How to add a time delay to a diagonal ss from a reside pole function?

Hi,

I have a list of poles (Ak) and residues(Ck) and one delay (tau) that corresponds to the following function:

I have created a ss representation of the rational function and also calculate the function analytically.

I have compared both functions using a bodeplot and they do not match when the time delay is present but match without it.

Can someone help me to find out how to add the time delay correctly?

% Delay problem
Ak = [-0.0633296088793117 + 0.00000000000000i
-0.188476918974608 + 0.00000000000000i
-0.592850411790702 + 0.00000000000000i];

Ck = [2.67546115276169e-05 + 0.00000000000000i
0.000235105989428637 + 0.00000000000000i
0.00132009482602487 + 0.00000000000000i];

D = 0;
tau = 0.001013802649678;

% Express as diagona form
F_A = eye(size(Ak,1),size(Ak,1)).*Ak;
F_B = ones(size(Ak));
F_C = Ck.';
F_D = D;

Fss = ss(F_A,F_B,F_C,F_D,'OutputDelay',tau);

% Analytic function
freq = logspace(-3,6,1000);
w = 2*pi*freq;

f = D;
for i = 1:length(Ak)
    f = f + Ck(i)./(1j*w - Ak(i));
end
f = f.* exp(-1j*w.*tau);

Fana = frd(f,w);

% Comparison bode plot
opts = bodeoptions;
opts.FreqUnits = 'Hz';
opts.FreqScale = 'Log';
opts.MagUnits = 'abs';
opts.MagScale = 'Linear';
opts.grid = 'on';

figure, bodeplot(Fss,Fana,w,opts)
title("Bode plot of H")
legend('SS model', 'Analitic')

Thanks and BR,

//JH

Answer 1

[Paul]

Hi Joan,

The problem appears to be with how bodeplot is unwrapping the phase.

Here is the original code

% Delay problem
Ak = [-0.0633296088793117 + 0.00000000000000i
-0.188476918974608 + 0.00000000000000i
-0.592850411790702 + 0.00000000000000i];
Ck = [2.67546115276169e-05 + 0.00000000000000i
0.000235105989428637 + 0.00000000000000i
0.00132009482602487 + 0.00000000000000i];
D = 0;
tau = 0.001013802649678;
% Express as diagona form
F_A = eye(size(Ak,1),size(Ak,1)).*Ak;
F_B = ones(size(Ak));
F_C = Ck.';
F_D = D;
Fss = ss(F_A,F_B,F_C,F_D,'OutputDelay',tau);

% Analytic function
freq = logspace(-3,6,1000);
w = 2*pi*freq;
f = D;
for i = 1:length(Ak)
    f = f + Ck(i)./(1j*w - Ak(i));
end
f = f.* exp(-1j*w.*tau);
Fana = frd(f,w);
% Comparison bode plot
opts = bodeoptions;
opts.FreqUnits = 'Hz';
opts.FreqScale = 'Log';
opts.MagUnits = 'abs';
opts.MagScale = 'Linear';
opts.grid = 'on';
figure, bodeplot(Fss,Fana,w,opts)
title("Bode plot of H")
legend('SS model', 'Analitic')

Zoom in on the end of the plot where things are different

xlim([1e4 1e6])

Now repeat the plot, but keep -180 <= phase <= 180

opts.PhaseWrapping = 'on';
figure, bodeplot(Fss,Fana,w,opts)
title("Bode plot of H")
legend('SS model', 'Analitic')
xlim([1e4 1e6])

Now both plots match.

I'm not sure why the phase unwrapping is a problem for Fana but not for Fss.

[Joan]

Thanks Paul. It seems the analyitic function phase gets wrapped when performing the operation exp(-1j*w.*tau).

I have tried the code and it works.