digital communications 2014/dr... · 2017-01-22 · 1. base band signal the simplest digital data...

Digital Communications

Computer Engineering Department

Dr. Eng. Riyadh J.S. Al-Bahadili

Fourth Year

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Dr. Riyadh Jabbar Digital Communications 4th IT class computer engineering

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Digital Communications

Chapter (1) Introduction

1. Base band signal

The simplest digital data signal contains a sequence of signal element (units or pulses of a data

signal) where each element is binary coded. Having the choice of two possible shapes that

correspond to the element values 0 or 1, each signal element has the same duration of T seconds.

So that the signal element rate is 1/T elements per sec (or bauds). The digital signal above is clearly

a 2-level or binary signal. In a sequence of M-level signal elements, where = 2 and n is the

number of bits, the M-level data symbol that determines the element value of a signal element can

be represented by a sequence of n bits. For example, if = 3 bits then = 2 = 8 levels and the

corresponding sequence of 3 bits are 000, 001, 010… 111.

• 2-level (binary) signal = , = 1 , = 2 1

= = …bits per second = = …symbol per second (baud) = • 8-level signal = , = 3 , = 2 = 8 1 0 1 = ….bits per second = = … symbol/sec

In general, = 2 , = log , = , = , =

?Exercise.1

For 2-level and 8-level systems, what is the difference between them (bit rate? noise?) for the same .




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2. Model of Digital communication system

?Exercise.2

Explain briefly the function of each component in the model

3. Measure of information

Consider M-level system with symbols ( , , … , ) .The information contents of a symbol , denoted by ( ) is defined by: ( ) = log ( ) = − log ( )

Where b is the radix of digits (2 for binary) and ( ) is the probability of symbol ( , , … , ) = ( ) + ( ) + ⋯+ ( ) The information measure (bits) of a message is equal to the minimum number of binary pulses required to encode that message.

Example.1

How many bits per symbol to encode 32 different symbols?

= 32, ( ) = =

( ) = log 32 = 5 / Example.2

The four symbols , , , occur with probability 1/2, 1/4, 1/8, 1/8 respectively. Find the info content in the message ( ) ( ) = ( ) + ( ) + ( ) = log 4 + log 8 + log 2 = 6

Note: log =

Message Source encoder

Channel encoder

Modulator

Demodulator Channel decoder

Source decoder

User

Noise

Channel

Modem Codec

Tx

Rx




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4. Bandwidth and Noise

There are two factors affecting the information transfer rate on a channel

• The bandwidth of the channel will determine how quickly the signaling states on the channel can be changed.

• The level of noise in the channel will impose a limit on the number of different unique

states that can be correctly decoded at Rx. In addition, the degree of distortion introduced

by the channel will also limit the number and rate of change of symbol states.

So, if we had a channel with infinite BW (or no noise and distortion), it would be possible

to send, say, one Mbits at the speed of light.

5. Bandwidth efficiency

It is a measure of how well a particular format (and coding scheme) is making use of the available

BW. The units of BW efficiency is bits/second/Hz.

ℎ efficiency =

For example, if a system requires 4 KHz of BW to send 8000 bps of information, then = / = 8000/4000 = 2 / /

6. Multi-level signaling (M-ary)

For transmission with high data rate, it is possible to use M-level signaling: = 2 symbol states

01 00

01

10

11

0

1

2

3

4

5

6

7

2-level 4-level 8-level




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• Advantage of M-ary signaling

A higher information transfer rate is possible for a given symbol rate and channel BW.

(Alternatively, reduced the required BW for a given information rate).

• Disadvantages of M-ary signaling

Low immunity in noise/interference compared with binary signaling, as it becomes more difficult

to distinguish between symbol states. More complex symbol recovery in the Rx. required for

linearity and reduced distortion in Tx/Rx hardware.

Example.3 A modem claims to operate with BW efficiency of 5 bits/sec/Hz when using 1024 symbol states.

a) How many bits are being encoded in each symbol, and what is the modem capacity if the

symbol rate is 4000 symbol/sec

b) How many symbol states should be employed if the user wishes to send his info in half

the time?

Solution

a) = log = log 1024 = 10 / = = = = 10 x 4000 = 40 Kbps

b) To send the info in half the time, it would be necessary to send data at bit rate = 80 Kbps hence we need 20 bits in each symbol, so symbol states = 220 = 1048576

7. Channel capacity (for baseband signals)

The minimum BW required for error-free transmission

= 0.5 , ℎ ∶ Since is the symbol rate ( ), then

= 2

The channel capacity is max but measured in bit/sec, so

ℎ ( ) = 2

8. Additive white Gaussian Noise channel (AWGN)

As M increases, the ability of the receiver to distinguish between symbols in the presence of

noise/interference/distortion decreases. Hence SNR (signal to noise ratio) will be an important




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factor in determining how many symbol states can be utilized and still achieve (error-free)

communication. Of each symbol is also key in determining the noise tolerance of a receiver system, with longer

symbols giving the receiver more time to average out the effects of noise than shorter symbols.

The capacity of AWGN channel (Shannon capacity) defined as

C = B log (1 + SNR) … bits/sec Where B is the channel BW, SNR = , S is the signal power, N is the noise power =No B, and No is PSD of the noise (watt/Hz) ?Note

• The channel is error-free if R ≤ C • For given C, the BW can be increased for decreased signal power

Example.4 Consider AWGN channel with B=4 KHz and noise PSD is 2x10-12 W/Hz, the signal power required at the modem receiver is 0.1 mW. Calculate the capacity of this channel. = 4000 , = (0.1)10 = No B = 2(10 )(4000) = (8)10 SNR= = . ∗ ∗ = (1.25)104 C= B log2 (1+SNR) = 4000 log2 [1+1.25(104)] ≈ 54.44 Kb/s Example.5 The specification of two telephone links are Link B SNR Class 1 300-3400 Hz 40 dB

Class 2 600-2800 Hz 30 dB A company has a requirement to send data over a telephone link at bit rate R= 20 Kbps without error. Would you advise the company to rent the more expensive class 1 service, or the cheaper class 2 service? Justify your decision. Solution: C = B log (1 + SNR) For class one line:

B= 3400-300 = 3100 Hz, SNR = 40 dB=10000 C = 3100 log (1 + 10000) = 41.2 Kbps




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For class two line: B = 2800-600 = 2200 Hz, SNR= 30 dB=1000 C = 2200 log2 (1+1000) = 21.9 Kbps

So both of links will meet the specification of R=20 Kbps error-free. However, the performance

of class 2 line is very close to Shannon bound, in practice, it is unlikely that a modem could be

realized that would give the desired result on the class 2 line.

?Exercise.3

A signal with 256 symbols is transmitted by 104 symbol per second.

a) What is the information rate R?

b) Can the output be transmitted without error over AWGN channel with B= 10 KHz and SNR=

100

c) Find the SNR required for error-free transmission for part (b)

d) Find the B required for AWGN channel for error-free transmission if SNR= 100

9. Power and bandwidth efficiency

For a system transmission at maximum capacity, the average signal power (S) at receiver input

can be written as S=C Eb, where Eb is the average received energy per bit.

The average noise power (N) is N=No B, then Shannon expression can be written as:

C = B log (1 + )

= log 1 + = log (1 + )

The ratio [ ] represents the bandwidth efficiency of the system (bits/sec/Hz).

The ratio [ ] is the power efficiency. The smaller the ratio, the less energy used by each bit.

