nten 216 data communications 5 digital modulation · 2006. 11. 20. · nten 216 data...

NTEN 216 Data Communications Digital Modulation

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5 Digital Modulation

5.1 Introduction to Digital Modulation

What is Digital Communications?

Communication using digital data

– Digital Data = bits, nibbles, bytes…1’s and 0’s

Two Broad Categories of Digital Communications:

1. Digital Transmission

2. Digital Radio

Digital Transmission

Communicating digital information in digital form

– Pulses representing 1’s and 0’s sent back and forth to each other

Baseband communication

– Not wirelessly transmittable

– Requires physical connection via conductor or fibre

Digital Radio

Communicating digital data using analog signals

Similar to analog communications – a carrier is modulated by the information

containing signal

– Difference is modulating signal is a digital one

– Amplitude, frequency and phase can all be modulated

Digital Radio Modulation

Carrier Signal:

Vc(t) = Vcsin(2πfct + θ)

Modulate Vc to get Amplitude Shift Keying

Modulate fc to get Frequency Shift Keying

Modulate θ to get Phase Shift Keying

Modulate Vc and θ to get Quadrature Amplitude Modulation

Digital Radio Advantages

Noise Reduction

– Data Repeaters remove noise easily

Easy Multiplexing in Time (TDMA), Frequency (FDMA), Code (CDMA)

– Analog can multiplex, but only FDMA is easy

Signal Processing capabilities – filtering, amplification and other effects

– All done in software. Limited only by imagination and technical skills

Becoming cheaper and easier to build

– Integrated circuits and digital systems

– System on a Chip (SOC)

NTEN 216 Data Communications: Digital Modulation

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Many forms of data/info already in digital form

– Internet, databases, email, video, radio

Claude Shannon: can pass digital data error free through any arbitrarily noisy

system if you decrease the data rate

Digital Radio Applications

1. Computer Modems

2. DSL

3. Microwave and satellite communications

4. PCS

5. Digital TV

6. IEEE 802.11

7. …

Bit Rate and Baud Rate

Bit Rate is the number of bits that can be sent per second over a channel

Baud Rate is the number of symbols that can be sent per second over a channel

If we designate N to be number of bits per symbol, and M to be number of

symbols necessary to represent N bits per symbol:

N = log2M or

M = 2N

Symbols and Bits

Example: if you want 3 bits per symbol, you need M = 23 = 8 symbols

For ASK, different symbols represented by different amplitudes

For FSK – different frequencies

For PSK – different phases

For QAM – combo different amplitudes and phases

Symbol # Bit Representation

0 000

1 001

2 010

3 011

4 100

5 101

6 110

7 111


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Binary vs M-ary Communications

Binary Communication:

– Sending only symbols representing a 1 or a 0

– Absolute Maximum Data Rate:

bps = 2*BW

– Typical Maximum Data Rate:

bps = BW

M-ary Communications:

– Sending symbols representing more than 1 bit each (i.e., there are M

symbols representing N bits each

– Absolute Maximum Data Rate:

Baud rate = 2*BW

Bit rate = 2*BW*log2M = 2*BW*N bps

– Typical Maximum Data Rate:

Baud rate = BW

Bit Rate = BW*log2M = 2*BW*N bps

5.2 Amplitude Shift Keying (ASK)

Simplest digital modulation technique

Carrier Signal:

vc(t) = Vccos(2πfct)

Modulating Signal is data sequence of 1s and 0s…m(t) = 1,0,0,1,1,1,0,1,1,1…

Modulated Signal:

vask(t) = m(t)*vc(t)

ASK Example

vask(t) = m(t)*vc(t)

m(t) = 1,0,1,1,0,1,1,1,0,1,0…


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ASK Example System

Switch turned on when input bit is a 1, and turned off when input bit is a 0

Rate of input bits must be constant so that each bit has a fixed, known length of

time.

ASK Demodulated Signal

Demodulation is fairly simple

– Incoming signal will have an envelope like analog AM

– Detect the envelope, and this will give first approximation of incoming

data

– Add threshold detector to determine whether data is a 1 or 0 to get

regenerated data

5.3 Frequency Shift Keying (FSK)

Frequency shift keying or FSK is another easy to implement but lower performance form

of digital communication. In FSK, different frequencies are used to transmit the digital

data. Just like with the other forms of digital modulation, the FSK can be binary, or it

can be M-ary, although binary FSK is the most common. The general expression for an

FSK modulated signal is:

}])([2cos{)( tftvfVtv mccfsk ∆+= π

Vfsk(t) is the FSK modulated signal.

