OFDM Example_final

# OFDM Example_final - OFDM EXAMPLE We are going to explain...

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Unformatted text preview: OFDM EXAMPLE We are going to explain the example given in the book in pages 30 to 33 covering equations 3.8 to 3.12. Transmitter The precondition of solving this example is that the source symbols, Ci have to be given. On these source symbols we perform IDFT to get the transmitted signal S. The classic definition of IDFT is: where, Ci are the source symbol S is the transmitted signal NSC is the number of subcarriers Here, for the sake of simplicity, the number of source symbols has been taken to be equal to the number of subcarriers, that’s why in the second summation we have written, i goes from 0 to NSC. The above classical definition is not used in the example in the book, but the author has used a matrix W(NSC)‐1 (the superscript ‐1 just signifies the operation of IDFT). The members of the W(NSC)‐1 matrix are nothing but different values of 1/NSC* evaluated for different values of i and k. The first row of the W(NSC)‐1 matrix is evaluated by keeping the value of k=0 and values of i going from 0 to NSC. The second row of the matrix is evaluated by keeping the value of k=1, and again the values of i go from 0 to NSC. Likewise, the other rows are generated. Later in this note, I have given a short algorithm to generate the W(NSC)‐1 matrix with the help of Matlab. In equation 3.9 of the book, the author has represented the row and column operators as ‘q’ and ‘k’ respectively, and hence the wqk‐1 are the elements of the W(NSC)‐1 matrix, whereas in the above paragraph, in the classical definition, we have represented the row and column operators as ‘k’ and ‘i’ respectively. The author has written as wqk‐1= exp(j2Πq(k‐1)/NSC), in equation 3.9. It means q’s are the rows of the W(NSC)‐1 matrix, running from 0 to NSC‐1 and k’s are the columns of the matrix running from 1 to NSC. Note that, the values of q are from 0 to NSC‐1 and the values of k are from 1 to NSC, i.e. one operator starts from 0 and runs until NSC‐1, and, another operator starts from 1 and runs until NSC. The initial and final values of q and k have no mathematical significance. In the algorithm that I have provided below, I have used q as the row operator and k as the column operator, like the author has used in the book. 1 The following block diagram shows us the transmitter, and the elementary functions the transmitter performs. Our area of interest in the blue box which is the one in which the IDFT takes place. As I have mentioned earlier, we can either use the classical method or matrix method to perform the IDFT. Let us do the example for NSC = 8. The algorithm which I used in Matlab to generate the W(NSC)‐1 matrix is as follows: for q = 0:N‐1 %N=NSC, which is 8 in our example for k = 1:N W(q+1,k) = exp(j*2*pi*q*(k‐1)/N); end end As I have mentioned earlier in the notes, ‘q’ and ‘k’ are the column and row operators used to generate this matrix. I think, now you can relate to equation 3.9. In this equation wqk‐1 are elements of the W(NSC)‐1 matrix, evaluated for different values of exp(j*2*pi*(k‐1)*(q)/N). With the help of the above algorithm, we generate the W(NSC)‐1 matrix, which is 8x8, since we are solving for NSC=8. The matrix looks like as follows: 1 1 1 1 ‐1 1 0.7+0.7i 0 + i ‐0.7+0.7i ‐1 ‐0.7‐0.7i 0 ‐ i 0.7‐0.7i 1 0 + i ‐1 0 – i 1 0 + i ‐1 0 ‐ i 1 ‐0.7+0.7i 0 ‐ i 0.7+0.7i ‐1 0.7‐0.7i 0 + i ‐0.7‐0.7i 1 ‐1 1 ‐1 1 ‐1 1 ‐1 1 ‐0.7‐0.7i 0 + i 0.7‐0.7i ‐1 0.7+0.7i 0 ‐ i ‐0.7+0.7i 1 0 ‐ i ‐1 0 + i 1 0 ‐ i ‐1 0 + i 1 0.7‐0.7i 