State Space Systems (applied to following circuit)

Discussion in 'Homework Help' started by u-will-neva-no, Dec 3, 2011.

1. u-will-neva-no Thread Starter Member

Mar 22, 2011
230
2
Hey everyone,

I have attached a diagram to show you what I am dealing with. My first question is:

Are the currents flowing the correct way for me to eventually solve the state space model of the circuit?

I have drawn in green(X and Y) where I would like to do my nodal current equations. They have to be where the capacitors are as they are storing devices (well that's what I was taught to do). I have also drawn in a black dot as I think this could be a possible place to do a current node equation.

Any help, as always, is much appreciated!

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2. Vahe Member

Mar 3, 2011
75
9
If you are trying to write the state space equations of this circuit, first you have to define your states. I see that you have marked and with arrows -- I guess this is the alternate way of marking voltage which I always get confused with. I always use the plus and minus signs -- it is always more clear to me. The capacitor voltages definitely need to be the state variables.

I think you should write nodal (KCL) equations at the inverting (-) terminals of all three opamps. Do not write any KCL equations at the output nodes of the opamps (since the output current of the opamp is unknown). I would mark the state variables (capacitor voltages), with the positive at the opamp output and the negative at the inverting terminal. This is because all non-inverting terminals are grounded ... and since , the capacitor voltages will just be equal to the output of the two opamps with capacitors in the feedback loop.

Your state state expressions should look like this

A few things to point out ... the nodes that you have marked as A and B should be and , so I think the arrow voltage notation denotes that the arrow points from + to - ... this is a guess. Also note that nodes that you have marked as X and Y are at ground potential. Also current marked as is zero (open circuit).

I need to work this out and will post the results.

I hope this helps.

Best regards,
Vahe

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3. Vahe Member

Mar 3, 2011
75
9
Here is what I am getting by writing the KCL equations that I mentioned in my earlier post. First let me setup the state variables. The output of the top opamp (point B) is , the output of the lower left opamp is and the output of the lower right opamp is . The signal is an output and not a state variable. Also note that since all noninverting opamp terminals are grounded, the voltage at all inverting terminals are at 0V.

Let's start with the easiest opamp to deal with -- the lower right device, which is just an inverting amplifier with a gain of . Writing the KCL equation at the inverting node gives (sum of currents in = 0)

This is the output equation ... it has to be a function of the state variables and the inputs ... and it is.

Consider the top opamp next. Writing the KCL equation at the inverting node gives (sum of currents in = 0)

Lastly, consider the lower left opamp. Writing the KCL equation at the inverting node gives (sum of currents in = 0)

Now you can put them in matrix form if you wish

I hope this helps you out -- hope there are not typos.

Best regards,
Vahe

Last edited: Dec 4, 2011
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4. u-will-neva-no Thread Starter Member

Mar 22, 2011
230
2
wow, that helped me out so much! I really cannot thank you enough!! Thank you for writing everything out also as that must have taken a bit of time to work out and type (it would for me).

You explained it really well, thank you once again.

5. u-will-neva-no Thread Starter Member

Mar 22, 2011
230
2
I have one small question and it involves an easier circuit diagram attached. The equation that I am trying to get is:

I tried to aply what you said before but the same analysis would not work in this instance as there is a capacitor across the op-amp (compared to the resistor previously). Don't worry about explaining the other equations at X and Y as I understand them.

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Last edited: Dec 4, 2011
6. steveb Senior Member

Jul 3, 2008
2,431
469
It looks to me that your choice of polarity on x_2 forces you to use the output equation

If you change the polarity of x_2, then you can get .

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