Vc is the carrier amplitude

Fc is the carrier frequency

Vm(t) is the modulating digital data.

∆f is the frequency shift from between bits, or from the carrier frequency to the bits.


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For binary frequency shift keying, the sequency of 1’s and 0’s usually becomes a

sequency of 1’s and -1’s so that you end up with these two values for the modulated

signal:

))(2cos()( 1 tffVtv ccvmfsk ∆+== π

))(2cos()( 1 tffVtv ccvmfsk ∆−=−= π

So the frequency for sending a 1 is (fc + ∆f) and the frequency for sending a 0 (-1) is

(fc – ∆f).

M-ary FSK systems will simply use more frequencies for transmitting the required

number symbols. Each frequency will be a different symbol. A 4 symbol M-ary system

would require 4 frequencies. The frequencies chosen would be something like:

(fc + 3∆f)

(fc + 1∆f)

(fc - 1∆f)

(fc - 3∆f)

We use 3 times ∆f and 1 times ∆f in order to give equal spacing between symbols.

Equal frequency spacing between symbols is not necessary but gives the system a

symmetry making it easier to analyze.

Turning our attention back to binary FSK, let’s look at an FSK waveform, and the

corresponding digital input:

The frequency of a 1 is often called the mark frequency, or fm, and the frequency of a 0 is

often called the space frequency, or fs.

Bandwidth


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The minimum bandwidth required for an FSK system is equal to:

bsmbbsmbsbm fffffffffffBW 2)()()()( +−=−−+−=−−+=

Since (fm-fs)=∆f,

bffBW 22 +∆=

where fb is the bit rate.

So the FSK signal from the previous figure would look like this in the time domain:

5.3.1 FSK Transmitter

Basic FSK transmitters are surprisingly simple. They consist of a voltage controlled

oscillator (VCO) which has as the input, the stream of digital data, and as the output, the

FSK modulated signal.

To understand how the system can work, let’s first look at oscillators and in particular

voltage controlled oscillators.


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5.3.1.1 Voltage Controlled Oscillator

Oscillators

An oscillator is an electronic component that provides an output that alternates between

two output voltages at some frequency. Typically this “oscillation” is a sinusoidal wave,

but it can be pretty much any simple periodic waveform (e.g., square, triangle, sawtooth).

An important feature of oscillators is that this oscillation occurs without any external

input.

One common type of oscillator is a crystal oscillator. A crystal oscillator consists of a

very thin and specially cut piece of quartz crystal sandwiched between two metallic

plates that are then connected to metallic leads.

Quartz crystal has a special property called the piezoelectric effect that makes it perfect

as an oscillator. When quartz is compressed or mechanically stressed in any manner, it

generates a voltage, and alternatively, when a voltage is applied across the crystal, it

changes shape.

Crystal oscillators have a specific frequency at which they oscillate, and this frequency is

determined by the size and geometry of the crystal. In order to begin oscillating, at the

startup (powerup) of the circuit the crystal oscillator is a part of, random electrical noise

is generated. Since this noise is found across the entire frequency spectrum, some of it is

at the same frequency at which the crystal is designed oscillate. This electrical noise

causes the crystal to deform, and when it returns to normal shape, it overshoots a small

amount which will then cause a voltage across the crystal. The electrical noise will again

cause the crystal to deform and when it returns to its normal shape, it overshoots a little

bit more, creating a slightly larger overshoot. After a short period of time, the crystal is

able to oscillate by itself. It is analogous to being swung on a swing. The electrical noise

is like someone giving you a push, until you reach a point where you can swing by

yourself. This frequency that the crystal naturally oscillates at is called the resonant

frequency.

VCO

A VCO is a special type of oscillator whose frequency of oscillation is determined by the

voltage applied to it.

So if we have a different voltage input for each symbol to transmit in an FSK system, we

can generated a different frequency for each symbol. The only other things to consider is

the voltage level for each symbol, and the frequency shift those changes cause.

VCO’s have a deviation sensitivity which determines how much the frequency changes

when the voltage changes. The deviation sensitivity is designated kl, and is in units of

Hz/volt.

Example.

