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Lecture 6: Simulation of Carrier and sampling frequency clock frequency error

March 28 – April 19 2008

Yuping Zhao (Doctor of Science in technology)

Professor, Peking UniversityBeijing, China

Yuping.zhao@pku.edu.cn

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Carrier frequency error model

Assume the transmitted OFDM signal is

Here fc is the carrier frequency of the transmitted signals.If the frequency of the receiver is

The received base band signal becomes

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The frequency difference is

If the sub-carrier separation is

The normalized frequency error is

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The sampled signal becomes

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frequency error model of OFDM systems

X k s n r n Y k

π N εn

Frequencydomain

Frequencydomain

Timedomain

⎟⎠⎞

⎜⎝⎛−

Kn

Nj επ2exp

K is the up sample rate, ε is the normalized frequency error, here is 0.05

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-3 -2 -1 0 1 2 30

0 .1

0 .2

0 .3

0 .4

0 .5

0 .6

0 .7

0 .8

0 .9

1

Amplitude Spectrum of OFDM

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Influence of carrier frequency error to OFDM

• Assume the transmitted signal is：

Assume the carrier frequency error exist, then it can be expressed

is the AWGN samples

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The received signal after FFT

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The signal can be expressed as

Then

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0 5 10-0.5

0

0.5

1

k

Re(

Y(k

))

The signal transmitted on sub-carrier 5 is 1+j

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0 5 10-1

-0.5

0

0.5

1

1.5

k

Im(Y

(k))

The signal transmitted on subcarrier 5 is 1+j

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0 5 100

0.5

1

1.5

k

ABS(

Y(k

))

The signal transmitted on subcarrier 5 is 1+j

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-5 0 5-5

-4

-3

-2

-1

0

1

2

3

4

5

Real

Imag

inar

y

ε = 5%

Figure.1

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-6 -4 -2 0 2 4 6-6

-4

-2

0

2

4

6

Real

Imag

inar

y

ε = 25%

Figure.2

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Frequency error correction method

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Model of sampling clock frequency errors

t

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0 2000 4000 6000 8000 10000 12000 14000-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

Sampling timing error

Part 1 Part 2

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Sampling timing error – part 1

300 400 500 600 700 800 900 1000 1100 1200

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

Figure 3

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1.18 1.19 1.2 1.21 1.22 1.23 1.24 1.25 1.26 1.27

x 104

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

Sampling timing error – part 2

Figure 4

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Clock frequency error problems

The correct sampling time

When the clock error exist, the received signal becomes

Here n’ is the practical sampling time

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采样时钟偏差Constellation with Sampling frequency error

Figure.5

Error = 50ppm, r=0.5, delay=5

-4 -3 -2 -1 0 1 2 3 4-4

-3

-2

-1

0

1

2

3

4

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采样时钟偏差Constellation with Sampling frequency error

Figure.6

-4 -3 -2 -1 0 1 2 3 4-4

-3

-2

-1

0

1

2

3

4

Error = 100ppm

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Exercises 3-1: Clock and frequency errors simulations

Build a OFDM simulation system to show• Simulate the figure1-6 give in this ppt

Note: for the sampling clock error, you may use MATLAB function “resample”

----------------------------Example: 50ppm clock error:

P = 20000;Q = 19999; resampI = resample(filteredI,P,Q);resampQ = resample(filteredQ,P,Q);

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Exercises 3-2: Fixed point simulations

Assume the synchronization of the received signal is perfect, the channel is AWGN. Convert the down sampled data into fixed value, examine how the system performance is affected by parameter A and K.

