ASK, FSK, PSK - Faculty of Computing, Engineering and Technology

Faculty of Computing,
Engineering & Technology
Digital modulation:
ASK, FSK, PSK
Communications
COMMS (CE700038-2)
Alison L Carrington
C203
[email protected]
www.fcet.staffs.ac.uk/alg1
2008/9
Digital modulation:
ASK, FSK, PSK
2
Overview
Why is modulation is needed?
What is modulation?
Digital modulation & demodulation techniques
Obtain SNR of system by:
Signal Power
Noise Power
Determine the performance of different Digital
Mod/Demod techniques with different amounts
of noise
COMMS (CE700038-2) 2008/9
3
Introduction
Digital modulation:
ASK, FSK, PSK
Often send information in a digital form
As binary words, bits represented by 1’s & 0’s
Could be from a keyboard, characters encoded as
1’s & 0’s
Or a microphone (speech)
Or a temperature sensor
converted to binary form by A/D converter.
Why do we need to modulate?
COMMS (CE700038-2) 2008/9
Digital modulation:
ASK, FSK, PSK
4
Why modulate?
The purpose of a communication system is to
transfer information from a source to a
destination
Problems arise in baseband transmissions:
Noise
external & circuit noise reduces the signal-to-noise (S/N)
ratio at the receiver, (Rx) reducing the quality of the
output.
Not able to fully utilise the available bandwidth:
telephone quality speech has a bandwidth ≃ 3kHz
co-axial cable has a bandwidth of 100's of Mhz.
COMMS (CE700038-2) 2008/9
5
What is Modulation? #1
Digital modulation:
ASK, FSK, PSK
A message signal containing information is used to
control parameters of a carrier signal
i.e. the information is embedded onto the carrier
The message can either be:
Analogue – m(t)
Digital – d(t), sequence of 1's and 0's
When message is
analogue = Analogue Modulation
digital = Digital Modulation
The carrier could either a 'sinusoidal wave' or a 'pulse
train‘
At the destination the carrier+message must be
demodulated so that the message can be Rx’d
COMMS (CE700038-2) 2008/9
6
What is Modulation? #2
Digital modulation:
ASK, FSK, PSK
If the message d(t) controls
amplitude = AMPLITUDE SHIFT KEYING ASK.
Frequency = FREQUENCY SHIFT KEYING FSK
Phase
= PHASE SHIFT KEYING PSK.
d(t) is a binary or 2 level signal representing 1's and 0's
This is binary or 2 level, e.g. Binary FSK, BFSK, BPSK,
etc. or 2 level FSK, 2ASK, 2PSK etc.
COMMS (CE700038-2) 2008/9
7
What is Modulation? #3
Digital modulation:
ASK, FSK, PSK
Multiple Levels:
A message signal could be multi-level or m levels
each level represents a discrete pattern of 'information' bits.
For example, m = 4 levels
Why increase M?
As the number of levels increase the amount of information
transmitted increases – but the probability of receiving an
error also increases
COMMS (CE700038-2) 2008/9
Digital modulation:
ASK, FSK, PSK
8
Baud Rate #1
The baud rate of a data communications
system is the number of symbols per second
transferred.
As a symbol may have > 2 states, it may
represent > 1 binary bit (a binary bit always
represents exactly two states).
Therefore the baud rate may not equal the bit
rate
COMMS (CE700038-2) 2008/9
9
Baud Rate #2
Digital modulation:
ASK, FSK, PSK
A Bell 212A modem uses QPSK modulation
each symbol has one of four phase shifts (of 0(deg), 90(deg),
180(deg), or 270(deg))
two bits represent four states (00, 01, 10, and 11),
the modem transmits 1,200 bits/s of information, using a
symbol rate of 600 baud.
Usually the baud rate of a modem will not equal the bit
rate and is of no interest to the end user--only the data
rate, in bits per second.
As modems transfer signals over a telephone line
the baud rate is limited to a maximum of 2400 baud.
This is a physical restriction of telephone lines
Higher data throughput achieved with 9600 or higher baud
modems is accomplished by using sophisticated phase
modulation, and data compression techniques
COMMS (CE700038-2) 2008/9
10
Baud Rate #3
Digital modulation:
ASK, FSK, PSK
Asymmetric Digital Subscriber Line (ADSL) is
used for Broadband Internet access
Allows faster data transmission over copper telephone lines
than a conventional voiceband modem
By using the frequencies that are not used by a voice
telephone call using Frequency Division Multiplexing (FDM).