Choosing a power-efficient modem is particularly important in cellular handsets (say to maximize

battery lifetime).

Example.6

A digital cellular telephone system is required to work at a BW efficiency of 4 bits/sec/Hz. What

is the min Eb/No that must be planned for in order to ensure that users on the edge of the coverage

area receive error-free communication? If the mobile telephone company wishes to double the




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number of users, how much more power must the base-station and handsets radiate in order to

maintain coverage and error-free communication?

Solution: =log2 (1+ ), = 4

4=log2 (1+4 ) = (24-1) = 3.75 = 5.74 dB

In order to double the number of users for the same B, then = 8 = (28-1) = 31.87 = 15.03 dB

Thus, the transmitted power must be increase by a factor 15.03 - 5.74 = 9.29 dB ?Exercise.4 Find the BW efficiency for a wireless communication system having a bit rate of 9.6 Kbps and B of 200 KHz with of 10 dB.

?Exercise.5 Data has to be transmitted which has B=3 KHz. If SNR at the receiver is 12 dB, determine for

data rates: 2.4 Kbps and 4.8 Kbps. Also, determine the BW efficiency. ?Exercise.6 Sketch roughly the relationship between and , then

a) Find the value of when = 1, 0, ∞

b) Mark the region on the graph that been considered error-free transmission c) What is the minimum (in dB) for error-free transmission

10. Inter Symbol Interference (ISI)

With any particle channel, the filtering effect will cause a spreading of data symbols through the

channel. For consecutive symbols, this spreading causes part of the symbol energy to overlap with

neighboring symbols, causing ISI. This degrade the ability of the detector to differentiate a current

symbol from diffused energy of adjacent symbols. Even with no noise present in the channel, this

can lead to defection errors.

It is possible to control ISI such that it does not degrade the system performance by pulse shaping

or using Nyquist filtering ( ≥ 2 ), where is the sampling frequency.




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Chapter (2) Waveform coding techniques

A major application of speech processing concerns digitally coding the speech signal for efficient

secure storage and transmission. From a communications viewpoint, speech is encoded into a bit

stream, transmitted over a channel, and then converted back into an audible signal.

1. Pules Code Modulation (PCM)

The term waveform coding is applied to source encoding methods that seek to digitize the

incoming analog waveform. PCM (or A/D conversion process) is one method of waveform coding

involves sampled of the input signal level then quantized this sampled value then encode this level

into a number.

Voice message: = 0.3 ~3.4 , = 3 , = 8 /

• Sampling

To digitize only the minimum number of samples necessary to represent the waveform on

reception. The minimum rate at which a waveform can be sampled without loss of information is

Sampler Quantizer Encoder Decoder Filter

A/D D/A

000 010 110 111 100 011

0

1

2

3

4 5

6 7

Δ




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in fact twice the high frequency (or B) of the input waveform (Nyquist sampling criterion) ≥ 2 ( ≥ 2 ).

• Quantization

Each sample is rounded to the nearest on of set of levels ( = 3 means there are only 8

levels, = 4 means 16 levels and so on ) = 2 , where M is the number of quantized

levels.

If represents the spacing between possible output values (step size), then ≥ 2 , where

is rms value of the noise.

• Uniform Quantizer

Assume analog signal has peak value and low value , then ∆=

?Exercise.1

If represents the quantization error between the instantaneous (actual) value and its quantized

equivalent, then prove that = ∆ , where (average power) is the variance of quantization

error. [±Hint: −∆/2 ≤ ≤ ∆/2]

• Bit rate and Bandwidth of PCM

The bit rate of PCM is = and =

Input X

Output Y

Δ 2Δ 3Δ 4Δ

Δ/2

3Δ/2

5Δ/2

7Δ/2 Midrise Q Uniform Quantizer

X Y




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Example.1

For voice message, = 0.3~3.4 = 8 , calculate: quantization levels, sampling

frequency, bit rate, and the bandwidth of PCM.

Solution:

Quantization levels = = 2 = 2 = 256

The message bandwidth = 3.4 − 0.3 = 3.1

Sampling frequency = ≥ 2 → ≥ 2(3.1) → ≥ 6.2 → ≈ 8 /

Bit rate of PCM = = = 8(8) = 64 /

Bandwidth of PCM = = = / = 32

• Signal to Quantization noise ratio (SNR) for PCM

Each additional coding bit, which doubles the number of quantize levels (M), halves Δ, decrease

the quantization error, and increases SNR.

= , = 1.8 + 6

?Exercise.2

Suppose we are using PCM for sinusoidal signal swings between + and − , prove

that: =

Example.2

An audio signal has bandwidth of 5.8 MHz encoded by PCM. Given that the total number of bits

to represent a level are 10 bits. Determine: total number of levels, bit rate, bandwidth, and SNR in

dB.

Solution:

Quantization levels = = 2 = 2 = 1024

Bit rate of PCM = = = (2 ) = 10(2)(5.8)(10 ) = 116 / = = / = 58 = 1.8 + 6 = 1.8 + 6 (10) = 61.8




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?Exercise.3

The bandwidth of an input signal to the PCM is restricted to 4 kHz. The required SNR is 20 dB.

(1). Calculate number of bits required per sample.

(2). Calculate total transmission bandwidth for 30 PCM coder that has been time multiplexed.

?Exercise.4

PCM system uses a uniform Quantizer followed by 7-bit encoder. The bit rate of the system is 5

Mb/s. Calculate the maximum message bandwidth. [R Check answer: 360 kHz]

2. Log PCM (Non-uniform Quantization)

One way to control the noise in PCM is by distribute the levels as log form. This can overcome

the range problem by making SNR less dependent on signal level. If step size (Δ) is not constant

for all input amplitudes but rather is proportional to input magnitude, then weaker signals use a

smaller Δ than more intense signals do. The quantization levels should be logarithmically spaced

using non-uniform quantizer:

Or equivalently, the logarithm of the input should be coded with a uniform quantizer.

Input

Output = Non-Uniform

Quantizer =




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At the receiver, the sample reconstruction uses the inverse, exponential operation. This log PCM

process is called companding, since the input signal range is compressed at the coder and expanded

at the decoder.

A standard logarithmic operation is not practical, since (0) = −∞ leading to an infinite

quantization range. However, two approximations to log quantization ( ) give

nearly constant percentage error and they have found wide use in multimedia fields.

• Quantization

The quantizer characteristic is linear for input values up to a certain threshold (for | | ≤ / )

= … 0 ≤ ≤ 1/

And logarithmic beyond that

= … 1/ ≤ ≤ 1

Where = 86 = 7 , = 87.56 = 8

Uniform Quantizer

Compress Expand ( ) ( )

( )

( )

( )

( )




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• Quantization

The quantizer characteristic is defined with one smooth function as:

= [ ( | | ) ( ) ( )

For small input values ( ≈ 0) the output is linear

≈ [ (1 + )]

For large values of x, the output y varies directly with the logarithm of |x|.

The values of specifies the relative input value near which the quantizer evolves from a linear to

a logarithmic characteristic.

Increasing makes the quantizer logarithmic over a wider range of input values and makes SNR

decreases slightly with large values of .

3. Adaptive Pules Code Modulation (APCM)

APCM is the simplest time-adaptive technique where Δ is varied in proportion to the short-time

average signal amplitude. In this method (variable step size), is to evaluate signal (speech) energy

over a time window and modify Δ accordingly. For loud sounds, Δ is large while weaker sounds

permit smaller Δ. Thus, a constant SNR can be maintained independent of the signal input variance

(σ ).

