Lecture 8: Orthogonal Frequency Division Multiplexing (OFDM)

Transcription

1 Communication Systems Lab Spring 2017 ational Taiwan University Lecture 8: Orthogonal Frequency Division Multiplexing (OFDM) Scribe: 謝秉昂 陳心如 Lecture Date: 4/26, 2017 Lecturer: I-Hsiang Wang 1 Outline 1 Eigenfunction of LTI system 2 Circular Convolution and Cyclic Prefix 3 Matrix View of OFDM 2 Recap 1 Last time we introduce wireline channel and its equivalent discrete-time baseband model input: x [n] output: y [n] = L 1 j=0 h lx [n l] + z[n] = h 0 x [n] + L 1 j=1 h lx [n l] + z [n] (desired signal + past interference + noise) L T d T : number of taps in the LTI channel Inter-Symbol Interference (ISI) is a new challenge 2 We can use Viterbi Algorithm to optimally solve the MLSD pattern However, the complexity is proportional to 2 L, L is typically 100 to 400 for wideband system Thus, even though Viterbi Algorithm is an optimal approach, L is still too large in most cases and hence we use OFDM 3 Eigenfunction of LTI: concept of OFDM 1 Basic Idea of OFDM ỹ k = h k x k is ISI-free, ie still ISI-free if noise-free Observation If you are willing to go back to the analog world, then there is a simple solution! Fact u [n] DT F T û (f) = n= v [n] DT F T ˆv (f) x [n] e j2πfn (u v) [n] DT F T û (f) ˆv (f) Ŷ (f) = Ĥ (f) ˆX (f) But we don t want to go back to analog Is there a digital solution? 1

2 Communication Systems Lab Spring 2017 ational Taiwan University 2 Discrete Fourier Transform (DFT) ( Discrete-Time Fourier Transform, DTFT) x [n] DF ˆx T ( k x [n] e j2π k k n = ˆx DT F T IDF T ˆx k x [n] 1 Can we achieve the same with DFT? o, it cannot be ISI-free Remark ˆx k e j2π k n For linear convolution, DFT and IDFT cannot translate into multiplication in the other domain, ie DF T (h x) DF T (h) DF T (x) 3 Linear Convolution vs Circular Convolution Recap Proof of the convolution-multiplication property: ŷ (f) = = ˆx (f) ĥ (f) n= n= y [n] = (h x) [n] = n= k= k= h [k] = ĥ (f) ˆx (f) x [n] e j2πfn h [n] e j2πfn k= h [k] x [n k] h [k] x [n k] e j2πf(n k) e j2πfk ( n k= x [n k] e j2πf(n k) ) ) e j2πfk Since the range is not from infinity to infinity for DFT, the function cannot be shifted as in DTFT Def (Circular Convolution) Fact (h x) [n] = x [(n k) mod ] h [k] DF T {h x} = DF T {h} DF T {x} If the channel is doing circular convolution instead of linear convolution, then the following architecture is ISI-Free! We need to pay a price for converting linear convolution into circular convolution But as long as is large enough, the price can be overlooked Redo convolution-multiplication property: ˆx k 1 x [n] e j2π k n ĥ k 1 h [n] e j2π k n 2

3 Communication Systems Lab Spring 2017 ational Taiwan University ŷ l = 1 = 1 h [k] y [n] = (h x) [n] = h [k] x [(n k) mod ] h [k] x [(n k) mod ] e j2π l [(n k) mod ] e j2π l k t=(n k) mod =0 = 1 ( ĥ k ) ( ˆxk ) x [(n k) mod ] e j2π l [(n k) mod ] e j2π l k 4 Implement Circular Convolution on the LTI Channel Original Channel (ignore noise) Linear Convolution Desired Channel Circular Convolution L 1 y [n] = h l x [n l], n = 0, 1,, 1 l=0 L 1 y [n] = h l x [(n l) mod ] 1 Add cyclic prefix (CP) to implement circular convolution on a linear convolution system! Original Sequence l=0 ew Sequence: numbers of symbols = + L 1 2 OFDM Architecture 5 A Linear Algebra Perspective Overhead = L 1 + L 1 1 if L y = y [0] y [1] y [ 1], x = x [0] x [1] x [ 1] 3

4 Communication Systems Lab Spring 2017 ational Taiwan University Linear Convolution Circular Convolution y = Hx, H = y = H C x, H = h h 1 h h 2 h 1 h 0 0 h 2 h 1 h 0 h h 2 h 1 h 1 h h 2 h 2 h 1 h h h 2 h 1 h 0 H C is a circulant matrix, ie row (i + 1) is the right-shift of row i For post-processing, For pre-processing, y = Hx + z ŷ = V y = V Hx + V z x = U x ŷ = (V HU) x + ẑ which is ISI-free iff (V HU) is a diagonal matrix and that {V, U} orthonormal We wish: D = diag (λ 1, λ 2,, λ ) D = V H U H = V 1 DU 1 Det: Circulant Matrix C is an n n matrix, and it is circulant if consecutive rows are circular shifts of the same vector Prop: Eigenvalue & Eigenvectors of Circulant Matrix Let ϕ k [n] e j2π k n ϕ k [0] ϕ k [1] ϕ k = ϕ k [ 1] For any circulant matrix, ϕ k is the eigenvector Cϕ k = λ k ϕ k for k = 0, 1,, 1 Hence C = ΦDΦ 1 where Φ = [ϕ 0, ϕ 1, ϕ ] λ D = λ c 0 c 1 c c c 0 c 2 C = c 1 c 2 c 0 4

5 Communication Systems Lab Spring 2017 ational Taiwan University determined by c = [c 0, c 1,, c ] Cϕ k = c T 0 ϕ k c T 1 ϕ k c T ϕ k l th component: (ϕ k [l]) 1 c T l ϕ k = This is true for all l Cϕ k = λ k ϕ k where λ k = DF T {c n } Back to OFDM, Let x = Φ x where D = ĥ 0 The IDFT Matrix is ĥ1 ĥ ϕ k [n] C (n l)mod ϕ k [l] = C (n l) mod e j2π k (n l) mod = DF T {cn } λ k = DF T {c} y = H C x + z H C = ΦDΦ H ŷ = Φ H y = Φ H H C x + Φ H z ŷ = Φ H ( ΦDΦ H) Φ x + ẑ = D x + ẑ Φ = 1 (Φ kn ) k,n {0,1,,} where Φ kn = ϕ k [n] = e j2π kn 6 Summary 5

6 Communication Systems Lab Spring 2017 ational Taiwan University Equivalent System: Parallel AWG Channels ( ) ˆv k = ĥ k û k + ẑ k, k = 0, 1,, 1 Remark How to choose? 1 If we want to use FFT to implement DFT/IDFT, we will pick = 2 p for some p 2 also depends on total bandwidth and the allowed frequency spacing f Example: DSL modem W = 6 MHz = f 12 khz L 1 10% 6