OFDM Modulation And 16-QAM Encoding

Parameters

  1. Coherent bandwidth of the channel

Now

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D(t)

0

0.1

0.2

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0.5

2

P

0

-10

0

-10

-30

This can be done as

D

0

0.1

0.2

0.5

2

P

0

-10

0

-10

-30            

Wc

inf

20

10

4

  

The basic system design can be given as below:

  1. The signal has total of the active sub-carriers count as 360. The size of FFT is 512. The differences in amplitude are due to fading of type frequency selective triggered through multipath transmissions.

let apply an FFT to the signal .

If we deliberate a 1300ms guard interval, this centrals to a 1ms symbol period and a 13.33 kHz frequency for the carrier spacing. This says that about 20000 carriers (W=20MHz) in the minimum bandwidth that will be required for the given one OFDM transmission.

Parameters

Values

Ts

50 nsec

Number of samples

80

Symbol period

4µs (80 samples)

Sample frequency fs

20 MHz

Type of modulation

64-QAM

Number of sub- channel N

64

The equation is given below The equivalent baseband signal can be given as With the 16 QAM; the value of symbol period is 3.6µs having a guard interval of 0.8µs. The type of demodulation adapted is coherent. OFDM clearly delivers purposely wide frequency band and a possible value of 54Mbit/s as bit-rate. When 256 QAM subsymbols are adapted to a treillis encoder and in that scenario it is being modulated through a 512 points IFFT.

  1.  

Advantages

  • We can have factor of subcarriers spacing which is directly connected to the valuable symbol time.
  • Subcarriers are designated in a way that they are altogether orthogonal to each another for that particular duration of the symbol, thus evading the essential to have non overlying subcarrier channels to remove intercarrier interference.
  • It means more subcarriers and lesser is the power loss and more is the bandwidth efficiency.

Disadvantages

  • It has the issues of Doppler shift, multipath and increase in the complexity and cost of the design.

Data rate (Mbps)

Modulation

Coding rate

(R)

Coded bits per OFDM symbol (NCBPS)

Coded bits each subcarrier (NBPSC)

Coded bits MIMO-OFDM symbol

(NMCBPS)

Data bits per OFDM symbol (NDBPS)

Data bits per MIMO-OFDM symbol

(NMDBPS)

216

64-QAM

3/4

576

6

1152

432

864

192

64-QAM

2/3

576

6

1152

384

768

144

16-QAM

3/4

384

4

768

288

576

96

16-QAM

1/2

384

4

768

192

384

Parameters

Value

NSD: Number of data subcarriers

96

TLONG: Long training sequence duration

13.6µs (TGI+ 4×TFFT)

TSHORT: Short training sequence duration

6.4µs (8× TFFT/4)

NSP: Number of pilot subcarriers

8

TSYM: Symbol interval

4µs (TGI+ TFFT)

TGI: GI duration

0.8µs (TFFT/4)

NST: Number of subcarriers, total

104 (NSD+ NSP)

TFFT: IFFT/FFT period

3.2µs (1/?F)

Modulation

Transmit Antenna Number

KMOD

16-QAM

2

  

4

  

For 16-QAM, b0b1 gives you the value of the I and b2b3 determines the Q value

16QAM Encoding Table

Input bit (b0b1)

I-out

00

-3

01

-1

11

1

10

3

Input bit (b2b3)

Q-out

00

-3

01

-1

11

1

10

3

Solution 2 CDMA

% Use rand function to generate Random Bits

r=round(rand(1,20));

% Consider station A, B and C and form the Chip Pattern

acdma1=[1 -1 -1 1 -1 1];

acdma0=-1*acdma1;

bcdma1=[1 1 -1 -1 1 1];

bcdma0=-1*bcdma1;

ccdma1=[1 1 -1 1 1 -1];

ccdma0=-1*ccdma1;

