Digital modulation

1
Analog TransmissionAnalog Transmission
of Digital Data:of Digital Data:
ASK, FSK, PSK, QAMASK, FSK, PSK, QAM
CSE 3213, Fall 2010
Instructor: N. Vlajic
Required reading:
Garcia 3.7

2
Why Do We Need Digital-to-Analog Conversion?!
1) The medium/channel is band pass, and/or
2) Multiple users need to share the medium.

3
Modulation of Digital Data
ModulationModulation – process of converting digital data or a low-pass
analog to band-pass (higher-frequency) analog signal
Digital-to-analog modulation.
Analog-to-analog modulation.
Carrier SignalCarrier Signal – aka carrier freq. or modulated signal - high freq.
signal that acts as a basis for the information signal
• information signal is called modulating signal
freq
bandpass channel

4
Modulation of Digital Data (cont.)
DigitalDigital--toto--AnalogAnalog
ModulationModulation
– process of changing one of the characteristic
of an analog signal (typically a sinewave) based
on the information in a digital signal
• sinewave is defined by 3 characteristics (amplitude,
frequency, and phase) ⇒ digital data (binary 0 & 1)
can be represented by varying any of the three
• application: transmission of digital data over
telephone wire (modem)
Types of DigitalTypes of Digital--toto--AnalogAnalog

5
Modulation of Digital Data: ASK
ASKASK – strength of carrier signal is varied to represent binary 1 or 0
• both frequency & phase remain constant while amplitude changes
• commonly, one of the amplitudes is zero
• demodulation: only the presence or absence of a sinusoid in a
given time interval needs to be determined
• advantage: simplicity
• disadvantage: ASK is very susceptible to noise interference –
noise usually (only) affects the amplitude, therefore ASK is the
modulation technique most affected by noise
• application: ASK is used to transmit digital data over optical fiber
⎩
⎨
⎧
=
⎩
⎨
⎧
=
1binaryt),fAcos(2
0binary0,
1binaryt),fcos(2A
0binaryt),fcos(2A
s(t)
cc1
c0
ππ
π
Is this picture,
from the textbook,
entirely correct?!
+A
-A

6
Modulation of Digital Data: ASK (cont.)
Example [ ASK ]
vd(t)
vc(t)
vASK(t)
How does the frequency spectrum of vASK(t) look like!?
fvd(f) vc(f)

7
Modulation of Digital Data: ASK (cont.)
ASKASK--Modulated Signal: Frequency SpectrumModulated Signal: Frequency Spectrum
t)cos(ωt)fcos(2(t)v ccc == π
⎥
⎦
⎤
⎢
⎣
⎡
−+−+⋅= ...tcos5ω
5π
2
tcos3ω
3π
2
tcosω
π
2
2
1
A(t)v 000d
Carrier signal: , where 2πfc=ωc
Digital signal:
(unipolar!!!)
ωc ωωd_max
Modulated signal:
( ) ( )[ ]
( ) ( )[ ] ...t3ωωcost3ωωcos
3π
1
-tωωcostωωcos
π
1
tcosω
2
1
...tcos3ωtcosω
3π
2
-tcosωtcosω
π
2
tcosω
2
1
...tcos5ω
5π
2
tcos3ω
3π
2
tcosω
π
2
2
1
tcosω
(t)v(t)v(t)v
0c0c
0c0cc
0c0cc
000c
dcASK
+++−−
++−+=
=+⋅⋅+=
=⎥
⎦
⎤
⎢
⎣
⎡
−+−+⋅=
=⋅=
( )B)cos(AB)-cos(A
2
1
cosBcosA ++=⋅
ωc ωωc+ωd_maxωc-ωd_max

8
Modulation of Digital Data: FSK
FSKFSK – frequency of carrier signal is varied to represent binary 1 or 0
• peak amplitude & phase remain constant during each bit interval
• demodulation: demodulator must be able to determine which of
two possible frequencies is present at a given time
• advantage: FSK is less susceptible to errors than ASK – receiver
looks for specific frequency changes over a number
of intervals, so voltage (noise) spikes can be ignored
• disadvantage: FSK spectrum is 2 x ASK spectrum
• application: over voice lines, in high-freq. radio transmission, etc.
⎩
⎨
⎧
=
1binaryt),fAcos(2
0binaryt),fAcos(2
s(t)
2
1
π
π
f1<f2
+A
-A

9
Modulation of Digital Data: FSK (cont.)
Example [ FSK ]
vd(t)
vc1(t)
vc2(t)
vFSK(t)
ω1 ωωd_max ω2
ω1 ωω1-ωd_max
ω2 ω2+ωd_max