Another method is the variable gain. A gain factor G inversely proportional to σ of the signal

input, multiplies the input, effectively compressing or expanding it to have uniform range for

different inputs.

Variable Quantizer

Δ estimation

σ

Δ

Δ ∝ σ

Input signal APCM

(Variable step size)




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APCM improves SNR performance and signal (speech) quality when compared to non-adaptive

PCM and log PCM systems.

4. Differential Pulse Code Modulation (DPCM)

In DPCM the error samples formed as the difference between the input samples and their estimates,

are quantized. There is a strong correlation between adjacent samples in many signals including

speech. DPCM is more efficient when compared to PCM and provide higher SNR at given

transmission rate.

In APCM, the quantization noise is proportional to Δ, which in turn is proportional to σ , i.e.

attempts to match Δ to σ using only information about signal value. In DPCM, instead of

quantizing signal samples directly, it quantizes the difference between a current sample and a

predicted estimate of that sample based on a weighted average of previous samples.

Where:

Fixed Quantizer

G estimation

Input signal

∝ 1

APCM (Variable gain)

Quantizer

P P

Inverse Quantizer

Input signal −

Coder Decoder

DPCM Codec




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: , : , :

: = − , : , : ( )

5. ADPCM

ADPCM refers to DPCM coders that adapt Δ and/or the predictor p. SNR can be increased in

DPCM if step size Δ may change dynamically. The improvements in SNR are additive: about 6

dB with the differential process and another 4-6 dB by adapting Δ to match the quantizer input.

6. Delta Modulation (DM)

An important subclass of DPCM system is DM, which uses 1-bit quantization ( = 1) with first-

order predictor. The main advantages of DM are simplicity and low cost.

The quantizer just checks the input sign bit. To compensate for large amount of noise arising from

such a sharp quantizer, the sampling rate is several times the Nyquist rate.

= 2

Where: D is oversampling ratio, and is the highest frequency of signal x (input speech).

Transmission rates are comparable for DM and DPCM, because D is equal to the number of bits

in typical DPCM quantizer. For example: in DPCM = 5 bits/sample, and = 8 Samples/sec,

so for DM: = 1 bit/sample, and = 2(5)(4000) = 40 / .

P P

−

Coder Decoder

DM Codec

Input signal

Δ

-Δ




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F The choice of Δ is very important in DM performance since the output value can change only

by ±Δ each sample interval (T), the step size Δ must be large enough to accommodate rapid

changes

≥ max | ( ) ( )|

Otherwise (when Δ is small) a form of clipping called slop overload results, with the output noise

exceeding Δ.

F If Δ is chosen too large, the noise becomes excessive since this noise is proportional to Δ (when

the speech input is silence, the output still oscillates with ±Δ)

F One way to improve DM is to let Δ changes dynamically with the input variance . This coder

is ADM and based on the previous 2-4 samples, to raise Δ when the recent transmitted samples

have the same sign, and to decrease it when the samples alternate in sign.




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Chapter (3) Digital Modulation Techniques

The previous chapter has been concerned with so-called baseband signaling where the channel

band is assumed to extend from 0 Hz upwards. In application where bandwidth encompassing 0

Hz is not available, band pass signaling is required. Here, the task is to centers the symbol energy

at a given frequency of operation, for example, 900 MHz for a typical cellular telephone channel

and 30 THz for an optical fiber link the process usually involves modulation the amplitude,

frequency, or phase of carrier sine wave. The carrier is commonly written as ( ).

The choice of modulation method affects the ease of implementation; the noise tolerance and

occupied channel bandwidth of the resulting band pass data modem.

1. Amplitude Shift keying (ASK) The simplest form of band-pass data modulation is ASK. Here, the symbols are represented as

various discrete amplitudes of a fixed carrier frequency .

BASK

BFSK

BPSK

m (t)




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• BASK

In binary ASK (2ASK), where only two symbol states are needed, the carrier is simply turned on

or off, and this process is called ON-OFF keying (OOK)

The spectrum of ASK signal can easily be determined if the spectrum of the baseband data symbol

is known, by viewing ASK modulation process as a mixing or multiplication of the baseband

symbol ( ) = ( ) with carrier ( )

( ) = ( ) , . . . 0 < <

Where A is a constant, ( ) = 1 0, is the carrier frequency, and T is the bit duration. It has

a power = /2, so that = √2

( ) = √2 = √ = √ Where = is the energy contained in a bit duration.

If we take [ ( ) = ] as the orthonormal basis function, the applicable signal space or

constellation diagram of the BASK signals is shown as:

The Fourier transform of the BASK signal ( ) is

( ) = 2 BASK ( ) = 0

BASK Modulator




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( ) = ( − ) + ( + ) The effect of multiplication by the carrier signal ( ) is simply to shift the spectrum of the

modulating signal ( ) to The following figure shows the amplitude spectrum of the BASK signals when ( ) is a periodic

pulse train.

Since we define the bandwidth as the range occupied by the baseband signal m(t) from 0 Hz to the

first zero-crossing point, we have B Hz of bandwidth for the baseband signal and 2B Hz for the

BASK signal. The following figure shows the coherent demodulator for BASK signals.




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• M-ary ASK

If more than two levels are used, then an M-ary ASK is adopted for high bit rate (4ASK for 2bits,

8ASK for 3bits and so on). 4ASK is shown here:

An M-ary amplitude-shift keying (M-ASK) signal can be defined by

( ) = ( ) , 0 < < 0 , ℎ

Where = [2 − ( − 1)], … for = 0,1 … − 1

( ) = 2 = =

MASK Modulator

Here is a 4ASK signal constellation diagram:




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(MASK coherent demodulator)

• Bandwidth efficiency and Capacity of ASK

ℎ = … ⁄ /

= … /

Example.1

ASK is used for transmitted data at = 28.8 Kbps over a telephone channel with bandwidth =300 3400

a. How many symbol states are required in order to achieve this level of performance?

b. What would be the equivalent number of symbol states needed if the channel pass band

extended from 0 Hz to 3100 Hz and baseband M-ary was used?

c. What is the maximum capacity for the ASK if the SNR on the telephone link is 33 dB.

Solution:

a. The capacity of band pass ASK is

= . , = 3400 − 300 = 3100 , =

28800 = 3100 → = 626.1 or M ≈ 1024 states

b. The capacity of baseband M-ary system is

= 2 , = 3100

4




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28800 = 2(3100) → = 25.02 or M ≈ 32 states

c. Shannon capacity is

= (1 + ) = (3400 − 300) (1 + 10 . ) = 33.996 Kbps

• Probability of Symbol error There are two types of waveform used in digital communications: unipolar and bipolar waveforms

For unipolar waveform, the energy per symbol is different depending on whether a logic 0 or 1

is sent, having a zero value for logic 0 case. The Probability of symbol error for unipolar is:

= …

Where is noise power density, and ( ) is the complementary error function:

( ) = √ ∫

And the Probability of symbol error for bipolar waveform is

= …

A

0

1 1 1 0 1 1 1 0 A

-A

0

Unipolar Bipolar




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It is often to draw which is called SNR rather than where = / and n is the number of

bits per symbol.

When the number of levels is increased (M>2), the ability of the receiver to distinguish between

symbols in the presence of noise will decrease. The Ps for M-ary bipolar baseband signaling is:

=

Example.2

A company wishes to increase the through put of a telephone modem product by changing from

2-level signaling to 8-level signaling and has set a design target of maintaining a performance of

no worse than one symbol error in every 10 000 symbols sent. By using the plot of symbol error




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vs. Eb/No for M-ary, determine the reduction in noise tolerance for the modem because of this

change. What is the theoretical minimum Eb/No required supporting the bandwidth efficiency

achievable by the 8-level modem?

