Bode plot with right half plane zero

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Anindya
Anindya 2025-11-10，8:02
回答： Andrew Ouellette 2025-11-10，14:31
Hi, while using the bode function in matlab script for a transfer function which has right half plane zero, I am finding that the phase is starting from 360 instead of starting from 0. How can this be corrected. The code is inserted.
clc
clear all
s=tf('s')
fs=1e6;
ws=2*pi*fs;
L=1.8e-6;
C=220e-6;
R=5;
Vin=9;
Kg=1/5;
Vcarr=1;
Vbase=5;
Ibase=5;
Zb=Vbase/Ibase
%% Plant TFs
Vo=5;
D=0.357;
wrz=((1-D)^2)*R/(D*L); %% position of the RHP zero
wc=(1-D)/sqrt(L*C);
Q=(1-D)*R*sqrt(C/L);
Gdo=Vo/(D*(1-D));
Gvd=Gdo*(1-(s/wrz))/(((s/wc)^2)+(s/(Q*wc))+1); % control to output
%% bode plots
wct=1:10:10e7;
bode(Gvd,wct);

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回答（4 个）

Paul
Paul 2025-11-10，13:05
Hi @Anindya,
The bodeplot function offers options on controlling the appearance of the plot, including to force the phase between +-180 deg.
A bigger hammer would be to use ctrlpref to set the default phase wrapping preference for bode and bodeplot (actually, I'm not 100% sure it applies to bodeplot, some experimentation may be in order).
  1 个评论
Mathieu NOE
Mathieu NOE 2025-11-10，13:53
hello @Paul
+1
see my suggestion above
you can force the phase between +-180 deg with the regular bode function

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Andrew Ouellette
Andrew Ouellette 2025-11-10，14:31
Ran in:
Hi @Anindya,
You are looking for the phase matching options in bodeplot, which you can interpret as "If PhaseMatchingEnabled, then keep phase close to PhaseMatchingValue at the PhaseMatchingFrequency". Here is an example using your code:
s=tf('s');
fs=1e6;
ws=2*pi*fs;
L=1.8e-6;
C=220e-6;
R=5;
Vin=9;
Kg=1/5;
Vcarr=1;
Vbase=5;
Ibase=5;
Zb=Vbase/Ibase;
%% Plant TFs
Vo=5;
D=0.357;
wrz=((1-D)^2)*R/(D*L); %% position of the RHP zero
wc=(1-D)/sqrt(L*C);
Q=(1-D)*R*sqrt(C/L);
Gdo=Vo/(D*(1-D));
Gvd=Gdo*(1-(s/wrz))/(((s/wc)^2)+(s/(Q*wc))+1); % control to output
%% bode plots
wct={1 10e7};
h = bodeplot(Gvd,wct);
h.PhaseMatchingEnabled = true;
●
h.PhaseMatchingValue = 0;
h.PhaseMatchingFrequency = 0;
Claire
Claire 2025-11-10，8:56
When you use the bode function to plot the Bode diagram of a transfer function that has a right-half plane (RHP) zero, you may notice that the phase starts from 360 degrees instead of 0 degrees. This is because MATLAB does not normalize the phase when calculating it; therefore, when the phase exceeds 180 degrees, it will be displayed as a value greater than 180 degrees.
Mathieu NOE
Mathieu NOE 2025-11-10，9:56
Ran in:
hello
you can do this way :
BTW , you pulstaion vector was huge and there is a better approach for Bode plots (as the x axis is log scaled ,use therefore a log spaced frequency / pulstaion vector)
here 500 values are enough to make a correct plot
clc
clear all
s=tf('s')
s = s Continuous-time transfer function.
fs=1e6;
ws=2*pi*fs;
L=1.8e-6;
C=220e-6;
R=5;
Vin=9;
Kg=1/5;
Vcarr=1;
Vbase=5;
Ibase=5;
Zb=Vbase/Ibase
Zb = 1
%% Plant TFs
Vo=5;
D=0.357;
wrz=((1-D)^2)*R/(D*L); %% position of the RHP zero
wc=(1-D)/sqrt(L*C);
Q=(1-D)*R*sqrt(C/L);
Gdo=Vo/(D*(1-D));
Gvd=Gdo*(1-(s/wrz))/(((s/wc)^2)+(s/(Q*wc))+1); % control to output
%% bode plot
freq=logspace(1,7,500);
[g,p] = bode(Gvd,2*pi*freq);
g = g(:);
p = p(:) - 360 ;
figure
subplot(2,1,1),semilogx(freq,20*log10(g))
ylabel('Modulus (dB)')
title('Bode Plot')
grid on
subplot(2,1,2),semilogx(freq,p)
xlabel('Frequency')
ylabel('Phase(°)')
grid on
  7 个评论
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Paul
Paul 2025-11-10，14:06
In some weird cases, unwrap can have problems, IIRC.
If I want to manually force the output of bode to -180 <= p <= 180, I just do
p = angle(exp(1j*p*pi/180))*180/pi;
Though with the more "modern" functions perhaps
p = atan2d(sind(p),cosd(p));

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