% For all the variables do the Random bit Allotment

seq_cdma=[];

for counter=1:20

    switch(randi(3,1,1))

        case(1)

            if r(1,counter)==0;

                seq_cdma=[seq_cdma acdma0];

            else

                seq_cdma=[seq_cdma acdma1];

            end

        case(2)

            if r(1,counter)==0;

                seq_cdma=[seq_cdma bcdma0];

            else

                seq_cdma=[seq_cdma bcdma1];

            end

        case(3)

            if r(1,counter)==0;

                seq_cdma=[seq_cdma ccdma0]

                seq_cdma=[seq_cdma ccdma1

end line_select=1:6:120

    iterval=iterval+1;

    tempval=[seq_cdma(1,line_select) seq_cdma(1,line_select+1) seq_cdma(1,line_select+2) …

            seq_cdma(1,line_select+3) seq_cdma(1,line_select+4) seq_cdma(1,line_select+5)];

    output1=dot(acdma1,tempval);

    output2=dot(bcdma1,tempval);

    output3=dot(ccdma1,tempval);

    if (output1==6)|(output1==-6)

        fprintf(‘nThe bit # %d is from Uaer 1’,iterval);

    else

        if (output2==6)|(output2==-6)

            fprintf(‘nThe bit # %d is from User 2’,iterval);

        else

            if (output3==6)|(output3==-6)

                fprintf(‘nThe bit # %d is from User 3’,iterval);

  1. The number of users are 3 which can be simultaneously served by the system. Below are the results

The bit # 1 is from User 3

The bit # 2 is from Uaer 1

The bit # 3 is from User 2

The bit # 4 is from User 2

The bit # 5 is from User 3

The bit # 6 is from Uaer 1

The bit # 7 is from User 3

The bit # 8 is from User 3

The bit # 9 is from User 2

The bit # 10 is from User 2

Advantages

The bit # 11 is from User 2

The bit # 12 is from Uaer 1

The bit # 13 is from User 2

The bit # 14 is from User 2

The bit # 15 is from User 3

The bit # 16 is from User 3

The bit # 17 is from User 2

The bit # 18 is from User 2

The bit # 19 is from User 3

The bit # 20 is from User 2

  1. To enhance the relability for the user 1 we will need to increase the probability of the user 1 coming inside the network, this can be done by changing the switch case in MATLAB

    switch(randi(2,1,1))

        case(1)

            if r(1,counter)==0;

                seq_cdma=[seq_cdma acdma0];

            else

                seq_cdma=[seq_cdma acdma1];

            end

        case(2)

            if r(1,counter)==0;

                seq_cdma=[seq_cdma bcdma0];

            else

                seq_cdma=[seq_cdma bcdma1];

            end

        case(3)

            if r(1,counter)==0;

                seq_cdma=[seq_cdma ccdma0];

            else

                seq_cdma=[seq_cdma ccdma1

  1. The required parameter can be estimated as follow

% Use rand function to generate Random Bits

r=round(rand(1,20));

% Consider station A, B and C and form the Chip Pattern

acdma1=[1 -1 -1 1 -1 1];

acdma0=-1*acdma1;

bcdma1=[1 1 -1 -1 1 1];

bcdma0=-1*bcdma1;

ccdma1=[1 1 -1 1 1 -1];

ccdma0=-1*ccdma1;

% For all the variables do the Random bit Allotment

seq_cdma=[];

for counter=1:20

    switch(randi(3,1,1))

        case(1)

            if r(1,counter)==0;

                seq_cdma=[seq_cdma acdma0];

            else

                seq_cdma=[seq_cdma acdma1];

            end

        case(2)

            if r(1,counter)==0;

                seq_cdma=[seq_cdma bcdma0];

            else

                seq_cdma=[seq_cdma bcdma1];

        case(3)

            if r(1,counter)==0;

                seq_cdma=[seq_cdma ccdma0];

            else

                seq_cdma=[seq_cdma ccdma1

%   Process of the signal Decoding

iterval=0;

for line_select=1:6:120

    iterval=iterval+1;

    tempval=[seq_cdma(1,line_select) seq_cdma(1,line_select+1) seq_cdma(1,line_select+2) …

            seq_cdma(1,line_select+3) seq_cdma(1,line_select+4) seq_cdma(1,line_select+5)];

    output1=dot(acdma1,tempval);

    output2=dot(bcdma1,tempval);

    output3=dot(ccdma1,tempval);

    if (output1==6)|(output1==-6)

        fprintf(‘nThe bit # %d is from Uaer 1 with symbol %d ‘,iterval, acdma1);

        if (output2==6)|(output2==-6)

            fprintf(‘nThe bit # %d is from User 2 with  symbol %d ‘,iterval, bcdma1);

        else

            if (output3==6)|(output3==-6)

