Parameters
- Coherent bandwidth of the channel
Now
|
D(t) |
0 |
0.1 |
0.2 |
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:
- 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.
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);
- 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
- 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
- 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.
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