Shift Keying
by Erol Seke
For the course “Communications”
ESKİŞEHİR OSMANGAZİ UNIVERSITY
Basic PAM
Binary 1 is represented by voltage A
Binary 0 is represented by voltage B
Amplitude Shift Keying (ASK)
carrier
If A and B has opposite signs then there will be a phase jumps at bit-value changes
Binary - ASK case
𝐴𝑆𝐾 𝑡 = 𝑥 𝑡 cos(2𝜋𝑓𝑐𝑡)
Spectrum of ASK
x(t)
t
......
T
f
1/T 2/T
Power spectral density
t
c(t)
f
-fc fc
y(t)
ffc-fc
psd of binary ASK
t
M=4
x(t)
t... ...
T
ffc-fc
psd of M-ary ASK
f
1/T 2/T
psd of 4-ary PAM
How? : 4-ary PAM can be thought of a sum of two 2-ary PAM
y(t)
t
Same argument follows when M=2k (k:integer)
Frequency Shift Keying (FSK)
Use different frequency values (finite number of) instead of different amplitudes
f2
f1
Example : Binary FSK
Binary 1 is represented by a sinusoid with frequency f1
Binary 0 is represented by a sinusoid with frequency f2
Note: Amplitude does not change, phase is not an issue
y0(t)
t
y1(t)
t
y(t)
t
2-ary FSK can be thought of the sum of two 2-ary ASK
f
fc0-fc0
f
fc1-fc1
fspectrum very much dependent on the choices of fc0 and fc1
Phase Shift Keying (PSK)
Use different phase values (finite number of), and we get PSK
Note: Amplitude and frequency do not change
Example : Binary PSK (BPSK)
Binary 1 is represented by a sinusoid with 0 phase
Binary 0 is represented by a sinusoid with phase π
Carrier with phase φ
Carrier with phase φ+π
1
00)(
binaryfor
binaryforts
Note2: BPSK is the same as Binary ASK when amplitudes are –A and +A
𝐵𝑃𝑆𝐾 𝑡 = 𝐴𝑐𝑜𝑠(2𝜋𝑓𝑡 + 𝜑 + 𝑠 𝑡 )
Spectrum of 2-ary PSK
y(t)
ffc-fc
psd of binary PSK
Spectrum of 4-ary PSK
That is, there are 4 phases (π/2 apart) instead of 2 (π apart)
Think of 4-ary PSK as the sum of two 2-ary PSK and verify the following
ffc-fc
psd of 4-ary PSK
It seems that this sinc spectrum will always be with us in communication
Hmw : Check the spectrum of PSK of a general M=2k (k:integer)
Note2: BPSK is the same as Binary ASK when amplitudes are –A and +A
Cosine and Sine are Orthonormal
t t
0)()( 21
dttt
)(2 t)(1 t
f1f1
same frequencies
A sinusoidal signal with any phase (at frequency f1) can be obtained by a weighted
sum of these basis waveforms and)(1 t )(2 t
cos
sin
I
Q
In phase component
Quadrature component
phasors𝑠 𝑡 = −2cos 2π𝑓1𝑡 + sin(2π𝑓1𝑡)
BPSK
S1
Symbol
S2
Binary
0
1
)2cos( tfc
)2cos( tfc
Signal I Q
1
-1
0
0
01
A binary stream
1 1 0 1 0 0 0 01 1 1 1
Phase changes
For easier drawing, example shows 1 carrier period per bit.
There need not be any relation between them.
2D constellation diagram
1/z
Binary BPSK mod.
DBPSK mod.
DBPSKDPCM
Differential BPSK
Advantage : Non-Coherent Detection is possible
0
1 0-1 change
1-0 change
Changes can be easily detected even
when there is no reference carrier
Disadvantage : A bit error affects detection of all remaining bits
)2cos( tfc
)2sin( tfc
I
Q
0111011...01
binary streamI-Q mod.
