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Does the order of a system always depend on the number of independent initial conditions?

Consider this single mesh containing a DC voltage source, one resistor, and two capacitances and a switch which closes at t = 0. Nothing is stopping me from having two different initial voltages on these two caps and satisfying KVL all the time.

schematic

simulate this circuit – Schematic created using CircuitLab

$$V_{c1}\:\textrm{and}\:V_{c2} \textrm{ are initial voltages on the caps.} $$

Now, I also know that since these two caps are in series writing a KVL around the mesh will give an equation with a single equivalent capacitor and hence deducing that order of the system is one.

Which can also be verified by writing its transfer function. (Obviously when there are no initial conditions.)

\begin{equation} \frac{V_o(s)}{V_i(s)} = \frac{C_1}{{C_1}+{C_2}} \frac{1 + s{C_2}R}{1 + s{C_{eq}}R} \end{equation}

where $$C_{eq} = \frac{{C_1}{C_2}}{{C_1} + {C_2}}$$

Clearly, there is only one pole and hence order is one.

This confuses me. Is this some degenerate case or the statement that the order of a system is equal to the number of independent initial conditions is somewhat flawed.

EDIT 1

Now if I take this circuit.

schematic

simulate this circuit

As far as the structure of the network is concerned nothing has changed since the input ideal current source can be taken as an open circuit and I know this for a fact that injection of the input and the pair of terminals across which we're taking the output doesn't alter the location of the poles as long as the structure remains invariant.

This circuit has the same structure as that of the earlier one.

But the order here is two in agreement with the statement that order of a system is equal to the number of independent initial conditions as shown below.

\begin{equation} \frac{V(s)}{I(s)} = \frac{1}{s(C_1 + C_2)} \frac{1+sC_{2}R}{1+sC_{eq}R} \end{equation}

These two examples are countering each other.

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  • \$\begingroup\$ Regarding your edit (second schematic): If your intention here was to do Ceq = C1C2/(C1+C2) then it's mathematically correct - however I recall that it is meaningless when referring to the association of capacitors in series. This is because your second circuit does not admits that. Also, it's incorrect to say: "...This circuit has the same structure as that of the earlier one...." Your first circuit is just a case of series association only (I repeat, decomposition / composition - depending on the point of view). I see no contradiction here. Please, revise again my EDIT2. \$\endgroup\$ – Dirceu Rodrigues Jr Sep 3 '18 at 23:01
  • \$\begingroup\$ I didn't get your point. I don't find any correspondence between this and your EDIT 2. And why is it incorrect to say that This circuit has the same structure as that of the earlier one. Also, what do you think about the number of initial conditions in the system since the order is two now? \$\endgroup\$ – Astatine Sep 4 '18 at 5:43
  • \$\begingroup\$ Why are you ignoring my edit 1 while saying it has only one independent initial condition. My second example is a second order case and has the same structure as that of the first one. Shorting the voltage source in the first example and opening the current source in the second example would give me the same network structure and should have the same number of poles. I don't find any relevance of your edit 2 to the questions that I am asking. As I said I knew those stuff and they are fairly clear to me. \$\endgroup\$ – Astatine Sep 5 '18 at 6:05
  • \$\begingroup\$ Keep calm. I really missed the fact that in your first circuit the output was taken between C2 and R. This changes a lot. So forget the composition / decomposition part (although the discussion in EDIT2 is valid) I apologize for this. Yes, both circuits seem to have the same structure - with respect to the ZERO INPUT RESPONSE (since the output is a voltage). I'm solving it again with more care and soon will report to you my results. Let's together solve this. Thanks for a good question! \$\endgroup\$ – Dirceu Rodrigues Jr Sep 5 '18 at 15:32
  • \$\begingroup\$ Regarding to your first circuit: There is a POLE-ZERO CANCELLATION IN ORIGIN. Experiment write down the complete system responses (through Laplace T.) for first and second circuits. The ZIR (Zero Input Response) for both circuits exhibit a second order form - with a pole in origin). Are equal. Then, the complete response needs the two initial conditions - for both circuits. I will write my equations later - no time now. \$\endgroup\$ – Dirceu Rodrigues Jr Sep 5 '18 at 17:56
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The order of the linear system is defined as the order ("n") of the largest derivative of the differential equation describing the system. In order to solve it, also are necessary "n" initial conditions. The configuration you showed results in an equation of order 1.

EDIT 1

The definition of order using differential equation is general. It is not always practical to verify, but works. The subject is formally treated in more advanced courses of circuit theory, usually approached using state variables (the number of states will be the order of the system). The number of states will normally be equal to the number of energy storage elements, except if there are capacitor loops and inductor cut-sets on network. This is called degeneration. Search the subject.

EDIT 2

Ok. Let's go to talk about independent initial conditions. See the first circuit shown below. There are only 2 INDEPENDENT initial conditions (capacitor voltages). The initial voltage on \$C_3\$, for example, can be determined by the initial voltages on \$C_1\$ and \$C_2\$. So, even having three energy storage elements (capacitors), the order of system is two. On the second circuit There are only 3 INDEPENDENT initial conditions. The initial current through \$L_3\$, for example, can be determined by the initial currents \$I_1\$ and \$I_2\$. So, even having four energy storage elements, the order os system is three. These are more generic cases. I think this is not applies exactly to your circuit. For more detailed discussion and generic cases, please (as I already mentioned, too) consult the books.