You need to select a VCO for a system that transmits a 1 with a frequency of 300.004

MHz, and a 0 with a frequency of 299.996 MHz. What deviation sensitivity do you

require if the voltage of a 1 is 1.8V and the voltage of a 0 is 0V?

∆f = 300.004MHz – 299.996MHz = 8kHz


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∆voltage = 1.8V

kl = 8kHz/1.8V = 4.4kHz/V

5.3.2 FSK Receivers

FSK receivers are also surprisingly simple (aside from the filters and amplifiers needed).

At the receiver end, you need to go from the received signal which looks like this:

To a bit stream that looks like this:

In other words, we have to convert the frequencies back into 1’s and 0’s.

The simplest (yet still effective) FSK receiver looks like this:

The first step just splits the input into two paths. The second step is a band pass filter.

The center frequency of the upper band pass filter is centered on the frequency used for a

1. The center frequency of the lower band pass filter is centered on the frequency used

for a 0. The following diagram shows the input and output signals for each path.


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The BPFs are not perfect, and therefore, there will still be a small amount of both

frequencies in both signals. The signal which was allowed through by the BPF will be

much stronger though.

The third step is an envelope detector which basically draws an envelope around the

signal, so when the signal is strong, the envelope will be higher, and when the signal is

weak, the envelope will be close to 0 volts.

The last step is a comparator which will output a 1 if the signal into the “1” input is

stronger than the signal into the “0” output and vice versa. The output of this stage is the

final output, and should match the original data (assuming no bit errors).

5.3.3 FSK Applications

FSK is more prone to bit errors, and has requires a higher bandwidth for any bit rate than

Phase Shift Keying and Quadrature Amplitude Modulation. FSK is therefore used in

systems where lower speeds and lower costs are required. Most systems that transmit

data over systems designed to transmit voice use FSK. The best examples are telephone

line computer modems which pass data between computers over telephone line and

Caller ID which passes the ID data over the telephone line.

Another FSK system which you probably use daily is the Dual-Tone Multi Frequency

(DTMF) system of touch tone phones. A total of 16 different symbols can be sent over

the phone line by hitting different keys on the key pad. Each key produces two different

tones at the same time. The receiving end decodes the tones to determine which key was

pressed. The tones were specially designed to minimize harmonics and interference

between tones. The following table1 shows the tones used for each key:

Tone (Hertz) 1209 1336 1477 1633

1 Source: http://en.wikipedia.org/wiki/DTMF


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697 1 2 3 A

770 4 5 6 B

852 7 8 9 C

941 * 0 # D

FSK is also used in proprietary wireless LANs, and in digital control systems such as

home automation systems.

5.4 Phase Shift Keying (PSK)

Phase shift keying, or PSK is an digital angle modulation system where symbols are

transmitted as signals with different phases. Like other forms of digital communications,

the number of bits sent with each symbol depends on the total number of symbols used:

NM 2= or

NN 2log=

The simplest PSK system sends only one bit per symbol; it sends either a 0 or a 1. This

kind of system is called binary PSK or BPSK because, obviously, it is a binary system.

There are however several different M-ary PSK systems, and in fact M-ary PSK systems

are in wider use than M-ary ASK or FSK systems because of performance and ease of

design.

5.4.1 Binary PSK

BPSK sends a 0 with a signal with a phase shift of 180°, and a 1 with a signal with a

phase shift of 0°. The following figure shows a binary input of a string of 1’s and 0’s and

the BPSK signal created from the binary input.


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The following block diagram, shows a BPSK system.

The binary input of 1’s and 0’s is modified so that it becomes a sequence of 1’s and -1’s

(formerly 0’s). This sequence designated vm(t) is then multiplied by the sine wave carrier

sin2πfct where fc is the carrier frequency. The output then becomes either:

1. tf cπ2sin1× or

2. tf cπ2sin1×−

Which you can see pictorially in the previous figure on the right hand side.

