Lecture 4: Make fixed point simulation using Matlab

March 28 – April 19 2008

Yuping Zhao (Doctor of Science in technology)

Professor, Peking UniversityBeijing, China

Yuping.zhao@pku.edu.cn

General explanation• For the floating point simulation, normally we have• A=3; B=5; C=A+B;• Here A, B, C are the floating point data, the number of

bits to represent the data is defined in MATLAB• However, for the hardware implementation, each data

should be represented by a certain number of “0”,”1”. Therefore all the operation is also done for those binary sequences.

• In the real word, the A/D converter will make the samples of the received signals and change it to the binary data with defined length

How does the A/D converter works?

• It should define the sampling frequency, normally it is 2-6 times of the signal bandwidth

• It should define the maximum and minimum value of samples. The signal value out of the range will be truncated.

• It should define the number of bits for representing each sample. This will define the quantization accuracy

Min Max

Example 1Assume the A/D range is[-10~10]，the number of bits for

each sample is 10 bit. The first bit normally represent the plus or minus value, and rest 9 bits represent the absolute value of sample.

10: -> “0111111111”-10: -> “1000000001” (using complementary data)The accuracy of quantization is(10+10)/(511+511)=0.0196，(about 0.02)Note: If the input to the A/D converter is larger than 10, the

out put will be 10. The sample is truncated.

Example 2

Assume the A/D range is[-1~1]，the number of bits for each sample is 10 bit.

1: -> “0111111111”-1: -> “1000000001” (using complementary

data)The accuracy of quantization is(1+1)/(511+511)=0.00196，(about 0.002)

Example 3

Assume the A/D range is[-10~10]，the number of bits for each sample is 5 bit.

1: -> “01111”-1: -> “10001” (using complementary data)The accuracy of quantization is(10+10)/(15+15)=0.6667，(about 0.7)

How to do the calculation using fixed point signals

Example 4

C=A+BAssume the range of A, B, C are [-10~10]，for the

fixed point calculation, we have to change the data into fixed one, it can be done as follows:

A=round(min(10,max(A,-10))/10*511)/511*10;B=round(min(10,max(B,-10))/10*511)/511*10;C=round(min(10,max(C,-10))/10*511)/511*10;

Example: convert A into Fixed point value YY = round(min(10,max(A,-10))/10*511)/511*10;

• “max(A,-10)”: to ensure the data always larger than -10• “min(10,max(A,-10))”: to ensure the data always smaller

than 10• “/10*511”: change the data into binary one• “round”” take the integer data• “/511*10” change back into decimal data

By doing so, Y becomes fixed point data, but shown in floating point valueExample: A = 9.5, Y = 9.4912

A = 13, Y = 10A = -7, Y = -7.0059

General functionC=round(min(10,max(C,-10))/10*511)/511*10

function Y = fixed(X,A,K)% X: input floating point data% A: The maximum value of the output, A>0% K: The total number bits for quantizationN = 2^(K-1) – 1;Y = round(min(A,max(X,-A))/A*N)/N*A

(In practice, normally we use “floor” rather than “round”)

Example 5A=1，B=2，C=A+B

Then we have:A=round(min(10,max(1,-10))/10*511)/511*10=0.998;B=round(min(10,max(2,-10))/10*511)/511*10=1.996;C=A+B=0.998+1.996=2.994;C=round(min(10,max(2.994,-10))/10*511)/511*10=2.994;

If using floating point, C=3.

Note: You cannot first calculate C=A+B, then take the fixed point value

Example 6A=11，B=-2，C=A+B

Correct one:• A=round(min(10,max(11,-10))/10*511)/511*10=10;• B=round(min(10,max(-2,-10))/10*511)/511*10=-1.996;• C=A+B=10-1.996=8.004;• C=round(min(10,max(8.004,-10))/10*511)/511*10=8.004;

Error one:• A=11, B=-2, C=A+B=9;• C=round(min(10,max(9,-10))/10*511)/511*10=9.002;

Notes:

• Normally the multiplication results require more bits than the original data– Example: c=a*b; a,b—10 bits; c -- 20 bits

• For division, the accuracy of divider is very important

• Normally, small number of bits make systems simple, large number of bits make systems accurate

How to decide parameters for A/D converter

• The sampling rate for A/D converter• The A/D range• The number of bits for each data

Example: Structure of the receiverBase-band System

Amplifier A/D Frame detection

Frequency error

compensation

Frequency error

estimation

(AGC)

synchronizationLength detectio

n

demodulation CRC

Local PN sequence

• To make received signal in the suitable level– Too large:

some signals are truncated

– Too small: the resolution is not enough

AGC should be used to adjust signals to the correct level

Consideration for the parameters

• Use the maximum and minimum values of the received signals as the signal range [-X, X]

• Decide the number of N bit to quantize the data

• Lots of simulation should be done for different N and X, to ensure the accuracy and good performance

Example: Received BPSK signals distribution

If the received signals is truncated by A/D converter, normally it will not cause big problems since the truncated signals are AWGN

Received OFDM time domain signal distribution

-10 -8 -6 -4 -2 0 2 4 6 8 100

50

100

150

200

250

300Central Limit theorem, N=60

The signals cannot be truncated in since the it is the useful signals

Fixed point data for OFDM

0 20 40 60 80 100 120 140 160 180 200-3

-2

-1

0

1

2

3maximun value= 1, number of bits = 4

floating pointFixed point

0 20 40 60 80 100 120 140 160 180 200-3

-2

-1

0

1

2

3maximun value= 1.5, number of bits = 4

floating pointFixed point

Fixed point data for OFDM

0 20 40 60 80 100 120 140 160 180 200-3

-2

-1

0

1

2

3maximun value= 1.5, number of bits = 3

floating pointFixed point

Fixed point data for OFDM