10
Modulation of Digital Data: FSK (cont.)
FSKFSK--Modulated Signal: Frequency SpectrumModulated Signal: Frequency Spectrum
(t)vd
Digital signal: - modulated with ω1 , and
- modulated with ω2
Modulated signal:
( ) ( )[ ]
( ) ( )[ ]
( ) ( )[ ]
( ) ( )[ ] ++++−+
++−−
++++−−
++−+=
=
=⎥⎦
⎤
⎢⎣
⎡
−−+−⋅+
+⎥
⎦
⎤
⎢
⎣
⎡
−+−+⋅=
=−⋅+⋅=
...t3ωωcost3ωωcos
3π
1
-tωωcostωωcos
π
1
tcosω
2
1
...t3ωωcost3ωωcos
3π
1
-tωωcostωωcos
π
1
tcosω
2
1
...tcos5ω
5π
2
tcos3ω
3π
2
tcosω
π
2
2
1
tcosω
...tcos5ω
5π
2
tcos3ω
3π
2
tcosω
π
2
2
1
tcosω
(t))vtcosω(t)vtcosω(t)v
0202
02022
0101
01011
0002
0001
d2d1FSK
...
1(
(t)v1-(t)'v dd =

11
PSKPSK – phase of carrier signal is varied to represent binary 1 or 0
• peak amplitude & freq. remain constant during each bit interval
• example: binary 1 = 0º phase, binary 0 = 180º (πrad) phase
⇒ PSK is equivalent to multiplying carrier signal by +1 when the
information is 1, and by -1 when the information is 0
• demodulation: demodulator must determine the phase of received
sinusoid with respect to some reference phase
• advantage: PSK is less susceptible to errors than ASK, while it
requires/occupies the same bandwidth as ASK
more efficient use of bandwidth (higher data-rate)
are possible, compared to FSK !!!
• disadvantage: more complex signal detection / recovery process,
than in ASK and FSK
⎩
⎨
⎧
+
=
0binary),tfAcos(2
1binaryt),fAcos(2
s(t)
c
c
ππ
π
Modulation of Digital Data: PSK
2-PSK, or
Binary PSK,
since only 2
different phases
are used.
+A
-A
⎩
⎨
⎧
=
0binaryt),fAcos(2-
1binaryt),fAcos(2
s(t)
c
c
π
π

12
Modulation of Digital Data: PSK (cont.)
Example [ PSK ]
vd(t)
vc(t)
vPSK(t)
ωc ωωd_max ωc ωωc+ωd_maxωc-ωd_max

13
PSK DetectionPSK Detection – multiply the received / modulated signal
by 2*cos(2πfct)
• resulting signal
• by removing the oscillatory part with a
low-pass filter, the original baseband signal
(i.e. the original binary sequence) can be
easily determined
[ ] 1binary,t)fcos(41At)f(22Acos cc
2
ππ +=
[ ] 0binary,t)fcos(41At)f(22Acos- cc
2
ππ +−=
t)fAcos(2 cπ±
( )2Acos1
2
1
Acos2
+=

14
1 1 1 10 01 1 1 10 0Information
Baseband
Signal
+A
-A
0 T 2T 3T 4T 5T 6T
+A
-A
0 T 2T 3T 4T 5T 6T
+A
-A
0 T 2T 3T 4T 5T 6T
+A
-A
0 T 2T 3T 4T 5T 6T
Modulated
Signal
x(t)
A cos(2πft) -A cos(2πft)
After multiplication
at receiver
x(t) cos(2πfct)
+A
-A
0 T 2T 3T 4T 5T 6T
+A
-A
0 T 2T 3T 4T 5T 6T
A {1 + cos(4πft)} -A {1 + cos(4πft)}
Baseband
signal discernable
after smoothing
+A
-A
0 T 2T 3T 4T 5T 6T
sender
receiver
Signal shifted above / below zero level.

15
B
fc+B
f
f
fc-B fc
• If bandpass channel has bandwidth Wc [Hz],
then baseband channel has Wc/2 [Hz] available, so
modulation system supports 2*(Wc/2) = Wc [pulses/second]
recall Nyqyist Law: baseband transmission system of bandwidth Wc [Hz]
can theoretically support 2 Wc pulses/sec
Facts from Modulation TheoryFacts from Modulation Theory
Wc
Wc/2
Baseband signal x(t) with
bandwidth Wc/2
If
then
Modulated signal
x(t)cos(2πfct) has
bandwidth Wc Hz
How can we recover the factor 2 in supported data-rate !?
Data rate = 2*Wc/2 = B [bps]

16
QPSK = 4QPSK = 4--PSKPSK – PSK that uses phase shifts of 90º=π/2 rad ⇒ 4
different signals generated, each representing 2 bits
• advantage: higher data rate than in PSK (2 bits per bit
interval), while bandwidth occupancy remains the same
• 4-PSK can easily be extended to 8-PSK, i.e. n-PSK
• however, higher rate PSK schemes are limited by the
ability of equipment to distinguish small differences in
phase
⎪
⎪
⎪
⎩
⎪
⎪
⎪
⎨
⎧
+
+
+
=
11binary),
2
3
tfAcos(2
10binary),tfAcos(2
01binary),
2
tfAcos(2
00binaryt),fAcos(2
s(t)
c
c
c
c
π
π
ππ
π
π
π