5.4.1.1 BPSK Bandwidth

To determine the bandwidth of a BPSK signal consider that the maximum rate of change

of the system will occur when you have alternating 1’s a 0’s. So if that BPSK signal has

a bit rate given by fb, the maximum rate of change will be equal to ½ the bit rate. This is

because a full cycle will come from a ‘1’ bit with a bit time of 1/fb followed by a ‘0’ bit

with a bit time of 1/ fb to give you one full cycle with a time of 2/fb. The inverse of that


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full cycle time of 2/ fb is fb /2. Let’s call this maximum data rate fa so that fa = fb /2. For

a pictorial representation of this, look at the following diagram:

If you have a sequence of alternating 1’s and 0’s, you can low pass filter this sequence to

get a sine wave, so when you multiply the carrier by vm(t) (which is now a sine wave)

you get:

tftftvtf acmc πππ 2sin2sin)(2sin ×=×

You end up with a result similar to what you get for analog AM signals, and that is in the

frequency domain, you have signal at fc, and then signal at fc - fa, and fc + fa as in this

frequency spectrum:

Therefore the bandwidth of the system extends from (fc - fa) up to (fc + fa), which gives:

baaaccacac ffffffffffBW ==−−+−=−−+= 2)()()(

So the bandwidth of a BPSK system is equal to fb which is simply the bit rate.

Example

What is the bandwidth of a 1 Mbps BPSK signal that is transmitted on a 100MHz carrier?

What would it’s frequency spectrum look like?

The bandwidth is 1MHz, and the frequency spectrum looks like this:


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5.4.1.2 BPSK Receiver

The following figure is a BPSK receiver:

The input to the receiver is the BPSK signal which will be the carrier signal alternating

between being in phase (1’s) and being 180° out of phase (0’s).

(1) The first block, the band pass filter, cleans up the signal so that only the

frequencies of interest from (fc - fa) up to (fc + fa) are allowed through.

(2) The second block multiplies the signal by sin(2πfct). Since the input signal will

be one of two values, sin(2πfct) and -sin(2πfct), the output of the multiplication

block will be either:

a. ( ) ]22cos[2

1

2

1]22cos[1

2

12sin2sin2sin 2 tftftftftf ccccc ×−=×−==× πππππ

or

b. ( ) ]22cos[2

1

2

1]22cos[1

2

12sin2sin2sin 2 tftftftftf ccccc ×+−=×−−=−=×− πππππ

(3) Uh, oh. This seems to be getting pretty complicated, but have no fear, the next

block is a low pass filter, set with a cutoff point considerably less than fc. What

does this mean? When the equations in parts 2(a) and 2(b) both have a

cos[2πx2fct] term which is at a frequency of 2fc and therefore will be eliminated

by the low pass filter.


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All that we are left with is either a ½ or a -½. If we receive a ½, then the value

sent was a 1. If we receive a -½ , then the value sent must have been a 0.

5.4.2 Quadrature PSK (QPSK or 4-PSK)

QPSK is another form of phase shift keying, but this one uses four different phases to

represent 4 different symbols with two bits each. BPSK uses 180° phase difference

between the two different symbols. What spacing would make sense for QPSK? Well,

you want something with maximum and equal spacing between each symbol, so the

spacing should be 360°/4 = 90°. As will soon be clear, we don’t use 0°, 90°, 180° and

270°. Instead we use 45°, 135°, 225°, and 315°:

Symbol # Phase Bits

1 45° 00

2 135° 01

3 225° (or -135°) 11

4 315° (or -45°) 10

5.4.2.1 QPSK Transmitter

The following figure shows a QPSK transmitter:


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(1) Bit Splitter: Takes the input bit stream, and creates two identical bit paths from it:

the I path, and the Q path. The I path is the in-phase path, and the Q path is the

out of phase path (out of phase path is also known as the Quadrature path).

(2) Balanced Modulator: This is just a multiplier that multiplies the I (or Q) channel

by the carrier (or the phase shifted carrier). As you can seee, there is a Reference

Carrier Oscillator which creates the sin(2πfct) signal. This signal is used directly

in the I, but in the Q channel, this signal is phase shifted by 90° to give a

cos(2πfct) signal instead.

I Path: What you end up with at the output of the Balanced Modulator in the I

path is 1*sin(2πfct) or -1*sin(2πfct).

Q Path: What you end up with at the output of the Balanced Modulator in the Q

path is 1*cos(2πfct) or -1*cos(2πfct).

(3) Band Pass Filters: These filters remove any harmonics in the signal, and any noise

outside the passband.

