How to convert from rad/sec to Hz in bode plot, and what can i understand from the bode plot shown?

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Salah 2023-1-9

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编辑： Paul 2023-1-10

采纳的回答： Paul

在 MATLAB Online 中打开

%% These are the results i get after using genetic algorithm to maximize

%% the performance of a boost converter with input filter

Cc=0.01;

Cf = 1.0559e-9;

Lc = 1e-9;

Lf = 1.0182e-9;

Vo = 500;

R = 5;

DutyC=0.76;

s=tf('s');

dp = 1-DutyC;

w1 = -Lc;

w2 = dp*R;

u1 = Lc*Cc*R;

u2 = Lc;

u3 = (dp^2)*R;

x1 = Lf*Cf*(Lc^2)*dp*R*Cc;

x2 = (Lf+Lc*dp*R-(dp^2)*R*Lf*Cf)*Lc*Cc+(Lc^2)*Lf*Cf*dp*R;

x3 = (Lf+Lc*dp*R-(dp^2)*R*Lf*Cf)*Lc+Lf*Cf*Lc*(dp^3)*(R^2);

x4 = (Lf+Lc*dp*R-(dp^2)*R*Lf*Cf)*(dp^2)*R-(dp^2)*R*Lc;

x5 = -(dp^4)*(R^2);

y1 = Lc*Cc*Lf*Cf;

y2 = Lc*Cc + Lc*Lf*Cf;

y3 = Lc+Lf+Lf*Cf*(dp^2)*R;

y4 = R*(dp^2);

Gvd = (Vo*(s*w1 + w2)*((s^4)*x1+(s^3)*x2+(s^2)*x3+s*x4+x5))/(dp*((s^2)*u1+s*u2+u3)*((s^3)*y1+(s^2)*y2+s*y3+y4));

bode(Gvd)

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Answer 1

Paul 2023-1-9

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编辑：Paul 2023-1-10

在 MATLAB Online 中打开

To plot in Hz, rather than rad/sec.

Option 1:

%[m,p,w] = bode(Gvd);  % w - rad/sec
%f = w/2/pi;           % f - Hz
%subplot(211);semilogx(f,db(squeeze(m)));
%subplot(212);semilogx(f,squeeze(p));

Option 2:

Use bodeplot with the appropriate bodeoptions input to specify frequency in Hz

If Gvd is a transfer function of a BIBO stable system (I didn't check it), then the Bode plot tells you the amplitude and phase of the steady state sinusoidal output of the system relative to a sinusoidal input at a given frequency. For example, for a sinusoidal input with frequency 10^5 rad/sec, the steady state output would be sinusoidal with output amplitude increased by ~60 dB and phase shifted by ~360 deg.

BTW, if Gvd represents an unstable, input-ouput system, its output in response to a sinusoidal input will still include a sinusoid at the same frequency as the input with relative amplitude and phase determined from the Bode plot, but the output will also contain other terms that grow towards infinity.

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Salah 2023-1-9

编辑：Salah 2023-1-9

Thank you @Paul, it worked!

if it is okay for my second part of the question, why the phase looks like this, shouldn't it reach -180?

Paul 2023-1-9

在 MATLAB Online 中打开

Think of the phase as an angle that identifies the location of point on the unit circle in the complex plane. 180 deg is the same as -180 deg. That is, the point at (-1,0) can written equally as exp(1j*deg2rad(180)) or exp(1j*deg2rad(-180))

format short e
exp(1j*deg2rad(180))  % not quite perfect because pi not exactly represented
ans = 
  -1.0000e+00 + 1.2246e-16i
exp(1j*deg2rad(-180))
ans = 
  -1.0000e+00 - 1.2246e-16i