t) = md (t)cos(2πfct + φ1) Transmitted Signal (3)
Dr. Varun Kumar (IIIT Surat) Bandpass Signal Reception 6 / 15

7. Continued–
Block Diagram of ASK Receiver:
y1(t) = md cos(2π∆fc t+φ1−φ2)
2 + md cos(2π(2fc +∆fc )t+2φ)
2
yR(t) = md cos(2π∆fc t+φ1−φ2)
2 ⇒ ∆fc < fcutoff
yR(t) = 0 ⇒ ∆fc > fcutoff ∀ φ1, φ2
It is noted that the recover received signal are not digital in nature,
when there arise a frequency non-synchronization in communication
system.
Due to non-synchronous phase, the recover signal remains in digital
and by proper phase calibration this detrimental effect can be nullified.
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8. Cause of Signal Non-synchronization
Note: For analyzing the effect of non-synchronization, only ASK bandpass
signal has been taken for consideration. We can also analyze the effect of
non-synchronization for other bandpass signal (PSK, FSK, QAM)
Cause of Signal Non-synchronization:
The main causes of signal non-synchronization are as follow,
Due to multiple path between transmitter and receiver causes phase
non-synchronization.
Due to relative velocity (Doppler) between transmitter and receiver
causes apparent change in frequency, thus frequency non-
synchronization occur in communication system.
Usage of improper hardware (multiple electronic components) used
across the transmitter and receiver unit.
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9. Error Performance of Bandpass Signal
Bandpass signal: ASK, PSK, FSK
Probability of error occur: 1 is transmitted and 0 is received.
Mathematical Relation:
SNR, γ2
max = 2
η
T
0 |e(t)|2dt
Pe = 1
2erfc{γ2
max
8 }1/2
where η → Noise Power Spectral Density
T → Bit Duration
e(t) → Error Signal
Pe → Probability of Error
γ2
max = Signal to Noise Ratio (SNR)
erfc{.} → Error Function
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10. ASK Signal and Probability of Error:
ASK
In case ASK digital modulation technique:
1 → x1(t) = Acos(2πfct) ∀ 0 < t < T
0 → x2(t) = 0 ∀ 0 < t < T
e(t) → x1(t) − x2(t) = Acos(2πfct)
γ2
max = 2
η
T
0 A2cos2(2πfct)dt = A2T/η
Pe = 1
2erfc{A2T
8η }
Since Es = A2T
2 = Energy of the symbol, hence Pe = 1
2erfc{Es
4η }
Note: It is advised that all student study about the details of error
function as well as Q-function from reference book.
Dr. Varun Kumar (IIIT Surat) Bandpass Signal Reception 10 / 15

11. PSK Signal and Probability of Error:
PSK
1 → x1(t) = Acos(2πfct) ∀ 0 < t < T
0 → x2(t) = −Acos(2πfct) ∀ 0 < t < T
e(t) → x1(t) − x2(t) = 2Acos(2πfct)
γ2
max = 2
η
T
0 4A2cos2(2πfct)dt = A2T/η
Pe = 1
2erfc{A2T
8η }
Since Es = A2T
2 = Energy of the symbol, hence Pe = 1
2erfc{Es
η }
Dr. Varun Kumar (IIIT Surat) Bandpass Signal Reception 11 / 15

12. FSK Signal and Probability of Error:
FSK
Let f1 > f2, also f1 = fc + Ω and f2 = fc − Ω, ∀ fc >> Ω
1 → x1(t) = Acos(2πf1t) ∀ 0 < t < T
0 → x2(t) = Acos(2πf2t) ∀ 0 < t < T
e(t) → x1(t) − x2(t) = −2Asin(2πfct)sin(2πΩt)
γ2
max =
2
η
T
0
4A2
sin2
(2πfct)sin2
(2πΩt)dt
=
2
η
T
0
4A2 1 − cos(4πfct) 1 − cos(4πΩt)
4
dt
=
2A2T
η
−
2A2sin(4πfcT)
4πfcη
−
2A2sin(4πΩT)
4πΩη
+
A2sin(4π(fc + Ω)T)
4π(fc + Ω)η
=
A2sin(4π(fc − Ω)T)
4π(fc − Ω)η
(4)
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13. Continued–
In communication system, fc, Ω, fc + Ω, fc − Ω >> 1, Since
−1 < sin(θ) < 1
Hence
γ2
max =
2A2T
η
(5)
Pe = 1
2erfc{A2T
4η } = 1
2erfc{Es
2η }
-10 -5 0 5 10 15
SNR in dB
10-6
10-5
10-4
10-3
10-2
10-1
100
ProbabilityofError
ASK
PSK
FSK
Dr. Varun Kumar (IIIT Surat) Bandpass Signal Reception 13 / 15

14. References
B. P. Lathi, Z. Ding et al., “Modern Digital and Analog Communication Systems /
BP Lathi, Zhi Ding.” 2010.
M. Borda, Fundamentals in information theory and coding. Springer Science &
Business Media, 2011.
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15. Thank You
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