(4) Linear Summer: This simply adds the signal from the I path with the signal from

the Q path. What you end up with is one of four possibilities:

i. )2cos()2sin( tftf cc ππ +

ii. )2cos()2sin( tftf cc ππ −

iii. )2cos()2sin( tftf cc ππ +−

iv. )2cos()2sin( tftf cc ππ −−

Each one of these four possiblities, is in essence a symbol representing two bits, and this

table shows the relationship between the summed signals, and bits:

I Channel Q Channel I + Q Bits

sin2πfct cos2πfct sin2πfct + cos2πfct 11

sin2πfct -cos2πfct sin2πfct - cos2πfct 10

-sin2πfct cos2πfct -sin2πfct + cos2πfct 01

-sin2πfct -cos2πfct -sin2πfct - cos2πfct 00

At the output of the transmitter will be a new signal made up of the sum of the I and Q

channel output. This new signal carries two bits in each symbol.

As an aside, and to help you visualize the 4 signals that are created, look at the following

sinewave graphs:


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5.4.2.2 Phasor Diagrams

Remember how I said it would become clear as to why we use 45°, 135°, -135°, and -45°,

instead of 0°, 90°, 180°, and 270°. Well the reason is that the resulting I+Q signal can be

drawn on a phasor diagram. This is the same phasor diagram that we saw with analog

angle modulation. It represents a signals strength (the length of the line) and its phase

(how many degrees from the positive x-axis it is. If we designate the x-axis to be the

strength of the sin2πfct part of the signal and the y-axis to be the strength of the cos2πfct

part of the signal, then we can create a graph that looks like this:

Phasor and constellation diagrams for a QPSK system. The phasor diagram is the one on

the left, and it shows the actual vectors, or lines, for each of the four points. The

constellation diagram is the one on the right, and it only shows the point on the cosine (I)

and sine (Q) axes.

There are 4 points on the graph, each with a length of 1, but with different phase angles.

This table shows the relationship between the points on the graph, and the I and Q signals

of the QPSK transmitter:

I Channel Q Channel Phase Dibit


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1 1 45° 11

1 -1 -45° 10

-1 1 135° 01

-1 -1 -135° 00

PSK and QAM signals can have many, many different symbols, phasor diagrams are

simple graphs that help us keep track of all of the different symbols, and what bits they

actually represent.

Before we go forward, let’s go back a bit and look at ASK and BPSK phasor diagrams:

For ASK, phase is not used at all, only strength, so for our on-off ASK system, one point

on the phasor diagram is at the origin (0 strength), and the other is at 0° with a length of

1.

BPSK uses two signals out of phase by 180°, and this is represented in the BPSK phasor

diagram.


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5.4.2.3 QPSK Receiver

A QPSK receiver looks very similar to a transmitter:

(1) The input signal comes in and it is split into two identical paths again, the I and

the Q path.

a. I = Asin2πfct + Bcos2πfct

b. Q = Asin2πfct + Bcos2πfct

c. A = +/-1 B=+/-1

(2) The I path is multiplied by the carrier (sin2πfct) and the Q path is multiplied by

the carrier phase shifted by 90°

a. I = sin(2πfct)[Asin2πfct + Bcos2πfct] = Asin2(2πfct) + Bsin(2πfct)cos(2πfct)

= A/2[1-cos(2π(2fct))] + B/2[sin(2π(fc+fc)t)] + B/2[sin(2π(fc+fc)t)]

= A/2 + A/2 cos(2π(2fct)) + B/2 sin(2π(2fc)t) + 0

b. Q=cos2πfct[Asin(2πfct) + Bcos(2πfct)]=Acos(2πfct)sin(2πfct)+Bcos2(2πfct)

= -A/2sin[2π(fc+ fc)t] – A/2sin[2π(fc- fc)t] + B/2(1 + cos[2π(2fc)t])

= -A/2 sin[2π(fc+ fc)t] – 0 + B/2 + B/2cos[2π(2fc)t]

(3) These two rather complicated formulae are easily reduced once the signals pass

through the low pass filter. Everything except the constant values are filtered out.

a. I = A/2 If A is 1 then I is ½. If A is -1, then I is -½.

b. Q = B/2 If B is 1 then I is ½. If B is -1, then I is -½.

