State Space representation

Discussion in 'Homework Help' started by m.ikhlas, Jul 4, 2014.

  1. m.ikhlas

    Thread Starter New Member

    Mar 28, 2014
    7
    0
    Hello Friends,
    I am looking to learn space state representation and was working on the attached figure and equations. If any one could help in setting up a space space representation..

    Thanks
     
  2. MrAl

    Well-Known Member

    Jun 17, 2014
    2,439
    492
    Hi,


    A quick question for you:
    Why do you have all the inductor values (Ln) dotted in your first set of equations, are you trying to state that the inductor values change with time too?
     
  3. m.ikhlas

    Thread Starter New Member

    Mar 28, 2014
    7
    0
    Hello,

    Thanks for your reply. Yes I am trying to state the inductor values change with time too. The inductor values (L1, L2 and L3) are also dependent on time. This is what making me confuse as I have voltage, current and inductor, all are changing with time.

    Thanks
     
  4. MrAl

    Well-Known Member

    Jun 17, 2014
    2,439
    492
    Hello again,

    Well if you have a time dependent inductor, wouldnt you represent that as L(t) and wouldnt you know what this function is?

    So it looks like to start with we would write (for example)
    V=I*R+L(t)*i

    where upper case "I" is the current, and lower case "i" is the time derivative of the current. Solving for "i" we would get:
    i=(V-I*R)/L(t)

    and then we might be able to work this out or do it numerically if needed, depending on the complexity of L(t).

    Usually non linear inductors are modeled as current controlled, ie as in: "L(I)" where I is the current through the inductor. This makes it time dependent, but only as I is time dependent.

    Does this seem logical to you?
     
    Last edited: Jul 7, 2014
  5. m.ikhlas

    Thread Starter New Member

    Mar 28, 2014
    7
    0
    Hello,

    Thanks a lot for your reply. It seems logical as well. The inductors are also sinusoidal function depending on time same as the current. If both current and inductance values are dependent on time should not voltage be dependent time derivative of both inductor and current. like:
    V = I*R + L * derivative (I) + I * derivative (L)

    May be I am confusing something.
     
  6. MrAl

    Well-Known Member

    Jun 17, 2014
    2,439
    492
    Hi,

    I'll have to give this some more thought. In the mean time, where did you first learn to put in the "I*derivative(L)" term?
     
  7. m.ikhlas

    Thread Starter New Member

    Mar 28, 2014
    7
    0
    Hey,

    Am I doing it wrong. The voltage across inductor is written as derivative of (L*i). If both inductor and current are time varying functions then should not it be opened as derivative by parts ?
     
  8. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    5,448
    782
    If you are thinking of linear inductance then the third term is not of interest. For a simple series R-L circuit carrying a time varying current i(t) the total series time varying voltage v(t) is simply

    v(t)=i(t)\times R+ L \times \frac{di(t)}{dt}
     
  9. MrAl

    Well-Known Member

    Jun 17, 2014
    2,439
    492

    Hello again,

    Is that your only reason for doing this? I ask because of the following argument which suggests that would not be the way to handle this.

    To start, the derivative of x*y using the product rule would be x*y'+y*x', however, that's the derivative of the entire expression x*y.
    With L and i we have:
    V=L*di/dt

    so clearly we are not taking the derivative of (L*i) we are just taking the derivative of i with respect to time, then multiplying by the constant L.

    Did you perhaps have another reason for starting to do it this way?

    Also, do you happen to know what the exact function L(t) is?
     
  10. m.ikhlas

    Thread Starter New Member

    Mar 28, 2014
    7
    0
    Hello,

    This is how I was doing it..
    Any change in the current through an inductor creates a changing flux, inducing a voltage across the inductor. By Faraday's law of induction, the voltage induced by any change in magnetic flux through the circuit is :

    V = time derivative (flux)

    Where Flux = L*i, so

    V = time derivative ( L*i)

    If L is constant we take it out of the derivative as

    V = L * time derivative (i)

    But if L is also a time dependent function then how can we just take it out of the derivative directly.. Also L is a sinusoidal function depending on time.

