plugin-212-matched_filter_2up.pdf

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Correlation FunctionsA.J.Wilkinson, UCT EEE3086F Signals and Systems II212 Page 1 March 19, 2012

EEE3086F

Signals and Systems II

2012

Andrew Wilkinson

[email protected]

http://www.ee.uct.ac.zaDepartment of Electrical Engineering

University of Cape Town


2.7 Matched Filter

1. Goal of the Matched Filter

2. Derivation of the Matched Filter

3. Examples

Contents:

Reference: F.G. Stremler, Introduction to Communication Systems, 3rd Edition


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Goal of the Matched Filter

A.J.Wilkinson, UCT Correlation Functions EEE3086F Signals and Systems II212 Page 4 March 19, 2012

Aim and Applications of the Matched Filter

The matched filter is a linear filter that is designed to detect the presence of a waveform(or pulse)x(t) of known structure buried in additive noise.i.e. within a signal

whereA and t0

are unknown constants, and n(t) is noise.

The output of a matched filter will typically exhibit a sharp peak in response to thepresence of the desired waveform (or pulse) at its input. From this peak may bedetermined time location t

0, and amplitudeA of the pulse within the received waveform.

E.g. 1. Radars transmit electromagnetic pulses which reflect off distance objects (ortargets). The echoes are captured by a receiving antenna. The echoes may be veryweak signals buried in additive noise (picked up by the antenna, and also created by thereceiver itself). Matched filters are used to process the received waveform, and henceoptimize the chance of detecting a target.

E.g. 2. In telecommunications systems using pulse code modulation, a bank of N filtersmay be constructed, with each filter specifically designed to respond to the presence of a

particular pulse code. Comparison of the responses at the outputs identifies the receivedcode.

v (t)=A x (tt0)+n(t)


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Model of Transmitted and Received Signal with Additive

Noise

Transmitted pulse

Received signal

where t0is the propagation time delay,

A is an amplitude scaling factor, and

n(t) is additive noise, described by its PSD Sn().

The SNR of the received pulse is defined as the ratio of the the

peak signal power to average noise power, i.e.

SNR=peak signal power

average noise power=

max {A x (tt0)2}

n(t)2=A x (0)2

n(t)2

x (t)


t0

0


Definition of the Matched Filter

Given an input signal

wherex(t) is a pulse of known structure,

t0is an unknown delay, andA is an unknown scaling factor.

A matched filter is a linear filterh(t) orH() that isdesigned to optimize the peak SNR at its output at some

specified instant relative to t0.

The output is

The aim is to find h(t) or equivalentlyH() that maximizes

(at some instant tx, usually tx>t0):


peakSNR=y (tx)2

no(t)2

vo(t)=v (t) h(t)=y (t)+no (t)


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Derivation of the Matched Filter


Simplified Signal Model

Let the input be:

Filter output:

Let

Output written as

vo(t)=v (t) h(t)=x (t) h(t)+n(t) h(t)

v (t)=x (t)+n(t)

vo(t)=y (t)+no(t)

y (t)=x (t) h(t)

no(t)=n(t) h(t)

(output signal component)

(output noise component)

(wherex(t) is a known waveform

plus stationary noise n(t) )


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Signal Model Illustrated

H()

noise

Input

signal

Output

output SNR=y (td)

2

no(t)2

=peak signal power

average noise power

td

t

t

peak

Maximize at t= td

vo(t)=y (t)+no(t)

Filter

t

v (t)

n(t)

x (t)


Derivation of the Matched Filter

output SNR=y (td)

2

no(t)2

=peak signal power at t=td

average noise power

y (t)=x (t) h(t)=1

2

+

X()H()ej td

y (td)2= 12

+

X()H()ej tdd2

Peak output

signal power

Average noise

power no(t)2

=1

2

+

Sno ()d =1

2

+

Sn()H()2d

Output (signal

component)(from inverse FT)


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Optimization Problem

The objective is to find a/the filterH() that optimizes the SNR at

the time instant t= td:

y(td)2

no(t)2=

12+

X()H()e j td d2

1

2

+

Sn()H()2

d

To do this, we make use of a (provable) mathematical relationship calledthe Schwartz (or Cauchy-Schwartz) inequality.


Schwartz Inequality for Integrals

The Schwartz inequality states that for two complex

functions, f(x) andg(x), integrable over [a,b]

The equality holds (i.e. LHS = RHS) if

where kis a real constant.

If then the left side is less than the right side.

ab

f(x)g(x)dx2

a

b

f(x)2 dx a

b

g(x)2 dx

g(x)=k f(x)

g(x)k f(x)Proof? Google it!


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To maximize SNR, apply Schwartz Inequality

y(td)2

no(t)2=

12+

X()H()e j td d2

1

2

+

Sn()H()2

d

y (td)2

no(t)2=

12 +

[ X()Sn() ] [Sn()H()ej t

d] d2

1

2

+

Sn()H()2

d

y (td)2

no(t)2

1

2

+

X()Sn()2

d1

2

+

Sn()H()ej td

2d

1

2

+

Sn()H()2

d

f(x)[ X()Sn() ] g(x )Sn()H()ej tdApply Schwartz Inequality with:

Notice that the term

on the right of the

numerator cancels

with the denominator.

SNR:

and

Rewrite expression as:


y (td)2

no(t)2

1

2

+ X()2

Sn()d

1

2

+

Sn (w)H()2d

1

2

+

Sn()H()2d

y (td)2

n0(t)2

1

2

+ X()2

Sn()d

Swartz inequality becomes an equality i.e. LHS = RHS if

or in our case, if:

g(x)=k f(x)

Sn()H()ej td=k[ X()Sn() ]

=> H()=kX()

Sn()ej td

The term on the right

of the numerator

cancels with the

denominator.