17
Modulation of Digital Data: QAM
QuadratureQuadrature
AmplitudeAmplitude
(QAM)(QAM)
– uses “two-dimensional” signalling
• original information stream is split into two sequences that
consist of odd and even symbols, e.g. Bk and Ak
• Ak sequence (in-phase comp.) is modulated by
Bk sequence (quadrature-phase comp.) is modulated by
• composite signal is sent
through the channel
• advantage: data rate = 2 bits per bit-interval!
B1 B2 B3A1 A2 A3 …
Ak x
cos(2πfct)
Yi(t) = Ak cos(2πfct)
Bk
x
sin(2πfct)
Yq(t) = Bk sin(2πfct)
+
Transmitted
Signal
Y(t) = Ak cos(2πfct) + Bk sin(2πfct)
t)fcos(2 cπ
t)fsin(2 cπ
t)fsin(2Bt)fcos(2A ckck ππ +
1 -1 1 1 -1 1 …
1 0 1 1 0 1 …

18
Modulation of Digital Data: QAM (cont.)
Example [ QAM ]
Bk
vd(t)
Ak
sin(ωct)
cos(ωct)

19
QAMQAM DemodulationDemodulation • by multiplying Y(t) by and then low-
pass filtering the resultant signal, sequence Ak is
obtained
t)fcos(22 cπ⋅
• by multiplying Y(t) by and then low-
pass filtering the resultant signal, sequence Bk is
obtained
t)fsin(22 cπ⋅
=+ t)fsin(2Bt)fcos(2A ckck ππ
( )cos(2A)1
2
1
(A)cos2
+=
( )cos(2A)1
2
1
(A)sin2
−=
(A)2sin(A)cossin(2A) =
Y(t) x
2cos(2πfct)
2Akcos2(2πfct)+2Bk cos(2πfct)sin(2πfct)
= Ak {1 + cos(4πfct)}+Bk {0 + sin(4πfct)}
Lowpass
filter
(smoother)
Ak
2Bk sin2(2πfct)+2Ak cos(2πfct)sin(2πfct)
= Bk {1 - cos(4πfct)}+Ak {0 + sin(4πfct)}
x
2sin(2πfct)
Bk
Lowpass
filter
(smoother)
smoothed to zero
smoothed to zero
Y(t) x
2cos(2πfct)
Lowpass
filter
(smoother)
Ak
x
2cos(2πfct)
Lowpass
filter
(smoother)
Ak
2Bk sin2(2πfct)+2Ak cos(2πfct)sin(2πfct)
= Bk {1 - cos(4πfct)}+Ak {0 + sin(4πfct)}
x
2sin(2πfct)
Bk
Lowpass
filter
(smoother)
x
2sin(2πfct)
Bk
Lowpass
filter
(smoother)
smoothed to zero
smoothed to zero

20
Signal Constellation
Constellation DiagramConstellation Diagram – used to represents possible symbols that may
be selected by a given modulation scheme as
points in 2-D plane
• X-axis is related to in-phase carrier: cos(ωct)
the projection of the point on the X-axis defines
the peak amplitude of the in-phase component
• Y-axis is related to quadrature carrier: sin(ωct)
the projection of the point on the Y-axis defines
the peak amplitude of the quadrature component
• the length of line that connects the point to
the origin is the peak amplitude of the signal
element (combination of X & Y components)
• the angle the line makes with the X-axis is the
phase of the signal element

21
QAM cont.QAM cont. – QAM can also be seen as a combination of ASK & PSK
( ) )
A
B
tantfcos(2BAt)fsin(2Bt)fcos(2AY(t)
k
k1-
ckkckck ++=+= πππ 2
1
22
Ak
Bk
(A, A)
(A,-A)(-A,-A)
(-A,A)
Ak
Bk
Ak
Bk
(A, A)
(A,-A)(-A,-A)
(-A,A)
4-level QAM

22
Ak
Bk
Ak
Bk
1616--level QAMlevel QAM – the number of bits transmitted per T [sec] interval
can be further increased by increasing the number
of levels used
• in case of 16-level QAM, Ak and Bk individually can
assume 4 different levels: -1, -1/3, 1/3, 1
• data rate: 4 bits/pulse ⇒ 4W bits/second
( ) )
A
B
tantfcos(2BAt)fsin(2Bt)fcos(2AY(t)
k
k1-
ckkckck ++=+= πππ 2
1
22
In QAM various combinations of amplitude and phase are employed
to achieve higher digital data rates.
Amplitude changes are susceptible to noise ⇒ the number of phase shifts used
by a QAM system is always greater than the number of amplitude shifts.
Ak and Bk individually
can take on 4 different values;
the resultant signal can take
on (only) 3 different values!!!

Digital modulation

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Digital modulation