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Conference rusure::math

Title:Mathematics at DEC
Moderator:RUSURE::EDP
Created:Mon Feb 03 1986
Last Modified:Fri Jun 06 1997
Last Successful Update:Fri Jun 06 1997
Number of topics:2083
Total number of notes:14613

666.0. "STUMPED!!" by BARNUM::JORGENSEN () Wed Feb 11 1987 10:47

    I am an engineer in need of some math help. I have been writing
    state (differential) equations to describe an electrical network
    and have run into a problem when it comes to reducing the coefficients
    of the variables in the equation. My problem is really just algebra.
    
    
    	dx3/dt= wz1*(x1-x3)
    
    	dx2/dt= wp1*(Vin-x1-x2)
    
    	dx1/dt= (1/(r1*c1))*(Vin-x1)+(1/(r2*c1))*(Vin-x1-x2)+(1/(r3*c1))*
    		
    		(x3-x1)
    
    
     	were:
    
    		wz1=1/(c3*r3)
    
    		wz2=1/(c2*(r1+r2))
    
    		wp1=1/(c2*r2)
    
    		wp2=(c2+c3)/(c1*c3*r3)
    
    The problem is getting the last equation to look like the first
    two. ie getting the coefficients to be some combination of wz1,wz2,
    wp1, and wp2. The coefficients are the 1/(__) terms in the third
    equation. I have been just plugging away trying to find the correct
    combination, and have had no luck... there must be a more elegant
    solution to this problem...any ideas?
    
    
    /Kevin
T.RTitleUserPersonal
Name		DateLines
666.1are the poles and zeroes expressed right?ENGINE::ROTHWed Feb 11 1987 11:5413
    It would be much easier if you could post your network, though I can
    guess at what it may be.

    Is it possible that the pole frequency, wp2 is expressed incorrectly?
    I'd almost guess it should be either

	wp2=(c1+c3)/(c1*c3*r3) or wp2=(c2+c3)/(c2*c3*r3)

    instead of

	wp2=(c2+c3)/(c1*c3*r3)

    - Jim
666.2FixedBARNUM::JORGENSENWed Feb 11 1987 12:3825
    
    yes, wp2 should read:
    
    	(c2+c3)/(c2*c3*r3)
    
    sorry, I knew I'd miss something somewhere. The network looks like
    this:
    
    
    		--- c2 -- r2 ---
                I              I
                I              I
            ________ r1 ____________________
                               I      I
    			       I      I
                              r3      I
                               I     c1
                              c3      I
    			       I      I
                               I      I
            ________________________________
    
    Hope you can make this one out. Thanks for correcting me Jim.
    
    /Kevin
666.3UghMODEL::YARBROUGHFri Feb 13 1987 09:3518
>    dx3/dt= wz1*(x1-x3)
>    dx2/dt= wp1*(Vin-x1-x2)
>    dx1/dt= (1/(r1*c1))*(Vin-x1)+(1/(r2*c1))*(Vin-x1-x2)+(1/(r3*c1))*(x3-x1)
>    
>    		wz1=1/(c3*r3)
>    		wz2=1/(c2*(r1+r2))
>    		wp1=1/(c2*r2)
>    		wp2=(c2+c3)/(c1*c3*r3)

Looks algebraically hopeless to me. dx1/dt has only c1's and no c2 or c3, 
but the w..'s have only c2 and c3 and no c1 at all, so I don't see where
you can make progress. I don't understand the circuit analysis, so I can
comment only from intuition; either this is indeed the simplest form of the
equations, or something has been inadequately analysed. 

This form of dx1/dt may be revealing:

	dx1/dt= ((Vin-x1)/r1 + (Vin-x1-x2)/r2 + (x3-x1)/r3) / c1
666.4Network has 3 poles, not just 2ENGINE::ROTHFri Feb 13 1987 14:3520
    You're having difficulty because the network shown has 3 poles, not
    just 2.  This follows because there are 3 energy storage elements...

    If you write out the open circuit transfer function it will
    come out in the form

	(1+sT1)(1+sT2)
	--------------------------------
	(1+sT1)(1+sT2)+sT3(1+sT4)(1+sT5)

    Where, for example, T1 and T2 are the time constants for the zeroes
    you gave.  The poles will all be real and interlace the zeroes
    on the negative real axis, but solving for them will involve
    taking the roots of a cubic equation in general, not just linear
    algebra.

    Note that T1*T2/(T3*T4*T5) = (R1+R2)/(R1*R2*C1), considering the
    behaviour of the network at high frequencies.

    - Jim
666.5Oops...ENGINE::ROTHTue Feb 17 1987 07:5312
    Re .4 -

�   The poles will all be real and interlace the zeroes
�   on the negative real axis...

    Actually it occurs to me that this is only true for the impedance
    of an RC network, and is not necessarily true of a transfer function,
    I'm a little rusty at this.

    Sorry if that may have caused any confusion...

    - Jim