A splitter - or microfilter - allows a single telephone
connection to be used for both ADSL service and voice
calls simultaneously
As phone lines vary in quality and were not originally
engineered with DSL in mind, it can generally only be used
over short distances, typically less than 4km.
COMMS (CE700038-2) 2008/9
11
Baud Rate #4
Digital modulation:
ASK, FSK, PSK
Data
+ve
Bi polar
NRZ
-ve
+ve
Bi polar
RZ
-ve
+ve
Bi polar
Bi Phase
-ve
Bit rate=1/τB
Baud rate=1/τE
‘1’
τB
‘0’
‘1’
‘1’
‘0’
‘0’
Baud rate=1/τE=1000Baud
τE
Baud rate=1/0.5m=2000Baud
τE
Baud rate=1/0.5m=2000Baud
τE
Advantage of a transition in every bit – self checking
e.g. if τB=1ms, Bit rate=1000bps
COMMS (CE700038-2)
2008/9
12
System Block Diagram
(S/N)
Channel
BW
MOD
Digital modulation:
ASK, FSK, PSK
d(t)
MODULATOR
ASK
FSK
PSK
Noise
vn(t)
B Hz p0
CHANNEL
EG. Txn line
radio
satellite
Filter
Bn
DEMOD
Noise
bandwidth
DEMODULATOR
Decision
Detector
d(t)
prob of
error
p
DETECTOR/
DECISION
COMMS (CE700038-2) 2008/9
Amplitude Shift Keying, ASK
13
Different amplitudes of carrier
Usually, one amplitude is ‘0’
Digital modulation:
ASK, FSK, PSK
i.e. presence and absence of carrier is used
Susceptible to sudden gain changes
Inefficient, up to 1200bps on voice grade lines
Used over optical fiber – due to low error
source
V1
V0
Amplitude
Shift Keying
ASK
1
T
0
destination
0
1
0
0
1
1
digital
signal
On off Keying,
OOK
COMMS (CE700038-2) 2008/9
14
ASK Modulation of data sequence d(t)
Amplitude Shift Keying
Digital modulation:
ASK, FSK, PSK
Analogue Multiplier
d(t)
V(t)=d(t).VCcos(ωCt)
ASK/OOK
d(t)=1 for ‘1’
d(t)=0 for ‘0’
Carrier
VCcos(ωCt)
Analogue Gate (4066)
Carrier: VCcos(ωCt)
V(t)=d(t).VCcos(ωCt)
d(t)=1 for ‘1’
d(t)=0 for ‘0’
0V
COMMS (CE700038-2) 2008/9
Digital modulation:
ASK, FSK, PSK
15
Phase Shift Keying, PSK
Phase of carrier signal is shifted data
Binary PSK: Two phases represent two binary
digits
Differential PSK: Phase shifted relative to
previous transmission rather than some
reference signal
COMMS (CE700038-2) 2008/9
16
PSK Modulation of data sequence, d(t)
d(t)
Digital modulation:
ASK, FSK, PSK
Phase Shift
Keying
V(t)=d(t).VCcos(ωCt)
d(t)= ‘1’
d(t)= ‘0’
Carrier
VCcos(ωCt)
(+1).VCcos(wCt)
(-1).VCcos(wCt)
d(t) ‘1’ or ‘0’
Osc
‘in-phase’
or
‘phase inverted’
VCcos(ωCt)
+VCcos(ωCt)
-1
or
-VCcos(ωCt)
COMMS (CE700038-2) 2008/9
Frequency Shift Keying, FSK
Digital modulation:
ASK, FSK, PSK
17
Two binary values represented by two different
frequencies (near carrier)
Less susceptible to error than ASK
Applications: Up to 1200bps on voice grade lines, High
frequency radio, even higher frequency on LANs using
co-ax
source
V1
V0
Frequency
Shift Keying
FSK
1
T
0
destination
0
1
0
0
1
1
digital
signal
Various types:
OFSK, FFSK
COMMS (CE700038-2) 2008/9
18
FSK Modulation of d(t)
Digital modulation:
ASK, FSK, PSK
Frequency Shift Keying
VIN
d(t)
fOUT
FSK
V/F
Osc
f1
f1=fC+αV
f0=fC
fC
‘1’
t
d(t) ‘1’ or ‘0’
‘0’
V
VIN
f1 or f0
Osc
f0
COMMS (CE700038-2) 2008/9
19
System Block Diagram – model for analysis
(S/N)
Channel
BW
MOD
Digital modulation:
ASK, FSK, PSK
d(t)
MODULATOR
ASK
FSK
PSK
Noise
