x1d=1/(Ld+Ls)(-Rx1+p(Lq-Ls)x1*x2)-(1/Ld+Lq)
x2d=1/(Lq+Ls)(-Rx2-p(Ld+Ls)x1x3+p*Phix3)
x3d=1/J(d1v^2+d2vx^3-p*Phix2)
y=x3
Need Experts Opinion.
Thanks
Means the direct coordinate transform and inverse coordinates transform.
Hi, do you mean that you want to invert the above equations analytically ?
Dear Richard Epenoy Sir,
Yes sir Analytically and confirmation of analytical equations from MATLAB tool box
xd=f(x)+g(x)*u
x1d=1/(Ld+Ls)(-Rx1+p(Lq-Ls)x1*x2)-(1/Ld+Lq)*u
x2d=1/(Lq+Ls)(-Rx2-p(Ld+Ls)x1x3+p*Phix3)*u
x3d=1/J(d1v^2+d2vx^3-p*Phi*x2)
y=x3
Hi, did you try the symbolic calculation toolbox of MATLAB or an equivalent tool as MAPLE or MATHEMATICA?
Hi Annas,
First, in a continuous-time LINEAR TIME-INVARIANT (LTI) system, the Laplace Transform manages to give you the frequency response of your system. First, you compute the Laplace Transform H(s) of its impulse response h(t) and for reasons that are shown below, this can be called Transfer Function (TF). Then, it can be shown that if you compute the Laplace Transform of the time-response of sin(wt), you get the product of the TF H(jw)=H(s=jw) with the Laplace Transform of sin(wt). This is why H(s) can also be called Frequency Response, because a simple product gives you the response of the linear systems for any input frequency. Moreover, next computation managed to show that, in linear systems, for any input u(t), the (transform of the) convolution integral that would give you the output y(t) in time-domain, becomes a simple product of H(s) with U(s) in Laplace domain, that can now be called Frequency domain. That’s why one one can separate the linear system H(s) from the input U(s) and get the output Y(s) as a simple product Y(s)=H(s)U(s).
In discrete linear systems, you get similar relations, only you use the z variable.
All that ends when you move away from LTI systems. While we still can, and may also want, to compute the transform of the input command and of the output signal, there is no simple relation with some transform of the nonlinear system.
That’s why, for example, stability analysis in linear systems is simply based on the Transfer Function (poles and zeros), while in nonlinear systems becomes some pretty complex analysis (Lyapunov, etc.)
Sorry if this is disappointing, yet nonlinear systems are complex and there is no use of s-transform or z-transform of the nonlinear SYSTEMS.
With best regards,
Itzhak
Thankyou Itzhak Barkana Sir
Actually i was asking about transformation that is used in feed back linearization of non linear control system
Sorry, misunderstanding, as the question that I saw was about nonlinear control systems.
(and, except for strictly official occasions, I am Itzhak for all, no need for any Prof., Dr., Mr., or Sir)
hi Annas
here is a simple example for using symbolic toolbox of matlab. you can write yours
syms x1 x2 x3 x4 p1 p2 p3 Ib Gb H n r t taw D
F=[-p1*(x1-Gb)-x1*x2+D;...
-p2*x2+p3*(x3-Ib);...
-n*(x3-Ib)+r*(x1-H)*t+x4;...
-x4/taw];
G=[0;0;0;1/taw];
T1=x1;
T2=[diff(T1,'x1') diff(T1,'x2') diff(T1,'x3') diff(T1,'x4')]*F;
T3=[diff(T2,'x1') diff(T2,'x2') diff(T2,'x3') diff(T2,'x4')]*F;
T4=[diff(T3,'x1') diff(T3,'x2') diff(T3,'x3') diff(T3,'x4')]*F;
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