ECS455: Chapter 5 OFDM

Transcription

1 ECS455: Chapter 5 OFDM 5.4 Cyclic Prefix (CP) 1 Dr.Prapun Suksopong prapun.co/ecs455 Office Hours: BKD Tuesday 9:30-10:30 Friday 14:00-16:00

2 Three steps towards odern OFDM 1. Mitigate Multipath (ISI): Decrease the rate of the original data strea via ulticarrier odulation (FDM) 2. Gain Spectral Efficiency: Utilize orthogonality 3. Achieve Efficient Ipleentation: FFT and IFFT Extra step: Copletely eliinate ISI and ICI Cyclic prefix 2

3 Cyclic Prefix: Motivation (1) Recall: Multipath Fading and Delay Spread 3

4 Cyclic Prefix: Motivation (2) OFDM uses large sybol duration T s copared to the duration of the ipulse response τ ax of the channel to reduce the aount of ISI Q: Can we eliinate the ultipath (ISI) proble? To reduce the ISI, add guard interval larger than that of the estiated delay spread. If the guard interval is left epty, the orthogonality of the sub-carriers no longer holds, i.e., ICI (inter-channel interference) still exists. To prevent both the ISI as well as the ICI, OFDM sybol is cyclically extended into the guard interval. 4

5 5 Cyclic Prefix

6 N n v v-1 N N-1 Recall: Convolution x Flip Shift Multiply (pointwise) Add h h = h 0 h 1 x hn xhn h n h N 1 h N h N + L 1 6

7 N n n v v-1 v v-1 N N-1 N N-1 Circular Convolution x (Regular Convolution) x h h h = h 0 h = h 0 h 1 h 1 h n h n h N 1 h N 1 7 Replicate x (now it looks periodic) Then, perfor the usual convolution only on n = 0 to N-1 h N h N + L 1

8 Circular Convolution: Exaples 1 Find

9 Discussion Regular convolution of an N 1 point vector and an N 2 point vector gives (N 1 +N 2-1)-point vector. Circular convolution is perfor between two equal-length vectors. The results also has the sae length. Circular convolution can be used to find the regular convolution by zero-padding. Zero-pad the vectors so that their length is N 1 +N 2-1. Exaple: In odern OFDM, we want to perfor circular convolution via regular convolution. 9

10 Circular Convolution in Counication We want the receiver to obtain the circular convolution of the signal (channel input) and the channel. Q: Why? A: CTFT: convolution in tie doain corresponds to ultiplication in frequency doain. This fact does not hold for DFT. DFT: circular convolution in (discrete) tie doain corresponds to ultiplication in (discrete) frequency doain. We want to have ultiplication in frequency doain. So, we want circular convolution and not the regular convolution. Proble: Real channel does regular convolution. Solution: With cyclic prefix, regular convolution can be used to create circular convolution. 10

11 0 0 Exaple Solution: ? Let s look closer at how we carry out the circular convolution operation. Recall that we replicate the x and then perfor the regular convolution (for N points) Goal: Get these nubers using regular convolution

12 Exaple ? Observation: We don t need to replicate the x indefinitely. Furtherore, when h is shorter than x, we don t even need a full replica. Not needed in the calculation

13 Exaple *? Copy the last saples of the sybols at the beginning of the sybol. This partial replica is called the cyclic prefix Try this: use only the necessary part of the replica and then convolute (regular convolution) with the channel Junk!

14 Exaple 2 We now know that * Cyclic Prefix Siilarly, you ay check that * Cyclic Prefix

15 Exaple 3 We know, fro Exaple 2, that [ ] * [3 2 1] = [ ] And that [ ] * [3 2 1] = [ ] Check that [ ] * [3 2 1] = [ ] and [ ] * [3 2 1] = [ ] 15

16 Exaple 4 We know that [ ] * [3 2 1] = [ ] [ ] * [3 2 1] = [ ] Using Exaple 3, we have [ ] * [3 2 1] = [ ] +[ ] * [3 2 1] = [ ] +[ ] = [ ] 16

17 Putting results together Suppose x (1) = [ ] and x (2) = [ ] Suppose h = [3 2 1] At the receiver, we want to get * [ ] * [ ] = [ ] We transit [ ]. [ ] [ ] = [ ] Cyclic prefix Cyclic prefix At the receiver, we get [ ] * [3 2 1] = [ ] 17 Junk! To be thrown away by the receiver.

18 Circular Convolution: Key Properties Consider an N-point signal x[n] Cyclic Prefix (CP) insertion: If x[n] is extended by copying the last saples of the sybols at the beginning of the sybol: xn Key Property 1: Key Property 2: x n, 0 n N 1 x n N, v n 1 h x n h* x n for 0 n N 1 FFT h x n H X k k 18

19 19 OFDM with CP for Channel w/ Meory We want to send N saples S 0, S 1,, S N-1 across noisy channel with eory. First apply IFFT: Then, add cyclic prefix This is inputted to the channel. The output is Reove cyclic prefix to get Then apply FFT: Sk IFFT s n,, 1, 0,, 1 s s N s N s s N,, 1, 0,, 1 rn hn snwn FFT y n p N p N r r N r n R k By circular convolution property of DFT, Rk HkSk W No ICI! k

20 OFDM Syste Design: CP A good ratio between the CP interval and sybol duration should be found, so that all ultipaths are resolved and not significant aount of energy is lost due to CP. As a thub rule, the CP interval ust be two to four ties larger than the root ean square (RMS) delay spread. 20 [Tarokh, 2009, Fig 2.9]

21 Reference A. Bahai, B. R. Saltzberg, and M. Ergen, Multi-Carrier Digital Counications: Theory and Applications of OFDM, 2nd ed., New York: Springer Verlag,