Lecture 4: Make fixed point simulation using Matlab

Transcription

1 Lecture 4: Make fixed point simulation using Matlab March 28 April Yuping Zhao (Doctor of Science in technology) Professor, Peking University Beijing, China

2 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

3 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

4 Min Max

5 Example 1 Assume 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: -> : -> (using complementary data) The accuracy of quantization is (10+10)/( )=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.

6 Example 2 Assume the A/D range is[-1~1],the number of bits for each sample is 10 bit. 1: -> : -> (using complementary data) The accuracy of quantization is (1+1)/( )= ,(about 0.002)

7 Example 3 Assume the A/D range is[-10~10],the number of bits for each sample is 5 bit. 1: -> : -> (using complementary data) The accuracy of quantization is (10+10)/(15+15)=0.6667,(about 0.7)

8 How to do the calculation using fixed point signals

9 Example 4 C=A+B Assume 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;

10 Example: convert A into Fixed point value Y Y = 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 value Example: A = 9.5, Y = A = 13, Y = 10 A = -7, Y =

11 General function C=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 quantization N = 2^(K-1) 1; Y = round(min(a,max(x,-a))/a*n)/n*a (In practice, normally we use floor rather than round )

12 Example 5 A=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= =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

13 Example 6 A=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= =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;

14 Notes: Normally the multiplication results require more bits than the original data Example: c=a*b; a,b 10 bits; c 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

15 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

16 Example: Structure of the receiver Base-band System Amplifier A/D Frame detection Frequency error compensation Frequency error estimation (AGC) synchronization Length detectio n demodulation CRC Local PN sequence

17 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

18 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

19 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

20 Received OFDM time domain signal distribution 300 Central Limit theorem, N= The signals cannot be truncated in since the it is the useful signals

21 Fixed point data for OFDM 3 2 maximun value= 1, number of bits = 4 floating point Fixed point

22 Fixed point data for OFDM 3 2 maximun value= 1.5, number of bits = 4 floating point Fixed point

23 Fixed point data for OFDM 3 2 maximun value= 1.5, number of bits = 3 floating point Fixed point

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