Solution:

From the plot of for M-ary signaling, at =10-4, it can be seen that an increase of about 8 dB is

required to maintain the same error rate. Therefore, the new modem will be approximately 8 dB

less tolerant to noise.

For 8-level modem, maximum bandwidth efficiency is 6 bits/sec/Hz, so

= 1 + 6 = 1 + 6 Therefore the minimum for error-free transmission is

( ) = = 10.5 or 10.2 dB

• Bit-error rate (BER) performances of ASK

The performance of digital communication systems is presented at the simplest level as a

probability of bit error or probability of symbol error , as a function of the received Eb/No

ratio. Binary ASK effectively uses a unipolar baseband modulation source.




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2. Frequency Shift keying (FSK) FSK has until recent years been the most widely used form of digital modulation, being simple

both to generate and to detect, and being insensitive to amplitude fluctuations in the channel. FSK

conveys the data using distinct carrier frequencies to represent symbol states. ? An important

property of FSK is that the amplitude of the modulated wave is constant.

( )2 = ( 1 ) , 1 ( 2 ) , 0 Consider the case of unfiltered 2FSK. This waveform can be viewed as two separate ASK symbol

streams summed prior to transmission.

• BFSK generation

FSK can be generated by switching between distinct frequency sources; however, it is likely that

there will be discrete phase jumps between the symbol states at the switching time. Any phase

discontinuity at the symbol boundary will result in much greater prominence of high frequency

terms in the spectrum, implying a wider bandwidth for transmission.




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Alternatively, FSK can be realized by applying the data signal as a control voltage to a voltage-

controlled oscillator (VCO). Here the phase transition between consecutive symbols states is

guaranteed to be smooth (continuous). FSK with no phase discontinuity between symbols is known

as a continuous phase (CPFSK).

• The vector modulator

The arrangement of mixer and a combiner forms an extremely useful building block in digital

communication systems. It achieves a linear frequency translation of all components in the input

signal (represented by its in-phase and quadrature components) by a carrier frequency component

(also represented by its in-phase and quadrature components). This block is often referred to as a

vector modulator or quadrature modulator, and can be used for both frequency up-conversion and

down-conversion. The output of the two mixing processes is given by




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cos . cos = [cos( − ) + cos( + ) ] sin . sin = [cos( − ) − cos( + ) ] When the above terms are summed, the result gives a down-converted component:

cos( − ) In addition, when subtracted from each other result in a signal up-converted component:

cos( + )

?Exercise.1

A vector modulator is fed with a perfect quadrature sine wave at the input, but there is a small

phase error of 5o between the notional quadrature inputs of the carrier signal. What will be the ratio

in dB between the sum and difference outputs of the vector modulator (ratio of the amplitude of

the wanted to unwanted output signal)? [±Hint: Sin ≈ 0 for small ], [RCheck answer: 27 dB]

• BFSK Vector modulator

BFSK requires the generation of two symbols, one at a frequency ( + ) and one at a

frequency ( − ). So to generate a shift of (+ ), I and Q inputs need to be fed with (− cos ) and (sin ) respectively. Generating a shift of (− ) requires inputs of cos and sin .

cos

sin

cos

sin( + )

sin

cos sin

±cos




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This approach is now frequency used to generate filtered CPFSK particularly in cellular handsets.

?Exercise.2

Draw the block diagram of vector modulator to generate 2ASK signal.

• Spectrum of BFSK

An approximation of BFSK spectrum can be obtained by plotting the spectra for two ASK streams

centered on the respective carrier frequencies.

Clearly, the overall bandwidth occupied by FSK signal depends on the separation between the

frequencies representing the symbol states. CPFSK system will have much lower side-lobe energy

than the discontinuous case.

• Coherent BFSK detection

This method is very similar to that for ASK but in this case there are two detectors tuned to the

two carrier frequencies.




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• M-ary FSK system

M-ary FSK (multi-level) is very much of interest for increasing the noise immunity of the

modulation format compared with BFSK, allowing a designer to achieve reliable data transmission

in the presence of high levels of noise. This is only possible by using a set of “orthogonal symbols”.

Two symbol states ( ) ( ) are said to be orthogonal over the symbol period if: ( ). ( ). 0

If the frequencies of M-FSK symbols are chosen to be of the form:

( ) = cos + = 1,2 … ( − 1)

Then these frequencies are orthogonal over a symbol period.

Example.3

For 8-FSK and Rs=1200, the required frequencies are 1000,1600,2200,2800,3400,4000,4600 and

5200 Hz.

? Orthogonal system gives better SNR at detector output, improving the probability of correct

symbol detection but required high bandwidth.

• M-ary FSK detection




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A typical M-ary FSK detector consists of a bank of “correlators” (mixers with coherent carrier

reference), followed by a decision circuit at the output determining which correlator has the largest

output and hence which symbol was sent.

• BER performance for M-ary FSK

? As the number of symbol states is increased, the BER improves but at the expense of BW.

• Advantage of FSK

v FSK is constant envelope modulation and hence insensitive to amplitude variations in

the channel.

v The detection of FSK is based on relative frequency changes between symbol states

and thus does not required absolute frequency accuracy in the channel.

v In deep space missions where the path loss is so great, M-ary FSK is very effective

modulation.

• Disadvantage of FSK

v FSK is less bandwidth efficient than ASK or PSK

v The bit/symbol error rate performance of FSK is worse than for PSK.




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3. Phase shift keying (PSK)

With PSK, the information is contained in the instantaneous phase of the modulated carrier.

Usually this phase is imposed and measured with respect to fixed carrier of known phase-coherent

PSK. For binary PSK (2PSK), phase states of 0o and 180o are used. It is also possible to transmit

data encoded as the phase change (phase difference) between consecutive symbols (Differentially

coherent PSK). There is no non-coherent detection for PSK.

For BPSK:

= cos( + 0) … 0

= cos( + ) … 1

Where Es is energy per symbol, T is symbol time and is the amplitude (A) of the signal.

In general, for MPSK:

= cos + … = 0,1 … , ( − 1)

Where = is the modulation angle.

The constellation mapping for MPSK can be shown below




32 | P a g e

? QPSK is quadrature PSK (4PSK).

• BPSK spectrum

The bandwidth of BPSK signal is identical to that of BASK. In fact, BPSK can be viewed as ASK

signal with the carrier amplitudes as + A and –A (rather than +A and 0 for ASK).

• PSK generation

The simplest means of realizing unfiltered BPSK is to switch the sign of the carrier using the data

signal, causing 0o or 180o phase shift.

BPSK QPSK QPSK 8PSK




33 | P a g e

? The square pulses for data signal are not practical to send. They are hard to create and required

a lot of bandwidth. The solution here is to send shaped pulses that convey the same information

but use smaller bandwidth and have other good properties such as ISI rejection.

There are some common pulse shaping methods that control the shape and the bandwidth of the

signal:

v Root raised cosine (used with QPSK)

v Half sinusoid (used with MSK (minimum shift keying))

v Gaussian (used with GMSK. This system is used in several mobile systems around the

world such as in GSM (global special mobile)

v Quadrature partial response (QPR)

• Detection of BPSK

There is no non-coherent detection for PSK, and various forms of coherent detection must be

employed. The ideal detector thus requires perfect knowledge of the unmodulated carrier phase at

the receiver (carrier recovery). As with ASK, any phase error of the locally generated carrier

reference reduces the signal level at the output of the detector by cos . This in turn degrades the

Es/No performance.