The following table summarizes the possible results:

Input Signal I Value Q Value Dibit

sin2πfct + cos2πfct 1 1 11

sin2πfct - cos2πfct 1 -1 10


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-sin2πfct + cos2πfct -1 1 01

-sin2πfct - cos2πfct -1 -1 00

5.4.2.4 QPSK Summary

• N (number of bits per symbol) = 2

• M (number of symbols) = 4

o 45°, 135°, 225°, 315°

• Two bits are sent per symbol, therefore the bit rate is twice the bit rate of BPSK

for the same bandwidth

o Example if carrier frequency is 100MHz, and the bit rate is 10Mbps, then

the bandwidth for a BPSK system is 10MHz, but the bandwidth for a

QPSK system is only 5MHz.

• QPSK in the time domain:

5.4.3 8 PSK

Eight different symbols, representing 3 bits each. The phase difference between symbols

is 360/8 = 45°. The following table shows the binary inputs and the phase outputs:

Binary Input Phase Output


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000 -112.5°

001 -157.5°

010 -67.5°

011 -22.5°

100 112.5°

101 157.5°

110 67.5°

111 22.5°

The following figure is the phasor diagram/constellation diagram for an 8-PSK system:

5.4.4 16-PSK

Sixteen different symbols, representing 4 bits each. The phase difference between

symbols is 360/16 = 22.5°. The following diagram is a constellation diagram for a 16-

PSK system:


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5.4.5 Bit Errors

The receiver receives the transmitted symbols, and converts them into the bit

representation of those symbols. That doesn’t mean that the receiver gets it right every

time. Noise in the system may cause the receiver to get detect a symbol that was

different from the one transmitted. If we look at the constellation diagram for a BPSK

system, we can see that the symbol transmitted will have a strength of 1, and a phase of

0° or 180°. However, the received symbol may have a different strength, and a different

phase, and therefore will appear elsewhere on the constellation diagram. The points

numbered 1-7 on the constellation diagrams are received symbols. If a 0 was transmitted

(strength 1, phase 180°), the receiver will detect the correct value as long as the received

symbol is to the left of the I=0 line (meaning the phase was greater than 90° and less than

270°. If a 1 was transmitted (strength 1, phase 0°), the receiver will detect the correct

value as long as the received symbol is to the right of the I=0 line (meaning the phase was

less than 90°, or greater than 270°. We can draw a line down the I=0 line to represent the

boundary for receiver detection.


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For M-ary systems, we can do the same thing, but we have to draw more boundaries for

the receiver. For QPSK, we draw two lines, one is the I axis and one is the Q axis to

form 4 sections, one in each quadrant. Each quadrant then represents the area for a

particular dibit (two bit sequence). If a received symbol falls in quadrant one, the

receiver interprets that as a ‘11’. Quadrant II means ‘01’. Quadrant III means ‘00’.

Quadrant IV means ‘10’. The following diagram shows the QPSK constellation diagram

with the quadrants clearly marked.

A received symbol without any noise will fall exactly on one of the four points which, for

a ‘11’. is at a phase angle of 45° and a length of 1. As long as the received symbol falls

between 0° and 90° the symbol will be received correctly; this means, the symbol can

have a phase error up to +/-45° without any bit errors.

As you increase the number of bits per symbol, the phase error that you can have

decreases. The general rule is that the maximum phase error will be have of the phase

difference between symbols:

)2/(360_max_ Merrorphase ×=

Where M is the number of symbols.

Draw the symbol boundaries for the 16-PSK constellation diagram below:


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What is the maximum phase error for a 16-PSK signal?

5.4.6 PSK in Use

The most prominent use of PSK communication is in 802.11, the different modulation

methods for 802.11 are shown in the following table2:

Communication Standard Modulation Method

802.11b (1Mbps) DBPSK

802.11b (2Mbps) DQPSK

802.11a,g (6,9 Mbps) BPSK

802.11a,g (12,18Mbps) QPSK

802.11a,g (24,36 Mbps) 16-QAM

802.11a,g (48,54 Mbps) 64-QAM

5.4.7 Differential PSK Systems

Differential PSK does not assign a fixed bit to the different phases of the system. Instead

it assigns bits to phase differences. You need to know the previous symbol phase and the

2 Source: http://en.wikipedia.org/wiki/Wi-Fi_Router


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current symbol phase to calculate the difference between phases, this difference is then

used to determine the symbol/bit(s) received. For a differential BPSK (DBPSK) system,

if the phase difference between previous and current symbol is 0 then this is interpreted

as a 1. If a phase difference of 180 is calculated, then this is interpreted as a 0.