vn(t)
B Hz p0
CHANNEL
EG. Txn line
radio
satellite
Filter
Bn
DEMOD
Noise
bandwidth
DEMODULATOR
Decision
Detector
d(t)
prob of
error
p
DETECTOR/
DECISION
COMMS (CE700038-2) 2008/9
20
Channel - Impairments
When signals are transmitted they are subject to noise
Signal y = cos(x)
Amplitude (V)
2
0
-2
-4
0
1
2
3
time (s)
4
5
6
0
1
2
3
time (s)
4
5
6
4
Amplitude (V)
Digital modulation:
ASK, FSK, PSK
4
2
0
-2
-4
Noise is assumed to be
ADDITIVE, WHITE, GAUSSIAN, AWGN – mean of zero
infinite bandwidth – hence a bandlimiting filter is required
p0 is the noise power spectral density (Watts/Hertz)
COMMS (CE700038-2) 2008/9
21
System Block Diagram – model for analysis
(S/N)
Channel
BW
MOD
Digital modulation:
ASK, FSK, PSK
d(t)
MODULATOR
ASK
FSK
PSK
Noise
vn(t)
B Hz p0
CHANNEL
EG. Txn line
radio
satellite
Filter
Bn
DEMOD
Noise
bandwidth
DEMODULATOR
Decision
Detector
d(t)
prob of
error
p
DETECTOR/
DECISION
COMMS (CE700038-2) 2008/9
ASK & PSK Demodulator
22
Digital modulation:
ASK, FSK, PSK
Optimum Demodulator, based on ‘correlation’:
Signal
+
Noise
p0
i.e. how similar is the received signal to one of the
possible transmitted signals:
(S/N)IN
VIN
Bn
cos(ωCt)
carrier
1
T
∫
T
0
VINdt
I&D
∆, ND
+
D
VREF -
Integrate & Dump,
T is the bit duration
ck
Data
Clock
Q
d(t)
prob of
error
p
COMMS (CE700038-2) 2008/9
23
Demodulator – FSK
Digital modulation:
ASK, FSK, PSK
Optimum Demodulator, based on ‘correlation’:
Signal
+
Noise
p0
1
T
(S/N)IN
cos(ω1t)
∫
T
0
dt
I&D
+
Bn
-
1
T
cos(ω0t)
∫
T
0
dt
I&D
∆, ND
+
D
VREF -
ck
Data
Clock
Q
d(t)
prob of
error
p
A requirement to receive the
carrier ωC or ω1 & ω0 & the
2008/9
‘bit rate’ COMMS
clock (CE700038-2)
- DATA CLOCK
24
Demodulator – Circuit – A/P/FSK
Digital modulation:
ASK, FSK, PSK
I&D
pulse
SIN
S+N
VIN
Bn
p0
NIN=p0Bn
+
R
cos(ωCt)
carrier
VREF
+V
∆, ND
-V
Input signal + bandlimited noise NIN=poBN
multiplied by a carrier
Cos(ωct) for ASK/PSK
Cos(ω0t) & Cos(ω1t) for FSK
COMMS (CE700038-2) 2008/9
25
PSK demodulation ex #1
Digital modulation:
ASK, FSK, PSK
If d(t) is a ‘1’, input is d(t)VCcos(ωCt)
VIN=V1=d(t)VCcos(ωCt)*cos(ωCt)+Noise
VIN=d(t)VCcos2(ωCt)+Noise
d(t)
d(t)
VIN =
VC +
VC cos(2ϖ C t) + Noise
2
2
If d(t) is a ‘0’, input is -d(t)VCcos(ωCt)
VIN=-d(t)VCcos2(ωCt)+Noise
d(t)
d(t)
VIN = −
VC −
VC cos(2ϖ C t) + Noise
2
2
VIN
COMMS (CE700038-2) 2008/9
26
PSK demodulation ex #2
The Integrate & Dump performs the ‘function’
Digital modulation:
ASK, FSK, PSK
where T is the duration of a bit
1
T
∫
T
0
T
VIN dt
1
0
1
1
T
The Integrator
VOUT
1 RC
=−
VIN dt
∫
0
RC
Design RC=T
VIN
+
R
0V
+V
VOUT
-V (CE700038-2) 2008/9
COMMS
PSK demodulation ex #3
Digital modulation:
ASK, FSK, PSK
27
I&D pulse shown, allows the integrator to integrate for
a time T, after which the integrator is cleared or
‘dumped’ by shorting out C & the integration recommences
T
1
0
1
1
I&D
∆ Volts
VREF
VOUT
Decision
‘1’ or ‘0’
‘1’
‘0’
COMMS (CE700038-2)
‘1’
‘1’
2008/9
28
PSK demodulation ex #4
Digital modulation:
ASK, FSK, PSK
Integrate & Dump Output:
Input
VIN
1
T
∫
T
0
VINdt
VOUT=
1
T
∫
T
0
VINdt
I&D
cos(ωCt)
Input = d(t)VCcos(ωCt)
1 T
2
d
(
t
).