0 ‐ i ‐0.7‐0.7i ‐1 ‐0.7+0.7i 0 + i 0.7+0.7i W (NSC)= 1 1 1 1 2 This matrix now has to be multiplied by the Ci matrix as has been given in equation 3.10 Ci = [ C1i , C2i , ……. , CNsc i]T Here I denotes a particular time instant, i.e., at time i, NSC bits are generated to form the first source symbol Ci. After the multiplication, a set of 8 transmitted signals, denoted by S, is generated. Suppose, Ci = [‐1 1 ‐1 1 ‐1 ‐1 ‐1 1]T We have to do the operation: 1/NSC * W‐1(NSC) * Ci to generate S. We get S as [0, 0.17678 ‐ 0.073223i, ‐0.25 ‐ 0.25i, ‐0.17678 + 0.42678i, ‐0.5, ‐0.17678 ‐ 0.42678i, ‐0.25 + 0.25i, 0.17678 + 0.073223i]T Thus, we get the transmitted signal from the matrix multiplication. We could have also got the above result from the classic definition of IDFT as At the Receiver The following block diagram may be considered as the structure of the receiver of an OFDM signal. From the air S Analog to Digital Thus from the above block diagram we can understand that the same thing as in the transmitter happens, but in the inverse order. Like before, we will be again be interested in the blue box, where the Discrete Fourier Transform takes place, to recover the symbols, Ci. Discrete Fourier Transform Ci=W(NSC)S Converter Recovered symbols, Ci 3 Hence, in this case we start off with transmitted signals analog transmitted signal, which is then digitalized into the digital transmitted signal S, and then goes through the DFT block, and the original symbols are recovered. We continue with our analysis with 8 subcarriers (NSC), which gave rise to our set of transmitted signals S. The classical definition of DFT is as follows: where Si stands for the set of transmitted signals. NSC stands for the number of subcarriers, in this case it is 8. Ci stands for the recovered source symbols As we saw in the transmitter case, in the receiver case too, we can solve the example with the above defined classical method, or with the matrix method. The author calls this matrix W (note, there is no subscript, ‐1 since this matrix signifies the operation of DFT). We can find out the W(NSC) matrix, very easily just by taking the complex conjugate of the W(NSC)‐1 matrix. Thus, the elements of the W(NSC) matrix are as follows: 1 1 1 1 1 0.7‐0.7i 0‐i ‐0.7‐0.7i ‐1 ‐0.7+0.7i 0+i 1 0‐i ‐1 0+i 1 0‐i ‐1 1 ‐0.7‐0.7i 0+i 0.7‐0.7i ‐1 0.7+0.7i 0‐i 1 ‐1 1 ‐1 1 ‐1 1 1 ‐0.7+0.7i 0‐i 0.7+0.7i ‐1 0.7‐0.7i 0+i 1 0+i ‐1 0‐i 1 0+i ‐1 1 0.7+0.7i 0+i ‐0.7+0.7i ‐1 ‐0.7‐0.7i 0‐i W(NSC)= 1 1 1 4 1 0.7+0.7i 0+i ‐0.7+0.7i ‐1 ‐0.7‐0.7i 0‐i 0.7‐0.7i The elements of W are nothing but the different values evaluated for We can find out the original source symbols just by the following matrix multiplication: Ci =W(NSC)*S (as given in equation 3.11). . Like we did in the transmitter case, we could use another algorithm to generate the W(NSC) matrix, but since there’s an easier way to do that, we don’t make use of equation 3.12. wqk are nothing but the elements of the W(NSC) matrix . Based on the values of S found in the transmitter part we find the values of Ci with the multiplication of W(NSC) and [0, 0.17678 ‐ 0.073223i, ‐0.25 ‐ 0.25i, ‐0.17678 + 0.42678i, ‐0.5, ‐0.17678 ‐ 0.42678i, ‐0.25 + 0.25i, 0.17678 + 0.073223i]T. The result we get is: Ci = [‐1 1 ‐1 1 ‐1 ‐1 ‐1 1]T , which is exactly similar to the one we used in the beginning of our analysis. 5 ...
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## This note was uploaded on 09/21/2010 for the course ECE 567 taught by Professor Attkin during the Spring '10 term at Academy of Design Tampa.

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