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:
x[n]
y[n]
y[n]  h[n]* x[n] 
h[n]

 h[
]x[n  ]

Excite the system with white noise and unit variance


Rxx[m]  E x[n]x*[n  m]   [m]
m
and compute the crosscorrelation between input and output
Ryx [m]  E  y[n]x*[n  m]


 h[

]E  x[ n  ] x [ n  m] 
*

 h[

] [ m  ]  h[ m]
In matlab:
y[n]
x[n]
?
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
x[n]
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:
1. Sampling frequency Fs
2. Upper bound on max Doppler Frequency FD max
1. Collect Data and partition in blocks of length N 
:
n0
nN
n  2N
x
y
N  Ts  1/ FDMAX

n  N

Fs
FD max

n  NB  N

Within each block the channel is
almost time invariant
X=reshape(x,N,length(x)/N);
Y=reshape(x,N,length(y)/N);
X,Y =
[

NB
]
N
2. Estimate impulse response in each block :
h(:,i)=xcorr(Y(:,i),X(:,i))/N;
h =
[

]
2N 1
NB
Take the transpose:
h’ =
[ ]

Each row is an impulse response
taken at different times
NB
2N 1
plot((-N+1:N-1)/Fs, abs(h(:,i)));

 N / Fs

 N / Fs
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):
time
time
t  n  NTs 
h’ =
  Ts
[ ]

NB
2N 1
H=fft(h’);
S=H.*conj(H);

Fs
F k
N  NB
Freq.
S =
time
[
  Ts

2N 1
]
NB
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 k
t  m / Fs
Time Resolution:
( Fs / N )
NB
t  1/ FS
Frequency Resolution:
F 
FS
1

Hz
N  N B totaldata length(sec)
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;
Bernoulli
Binary
Rectangular
QAM
Bernoulli Binary
Generator
Rectangular QAM
Modulator
Baseband
% sampling freq. In Hz
% attenuations in dB
% time delays in sec
%doppler shift in Hz
Rayleigh
Fading
y
To Workspace1
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
0
0.02
0.04
0.06
0.08
0.1
0.12
0.14
0.16
0.18
0.2
t (sec)
sum(S)/NB; sum of each column
-3
9
Time Spread
x 10
8
7
Time Spread
6
5
4
3
sum(S’)/(2N-1); ave. of each row
2
-3
1
0
-1.5
1.2
-1
-0.5
0
0.5
time (sec)
Frequency Spread
x 10
1
-4
x 10
1
15 sec
0.8
Frequency Spread
0.6
0.4
0.2
0
-1000
-800
-600
-400
-200
0
200
frequency (Hz)
 70 Hz
400
 70 Hz
600
800
1000