V
cos
(ϖ C t)dt
C
∫
0
T
V d(t)
= C
2
VOUT =
VOUT
COMMS (CE700038-2) 2008/9
29
V0
VREF
V1
Digital modulation:
ASK, FSK, PSK
-
∆
VREF
D
+
ck
Data
Clock
Q
d(t)
prob of
error
p
V0 is the voltage when a ‘0’ was transmitted
V1 is the voltage when a ‘1’ was transmitted
The peak-to-peak input to the comparator is
Change in Volts ∆ = V0-V1
This difference between a ‘1’ & a ‘0’
The reference VREF is set midway between V1 & V0.
VREF
∆
∆ V1 + V0
= V0 − = V1 + =
2
2
2
A HALF LEVEL THRESHOLD DETECTOR
If V1=10V, V0=5V: ∆=2.5 & VREF=7.5VCOMMS (CE700038-2)
2008/9
PSK demodulation ex #6
30
Digital modulation:
ASK, FSK, PSK
Integration in presence of Noise
i.e.
Comp
output
which is clocked at the Data Clock Rate
This makes a final decision if a ‘1’ or a ‘0’ is received
1
Noise
1
Noise
0
1
1
∆ Volts
VREF
Noise
causes
this
0
1
0
1
0
1
Data
CLK
Q output
0, error
Noise causes the integrator to fluctuate & causes errors
COMMS (CE700038-2) 2008/9
31 Modulation & Demodulation
FSK
f
f
f
f
V1
1
0
Digital modulation:
ASK, FSK, PSK
V0
1
0
0
1
V1
1
d(t)
fOUT
f1
0
1
MOD
V0
DEMOD
V/F
F/V
VOUT
gdt=α KHz/V
V1
0
0
1
gdt=K V/KHz
fC
V0
f0
VIN
VDC
TB
f0
f1
fIN
‘1’
‘0’
V0
t
V1
K is usually designed to be 1/α
COMMS (CE700038-2) 2008/9
32
FSK Demodulation #2
gdt= α KHz/V
fOUT
Digital modulation:
ASK, FSK, PSK
fC
f0
VDC
TB
‘1’
‘0’
t
V1
V0
Bit rate=Baud rate
f1=fC+αV1
Peak-to-peak deviation =f1-f0
fC=αVDC
y=mx+c
f0=fC-αV0 (V0 is –ve)
fOUT=αVIN+f0
VIN
VIN=VDC+d(t)
VIN=VDC+V1 for 1’s
VIN=VDC+V0 for 0’s
f1=αVDC+αV1+f0 for 1’s
fC=αVDC, TC=1/fC
Normalised frequency Deviation ratio, h =
f1 − f 0
Rb
Rb=1/τB
i.e. modulus of f1-f0
COMMS (CE700038-2) 2008/9
33
FSK Demodulation #3
Digital modulation:
ASK, FSK, PSK
The spectrum of the FSK signal depends on h.
