### 8-Channel Model Estimation.ppt

```Channel Estimation from Data
1. Recall Impulse Response Identification from
Correlation
2. Estimation of Time Spread and Doppler Shift
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
x[n]
y[n]
Rayleigh
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
y
To Workspace1
Multipath Rayleigh
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
x 10
8
7
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)
x 10
1
-4
x 10
1
15 sec
0.8
0.6
0.4
0.2
0
-1000
-800
-600
-400
-200
0
200
frequency (Hz)
 70 Hz
400
 70 Hz
600
800
1000
```