The purpose of DPSK is to simplify the receiver.

5.5 Quadrature Amplitude Modulation (QAM)

QAM systems encode data in both the amplitude and phase of a carrier, and as such,

these systems are sometimes called Amplitude Phase Shift Keying (APSK). QAM is

used in multi M-ary systems, meaning systems where many different symbols (and

therefore many bits per symbol) are used for transmitting data. QAM is the best choice

for this type of system because varying phase and amplitude provides the largest distance

between symbols. This, of course means that the receiver will have a better chance of

discerning the correct transmitted symbol.

It would be possible to have large distance between symbols for ASK systems, by setting

the amplitude difference to be very large, e.g. 0 volts for a ‘0’ and 100 volts for a ‘1’ but

this extreme example is obviously impractical because it would use too much power.

FSK systems also could have a large distance between symbols by setting the frequency

separation between symbols very high (e.g. 1 GHz for a ‘0’ and 10 GHz for a ‘1’). But

this is also impractical because it will use a bandwidth of at least 9GHz.

What does “the largest distance between symbols” mean? The best way to think about

this is to look at the physical distance between symbols on a constellation diagram. For

PSK, the symbols are fixed on a circle centered at the origin because the amplitude is

constant. QAM is not limited to being on the circle because the amplitude can change

too. So, QAM can have more symbols for a given bandwidth and power than any of the

other types of modulation.

5.5.1 8-QAM

8-QAM uses 8 different symbols, and therefore transmits log28 = 3 bits per symbol.

Among the 8 different symbols, 4 different phases and 2 different amplitudes are used

(4x2=8). This is the constellation diagram for 8-QAM:


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This table outlines the features (i.e. phase, amplitude and bit value) of each symbol:

Symbol Bit Value Phase Amplitude

1 000 -135 0.765V

2 001 -135 1.848V

3 010 -45 0.765V

4 011 -45 1.848V

5 100 135 0.765V

6 101 135 1.848V

7 110 45 0.765V

8 111 45 1.848V

In the time domain, an 8-QAM message might look something like this:

5.5.1.1 8-QAM Receiver

At the receiver, the symbols will not be at the exact same phase and amplitude as the

transmitted symbols due to noise in the system. The receiver has decision boundaries

which can be drawn on the constellation diagram to show the regions for each symbol.

When a received symbol is received, the receiver interprets the bit value of that symbol

based on which region it falls in. If there was so much noise that the transmitted symbol

was from one region but received in another, we get a bit error or errors. You will notice

on the diagram that where there are large boundary lines between symbols, the bit values

for those symbols on each side of the boundary line are only different by 1 bit. Therefore

if there is a symbol error, you will likely only get a single bit error.


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5.5.1.2 8-QAM Bandwidth

8-QAM transmits 3 bits per symbol, and since the bandwidth is equal to the baud rate, the

bandwidth is equal to 1/3 of the bit rate:

3_8

BitRateBandwidthQAM =

5.5.2 16-QAM

16-QAM has 16 different symbols; there are 3 different amplitudes and 12 different

phases used (obviously, each amplitude does not use all of the phases). The following

table shows the features of each symbol:

Symbol Bits Phase Amplitude

1 0000 -135 0.311V

2 0001 -165 0.850V

3 0010 -45 0.311V

4 0011 -15 0.850V

5 0100 -105 0.850V


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6 0101 -135 1.161V

7 0110 -75 0.850V

8 0111 -45 1.161V

9 1000 135 0.311V

10 1001 165 0.850V

11 1010 45 0.311V

12 1011 15 0.850V

13 1100 105 0.850V

14 1101 135 1.161V

15 1110 75 0.850V

16 1111 45 1.161V

This next diagram shows the constellation diagram for a 16-QAM system, you can also

see a grid showing the approximate regions for each symbol. Again, when a received

symbol falls in that region, it is decoded as the bit value for that region.


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5.5.3 Multi-Multi M-ary QAM Systems

As mentioned before, QAM systems are the ones used when many symbols need to be

transmitted. The following diagrams show the constellation diagrams for such systems.

As you can imagine, since the symbols are so close together, there are many tricks used

by designers to ensure that symbols pass through the system without errors.

128 Point QAM Constellation Diagram


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One Quarter of a 960 Point QAM Constellation

nten 216 data communications 5 digital modulation · 2006. 11. 20. · nten 216 data...

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