Waveforms
V1
V0
1
0
0
1
0
0
1
Bit rate=Baud rate
1 Rb=1/τB
VDC
TB
t
Frequency
Shift Keying
FSK
t
f1
f0
f1
f0
f1
COMMS (CE700038-2) 2008/9
34
System Block Diagram – model for analysis
(S/N)
Channel
BW
MOD
Digital modulation:
ASK, FSK, PSK
d(t)
MODULATOR
ASK
FSK
PSK
Noise
vn(t)
B Hz p0
CHANNEL
EG. Txn line
radio
satellite
Filter
Bn
DEMOD
Noise
bandwidth
DEMODULATOR
Decision
Detector
d(t)
prob of
error
p
DETECTOR/
DECISION
COMMS (CE700038-2) 2008/9
35
Performance of Digital Systems #1
Digital modulation:
ASK, FSK, PSK
Performance of Digital Data Systems is
dependent on the bit error rate, b.e.r.*
i.e. the probability of a bit being in error
p=
no. of errors, E
as N → ∞
Total no. of bits N
p depends on several factors
The
The
The
The
modulation type: ASK, FSK or PSK
demodulation method
noise in the ‘system’
signal-to-noise ratio
*p is a prediction of how the system will perform
b.e.r. is a measure of how the system actually performs
COMMS (CE700038-2) 2008/9
36
Performance #2
Digital modulation:
ASK, FSK, PSK
Obtain the received
Signal Power
Noise Power
Calculate the SNR based on a particular
modulation method
Predict the theoretical performance using Error
Function Tables
COMMS (CE700038-2) 2008/9
37
Performance #3
Digital modulation:
ASK, FSK, PSK
(S/N)IN
Filter
Bn
DEMOD
∆, ND
VREF
+
D
-
ck
Q
d(t)
prob of
error
p
Data
Clock
Find ∆, ND in terms of SIN & NIN
p as a function of (S/N)IN
COMMS (CE700038-2) 2008/9
38
Performance #4 - Signal Power, SIN
Digital modulation:
ASK, FSK, PSK
Usually assume an un-modulated carrier
i.e. a signal=VCcos(ωCt) - ALWAYS
VC
Normalised Average Power=(VRMS)2
VC
VRMS =
Vpeak
2
=
VC
2
2
2
Normalised Average Power=(VRMS)2= ⎛⎜ VC ⎞⎟ = VC = S IN
2
⎝ 2⎠
2
S IN
VC
=
Watts
2
Similarly for FSK & PSK
VC
COMMS (CE700038-2) 2008/9
Digital modulation:
ASK, FSK, PSK
39
Performance #4 - Noise Power, NIN
Channel
BW
vn(t)
Signal
+
Noise
B Hz
VIN
Bn
p0
‘Peak’ value of noise,
Normalised Average Noise Power,
NIN=p0Bn
vn(t)IN
= 2p0 B
∴ N IN
∫
T
0
VINdt
Dec
Det
I&D
cos(ωCt)
carrier
⎛ N pk
= ⎜⎜
⎝ 2
1
T
Contrived to be
equivalent to
noise power
⎞
⎟⎟
⎠
2
COMMS (CE700038-2) 2008/9
40
Performance #5 – SNR & p
Digital modulation:
ASK, FSK, PSK
SIN/NIN
2
S IN
2
N pk
S IN VC
=
/
N IN
2
2
2
VC
=
Watts
2
N IN
⎛ N pk
= ⎜⎜
⎝ 2
⎞
⎟⎟
⎠
2
2
2
S IN
VC
=
N IN N pk 2
VC
=
2
N pk
Substitute into the correct equation to obtain p
B - ASK
OOK
B − FSK
B − PSK
PRK
⎞
⎟
⎟
⎠
S IN ⎞⎟
1 ⎛⎜
p = ⎜1 − erf
2⎝
2 N IN ⎟⎠
S IN ⎞⎟
1 ⎛⎜
p = ⎜1 − erf
2⎝
N IN ⎟⎠
COMMS (CE700038-2)
1 ⎛⎜
p = ⎜1 − erf
2⎝
S IN
4 N IN
2008/9
41
Performance #6
B - ASK
OOK
Digital modulation:
ASK, FSK, PSK
B − FSK
B − PSK
PRK
1 ⎛⎜
p = ⎜1 − erf
2⎝
⎞
⎟
⎟
⎠
S IN ⎞⎟
1 ⎛⎜
p = ⎜1 − erf
2⎝
2 N IN ⎟⎠
S IN ⎞⎟
1 ⎛⎜
p = ⎜1 − erf
2⎝
N IN ⎟⎠
S IN
4 N IN
1
10-1
p
PSK is ‘best performance’
FSK is next
ASK is worst
10-2
10-4
10-5
PSK FSK ASK
5dB
(S/N)IN dB
COMMS (CE700038-2) 2008/9
Error Function
42
Digital modulation:
ASK, FSK, PSK
The "error function" encountered in integrating the normal distribution (which is a
normalized form of the Gaussian function).