M-ary PSKIm
Qm
Generation of M-PSK
M=2k k: number of bits in a symbol
coefficient look-up table
I
Q
(instead of vectors, we just use points at tip of the vectors)
𝑀 = 𝐼2 + 𝑄2 = 1
𝜑𝑖 = tan−1𝐼
𝑄
Remember the efficiency statement in the baseband receiver block diagrams
I
Q
I
Q
BPSK
QPSK
I
Q QPSK
000
001
010
011
100
101
110
111
01
00
11
01
10
Q
I
Q
I8-PSK
16-QAM
In QAM, both amplitude and phase
are used to represent a symbol
(to be continued)
0011
01
10
Quadrature PSK
S1
Symbol
S2
Binary
00
11
)2cos( tfc
)2cos( tfc
Signal I Q
1
-1
0
0
S3
S4
01
10
)2/2cos( tfc
)2/2cos( tfc
0
0
1
-1
00
11
01
10
QPSK
S1
Symbol
S2
Binary
00
11
)4/2cos( tfc
)4/52cos( tfc
SignalI Q
0.707 0.707
S3
S4
01
10
)4/32cos( tfc
)4/32cos( tfc
-0.707 -0.707
-0.707 0.707
0.707 -0.707
Binary value
I channel
Q channel
I
Q
I+Q channel
QPSK (sum of two BPSKs)
Modulated
In-Phase carrier
Modulated
Quadrature-Phase
carrier
QPSKIr = [ 0.7071 -0.7071 -0.7071 0.7071 ]
Qr = [ 0.7071 0.7071 -0.7071 -0.7071 ]
Ir = [1 0 -1 0]
Qr = [0 1 0 -1]QPSK
Binary
000
001
)2cos( tfc
)4/2cos( tfc
Signal I Q
1 0
011
010
)2/2cos( tfc
)4/32cos( tfc
0.707 0.707
0 1
-0.707 0.707
110
111
)8/52cos( tfc
)8/72cos( tfc
-1 0
101
100
)8/92cos( tfc
)8/112cos( tfc
-0.707 -0.707
0 -1
0.707 -0.707
8-PSK
8-PSK
I
Q
000
001
010
011
100
101
110
111
I
Q
000
001010
011
100
101110
111
4 different I and Q values
000
001
010
011
100
101
110
111
Gray-coded
Ir = [1 0.7071 0 -0.7071 -1 -0.7071 0 0.7071 ]
Qr = [0 0.7071 1 0.7071 0 -0.7071 -1 -0.7071 ]
8-PSK
(bit assignments are different than shown in previous slide)
Ir = [0.9239 0.3827 -0.3827 -0.9238 0.9238 0.3827 -0.3827 -0.9238]
Qr = [0.3827 0.9238 0.9238 0.3827 -0.3827 -0.9238 -0.9238 -0.3827]
8-PSK
)2cos( tfc
)2sin( tfc
I
Q
0111011...01
binary streamI-Q mod.
M-ary QAMIm
Qm
coefficient look-up table
Q
I
16-QAM
QAM
Both phase & amplitude are used
𝐴𝑖 = 𝐼2 + 𝑄2
𝜑𝑖 = tan−1𝐼
𝑄
1024-QAM
2048-QAM
Now you know the sky is the limit.
A symbol is 10 bits.
Spectrum is the same as BPSK. So, why not use
64536-QAM and transmit 16 bits with each symbol?
Caveat : The receiver must distinguish each symbol,
and symbols will be closer, unless the voltage is
increased.
Q
I
16-QAM
64-QAM (from IEEE-802.1a-1999)
5180
c0
20 MHz
c1c-1c8c-22 c6
c20 c22 c26
c27 c32
c-6c-8c-20c-26
c-31 c-27 f (MHz)
312.5 kHz
Placement of 64 Carriers in IEEE-802.1a-1999)
48 carriers use BPSK, QPSK, 16-QAM or 64-QAM
How these signals are generated will be discussed in OFDM
END