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IV. Orthogonal Frequency Division Multiplexing (OFDM)

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IV. Orthogonal Frequency Division Multiplexing (OFDM)

- Frequency Equalization:
- Divide by received signal Y[i] by H[i] for all sub-carriers
- Requires channel estimation
- For low values of H[i] equalization results in noise amplification

- Precoding
- Divide transmitted signal X[i] by H[i] for all sub-carriers
- Requires channel estimation knowledge at transmitter
- Does not result in any noise amplification at the receiver
- For low values of H[i] excessively high transmission power might be needed at the transmitter

Problem Definition:

N Sub-carriers, Maximum Transmission Power P

How to divide the available power P over sub-carriers so as to maximize the total rate

Problem Formulation:

16

8

4

2

1

1/2

1/4

1/8

1/16

-4

-2

0

2

4

6

8

10

12

14

16

18

20

22

24

26

28

30

32

34

36

38

40

Solution:

In this region a increase in power corresponds to minimal increase in rate

Intuitively in order to maximum the total rate the following partial differentiation equation should be satisfied where C is some constant

Ri

SNRi

In this region a small increase in power corresponds to a significant increase in rate

16

8

4

2

1

1/2

1/4

1/8

1/16

-4

-2

0

2

4

6

8

10

12

14

16

18

20

22

24

26

28

30

32

34

36

38

40

Solution:

In this region a increase in power corresponds to minimal increase in rate

Intuitively in order to maximum the total rate the following partial differentiation equation should be satisfied

Ri

SNRi

In this region a small increase in power corresponds to a significant increase in rate

16

8

4

2

1

1/2

1/4

1/8

1/16

-4

-2

0

2

4

6

8

10

12

14

16

18

20

22

24

26

28

30

32

34

36

38

40

Solution:

In this region a increase in power corresponds to minimal increase in rate

Intuitively in order to maximum the total rate the following partial differentiation equation should be satisfied

Ri

Constant

SNRi

In this region a small increase in power corresponds to a significant increase in rate

Bandwidth for all sub-carriers is usually equal

16

8

4

2

1

1/2

1/4

1/8

1/16

-4

-2

0

2

4

6

8

10

12

14

16

18

20

22

24

26

28

30

32

34

36

38

40

Solution:

In this region a increase in power corresponds to minimal increase in rate

Water-Filling Algorithm

Ri

K

SNRi

In this region a small increase in power corresponds to a significant increase in rate

0

1

N-1