Order examples

EDIT3

Some comments on circuits provided by OP. For the discussion that follows is used \$\left |vo_{(0-)} \right | = \left |vc1_{(0-)} \right |\$. Pay attention for the polarity in each case.

Circuit 1

Differential equation: $$ RC_1C_2\frac{dv_o^2}{dt^2}+(C_1+C_2)\frac{dv_o}{dt}=RC_1C_2\frac{dv_i^2}{dt^2} + C_1\frac{dv_i}{dt} \tag{1}$$ The complete response through L.T.: $$ Vo(s)= \frac{\left(RC_2s^2+s\right)C_1Vi(s)+\left(RC_2s+1 \right)C_1vo_{(0-)}+C_2vc2_{(0-)}}{s\left( RC_1C_2s + C_1+C_2\right )} $$

where \$vo_{(0-)} = -vc1_{(0-)}\$.

Separating in ZSR (Zero State Response) and ZIR (Zero Input Response):

$$ Vo(s)= \underbrace{\frac{\left(RC_2s^2+s\right)C_1}{s\left( RC_1C_2s + C_1+C_2\right )}Vi(s)} +\underbrace{\frac{\left(RC_2s+1 \right)C_1vo_{(0-)}+C_2vc2_{(0-)}}{s\left( RC_1C_2s + C_1+C_2\right )}} \tag{2}$$

Clearly indicating a second order response. But note that in the ZSR, there is a common \$s\$ factor between numerator and denominator. Note: If they are cancelled (POLE-ZERO CANCELLATION AT ORIGIN), it's obtained the first order transfer function \$\frac{\left(RC_2s+1\right)C_1}{RC_1C_2s + C_1+C_2}Vi(s)\$, which we would normally expect to see in such circuit. I think the reason for this is that the equation (1) is composed exclusively by derivatives of input and output. In State Space Model (an internal representation) the system is considered NON CONTROLLABLE. The transfer function is an external representation. In other hand, such cancelling cannot be applied on ZIR. This response exhibits a second order behavior, depending on two capacitor initial voltages (two initial conditions are required).

Circuit 2

Differential equation: $$ RC_1C_2\frac{dv_o^2}{dt^2}+(C_1+C_2)\frac{dv_o}{dt}=RC_2\frac{di}{dt} + i \tag{3}$$ The complete response through L.T.: $$ Vo(s)= \frac{\left(RC_2s+1\right)I(s)+\left(RC_2s+1 \right)C_1vo_{(0-)}+C_2vc2_{(0-)}}{s\left( RC_1C_2s + C_1+C_2\right )} $$ This time, \$vo_{(0-)} = vc1_{(0-)}\$.

Separating in ZSR (Zero State Response) and ZIR (Zero Input Response):

$$ Vo(s)= \underbrace{\frac{RC_2s+1}{s\left( RC_1C_2s + C_1+C_2\right )}I(s)} +\underbrace{\frac{\left(RC_2s+1 \right)C_1vo_{(0-)}+C_2vc2_{(0-)}}{s\left( RC_1C_2s + C_1+C_2\right )}} \tag{4}$$

Clearly indicating a second order response, too. But note that in the ZSR, there is no common factor to be cancelled. Note: The ZIR response also exhibits a second order behavior, depending on two capacitor initial voltages (two initial conditions are required). More than this: The ZIR responses (no input dependent) for both circuits are equal.

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  • \$\begingroup\$ So if I have to find the order of a system just by mere inspection, what should I do. Writing a differential equation (even for a simple three mesh network) all the time doesn't seem intuitive at all. \$\endgroup\$ – Astatine Sep 3 '18 at 17:15
  • \$\begingroup\$ @Astatine: Please, see my edit. \$\endgroup\$ – Dirceu Rodrigues Jr Sep 3 '18 at 18:57
  • \$\begingroup\$ Thanks for the edit. But I already knew that. I am not arguing over the fact that the order of the system in my question is two instead of one. As clearly evident from the Laplace transform, the order is one. I want to know why this well-known statement that the order of a system is equal to the number of independent initial conditions doesn't work here. \$\endgroup\$ – Astatine Sep 3 '18 at 20:00
  • \$\begingroup\$ @Astatine: Please, see my edit 2. \$\endgroup\$ – Dirceu Rodrigues Jr Sep 3 '18 at 20:20
  • \$\begingroup\$ Thanks for revising my concepts. I am aware of all these points. Could you please take a look at my EDIT 1. \$\endgroup\$ – Astatine Sep 3 '18 at 21:12
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Two or more capacitors in series or in parallel are equivalent to one capacitor, so there's one initial condition and the order of the system is one. It doesn't really matter that you can charge the series capacitors with different voltages.

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  • \$\begingroup\$ Are you acquainted with the statement that order of a system is equal to the number of independent initial conditions? \$\endgroup\$ – Astatine Sep 3 '18 at 17:37

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