Sound from a plucked string
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Can anyone debug this program for me?? I will be very grateful to you.
l = 0.6;
x = 0.3*l; % pluck position is 30% of total length
xp = x ; %pickup point same as pluck position
u = 1:100;
Ts = 60.97;
T = 1/8000;
I = 1.71e-13;
E = 7.4000e+9;
rho = 1140;
A = 5.188e-7;
d1 = 8.0e-005;
d3 = -1.4e-005;
stl = (u.*pi/l);
e1 = d1-(d3*(stl.^2));
e2 = (E*I*(stl.^4))+(Ts*(stl.^2));
Kux = sin(stl*x);
Nu = l/2;
den = rho*A;
Wu = (stl.^4*((E*I/den)-(d3^2/(2*den)^2)))+
(stl.^2*((Ts/den)+((d1*d3)/2*(den^2))))-(d1^2/(4*(den^2)));
Wu = sqrt(Wu);
sg = (-d1+(d3*(stl.^2)))/(2*den);
b0= 1/den;
b1= (b0*sin(Wu.*T))./(Wu);
c1 = -2*exp(sg.*T).*cos(Wu*T);
c0 = exp(2*sg*T);
% from Fe_mu.m :
m = 100;
u = 1:100;
x = 0.3;
xp = x;
% ******* recursive system implementation ******
for n = 1:1:m;
Ge(n) = dfilt.df1([0 b1(n)],[1 c1(n) c0(n)]); % Discrete Filter Ge^d(mu,z)
a(n) = dfilt.scalar((fe1_mu(n)*Kux(n))/Nu); % Gain factor
a(mu,xp)
hd(n) = cascade(Ge(n),a(n)); % series connection of Ge^d and a
end
H =
dfilt.parallel(hd(1),hd(2),hd(3),hd(4),hd(5),hd(6),hd(7),hd(8),hd(
9),hd(10),hd(11),hd(12),hd(13),hd(14),hd(15),hd(16),hd(17),hd(18),
hd(19),hd(20)...,hd(100)));
% display response of system
fe2k=[1 ; zeros(10000,1)];
g=conv(impz(H),fe2k); %convolution with fe^d_2(kT)
fvtool (g)
sound(g)
2 commentaires
Réponse acceptée
Walter Roberson
le 2 Nov 2011
There are two possibilities. I suspect the easier one will not work, but it is easier:
1) Change the line
dfilt.parallel(hd(1),hd(2),hd(3),hd(4),hd(5),hd(6),hd(7),hd(8),hd(
9),hd(10),hd(11),hd(12),hd(13),hd(14),hd(15),hd(16),hd(17),hd(18),
hd(19),hd(20)...,hd(100)));
to
dfilt.parallel(hd(1),hd(2),hd(3),hd(4),hd(5),hd(6),hd(7),hd(8),hd(
9),hd(10),hd(11),hd(12),hd(13),hd(14),hd(15),hd(16),hd(17),hd(18),
hd(19),hd(20),...
hd(100)));
But I suspect that will not work in practice.
2)
Change the line
hd(n) = cascade(Ge(n),a(n)); % series connection of Ge^d and a
to
hd{n} = cascade(Ge(n),a(n)); % series connection of Ge^d and a
and change the line
dfilt.parallel(hd(1),hd(2),hd(3),hd(4),hd(5),hd(6),hd(7),hd(8),hd(
9),hd(10),hd(11),hd(12),hd(13),hd(14),hd(15),hd(16),hd(17),hd(18),
hd(19),hd(20)...,hd(100)));
to
dfilt.parallel(hd{:});
I believe this is the solution you need.
8 commentaires
Walter Roberson
le 5 Nov 2011
Yes, that one -- put the Fe_mu invocation at the top of the routine that needs to use the value of the fe1_mu variable.
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