It is an entire function defined by :
erf ( z ) =
2
π
∫
z
0
e −t dt
2
erfc( z ) = 1 − erf ( z )
x
erf(x)
erfc(x)
0
0.000
1.000
0.1
0.112
0.888
0.2
0.223
0.777
0.3
0.329
0.671
0.4
0.428
0.572
0.5
0.520
0.480
0.6
0.604
0.396
0.7
0.678
0.322
0.8
0.742
0.258
0.9
0.797
0.203
1
0.843
0.157
1.1
0.880
0.120
1.2
0.910
0.090
1.3
0.934
0.066
x
erf(x)
erfc(x)
1.4
0.952
0.048
1.5
0.966
0.034
1.6
0.976
0.024
1.7
0.984
0.016
1.8
0.989
0.011
1.9
0.993
0.007
2
0.995
0.005
2.1
0.997
0.003
2.2
0.998
0.002
2.3
0.999
0.001
2.4
0.999
0.001
2.5
1.000
0.000
2.6
1.000
0.000
2.7
1.000
0.000
x
erf(x)
erfc(x)
2.8
1.000
7.50E-05
2.9
1.000
4.11E-05
3
1.000
2.21E-05
3.1
1.000
1.16E-05
3.2
1.000
6.03E-06
3.3
1.000
3.06E-06
3.4
1.000
1.52E-06
3.5
1.000
7.43E-07
COMMS (CE700038-2) 2008/9
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Tutorial 1: find SNRdB for each case
Case B
Case A
Signal y = cos(x)
Signal + Noise: mean=0 & variance=0.1
Digital modulation:
ASK, FSK, PSK
4
4
2
0
0
-2
-4
0
Case C
-2
SNR = ?
1
2
3
4
5
6
-4
0
Case D
Signal + Noise: mean=0 & variance=0.5
4
1
2
3
4
5
6
Signal + Noise: mean=0 & variance=1.0
4
Npk = 1.76V
2
0
-2
-2
1
2
3
4
Npk = 3.16V
2
0
-4
0
Npk = 0.302V
2
Npk = 0V
5
6
-4
0
1
2
3
4
5
6
COMMS (CE700038-2) 2008/9
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Solution 1
SNR for case A:
Digital modulation:
ASK, FSK, PSK
2
S IN
VC
1
=
= =∞
2
0
N IN N pk
SNR for case B:
S IN
1
=
= 10.96
2
N IN (0.302 )
⎛ Sin ⎞
⎟⎟
SNRdB = 10 log10 ⎜⎜
⎝ N in ⎠
SNRdB = ∞dB
SNRdB = 10 log10 (10.96)
SNRdB = 10.40dB
SNR for case C: -4.91dB
SNR for case D: -10.00dB
COMMS (CE700038-2) 2008/9
Digital modulation:
ASK, FSK, PSK
45
Tutorial 2
With Reference to the following equations and
error function tables, what is the performance
of Binary ASK, FSK & PSK at 0, 4 & 8dB?
Plot your results on logarithmic / linear paper
Comment on the performance at 4dB, rank the
modulation schemes and comment
COMMS (CE700038-2) 2008/9
Solution 2
SNR (dB)
0
1
2
3
4
5
6
7
8
9
10
1.E+00
probability of error, p
Digital modulation:
ASK, FSK, PSK
46
1.E-01
1.E-02
1.E-03
1.E-04
1.E-05
1.E-06
ASK
FSK
PSK COMMS (CE700038-2) 2008/9
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Tutorial 3
Digital modulation:
ASK, FSK, PSK
Explain what is meant by ‘Baud Rate’
If data signal is represented as
h(t) = +V for 1’s
= -V for 0’s
h(t)
v(t)
cos(ωCt)
i) What modulation scheme is this?
ii) Sketch the output waveform, v(t)
iii)Obtain Vout(t) stating any assumptions
h(t)[cos(ωCt)]
LPF
cos(ωCt)
VOUT(t)
COMMS (CE700038-2) 2008/9
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Solution 3
Digital modulation:
ASK, FSK, PSK
i) Phase reversed keying
ii)
iii) Vout(t) :
VOUT (t ) = h(t ) cos(ϖ C t ). cos(ϖ C t )
VOUT (t ) =
h(t )
cos(0 ) +
2
h(t )
cos(2ϖ C t )
1244244
3
use LPF to filter out this component
VOUT (t ) =
h(t )
2
COMMS (CE700038-2) 2008/9