Thus, we need zero phase error for optimum detection. Note that if the phase error reaches 90o,

the output falls to zero.




34 | P a g e

• Quadrature phase-shift keying (QPSK)

QPSK uses the orthogonality between cosine and sine carrier. This would imply that if we send

BPSK on the cosine of a carrier, and simultaneously send a second BPSK using the sine of a

carrier, then it would be possible to detect each one independently of the other. ? Orthogonality

property of QPSK means that it can be used to send information at twice the speed of BPSK in the

same bandwidth. The block diagram of QPSK modulator is simply two BPSK using quadrature

carriers summed in parallel. The source data is first split into two data streams, with each data

stream running at half the rate of input data.

• Performance of MPSK

Increasing M allows further improvements in bandwidth efficiency but requires more for same

Ps.

ℎ = … / /




35 | P a g e

4. Quadrature Amplitude Modulation (QAM) So far, we have considered only signal property modulators using amplitude, frequency, or phase

symbols to conveying the data. We can combine two or more symbols types, which gives improved

performance (trade-off between bandwidth efficiency and noise performance).

The simplest form of QAM is in fact the QPSK symbol set, which Can be viewed as two quadrature

amplitude modulated carriers, with amplitude levels of +A and -A .

Increasing the number of amplitude levels on each carrier to 4 (for example ±A, ±3A) gives 16

possible combinations of symbols at the output, each equally spaced on the constellation diagram,

and each represented by a unique amplitude and phase.

• QAM generation

The modulator is making use of orthogonality of the sine and cosine carriers to allow independent

detection of the two ASK data.

16-QAM constellation




36 | P a g e

Pulse shaping is performed by filtering the multi-level baseband input symbol streams as in ASK.

• QAM detection

QAM can be decoded using coherent detection just as for PSK (requires carrier recovery). The

output of each demodulator is a baseband multi-level symbol set; this should undergo matched

filtering for optimum performance in noise. The aim of comparator is to determine the level at the

sampling instant, and hence decode the corresponding bit pattern.

• M-ary QAM vs. M-ary PSK

Comparing the constellation diagrams of 16 QAM with 16 PSK, we can see that the spacing

between symbol states for QAM is greater than that for PSK, which means that the detection

process in QAM should be less susceptible to noise. However, the power for QAM is greater than

that for PSK and this must be taken into account if the transmission process is power limited.




37 | P a g e

Example.4

A digital TV has a source analogue video signal with BW from 0 Hz to 2 MHz. This signal is

sampled at four times the highest frequency using 16-bit ADC. The resulting data signal is sent

over the air using 16QAM modulation. Assume ideal pulse-shaping filter, what is the bandwidth

occupied by the transmitted digital video signal?

Solution:

Sampling rate at ADC = 4 = 4 (2) = 8 /

Bit rate at ADC output = 16 bits (8M) = 128 Mbps

16 QAM uses = 4 , so = 4 bit/sec/Hz

Hence = = 32




38 | P a g e

Example.5

A transmitter for digital radio system is peak power limited to 150 W with 50 Ω antenna. Determine

the average power that can be supported for both 16 PSK and 16 QAM transmission if each point

in the constellation has an equal probability of transmission.

Solution:

With reference to one quadrant of the 16 QAM constellation, the average power developed by each

of the vectors A, B, C, D is as follows:

A2 = (3a)2+(3a)2=18 a2 , B2 = (3a)2 + (a)2 = 10 a2

C2 = (a)2 + (a)2 = 2 a2 , D2 = (3a)2 + (a)2 =10 a2

Average power = ( )/ =

The maximum vector power is 150 w, so =150 = = → a = × = 20.4

The average power for all symbol states is: Pav (QAM) = = 83.33 w

Pav for 16 PSK is the same for all symbol states Pav (PSK) = = 150 w

?Exercise.3

If the maximum vector length in 16 QAM is 100 v rms, determine the average power that would

be delivered into R=50 Ω antenna load if each point in the constellation has an equal probability

of transmission. [R Check answer: Pav = 111 w]

?Exercise.4

If the peak symbol power for 16QAM is 200 w, measured in R=50 Ω antenna load. What are the

amplitudes of the different symbol vectors in the transmitted waveforms?




39 | P a g e

?Exercise.5

Orthogonal 4FSK modem has = 2400 / . If the lowest symbol frequency is 8 kHz,

what will be the other three symbol frequencies?

?Exercise.6

64QAM data link operates at 256 kbps. What is the symbol rate on the channel, and what is the

occupied bandwidth?

?Exercise.7

What is the minimum bandwidth required to support 256 kbps data stream using BPSK, QPAK,

and 64QAM?

?Exercise.8

A customer requires a microwave radio link to provide a bit rate of 2 Mbps in a bandwidth of 400

kHz. The minimum SNR on the channel is 30 dB. Can the channel support the required capacity?

Moreover, how many symbol states would be required?




40 | P a g e

Chapter (4) Channel Coding

The goal of channel coding is to detect the data digits with minimum probability of error. This is

an effective way of increasing the channel capacity. The basic idea of coding is to add a group of

check digits to the message and transmit the entire block through the channel. The check digits

may then provide the receiver with sufficient information to either detect or/and correct channel

errors. There are many techniques used in channel coding: block coding, convolutional coding,

and combined coding and modulation (TCM).

= +

k: number of message digits

r: number of check digits

n: code word

Single parity check

A simple one-error detection with = 1 bit

Hamming code

A class of linear codes that can correct all patterns of single error in received word.

= 2 − 1

Block coding

Let the encoding word

= [ … … ]

Message check digits

Then

= ( , , … , )

= ( , , … , )

xor

k r




41 | P a g e

…

= ( , , … , )

For = 3, = 3: = [ ] With the following functions for check digits:

= ⨁

= ⨁

= ⨁

(? Note: the operator ⨁ is modulo-2 addition)

Then

= 0. ⨁ 1. ⨁ 1.

= 1. ⨁ 0. ⨁ 1.

= 1. ⨁ 1. ⨁ 0.

Alternatively, in matrix form:

= 0 1 11 0 11 1 0 0 1 1 1 0 01 0 11 1 0 0 1 00 0 1 = 0

In general:

= 0

Where H is matrix called parity check matrix.

Decoding process

Let the error vector be

= [ … ] , where = 0 … 1 … The received word is

= [ … ] And = ⊕




42 | P a g e

The decoder begins computing the syndrome S

=

Now we have two cases:

• If = 0 (no error), then = 0, i.e. = = • If ≠ 0 (there’s an error), then:

= = ( + ) = +

But = 0, then =

It means represents a column of H matrix.

Example.1

For = 3, = 6 and the parity check matrix is:

= 0 1 1 1 0 01 0 11 1 0 0 1 00 0 1

If the received word is R= [0 1 0 0 1 1], check if there’s an error occurred in R, then find the correct transmitted word C.

Solution:

= = 0 1 1 1 0 01 0 11 1 0 0 1 00 0 1 ⎣⎢⎢⎢⎢⎡010011⎦⎥⎥⎥⎥⎤ = 110

Since S≠0 (an error occurred)

E= [0 0 1 0 0 0]

= ⊕ = [0 1 0 0 1 1] ⊕[0 0 1 0 0 0] = [0 1 1 0 1 1]

? Exercise.1

For = 4, = 7, if the received word is R= [1 1 1 1 0 1 0], check if there’s an error occurred in R, then find the correct transmitted word C for the flowing functions.

