Channel equalization with transmitter redundancy - Signal Processing and Optimization for Transceiver Systems - page 289

3. Since the channel has order L , at most L of its N -point DFT coe -
cients are zero. So there exist at least M nonzero coe cients, since
N = M + L. Let

C ( e j 2 πk i /N ) ,

0 ≤ i ≤ M − 1 ,

be the M nonzero coe cients with the largest magnitude. The “selec-
tor” at the receiver picks these M components, that is, u i ( n )= v k i ( n ) .
What are the elements of the diagonal equalizer Λ e which ensure that

r i ( n )= r k i ( n ) , 0 ≤ i ≤ M − 1? Are these diagonal elements finite

and nonzero?
4. Assume that the N

M matrix Q is chosen such that any set of
M rows is linearly independent. Then show that regardless of what
the M integers

×

above are, there exists a matrix R such that
its output is the original vector s ( n ) (in absence of channel noise, of
course).

{ k i }

Thus the method provides perfect equalization even if the channel has nulls
in frequency. The use of the integer N = M + L simply avoids these nulls
in the equalization process. The system is therefore referred to as a null-
resistant system. The matrix Q with the property that any set of M rows
is linearly independent is called a valid null-resistance matrix.

p ( n )
1

x ( n )
1

y ( n )
1

s ( n )

r ( n )

s ( n )
1

s ( n )

M

C ( z ) N

W − 1

Q

channel

s ( n )
1 blocking inverse

DFT

cyclic
prefixing

unblocking

L

ignore

DFT domain
equalizers

v ( n ) r ( n ) s ( n )

u ( n )

s ( n )

N

Λ e

W

R

y ( n )
1

blocking unblocking

DFT selector

Figure P7.23 .