bode plot
I sent in a function
$$T = -0.1 \frac{s+400}{(s+20)*(s+2000)}$$
> T = -0.1*(s+400)/((s+20)*(s+2000));
> bode(T)
> grid on
I was expecting the phase to start at -180 deg, but instead the MATLAB is showing +180 deg.
Is this an inconsistency or a conceptual thing?
Partial answer only:
Check the period associated with the frequency of each notch in the frequency response. You will probably find they are integer multiples of the shortest period. This is typical of a "comb filter", often implemented as the sum of a signal and a delayed version of itself (the notch corresponds to delay=half a cycle). So look for a delay line in the block diagram.
Any complex number \$Z\$ can be represented in rectangular coordinates or polar coordinates $$Z = x + jy = A\angle\phi$$ where, \$A = \sqrt{x^2+y^2}\$ and \$\phi = atan2(y,x)\$ where, \$atan2(y,x)\$ is defined as
As you can see your conversion formula is valid only if \$x>0\$.
Hence \$\phi\$ for \$a+jb\$ and \$\phi\$ for \$-a-jb\$ will be different.
I think this solves your problem.
Best Answer
You can add or subtract 360 deg (2*pi rad) as many times as needed in Bode phase graph. Right click in your graph, Properties > Options > Phase Response > Adjust phase offsets [on] > Keep phase close to [-pi] (in your case), At frequency [0.000]. Done in MATLAB R2015a.
Edit: still works in MATLAB R2020b.