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- Digital Communications I: Modulation
and Coding Course
Period 3 - 2007
Catharina Logothetis
Lecture 4
- Last time we talked about:
Receiver structure
Impact of AWGN and ISI on the transmitted
signal
Optimum filter to maximize SNR
Matched filter receiver and Correlator receiver
Lecture 4 2
- Receiver job
Demodulation and sampling:
Waveform recovery and preparing the received
signal for detection:
Improving the signal power to the noise power (SNR)
using matched filter
Reducing ISI using equalizer
Sampling the recovered waveform
Detection:
Estimate the transmitted symbol based on the
received sample
Lecture 4 3
- Receiver structure
Step 1 – waveform to sample transformation Step 2 – decision making
Demodulate & Sample Detect
z (T ) Threshold ˆ
mi
r (t ) Frequency Receiving Equalizing
comparison
down-conversion filter filter
For bandpass signals Compensation for
channel induced ISI
Received waveform Baseband pulse
Baseband pulse Sample
(possibly distored)
(test statistic)
Lecture 4 4
- Implementation of matched filter receiver
Bank of M matched filters
z1 (T )
s (T − t ) ⎡ z1 ⎤
*
1
Matched filter output:
r (t ) ⎢ M ⎥=z z Observation
⎢ ⎥ vector
⎢zM ⎥
⎣ ⎦
sM (T − t )
*
zM (T )
zi = r (t ) ∗ s ∗i (T − t ) i = 1,..., M
z = ( z1 (T ), z 2 (T ),..., z M (T )) = ( z1 , z 2 ,..., z M )
Lecture 4 5
- Implementation of correlator receiver
Bank of M correlators
s ∗1 (t )
T z1 (T )
∫0
⎡ z1 ⎤ Correlators output:
r (t ) ⎢M⎥ =z
z Observation
⎢ ⎥
∗ vector
s M (t )
T ⎢zM ⎥
⎣ ⎦
∫0 z M (T )
z = ( z1 (T ), z 2 (T ),..., z M (T )) = ( z1 , z 2 ,..., z M )
T
zi = ∫ r (t )si (t )dt i = 1,..., M
0
Lecture 4 6
- Today, we are going to talk about:
Detection:
Estimate the transmitted symbol based on the
received sample
Signal space used for detection
Orthogonal N-dimensional space
Signal to waveform transformation and vice versa
Lecture 4 7
- Signal space
What is a signal space?
Vector representations of signals in an N-dimensional
orthogonal space
Why do we need a signal space?
It is a means to convert signals to vectors and vice versa.
It is a means to calculate signals energy and Euclidean
distances between signals.
Why are we interested in Euclidean distances between
signals?
For detection purposes: The received signal is transformed to
a received vectors. The signal which has the minimum
distance to the received signal is estimated as the transmitted
signal.
Lecture 4 8
- Schematic example of a signal space
ψ 2 (t )
s1 = (a11 , a12 )
ψ 1 (t )
z = ( z1 , z 2 )
s 3 = (a31 , a32 )
s 2 = (a21 , a22 )
s1 (t ) = a11ψ 1 (t ) + a12ψ 2 (t ) ⇔ s1 = (a11 , a12 )
Transmitted signal
s2 (t ) = a21ψ 1 (t ) + a22ψ 2 (t ) ⇔ s 2 = (a21 , a22 )
alternatives
s3 (t ) = a31ψ 1 (t ) + a32ψ 2 (t ) ⇔ s 3 = (a31 , a32 )
Received signal at
z (t ) = z1ψ 1 (t ) + z2ψ 2 (t ) ⇔ z = ( z1 , z 2 )
matched filter output
Lecture 4 9
- Signal space
To form a signal space, first we need to know
the inner product between two signals
(functions):
Inner (scalar) product:
∞
< x(t ), y (t ) >= ∫
−∞
x(t ) y * (t )dt
= cross-correlation between x(t) and y(t)
Properties of inner product:
< ax (t ), y (t ) >= a < x(t ), y (t ) >
< x(t ), ay (t ) >= a * < x(t ), y (t ) >
< x(t ) + y (t ), z (t ) >=< x(t ), z (t ) > + < y (t ), z (t ) >
Lecture 4 10
- Signal space …
The distance in signal space is measure by calculating
the norm.
What is norm?
Norm of a signal:
∞
x(t ) = < x(t ), x(t ) > = ∫ x(t ) dt = E x
2
−∞
= “length” of x(t)
ax(t ) = a x(t )
Norm between two signals:
d x , y = x(t ) − y (t )
We refer to the norm between two signals as the
Euclidean distance between two signals.
Lecture 4 11
- Example of distances in signal space
ψ 2 (t )
s1 = (a11 , a12 )
E1 d s1 , z
ψ 1 (t )
E3 z = ( z1 , z 2 )
d s3 , z E2 d s2 , z
s 3 = (a31 , a32 )
s 2 = (a21 , a22 )
The Euclidean distance between signals z(t) and s(t):
d si , z = si (t ) − z (t ) = (ai1 − z1 ) 2 + (ai 2 − z 2 ) 2
i = 1,2,3
Lecture 4 12
- Orthogonal signal space
N-dimensional orthogonal signal space N characterized by
is
N linearly independent functions { j (t )}j =1 called basis
ψ
functions. The basis functions must satisfy the orthogonality
condition
0≤t ≤T
T
< ψ i (t ),ψ j (t ) >= ∫ψ i (t ) * (t )dt = K iδ ji
ψj
0
j , i = 1,..., N
where ⎧1 → i = j
δ ij = ⎨
⎩0 → i ≠ j
If all K i = 1, the signal space is orthonormal.
