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