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A signal $x[n]$ is to be transmitted over a communication channel. The communication channel is described by an FIR filter of length $M$ such that the received signal is given by $$r[n]=\sum_{k=0}^{M-1} x[k]h[n-k]\text.$$

The data are transmitted in blocks where each block contains N samples of a data stream $d[n]$, and in addition each transmitted block also contains a so-called cyclic prefix where the last values of the block are duplicated at the beginning of the block. Consider for example the first block of data ${d[0], d[1], . . . , d[N − 1]}$.

The corresponding block to be transmitted with the cyclic prefix is ${x[n]}_{k=0}^{N+K-1}= {d[N − K], d[N − K + 1], . . . , d[N − 1], d[0], d[1], . . . , d[N − 1]}$ $i.e. x[0] = d[N − K], x[1] = d[N − K + 1], . . ..$ At the receiver end, the first $K$ values are discarded and the next $N$ values are saved in a buffer ${y[n]}_{k=0}^{N+K-1}={r[K], r[K + 1], . . . , r[K + N − 1]}, i.e. y[0] = r[K], y[1] =r[K + 1], . . ..$ a) Show that at the receiver end the original data stream d[n] can be recovered using the following algorithm if the cyclic prefix is sufficiently long.

  1. $Y [k] = FFT({y[n]})$
  2. $D'[k] = Y [k]/H[k]$
  3. $d'[n] = IFFT (D'[k])$

Show all workings, and in particular determine the length $K$ of the cyclic prefix such that $d'[n] = d[n]$,$ n = 0, . . . , N − 1,$ and show exactly how $H[k]$ should be computed from $h[n]$.

To solve this question for the given info do we have to convert the sequence into odd and even sequence respectively and then arrive at the FFT or is there any alternative? I don't need the solution. A hint would be good to start.

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    $\begingroup$ Homework? or mid term question? $\endgroup$ – Stanley Pawlukiewicz Oct 12 at 15:13
  • $\begingroup$ Homework @StanleyPawlukiewicz $\endgroup$ – A M Ankit Kalluraya Oct 13 at 5:16
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don't need the solution. A hint would be good to start.

This is bog-standard cyclic-prefix OFDM, and that's really well-covered in literature.

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  • $\begingroup$ What does bog-normal mean? $\endgroup$ – Laurent Duval Oct 17 at 17:48
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    $\begingroup$ Looks like a genuine hint, then;) $\endgroup$ – Laurent Duval Oct 17 at 17:53

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