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derivation of matched filter is explained very nicely to help understand the topic
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Baseband Data Transmission I
After this lecture, you will be able to
destination
the receiver
of thermal noise
Reference
Reference
D.
Introduction (1)
Digits
Quantizationnoise
t
Introduction (2)
Digits 0 2 3 4 5 6 7 8
Binary code 000 001 010 011 100 101 110 111
Return-to-zero
Introduction (4)
Channel
receiver
Receiver
Example: 1 or 0 for a binary system
Destination
D.
Signaling Rate
Digital message
drawn from an alphabet of
finite size
μ
Binary source:
μ
=2 for alphabet 0,1 where 0 and 1 are
symbols ¾
A 4 level signal has 4 symbols in its alphabet such as
Signaling Rate
of signal-elements. Each signal-element has the same duration of
second and is transmitted immediately one after another, so that thesignal-element rate (signaling rate) is 1/
elements per second
(bauds).
Bit Rate
Bit Rate
bit/symbol.
2 (4) = 4800 bits/s (bps)
D.
Matched Filter (1)
communication systems is that of detecting a pulsetransmitted over a channel that is corrupted by channel noise
t
t
Square pulse
Signal at the receiving end
D.
Matched Filter (3)
h
t )
of the filter such that the output signal-to-noise ratio
is
maximized. Let
g
f ) and
h
f ) denoted the Fourier Transform of
g
t ) and
h
t
df ft
j f G f H t g
exp( ) ( ) ( ) ( 0
2
2
0
exp( ) ( ) ( ) (
∞ ∫ ∞−
df ft
j f G f H t g
Signal Power
Matched Filter (4)
Noise Power
w
t ) is white with a power spectral density
, the
spectral density function of Noise is
0 N
2
0
f
f
N
df
f H N t n E
2
0
2
∞ ∫ ∞−
D.
Matched Filter (6)
f ), the particular form of the
transfer function
f ) of the filter that makes
η
at maximum.
Schwarz’s inequality
:
If
∞ <
∞ ∫ ∞−
dx
x
2
1
) ( φ
and
∞ <
∞ ∫ ∞−
dx
x
2
2
) (
φ
,
2
2
1
) (
) (
∞ ∫ ∞−
dx x
x
φ
φ
dx
x
∞ ∫ ∞−
2
1
) (
φ
dx
x
∞ ∫ ∞−
2
2
) (
φ
This equality holds, if and only if, we have
) (
) (
1
x
k
x
φ
φ
=
where
k
is an arbitrary constant, and *
denotes complex
conjugation.
Matched Filter (7)
Applying the schwarz’s inequality to the numerator ofequation (1), we have
≤
2
)
2
exp( )
(
)
(
df
fT
j
f
G f
H
df
f
H
2 )
(
df
f
G
2 )
(
……(2)(Note:
1
2
=
fT j e
π
)
Matched Filter (9)
N
otice that the S/N ratio does not depend on the transfer function H(f) of the filter but only on the signal energy.The optimum value of H(f) is then obtained as
)
2
exp( )
(
)
(
fT
j
f
kG
f
H
π
−
=
Matched Filter (10)
Taking the inverse Fourier transform of
H
(
f ) we have
h
(
t )=
k
df
t T f j f G
)]
(
2
exp[ )
(
−
−
∞ ∫ ∞−
π
and
)
(
)
(
f
G
f
G
−
=
for real signal
) (
t
g
h
(
t )=
k
df
t T f j f G
)]
(
2
exp[ )
(
−
−
−
∞ ∫ ∞−
π
h
(
t )=
kg
(
T
t )
…..(4)
Equation (4) shown that the impulse response of the filter isthe time-reversed and delayed version of the input signal g
(
t )
. ”Matched with the input signal”