                fprintf(‘nThe bit # %d is from User 3  with  symbol %d ‘,iterval,ccd

The bit # 1 is from User 2 with  symbol -1

The bit # -1 is from User 2 with  symbol 1

The bit # 1 is from User 2 with  symbol

The bit # 2 is from Uaer 1 with symbol 1

The bit # -1 is from Uaer 1 with symbol -1

The bit # 1 is from Uaer 1 with symbol -1

The bit # 1 is from Uaer 1 with symbol

The bit # 3 is from User 3  with  symbol 1

The bit # 1 is from User 3  with  symbol -1

The bit # 1 is from User 3  with  symbol 1

Disadvantages

The bit # -1 is from User 3  with  symbol

The bit # 4 is from User 3  with  symbol 1

The bit # 1 is from User 3  with  symbol -1

The bit # 1 is from User 3  with  symbol 1

The bit # -1 is from User 3  with  symbol

The bit # 5 is from User 3  with  symbol 1

The bit # 1 is from User 3  with  symbol -1

The bit # 1 is from User 3  with  symbol 1

The bit # -1 is from User 3  with  symbol

The bit # 6 is from Uaer 1 with symbol 1

The bit # -1 is from Uaer 1 with symbol -1

The bit # 1 is from Uaer 1 with symbol -1

The bit # 1 is from Uaer 1 with symbol

The bit # 7 is from User 2 with  symbol 1

The bit # 1 is from User 2 with  symbol -1

The bit # -1 is from User 2 with  symbol 1

The bit # 1 is from User 2 with  symbol

The bit # 8 is from User 2 with  symbol 1

The bit # 1 is from User 2 with  symbol -1

The bit # -1 is from User 2 with  symbol 1

The bit # 1 is from User 2 with  symbol

The bit # 9 is from User 3  with  symbol 1

The bit # 1 is from User 3  with  symbol -1

The bit # 1 is from User 3  with  symbol 1

The bit # -1 is from User 3  with  symbol

The bit # 10 is from User 2 with  symbol 1

The bit # 1 is from User 2 with  symbol -1

The bit # -1 is from User 2 with  symbol 1

The bit # 1 is from User 2 with  symbol

The bit # 11 is from User 3  with  symbol 1

The bit # 1 is from User 3  with  symbol -1

The bit # 1 is from User 3  with  symbol 1

The bit # -1 is from User 3  with  symbol

The bit # 12 is from User 2 with  symbol 1

The bit # 1 is from User 2 with  symbol -1

The bit # -1 is from User 2 with  symbol 1

The bit # 1 is from User 2 with  symbol

The bit # 13 is from User 2 with  symbol 1

16QAM Encoding Table

The bit # 1 is from User 2 with  symbol -1

The bit # -1 is from User 2 with  symbol 1

The bit # 1 is from User 2 with  symbol

The bit # 14 is from User 3  with  symbol 1

The bit # 1 is from User 3  with  symbol -1

The bit # 1 is from User 3  with  symbol 1

The bit # -1 is from User 3  with  symbol

The bit # 15 is from Uaer 1 with symbol 1

The bit # -1 is from Uaer 1 with symbol -1

The bit # 1 is from Uaer 1 with symbol -1

The bit # 1 is from Uaer 1 with symbol

The bit # 16 is from Uaer 1 with symbol 1

The bit # -1 is from Uaer 1 with symbol -1

The bit # 1 is from Uaer 1 with symbol -1

The bit # 1 is from Uaer 1 with symbol

The bit # 17 is from Uaer 1 with symbol 1

The bit # -1 is from Uaer 1 with symbol -1

The bit # 1 is from Uaer 1 with symbol -1

The bit # 1 is from Uaer 1 with symbol

The bit # 18 is from User 2 with  symbol 1

The bit # 1 is from User 2 with  symbol -1

The bit # -1 is from User 2 with  symbol 1

The bit # 1 is from User 2 with  symbol

The bit # 19 is from User 3  with  symbol 1

The bit # 1 is from User 3  with  symbol -1

The bit # 1 is from User 3  with  symbol 1

The bit # -1 is from User 3  with  symbol

The bit # 20 is from User 3  with  symbol 1

The bit # 1 is from User 3  with  symbol -1

The bit # 1 is from User 3  with  symbol 1

The bit # -1 is from User 3  with  symbol

Solution 3: Multi-antenna system

Its squared magnitude is exponentially distributed and hence Z has PDF as                                        

For Z the value of mean is derived as SNR at the receiver is

Its squared magnitude is exponentially distributed. PDF can be given as below

The mean of Z is easily derived as Output SNR is then equivalent toow we can have

The post-detection SNR conforming to the kth sub stream.