Lecture 4 13
- Example of an orthonormal bases
Example: 2-dimensional orthonormal signal space
⎧ 2
⎪ψ 1 (t ) = cos(2πt / T ) 0≤t
- Signal space …
Any arbitrary finite set of waveforms {si (t )}iM1
=
where each member of the set is of duration T, can be
expressed as a linear combination of N orthonogal
waveforms ψ { j (t )}Njwhere N ≤ M .
=1
N
si (t ) = ∑ aijψ j (t ) i = 1,..., M
j =1 N≤M
where
j = 1,..., N
T
1 1
aij = < si (t ),ψ j (t ) >= ∫ si (t )ψ j (t )dt 0≤t ≤T
*
Kj Kj 0 i = 1,..., M
N
Ei = ∑ K j aij
2
s i = (ai1 , ai 2 ,..., aiN )
j =1
Vector representation of waveform Waveform energy
Lecture 4 15
- Signal space …
N
si (t ) = ∑ aijψ j (t ) s i = (ai1 , ai 2 ,..., aiN )
j =1
Waveform to vector conversion Vector to waveform conversion
ψ 1 (t ) ψ 1 (t )
T ai1
∫
ai1
⎡ai1 ⎤ ⎡ai1 ⎤
⎢M⎥
0 sm
si (t ) ⎢ M ⎥ = sm si (t )
ψ N (t ) ⎢ ⎥ ⎢ ⎥ ψ N (t )
T ⎢aiN ⎥
⎣ ⎦ ⎢aiN ⎥
⎣ ⎦
∫0 aiN aiN
Lecture 4 16
- Example of projecting signals to an
orthonormal signal space
ψ 2 (t )
s1 = (a11 , a12 )
ψ 1 (t )
s 3 = (a31 , a32 )
s 2 = (a21 , a22 )
s1 (t ) = a11ψ 1 (t ) + a12ψ 2 (t ) ⇔ s1 = (a11 , a12 )
Transmitted signal
alternatives s2 (t ) = a21ψ 1 (t ) + a22ψ 2 (t ) ⇔ s 2 = (a21 , a22 )
s3 (t ) = a31ψ 1 (t ) + a32ψ 2 (t ) ⇔ s 3 = (a31 , a32 )
T
aij = ∫ si (t )ψ j (t )dt j = 1,..., N i = 1,..., M 0≤t ≤T
0 Lecture 4 17
- Signal space – cont’d
To find an orthonormal basis functions for a given
set of signals, Gram-Schmidt procedure can be
used.
Gram-Schmidt procedure:
Given a signal set {si (t )}i =1 compute an orthonormal basis { j (t )}j =1
ψ
M N
,
1. Define ψ 1 (t ) = s1 (t ) / E1 = s1 (t ) / s1 (t ) i −1
2. For i = 2,..., M compute d i (t ) = si (t ) − ∑ < si (t ),ψ j (t ) > ψ j (t )
If d i (t ) ≠ 0 let ψ i (t ) = d i (t ) / d i (t )
j =1
If , (t = 0
d ido) not assign any basis function.
1. Renumber the basis functions such that basis is
{ 1 (t ),ψ 2 (t ),...,ψ N (t )}
ψ
This is only necessary if d i (t ) = 0 for any i in step 2.
Note that N ≤ M
Lecture 4 18
- Example of Gram-Schmidt procedure
Find the basis functions and plot the signal space for the following
transmitted signals:
s1 (t ) s2 (t )
A
T 0 T t
−A
0 T t T
Using Gram-Schmidt procedure:
T ψ 1 (t ) s1 (t ) = Aψ 1 (t )
1 E1 = ∫0 s1 (t ) dt = A
2 2
1 s2 (t ) = − Aψ 1 (t )
ψ 1 (t ) = s1 (t ) / E1 = s1 (t ) / A T
s1 = ( A) s 2 = (− A)
T
2 < s2 (t ),ψ 1 (t ) >= ∫ s2 (t )ψ 1 (t )dt = − A 0 T t
0
s2 s1
d 2 (t ) = s2 (t ) − (− A)ψ 1 (t ) = 0 ψ 1 (t )
-A 0 A
Lecture 4 19
- Implementation of matched filter receiver
Bank of N matched filters
z1
ψ (T − t )
∗
⎡ z1 ⎤
1 Observation
vector
r (t )
⎢ ⎥=z z
⎢ ⎥
ψ ∗ N (T − t ) ⎢zN ⎥
⎣ ⎦
zN
N
si (t ) = ∑ aijψ j (t ) i = 1,..., M
j =1
z = ( z1 , z 2 ,..., z N ) N≤M
z j = r (t ) ∗ψ j (T − t ) j = 1,..., N
Lecture 4 20
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