= ⨁ ⨁

= ⨁ ⨁

= ⨁ ⨁




43 | P a g e

Galois Fields Algebra

When we deal with only two symbols 0 and 1, this process is called (2). In coding theory, the

computations are frequently using polynomials whose coefficient are 0 or 1, with modulo-2

operation. i.e:

• Coefficients of polynomial just 1 or 0, any odd coefficient reduced to 1, and any even

coefficient reduced to 0.

• Both plus and minus operations are just modulo-2 addition (i.e. xor)

( )

In order to construct the set of field elements, a table of powers of α is developed:

0, 1, , , …

All these elements can be expressed as sum of

, , …

Example.2

Construct the elements of (2 ), given that the polynomial for this type is

( ) = + + 1

Solution:

We have = 4, so the last element is = =

The set of elements are (expressed as sum of 1, , 2, 3):

0

1

= + 1

= +

= +

= + = + + 1




44 | P a g e

= + + = + 1

= +

= + = + + 1

= + +

= + + = + + + 1

= + + + = + + 1

= + + = + 1

= + = 1

=

The polynomial ( ) = 4 + + 1 is called a primitive polynomial that gives the complete

table with 2 symbols.

In fact; 1, , , in (2 ) are just the weights of the symbols, where 1 is the weight of LSB

and is the weight of MSB in each symbol.

For example, with (2 )

1 value

0 0 0 0 0

0 0 0 1 1

0 0 1 0

0 0 1 1 + ⋮ ⋮ 1 0 1 1 + + ⋮ ⋮ 1 1 1 1 + + +

0

1




45 | P a g e

Computations in GF There are two rules in computation in GF:

• = 1

i.e. (for = 4) = = 1

• = ( ) … > (2 − 1)

i.e. (for = 4) = =

Example.3

For (2 ):

. =

. = =

=

= =

+ = ( + 1) + ( + ) = + + + 1 =

? The object of coding theory (error detection and correction) is to find the roots of a polynomial

whose coefficients are elements of GF.

Example.4

Find the roots of ( ) = + + = 0 , and specify which symbols they represent.

We have

=

= + + 1 = + + 1 + 2 = + + + + 1

= 6 + 10

Hence: ( ) = + + = + ( + ) + = 0

= ( + 6)( + 10)




46 | P a g e

So and are the roots of ( )

= + = [1100]

= + + 1 = [0111]

Minimal polynomial M(x)

Let be any element of (2 ), then the minimal polynomial ( ) of smallest degree such

that ( ) = 0 is called the minimal polynomial. [ ( )] = = 0

i.e. if is a root of ( ) , then is also root, for = 0,1,2, … . If is the degree of ( ), then , , … are all roots of ( ) In general if:

= ℎ ( )( )

Example.5

Find the minimal polynomial for (2 ) when = .

Solution:

The minimal polynomial is ( )( ), and the roots are

= , = , = , = , = =

So the roots for = are: , 2 , 4 , 8

( )( ) = ( + )( + )( + )( + ) = + + 1

So ( )( ) = + + 1

? Exercise.2

Find the minimal polynomial for (2 ) when = .

[R Check answer: ( )( ) = + + + + 1]




47 | P a g e

Two errors detection and correction algorithm

The hamming code can correct only signal error of length n, where = 2 − 1. In order to correct

two or more errors, more check digits are required, i.e.

= 2 − 1

Where m in hamming code is r and n is the number of columns of H matrix. For two-error

correction, we need

= 2

Then H matrix becomes (2 ) , the first m rows are made to correspond to the hamming

matrix. The second m rows must tell the decoder something new.

~ As example, for a message that has = 15,

= 2 − 1 , = 4 check bits

The H matrix becomes:

= … ( ) ( ) … ( ) Where represents the th location number (column), i.e.

= 1

=

= + 1

……

= + + + 1

And

( ) = (1)

( ) = ( )

……

( ) = ( + + + 1)

So, for = 15 , = 4, H matrix becomes:

← First m rows ← Second m rows




48 | P a g e

= ⎣⎢⎢⎢⎡ 0 0 00 0 00 1 1 … 1 … 1 … 1 1 0 1 1 2 3 … 1… 15 ⎦⎥⎥

⎥⎤ Now, how can we select ( )?

? In general, ( ) is taken to be some power of , i.e. ( ) = ℓ

The error locator polynomial Suppose that two errors occurred in the th and th locations of = [ … . ] . These errors

will correspond to the location numbers (column) and . Let S be the sum of these columns of H

= [ … … ]

Or

= [ ] , ℎ = + = ( ) +

? The decoder problem is therefore to determine and given the S components Ψ1 and Ψ2.

Case 1

Let ( ) =

= + , = +

= ( + ) = + 2 + = + =

Hence = … ℎ !

Case 1

Let ( ) =

= + , = + = ( + )( + 2 + )

= ( + )

1

First 4 rows

Second 4 rows




49 | P a g e

i.e. = +

But = + , then 1 + = 12 + 2 1

Also = + , then 1 + = 12 + 2 1 We have same equation but different root (i.e. or )

Let x be unknown root ( or ), then ( + ) = +

In general: + + + = 0

This is called error location polynomial for two errors. If we solve this equation to find the roots,

then: = =

In addition, this equation has the following properties:

→ 1 = 2 = 0

1 → 1 ≠ 0 , 2 = 13

2 → 1 ≠ 0 , 2 ≠ 13

Decoding Algorithm

1. Compute = and obtain and

2. Find the roots of error locator polynomial.

3. Change the corresponding bits of R.

2n Buffer

Compute ,

Solve + + + = 0

Find Roots ,

R Corrected

C




50 | P a g e

Computing = takes long time, so we compute S as follows:

= [ 1 2] = ( )( ) = ( ) ( )( )

= ( )( ) = ( ) ( )( )

Where: ( ) is the received word which has the form [1 … . ] ( )( ) is the minimal polynomial ( ) is the remainder of dividing ( ) by ( )( ) and evaluating it at =

Example.6

The message is encoded using block coding with each block has = 15, the received block is

= [0 1 0 1 0 1 1 1 1 0 1 1 0 0 1] Find and , then comment how to obtain the locations of two errors and .

Solution:

( ) = [1 ] = [0 1 0 1 0 1 1 1 1 0 1 1 0 0 1 ]

= + + + + + + + + We have (see Example.5 and Exercise.2)

( )( ) = + + 1

( )( ) = + + + + 1

= ( )( ) = ( ) ( )( )




51 | P a g e

Using the long division:

So ( )( ) = + + = ( )( ) = + + In the same way we obtain

= ( )( ) = ( ) ( )( )

( )( ) = + 1 → ( )( ) = + 1 = + + 1

So

= + + = + + 1

Now we can find the locations of errors and by solving the error locator polynomial

+ + + = 0

( ) ( )( )

+ + + + + + + + + + 1

+ + + + +

+ +

+ + + + + + + + + + + + + + + +




52 | P a g e

Encoding Process

Encoding process is simpler than decoding process and it just uses a generator polynomial ( ) depending on the number of errors.

( ) uses the minimial polynomials depends on the number of error correction that the system

uses.

