Understanding Channel Estimation and Impulse Response Identification

Channel Estimation from Data 1. Recall Impulse Response Identification from Correlation 2. Estimation of Time Spread and Doppler Shift 3. Simulink/Matlab Example 4. Stanford University Interim (SUI) Channel Models

Estimation of Channel Characteristics from Input - Output data. 1. For Linear Time Invariant (LTI) systems: [n y ] + [n x ] = = [ ] y n [ ]* [ ] h n [ ] [ h ] x n x n [n ] h = Excite the system with white noise and unit variance = = * [ ] [ ] [ ] [ ] Rxx m E x n x n m m m and compute thecrosscorrelation between input and output [ ] h E x n = = * [ ] m [ ] [ ] R E y n x n m yx + + = = = * [ ] [ x n m ] [ ] [ h ] [ ] h m m =

In matlab: [n y ] [n x ] ? 1. Get data (same length for simplicity): y x 2. Compute crosscorrelation between input and output: h=xcorr(x,y); If x[n] is white noise, h[n] is the impulse response.

2. For a Linear Time Varying Channel: Multipath Rayleigh Fading Channel [n x ] [ ] y n Rayleigh Fading + = [ ] y n [ , ] [ h n ] x n = [ , ] h n The impulse response changes with time Goal: estimate time and frequency spread. Known: s F 1. Sampling frequency F 2. Upper bound on max Doppler Frequency max D

F N s 1. Collect Data and partition in blocks of length : 0 = n N n = N n 2 = F max D = = B n N n N N x y N T 1/ MAX D F s Within each block the channel is almost time invariant [ ] X=reshape(x,N,length(x)/N); N X,Y = Y=reshape(x,N,length(y)/N); N B

2. Estimate impulse response in each block : h =[ N] h(:,i)=xcorr(Y(:,i),X(:,i))/N; N 2 1 B Take the transpose: Each row is an impulse response taken at different times h =[ ] N B N 2 1 plot((-N+1:N-1)/Fs, abs(h(:,i))); + N / / s F N F s

3. Compute Power Spectrum on each column of h (each row of h) , to determine time variability of the channel (If the channel is Time Invariant all columns of h are the same): = ( ) s t n NT = sT time time h =[ ] N B N 2 1 F = s F kN = sT time N B [ N ] 1 N Freq. H=fft(h ); S=H.*conj(H); B S = 2

4. Take the sum over rows for Doppler Spread and sum over columns for Time Spread (fftshift each vector to have zero term (sec or Hz) in the middle Sf St ( / ) F N = s N f k = / t m s F B / 1 = t S F Time Resolution: 1 F = = S F Hz Frequency Resolution: total data length(sec ) N N B Therefore if we want to a resolution in the doppler spread of (say) 1Hz, we need to collect at least 1 sec of data.

Example: % channel Fs=10^6; P=[0,-2,-3]; T=[0, 10, 15]*10^(-6); fd=70; % sampling freq. In Hz % attenuations in dB % time delays in sec %doppler shift in Hz Rayleigh Fading Bernoulli Binary Rectangular QAM y To Workspace1 Bernoulli Binary Generator Rectangular QAM Modulator Baseband Multipath Rayleigh Fading Channel x To Workspace test_scattering.mdl

Channel Output (Magnitude) with a QPSK Transmitted Signal: 3 2.5 2 1.5 1 0.5 0 t (sec) 0 0.02 0.04 0.06 0.08 0.1 0.12 0.14 0.16 0.18 0.2

sum(S)/NB; sum of each column -3 Time Spread 9x 10 8 7 Time Spread 6 5 4 3 sum(S )/(2N-1); ave. of each row 2 -3 Frequency Spread 1 1.2x 10 0 -1.5 -1 -0.5 0 0.5 1 time (sec) -4 1 x 10 15 sec 0.8 Frequency Spread 0.6 0.4 0.2 0 -1000 -800 -600 -400 -200 0 200 400 600 800 1000 frequency (Hz) Hz + 70 70 Hz