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Matched Filter

H()=kX

()Sn()

ej td

For the special case of white noise,

H()=k X()ej td

h(t)=k x(t+td)

The filter that maximizes

the peak SNR at time td

is:

kis a real constant, usually positive, and is often set to 1.

Sn()=constant

Inverse transforming gives:

Apply Fourier: X()x(t)

and Y()ej td y(ttd)


Matched Filter (additive white noise)

H()=X()e j td

h(t)=x(t+td)

t

x (t)

tx (t)

ttd

ttd

y(t)=x (t) h(t)c2 2BSa (2B (ttd))

X()

Y()=X()2 ej td

2B 2B

Flip h(t), slide, mult. and integrate

c

2B 2B

c2 rect( /4 B)


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Interpretation in Frequency Domain:

In the frequency domain, the matched filter magnitude boosts the

frequency components for which the signal is large compared to

the noise and suppresses those that are weak compared to thenoise i.e.

The phase of the filter cancels the phase ofX(),

and hence the inverse Fourier transform integral gives a strong

peak at t= td. In the white noise case

H()X()Sn()

H()X()

y (t)=1

2

+

X()H()ej td=1

2

+

X()2 ej tdej td

H()=kX

()Sn()

ej td

H()=X()ej td

y (td)=Ex (by Parseval)


Interpretation in Time Domain

In the time domain, the matched filter for white noise is seen as

the conjugate of a time-reversed (flipped) version ofx(t).

It is easiest to consider the case ofk=1 and td = 0, which is a

matched filter with zero-delay. Performing a graphical convolution, flips h(t), which re-flips

such that it will now match (or correlate well) with the

x(t) contained in the received signal v(t) as one slides from left

to right and integrates the product.

The width of the peak in the time domain depends on the

bandwidthB ofX(). To gety(t), we inverse transform:

h(t)=x

(t)

x(t)

v (t) h(t)=x (t) h(t)+n(t) h(t)

X()H()X()2


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Interpretation in Time Domain:

The width of the peak in the time domain depends on the

bandwidthB ofX(), since we are inverse transforming

If has an approximately rectangular bandwidth,then the output will have a Sa( ) shape.

The 3dB width of the lobe will be [s].

X()2

X()2

t1

2B

H()

noise

Bandlimited

input signal

Output

td

tt

peak

Matched Filter

x (t) t 12B


Physically Realizable Filters

The impulse response of the matched filter for waveformx(t) with

additive white noise is

If this filter is to be constructed as a filter operating in real-time,

then the output cannot respond before the input arrives.

Thus for physical realizability, the impulse response h(t) must

begin aftert= 0. i.e. h(t) = 0 fort


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Output SNR

The output SNR of the matched filter is

For white noise case,

Thus the output SNR is proportional to the energy of the

received signal (and hence the energy of the transmitted

signal).

y (td)2

n0(t)2=

1

2 + X()

Sn()

2

d

(previously derived

general case)

no(t)2 = 1

2

+ 2H()2 d= =

2

1

2

+

X()2 d=2

Ex

H()=X()e j td

y (td)

2

n0(t)2=

Ex2

2

Ex=

Ex

/2

y (td)2= 12

+

X()H()ej tdd2

=Ex2

Sn()=/2


Example of Matched Filter: detection of a received pulse

x (t)= A rect ( tT/2T ) T0A

t

td+T

td

kA

t

Transmitted

pulse.

Noisy received

waveform.

td

kA2T

Note: output peak is proportional to energy in pulse (A2T)

t

Delayed, scaled pulse.

Output of matched filter;

SNR at peak is optimised.

Peak SNR

h

(t

)=x

(t

)Matched Filter.T

A

t

h(t) v (t)

k x ( ttd)

v (t)=k x (ttd)+n(t)

td-Tt


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Example of Matched Filter: detection of a received pulse

x (t)T0

A

t

td+T

td

kA

t

Transmitted

pulse.

Noisy received

waveform.

tdNote: output peak is proportional to energy in pulse

t

Delayed, scaled pulse.

Output of matched filter;

SNR at peak is optimised.

Peak SNR

h(t)=x(t)Matched Filter.T

A

t

h(t) v (t)

k x ( ttd)

v (t)=k x (ttd)+n(t)

td-Tt

k Ex


Applications of Matched Filter

The matched filter is very useful for detecting known waveforms in thepresence of additive noise .

E.g. 1. Radars use correlation receivers to detect weak echoes from distanttargets. A replica of the transmitted pulse is correlated with the received

waveform. The radar resolution (ability to separate close targets) dependson the bandwidth of the transmitted pulse.

E.g. 2. In telecommunications systems using pulse amplitude modulation, amatched filter, matched to the shape of the pulse, can be used to estimateoptimally the pulse amplitude, since the height of the peak at the output ofthe matched filter is proportional to the amplitude of the received pulse.

h(t) v (t)

v (t)t

t

Received

Output of MF


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Drill Problem 2.14 (2012)

Design the impulse response of a physically realizable filter

matched to the waveform below:

The input noise is bandlimited white noise with a power

spectral density of

Sketch the impulse response h(t) of the matched filter. Plot the outputy(t) of the matched filter (excluding noise).

Calculate the peak signal to noise ratio at the input and at the

output.

Sn()=0.5rect ( /20)W/Hz into 1 ohm

2s

1V

2V

1st


Drill Problem 2.15 (2012)

By application of properties of the inverse Fourier transform,

determine the inverse transform of:

in terms of

Show each step carefully.

X()ej td

x (t)=F1{X()}


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EEE3086F

Signals and Systems II

End of handout

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