( ) = ( )( )

( ) = ( )( ) ( )( )

( ) = ( )( ) ( )( ) ( )( )

( ) = ( )( ) ( )( ) ( )( ) … ( )( )

? Exercise.3

Find the generator polynomial ( ) if the system is designed to detect and correct 2 errors.

[ R Check answer: ( ) = + + + + 1 ]

( )

Message (Blocks)

Transmitted Block

C

← For 1 error

← For 2 errors

← For 3 errors

← For t errors




53 | P a g e

Chapter (5) Spread Spectrum and Multi-User Modulation

Spread spectrum (SS) communication systems are widely used today in a variety of applications

for different purposes such as:

• Multiple access: access of same radio spectrum by multiple users.

• Anti-jamming capability: signal transmission cannot be interrupted or blocked by spurious

transmission.

• Interference rejection

• Secure communications

• Multi-path protection

Irrespective of the application, all SS systems satisfy the following criteria:

1. Bandwidth of the transmitted signal is much greater than that of the message that modulates

a carrier.

2. The transmission bandwidth is determined by a factor independent of the message

bandwidth.

So the power spectral density (PSD) of the modulated signal is very low and usually comparable

to noise and interference at the receiver.

The baseband signal bandwidth is intentionally spread over a larger bandwidth by injecting a

higher frequency signal (PN-generator). As a direct consequence, energy used in transmitting the

signal is spread over a wider bandwidth and appears as noise. The ratio (in dB) between the spread

baseband and the original signal is called processing gain (PG). Typical spread spectrum PG ran

from 10 dB to 60 dB.




54 | P a g e

Advantages of spread spectrum techniques

1) Resistance to interference and jamming

Intentional or unintentional interference and jamming signal are rejecting because they do not

contain the PN key. Only the desired signal, which has the key, will be seen at the receiver. That

rejection also applies to other SS signals that do not have the right key (other users). Thus different

SS communications can be active simultaneously in the same band (such as CDMA). Note that SS

is a wideband technology.

2) Resistance to interception

Because non-authorized listeners do not have the PN key used to spread the original signal, those

listeners cannot decode it. Without the right key, the SS signal appears as noise or as interferer.

3) Resistance to fading (multipath effects)

Wireless channels often include multiple-path propagation in which the signal has more than one

path from the Tx to the Rx. such multi-paths can be caused by atmospheric reflection and by

reflection from the ground or objects such as buildings. The reflected path R can interfere the

direct path D in a phenomenon called fading. Because the Rx synchronizes to signal D, signal R

is rejected even though it contains the same key.




55 | P a g e

4) Multiple access communications

Spread spectrum is not a modulation scheme, and should not be confused with other types of

modulation. We can use SS techniques to transmit a signal modulated by FSK or PSK. Spread

spectrum can also be used for implementing multiple access (i.e. multiple and simultaneous

communication links on the same physical media) as in TDMA, FDMA and CDMA.

Types of spread spectrum

• Direct sequence (DS)

• Frequency hopping (FH)

• Time hopping (TH)

• Hybrid systems (DS+FH)

Pseudo-noise (PN) generator

Several spreading codes are popular for use in practical SS systems. Some of these are maximal

sequence (m-sequence) codes, and Gold codes. A shift register can generate these longest codes.

An N-stage shift register with Xor gate can be used to generate m-sequence of length 2N-1. Note

that the sequence will repeat itself after 2N-1 bits.

Example.1

Direct sequence (DS)

With DS technique, the PN generator is applied directly to data entering the carrier modulator. The

modulator, therefore, sees a much large bit rate, which corresponds to the chip rate of PN generator.

1 1 1 Clock Rc Chip rate

O/P

1 1 1 0 1 1 1 0 1 0 1 0 0 0 1 1 0 0 1 1 0 = 2 − 1 = 2 − 1 = 7 ℎ

7 ℎ




56 | P a g e

The spectrum of the DS output is ( ) centered at the carrier frequency. The main lobe of this

spectrum (null to null) has a bandwidth 2Rc, and the side lobes have null-to-null bandwidths equal

to the Rc.

? An important feature of DS is its ability to operate in presence of strong co-channel interference.

• Processing gain

The ratio of the signal bandwidth to the message bandwidth (or the number of chips per

information bit) is just the bandwidth expansion factor. This ratio is called the “processing gain” = =

Where Tb is the message bit interval, and Tc is the chip interval (Tc = 1/Rc).

Example.2

For DS in which the transmitted signal bandwidth is 20 MHz and message bit rate is 10 Kb/s,

= 10 log × = 33

Which means that system would offer 33 dB improvement in SNRo at receiver output.




57 | P a g e

Also represents the advantage gained over the jammer that is obtained by expanding the

bandwidth of the transmitted signal.

• Jamming margin

The amount of interference (or jamming) that the receiver can withstand while produces an

acceptable SNR is called jamming margin , where

=

Where Jav is the jamming signal power and Pav is the signal power.

• Signal to Noise Ratio

The SNR required by the receiver to achieve a specified level of performance is

= − − …

Where is system implementation losses.

For DS which has = 33 , = 10 and = 2 , then

= 33 − 12 = 21

? The major disadvantage of DS is the ‘near-far’ effect. This effect is prominent when an

interfering transmitter is close to the receiver than the intended transmitter.

? DS uses BPSK and QPSK modulation techniques.




58 | P a g e

Frequency Hopping (FH)

In FH method, the carrier hops from frequency to frequency over a wide band according to a

sequence defined by PN generator. The frequency is constant in each time chip but changes from

chip to chip.

FH spectrum is flat over the band of frequencies and the bandwidth of FH signal is N times the

number of frequency slots, where N is the BW of each hop channel. FH systems can be divided

into fast-hop or slow-hop

• Fast-hop FH system is the kind in which hopping rate Rh in greater than the message bit

rate

• Slow-hop uses Rh that smaller than the message bit rate.




59 | P a g e

? FH system is usually using FSK modulation, and the receiver is non-coherent and needs FEC

(forward error correction)

? The advantages of FH system are:

• Less affected by near-far problem

• Better for avoiding jamming

• Less affected by multi-access interference.

Applications of spread spectrum technology are used in CDMA, satellite communications and

wireless LAN.

Multi-user digital modulation

An important use of the concept of SS in wireless communication systems is to allow multiple

users occupy the same transmission band for simultaneous transmission of signals without

considerable interference. In telecommunications and computer networks, a channel access

method or multiple access method allows several terminals connected to the same multi-point

transmission medium to transmit over it and to share its capacity. Examples of shared physical

media are wireless networks, bus networks, ring networks, hub networks and half-duplex point-

to-point links.




60 | P a g e

• Frequency division multiple access (FDMA) This classical technique has been in use in conventional telephone systems and satellite.

Communication systems. Every user gets a certain frequency band assigned and can use this part

of the spectrum to perform its communication. If only a small number of users is active, not the

whole resource (spectrum) is used.

• Time Division Multiple Access (TDMA)




61 | P a g e

Every user is assigned one “time slot” within a “time frame”. A transmitting user sends its own

data only in the designated time-slot, and waits for the remaining time-frame duration until it gets

another time-slot in the next time-slot in the next time-frame. The synchronization among all users

is an important and necessary feature of TDMA.

• Code Division Multiple (CDMA)




62 | P a g e

One unique spreading code is assigned to each user for accessing the RF bandwidth simultaneously

for transmission and reception of signals. The spreading codes, assigned to all participating users,

are carefully chosen to ensure very low cross-correlation among them. This ensure that the signal

from undesired transmitters appear as noise. CDMA does not need precise time synchronization

among the users.

