I just thought I would send the Butterworth algorithm to make sure you get
it correct.  Note that the scaling at the end will be explained on Friday.

------------------------------------------------------------------------------
INPUT: N (filter order), omega_c (cutoff frequency)
OUTPUT: b = [b(0) b(1) ... b(N)], a = [a(0) a(1) ... a(N)] where H(z) =
        B(z)/A(z) where B(z),A(z) are z transforms of b,a respectivly.

ALGORITHM:
------------------------------------------------------------------------------
% Prewarp
OMEGA_C = tan(omega_c/2)

% Compute b coefficients
b = 1
for n = 1 to N
   b = convolve(b, [OMEGA_C OMEGA_C])
end

% Compute a coefficients
if (N is odd)
   a = [(1+OMEGA_C) (-1+OMEGA_C)]
else
   a = 1
end;
for n = 1 to floor(N/2)
   a = convolve(a, [(1 - 2*OMEGA_C*cos((Pi/2)*(1 + (2*n - 1)/N)) + OMEGA_C^2)
                    (-2 + 2*(OMEGA_C^2))
                    (1 + 2*OMEGA_C*cos((Pi/2)*(1 + (2*n - 1)/N)) + OMEGA_C^2)])

end

% Scale H(z) so a(0) = 1
for n = 0 to N
   b(n) = b(n) / a(0)
   a(n) = a(n) / a(0)
end


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Q: From taking a look at your Chebyshev algorithm and the book's algorithm,
they seem to be different.  Is there any way to check the program from the
table in the book like the Butterworth tool?

A: I was unaware the book had an algorithm for digital Chebyshev LP filter
design.  Can you provide me with a page number?  The easy way to verify
your tool is working is to look at the cutoff and the ripple amount and
check against your input parameters.  Sometimes the cutoff may not be exact
if you require small ripple but do not give enough poles.  Anyhow, here are
results from two filters I designed using my chebyshev routine.  Verify
your filter with these coefficients.

Ex. [b,a] = chebyshev(10,0.3,pi/4)

b =

   1.0e-03 *

    0.0004
    0.0040
    0.0179
    0.0478
    0.0837
    0.1004
    0.0837
    0.0478
    0.0179
    0.0040
    0.0004


a =

    1.0000
   -7.6076
   27.2543
  -60.3043
   91.0218
  -97.7535
   75.5649
  -41.4922
   15.4868
   -3.5495
    0.3797


Ex. [b,a] = chebyshev(11,0.3,3*pi/4)

b =

    0.0187
    0.2054
    1.0268
    3.0805
    1.0268
    3.0805
    6.1610
    8.6255
    8.6255
    6.1610
    3.0805
    1.0268
    0.2054
    0.0187


a =

    1.0000
    3.4396
    6.9904
    8.9453
    8.4762
    5.6626
    2.9428
    0.9168
    0.1595
   -0.1527
   -0.0894
   -0.0555

-----------------------------------------------------------------------------

FILTER

The algorithm has some typos on the limits of both summations:

1) the first summation (feed-forward terms) should be from k = 0 to N - 1
2) the second summation (feed-back terms) should go from k = 1 to M - 1

MAGNITUDE_RESPONSE

The for loop should go from:

   for omega_0 = 0 to (2Pi - 2Pi/P) step 2Pi/P

so that this tool returns exactly P data points.

PHASE_RESPONSE

(same as above)

MAGNITUDE_RESPONSE

The input will have either 2 arguments (b,a) or 4 arguments
(b,a,fs,plot_type).  If 2 arguments are specified, assume fs = 2Pi and
plot_type = 0.

EE495: For this and all future projects, if you wish to do the EE545
problems they will be considered bonuses.  Please state for my grader that
the problem you are doing is a bonus.


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