    Isn't this right ?

    Thanks
     
  11. MrAl

    Well-Known Member

    Jun 17, 2014
    2,439
    492
    Hi,

    Oh yes, that makes a lot more sense now. So the product rule should apply then.

    This also then seems to introduce a new state variable, L.

    Do you know the function L(t) or not? For example, it is sinusoidal but does it have the same period as the AC source voltage itself? I would like to know more about this function.

    It does look clear that each branch of the delta is independent in voltage, however the currents into each node will depend on two out of three branches. This is probably how you solve the entire set.

    Also, do you just want the state equations or do you also want to solve them?
     
  12. m.ikhlas

    Thread Starter New Member

    Mar 28, 2014
    7
    0
    Hello,

    Yes the L(t) will have same period as the source itself. I would like to solve this as well using state space model. Once the A, B and C matrices are developed I can use MATLAB for its different analysis.

    Thanks.
     
  13. MrAl

    Well-Known Member

    Jun 17, 2014
    2,439
    492
    Hi again,

    This is very interesting because the inductor is usually non linear with its current, not explicitly with time, so we usually have L(i) not L(t), even though L also happens to change with time when we have L(i) because i changes. Most physical inductors are modeled this way because that's the way they behave. The inductance changes based on the current through the inductor and also the history of the current through the inductor. That's probably why we see all over the web L(i) and not L(t).

    Ok lets see if we have this right now...

    Vab(t)=Vp*sin(w*t)
    L(t)=L0*sin(w*t)
    where
    Vp is the peak of the source sine, and L0 is the inductor peak value.

    We have so far:
    Vab(t)=R*i(t)+L(t)*di(t)/dt+i(t)*dL(t)/dt

    and after substituting the functions for Vab and L and taking the derivative we get:
    Vp*sin(t*w)=I*R+w*cos(t*w)*I*L0+i*sin(t*w)*L0
    where
    I is the current i(t), and i is the derivative=di/dt.

    Solving for i we get:
    i=Vp/L0-(I*R)/(sin(t*w)*L0)-(w*cos(t*w)*I)/sin(t*w)

    and after reducing the equation using trig we get:
    i=Vp/L0-(csc(t*w)*I*R)/L0-w*cot(t*w)*I

    To get into state space form, factor out the I where possible:
    i=Vp/L0-I*(csc(t*w)*R/L0+w*cot(t*w))

    This is now in state space form where the A and B are actually single dimensional matrixes:
    A=-(csc(t*w)*R/L0+w*cot(t*w))
    B=1/L0

    We could rewrite this as:
    i=A*I+B*Vp

    or using the more traditional variable x:
    x'=A*x+B*Vp

    where x is the state variable and x' is its time derivative.

    If your L(t) is not exactly as that, then you'll have to adjust the equations accordingly.
    You may also be able to change the trig form a little or perhaps even reduce it further.

    LATER:
    I should also add that the inductor used in this example had equation:
    L(t)=L0*sin(w*t)
    however this would cause the inductance to go though zero and also become negative so a more likely candidate would be something like:
    L(t)=Lp*sin(w*t)+L0

    where
    Lp is the peak inductance of the sinusoidal part, and
    L0 is the inductance at t=0.
    Also, most likely L0>=Lp to keep L(t) from going negative, and even more likely L0>Lp to keep L(t) from going to zero.
    It would be good if you could specify this better.
     
    Last edited: Jul 10, 2014
  14. m.ikhlas

    Thread Starter New Member

    Mar 28, 2014
    7
    0
    Hi,

    Thanks a lot. this seems much logical and helped a lot in understanding how it will be done. Now I am working on the complete circuit.

    Thanks again.
     
Loading...