The first point is correct. As a general rule of thumb, the loop crossover frequency should be a fraction of the converter switching frequency. I've heard both one-fifth and one-sixth cited as the maximum.
Crossing over at -1 is not a firm requirement for stability. If the closed-loop response at the crossover frequency has sufficient phase margin (typically 45 degrees or more) and gain margin (-10 dB or more when the phase is at 0 degrees), the converter is stable.
The reason -1 slope is cited as the target for gain crossover is that, in general, the phase is small and isn't changing rapidly during that part of the curve. When the gain slope is -2, the phase shift is greater and changes rapidly, making it difficult to ensure phase margin if the crossover frequency shifts (due to component tolerances, etc.) That doesn't make it impossible to choose a compensation network, but it does make it more sensitive to changes.
For starters, can you tell the dimensions of each matrix ? It will make it easier to think further.
Assuming that u(t) and c are both of same dimensions. If they are, you can just make it :
der(x) = F*x(t)+G*[u(t) + c] , because u(t) is an input ( usually the one we control ) and c is also an input, only it's an external one. u(t) suggests it's time varying input, but that's just a general case, it doesn't have to be.
Edit after your comment:
Now that you have explained that u(t) is a 1x1 scalar, I see a solution, that may be too simple to work. You said you are using lsim command in Matlab, so I suppose you pass state space (ss) system model to it. In that case, since according to your comment u(t) is a scalar, G*u(t) is a constant 6x1 vector. Therefore, you can make a substition and say [G*u + c] is your new G matrix. For now I assume you are familiar with this part of using Matlab, but just in case you need a kickstart:
Matlab documentation on this website:
LINK
Says you can pass systems to lsim in this fashion :
[y,t] = lsim(sys,u,t)
[y,t,x] = lsim(sys,u,t) % for state-space models only
[y,t,x] = lsim(sys,u,t,x0) % with initial state
Best Answer
It's usually no problem at all. Take as a simple electronic example an op-amp precision rectifier. It has to cope with severe non-linearities i.e. the diode in the output circuit. It has a formula shown below: -
Where I is the current thru the device and V is the applied voltage to the terminals. It's exponential like in the question but the circuit below copes well: -
Vout has is produced and is perfectly linear with respect to the peak input voltage i.e. negative feedback has overcome the difficulties. The same is ture for a push pull amplifier with negative feedback. You can make one without final transistor biasing (to overcome the distortion arising when changing from one transistor to another) by choosing an op-amp that is high-speed.
In effect, when the transistors are at the crossover-point the op-amps's high speed almost perfectlyy compensates and "rushes" the input voltage to the transistors past this cross-over point. Again, negative feedback comes to the rescue: -
Real difficulties arise in control systems with complex frequency responses, hysterisis and deadband. Now these can cause major headaches.