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Increased Productivity Example #3
Optimum Matched Filter (Transfer
Function)
-
(Nested Processes ... Each Process
controlled by a Solver)
- The transfer function H(s) is the
Laplace transform of the output signal Yout(s) divided by the Laplace
transform of the input signal Yin (s): that is H(s) =
 where each signal's transform is assumed to be a ratio
of polynomials. Thus, H(s) can likewise be stated in the form:
-
- Assuming the numerator and denominator can be
factored, yields H(s) in the general form:
-
- where each Zi is known as a "zero"
and the Pi as a "pole" of the transfer function. Zi
and Pi are complex points in the Laplace domain.
- A realizable transfer function must have poles
and zeros with their conjugate point. That is, poles and zeros come in
pairs. If a pole or zero is located at the complex point si
+ jwi, then its conjugate is located at si - jwi.
Thus, a generalized transfer function is stated as
- Given n-data points from a Bode plot (see
drawing below) that define the mainlobe of the desired transfer
function, find the optimal Pole/Zero constellation such that H(s) has
equal sidelobe peak amplitudes in a Bode plot and curve fits the
following data in the mainlobe.
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Increased Productivity Example
#3 Source Code:
- A Calculus-level program for this optimal matched
filter transfer function is as follows:
Problem .Matched.Filter.Transfer.Function
execute .Setup
for i = 0 to 1
sidelobes = i [Include Zeros On Omega Axis ? ]
Find gain, p.real, p.imag
In .Laplace.Domain To Match error
repeat
End
Model .Laplace.Domain
if sidelobes gt 0 AND omega.zeros gt 0 then
for ij = 1 to omega.zeros do
side.limits( ij) = x.zeros( ij) * (1 + move( ij)) * (2**(ij-2))
old.zeros( ij) = x.zeros( ij)
repeat
Find x.zeros In .Stopband By HERA
With Bounds side.limits
To Minimize peak.diff
for ij = omega.zeros, move( ij) = x.zeros( ij) - old.zeros( ij)
close
for ij = 1 to npoints do [ --- Calculate Transfer Function ---- ]
x2 = freq( ij) ** 2
execute .Transfer.Function
if den eq 0, den = 1e-8
h( ij) = gain * num / den
error( ij) = y.out( ij) - h( ij) * y.in( ij)
error( ij) = error( ij) / y.out( ij) [relative error ... line optional]
repeat
End
Model .Stopband [locate sidelobe peaks]
peak.diff = 0
for ijk = 1 to omega.zeros do
step.limit = side.limits( ijk)
Find x.peak In .Sidelobes By HERA
With Bounds step.limit
To Maximize y.peak
peaks( ijk) = x.peak peak.ampl( ijk) = y.peak
if ijk gt 1, peak.diff = peak.diff + (y.peak - peak.ampl( ijk-1))**2
repeat
peak.diff = peak.diff + (y.peak - peak.ampl( omega.zeros))**2
End
Model .Sidelobes [calculate sidelobe amplitude at frequency 'x.peak']
x2 = x.peak**2
execute .Transfer.Function
y.peak = num / den
End
Model .Transfer.Function
num = 1 den = 1
for ii = 1 to p.pairs, den = den * .Factor( x2, p.real( ii), p.imag( ii))
if omega.zeros gt 0 then
for ii = 1 to omega.zeros, den = den * .Factor( x2, 0, x.zeros(ii))
close
End
Function .Factor( x.sq, sigma, omega)
real.sq = sigma**2 Imag.sq = omega**2
sum = real.sq + imag.sq - x.sq
if omega eq 0, exit with sum / real.sq
temp.f = sum * sum - 4 * x.sq * imag.sq
End with temp.f / (real.sq + imag.sq)**2
Procedure .Setup
freq = .data( 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12,
13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24)
y.in = .data( 51.31, 37.79, 28.26, 21.1, 15.37, 11.32, 8.06, 5.83, 3.91, 2.69,
1.78, .96, .52, .31, .25, .21, .18, .15, .12, .10, .08, .07, .06, .06, .05)
[y.out(w) = optimum response in the linear Van der Maas sense ]
y.out = .data(1, .981, .926, .836, .73, .609, .484, .365, .26, .173,
.105, .058, .027, .012, 7.2e-3, 4.4e-3, 2.7e-3, 1.7e-3,
1e-3, 6.2e-4, 3.8e-4, 2.3e-4, 1.4e-4, 8.8e-5, 5.4e-5)
fmax = freq( npoints)
npoints = 25 gain = 1 p.pairs = 5 omega.zeros = 3
allot h( npoints), error( npoints)
allot p.real( p.pairs), p.imag( p.pairs)
for i = 1 to p.pairs do [initial guess]
p.imag( i) = ((i-1) / p.pairs + .11) * fmax
p.real( i) = fmax
close
if omega.zeros gt 0 then
allot x.zeros( omega.zeros), old.zeros( omega.zeros), move( omega.zeros),
side.limits( omega.zeros), peaks( omega.zeros), peak.ampl( omega.zeros)
for i = 1 to omega.zeros, x.zeros(i) = (1 + 2**(i-1) /10)*fmax [initial guess]
close
End
- Note: This is a multi-level optimization or
nesting of optimizers example problem. Each 'find' statement starts a
solver and at times has up to three Find statements
executing at once (ie. nested). This nesting power should allow
companies to optimizes at many levels and combine all optimizations for
a true optimize company profits when necessary. Nice!
- Note:
Match-n-Freq application is exactly this multi-level optimization
problem. Download Match-n-Freq application file. Give it a test run and
see the nested solvers at work.
Increased Productivity Example
#3 More Individual Process Examples:
- This Filter Designing problem is another
increased productivity example do to using Calculus (level) programming.
- Kost, R.E. and Brubaker, P.B., Arbitrary Equalization with
Simple LC Structures, IEEE Transactions on Magnetics, Vol.
MAG-17, No. 6, November, 1981
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<strong>Optimum Filter Design</strong> </a>, Bode
Diagram, Transfer Function, Matched Filter, Modeling & Simulation.
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