Case 1: K=5,8,13, the value A should not make truncation of the received signals

Case 2: K=10, the value A should chosen as 1) just suitable value; 2) with truncation; 3) too large compare with received signal value

------------------------------------------------For exercise 3-2, you may chose the parameters yourself

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Instruction 3-1: carrier frequency error model

16QAM modulation

N-points IFFT

Guard insertion

Lowpass filtering

Binarydata

Lowpass filteringA/D converting

Guard deleting

N-points FFT

16QAM demodulation

Binarydata

X(k) x(n) x’(n) s(t)

y’(k) y(n) y’(n) r(t)

Up sampling

Down sampling

K is Up sampling rate, N is the FFT points,

show Signal constellation

⎟⎠⎞

⎜⎝⎛−

Kn

Nj επ2expX

is normalized frequency offset, its value is 5% and 25% in Figure 1 and Figure 2

⎟⎠⎞

⎜⎝⎛−

Kn

Nj επ2exp :

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Instruction 3-2: clock frequency error model

16QAM modulation

N-points IFFT

Guard insertion

Lowpass filtering

Binarydata

Lowpass filteringA/D converting

Guard deleting

N-points FFT

16QAM demodulation

Binarydata

X(k) x(n) x’(n) s(t)

y’(k) y(n) y’(n)r(t)

Up sampling

Down sampling

show time domain Signal

resample

using MATLAB function “resample” to get the clock frequency error

(see next page)

r(n)

show Signal constellation

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using MATLAB function “resample” to get the clock frequency error----------------------------Example: for 50ppm clock error:

P = 20000;Q = 19999;

real(y’) = resample(r_I,P,Q); % r_I is the real part of r(n) in the figureimag(y’) = resample(r_Q,P,Q); %r_Q is the imaginary part of r(n) in the figure

Note: The resample functionClock error = (P-Q)/P, however P and Q cannot be too big value, otherwise Matlabfunction cannot work

Instruction 3-2: clock frequency error model

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Instruction 3-3: Fixed point simulation

16QAM modulation

N-points IFFT

Guard insertion

Lowpass filtering

Binarydata

Lowpass filteringA/D converting

Guard deleting

N-points FFT

16QAM demodulation

Binarydata

X(k) x(n) x’(n) s(t)

y’(k) y(n) y’(n)r(t)

Up sampling

Down sampling

show histogram

Fixed point model

Using fixed point function (see next page) for the fixed point simulation

r(n)

show signal constellation

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Instruction 3-3: Fixed point simulation

Here is the fixed function (you also can make yourself)

----------------------------------------function Y = fixed(X,A,K)% X: input floating point data% A: The maximum value of the output, A>0% K: The number of bits for quantizationN = 2^(K-1) – 1;Y = round(min(A,max(X,-A))/A*N)/N*A-------------------------------------------------------For exercise 3-2, you may chose the parameters yourself. The following

figures are examples with different parameters.

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-0.5 -0.4 -0.3 -0.2 -0.1 0 0.1 0.2 0.3 0.4 0.50

200

400

600

800

1000

1200Real part, floating point

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-0.5 -0.4 -0.3 -0.2 -0.1 0 0.1 0.2 0.3 0.4 0.50

200

400

600

800

1000

1200Real part, fixed point, A = 0.4; K = 14

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-4 -3 -2 -1 0 1 2 3 4-4

-3

-2

-1

0

1

2

3

4Real part, fixed point, A = 0.4; K = 14

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-0.5 -0.4 -0.3 -0.2 -0.1 0 0.1 0.2 0.3 0.4 0.50

200

400

600

800

1000

1200Real part, fixed point, A = 0.25; K = 14

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-4 -3 -2 -1 0 1 2 3 4-4

-3

-2

-1

0

1

2

3

4Real part, fixed point, A = 0.25; K = 14

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-0.5 -0.4 -0.3 -0.2 -0.1 0 0.1 0.2 0.3 0.4 0.50

200

400

600

800

1000

1200Real part, fixed point, A = 0.2; K = 14

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-4 -3 -2 -1 0 1 2 3 4-4

-3

-2

-1

0

1

2

3

4Real part, fixed point, A = 0.2; K = 14

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-0.5 -0.4 -0.3 -0.2 -0.1 0 0.1 0.2 0.3 0.4 0.50

200

400

600

800

1000

1200Real part, fixed point, A = 0.4; K = 8

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-4 -3 -2 -1 0 1 2 3 4-4

-3

-2

-1

0

1

2

3

4Real part, fixed point, A = 0.4; K = 8

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-4 -3 -2 -1 0 1 2 3 4-4

-3

-2

-1

0

1

2

3

4Real part, fixed point, A = 0.4; K = 5

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