• Cellular CDMA

Cellular CDMA is a promising access technique for:

• Supporting multimedia services in a mobile environment as it helps to reduce the multi-

path fading effects and interference.

• Supporting universal frequency reuse, which implies large tele traffic capacity to

accommodate

? The quality of received signals degrades with increase in the number of active users at a given

point of time.

In fact, the number of users who can use the RF band is limited by the amount of interference

generated in the air interface.

? Fading is a major factor degrading the performance of CDMA system.




63 | P a g e

Chapter (6) Satellite Communication Systems

Satellite communication systems (SCS) are the outcome of research in the field of radio

communication with the aim of achieving the greatest coverage and capacity at lowest cost. SCS

are divided into two parts:

(1) Space segment: includes the sat and the means on earth necessary for launching and station

keeping.

(2) Earth segment: contains Tx and Rx for transmission and reception of signal from sat.

SCS services

1) Fixed-Sat service: For communication between earth stations at specified fixed points via one

more satellites.

2) Mobile-Sat service: Provides communication between mobile earth Stations and one or more

space stations.

3) Broadcasting-Sat service: Allows audio and video to be received by Individuals via sat.

4) Earth exploration-Sat service: Involves observation of earth for various purposes

(meteorological, earth resources, data collection)

5) Space research service: Spacecraft or other objects in space are used for scientific or technical

research.

6) Space operation service: Concerned exclusively with the operation of spacecraft (tracking,

telemetry , telecommand)

7) GPS-Sat service: For determining the position and velocity of an object.

SCS Frequency band Communication satellites have worked by receiving and “up-link” signal from an earth station,

shifting its frequency to the “down-link” signal, amplifying the shifted signal, and retransmitting

it back to the ground.

Frequency translation is necessary because it is impossible to achieve the necessary isolation

between Tx output and Rx input. Without sufficient isolation, the sat transmitted signals would

block the up-link received signals.




64 | P a g e

A. Commercial sat communications use a frequency band of 500 MHz bandwidth near 6 GHz for up-

link and 4 GHz for downlink.

B. Advanced technology sat are moving to higher frequency bands. 1000 MHz bandwidth has been

allocated near 12 GHz for downlink with corresponding up-link near 14 GHz.

C. Another band where extremely high capacities are potentially available is 30/20 GHz, where 2.5

GHz bandwidth has been allocated.

? FDMA, TDMA, and CDMA are most modulations used in SCS with 50-100 Mbps data rates.

Fundamental of SCS

• There are two types of SCS: Active and passive, a passive sat only reflects received signal

back to earth. An active sat acts as a “repeater”, it amplifies signal received and retransmits

it back to earth.

500 MHz

40 MHz

Ch.1 Ch.2 Ch.12




65 | P a g e

• A typical operational link involves an active sat and two or more earth terminals. One

station terminal transmits to the sat on a frequency called up-link frequency. The sat then

amplifies the signal, converts it to the downlink frequency, and transmits it back to earth.

• The basic design of a sat system depends to a great degree upon the characteristics of the

orbit of the sat. An orbit is either elliptical or circular in shape. A special of orbit is

“synchronous” orbit which has period (time required for one revolution) of the orbit the

same as that of the earth. An orbit that is not synchronous is called “asynchronous”.

Synchronous sat can provide coverage to almost half the surface of the earth.

• Three parameters are used to describe orbital data of sat: perigee, apogee, and angle of

inclination. The inclination of the orbit determines the area covered by the path of the sat.

Three of satellites can provide coverage over most of the earth (except for the extreme

north and south Polar Regions)




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• Early SCS were limited by the lack of suitable power sources. The only source of power

available within early weight restrictions was a very inefficient panel of solar cells without

battery backup. Also early SCS have been limited by low-gain antennas.

• Satellite orientation in space is impotent for continuous solar cell and antenna orientation.

Maximum number of the solar cells must be exposed to the sun at all time. The satellite

antenna must also be pointed at the appropriate earth terminals.

• Earth terminal antennas are directional, high gain antenna capable of transmitting and

receiving signal simultaneously. Generally, large, high-gain, parabolic antennas are used.

• All earth terminals are highly sensitive receivers. These receivers are designed to overcome

downlink power losses and to permit extraction of the desired information from the weak

received signal.

• All earth terminals transmitters generate high-power signal for transmission to overcome

up-link limitations and to ensure that the signals received by the sat are strong enough to

be detected by the satellites. Tx used in earth terminals have output power from 10 watts to

20 kW.




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• An essential operation in communicating by sat is the acquisition (locating) of the sat by

the earth terminal antenna and tracking of the sat. This process is important to ensure earth

terminal antennas are always pointed towards the sat.

• Limitations of SCS are determined by the technical characteristics of the sat and its orbital

parameters. Two things limit active SCS: Satellite transmitter power on the downlinks, and

satellite receiver sensitivity on the uplinks.

Link Budget Analysis

In the design of radio communication systems that transmit over line-of-sight microwave channels

and satellite channel, the system designer must specify the size of the transmit and receive

antennas, the transmitted power ( ), and the SNR required to achieve a given level of

performance at some desired data rate (R).

The transmit antenna radiates isotropically in free space at a power level of watts. The power

density at a distance d from the antenna is …watt/m2

If the transmitting antenna has some directivity in a particular direction, the power density in that

direction is increased by antenna gain ( ). Therefore, the power density now is … watt/m2

The product ERP = is called the effective radiated power.

Antenna

d




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The received power ( ) that has :

=

Where is the effective area of the received antenna

= … m2

Where = / is the wavelength of the transmitted signal, c is the speed of light (3x108 m/s), and

f is the frequency of the transmitted signal. So

= ( )

The factor

( ) =

is called the free-space path loss. So

=

The received antenna gain for a parabolic antenna of diameter D is

= ( )

Where is the efficiency factor for received antenna.

In general:

= + + + … all in dB

Example.1

A satellite radiates 100 watt and has antenna gain of 17 dB. The down-like signal has been received

by an earth station, which has 3m parabolic antenna and frequency of 4 GHz with efficiency factor

of 0.5. Calculate the received power by this earth station ( = 36000 )




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Solution:

= 100 = 20

ERP = = [ ] + [ ] = 20 + 17 = 37

= ( ) = 0.5 ( ) ( ) = 39 = ( ) = ( ) ( ) ( × ) = −195.6

= + + + = 20 + 17 + 39 − 195.6 = −119.6 dB = 1.1 × 10−12 watt

Maximum Data Rate

The performance of the digital communication system is specified by required to keep the error

rate performance below some given value.

Earth Station = 3 = 0.5

= 100 = 17




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We have

= . → = .

But = → = → = / /

Or = −

Where R is the maximum data rate for required SNR per bit.

Example.2

For previous example, find the maximum data rate if = 4.1x10 w/Hz and the required

SNR is 10 dB per bit.

Solution:

= −119.6 , = 4.1x10−21 = −203.9 dB = −119.6 + 203.9 = 84.3 dB

= 84.3 − 10 = 74.3 = 26.9 /

Exercise.1

A satellite in geosynchronous orbit (d=36000 km) is used as repeater in a digital communication

system. Consider the satellite-to-earth link in which the satellite antenna has a gain of 6 dB and

the earth station antenna has a gain of 50 dB. The down link is operated at frequency of 46 GHz,

and the data rate is 1 Mb/s. if the required is 15 dB, determine the transmitted power for the

satellite down link. Assume that = 4.1x10 w/Hz.



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