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How to find the time constant?

Discussion in 'Homework Help' started by Heidi, Sep 27, 2013.

  1. Ratchit

    Ratchit Well-Known Member

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    It means that there are two storage elements could have different energy storage values. What does the TC of a system mean, anyway. The TC refers to the storage component, not the system. Yes, series/parallel resistors/caps can be combined if the resistors have the same currents and the caps are supplied with the same voltage. But, that is not the case in this circuit.

    Ratch
     
  2. steveB

    steveB Well-Known Member Most Helpful Member

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    OK, that's fine. I can see your logic there. In this case the two caps are exchanging energy between each other. That's the dynamics of what is happening and one time constant value describes the interchange. If you want to say that means two equal times constants, I can't really argue against that. But, the dynamic system is a first order with one relevant time constant. This means you can find an equivalent system with one energy storage component, which I believe is what MikeMl did above. Still, you may prefer to say the real implementation has two time constants. Again, I don't see a lot of usefulness or meaning to it, but you do and perhaps others would too.
     
  3. MrAl

    MrAl Well-Known Member Most Helpful Member

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    Hi,

    We really need more input from the questioner for this. Questions like this usually come from a book that has been assuming lots of other stuff all along and so it is not stated repeatedly for each individual problem.

    If we assume small source series resistance Rs like 1uOhm it will be very difficult to see the response assuming R1 and R2 are much higher. So if we choose Rs=1 Ohm that might be a little better.

    The two basic mechanisms here are:
    1. The Rs and C1 and C2 time constant during start up.
    2. The C1 and R1 and C2 and R2 time constants some time just after start up.

    Rs being small means the initial time response is dominated by Rs*C where C is the parallel combo of C1 and C2. That simply means we'll see a fast rising exponential at the junction of C1 and C2. The resistors R1 and R2 wont play much of a role at this point in time because C1 and C2 set the short time voltage level at the junction.
    So the voltage output starts out at 0 and stops rising after maybe 5*Rs*C. At that time it will reach a voltage of C1/(C1+C2) times the input voltage.

    Next, the other time constants come into play if the resistor values do not match the capacitor values properly. If the ratio of the two resistors match the inverse of the ratio of the two capacitors then nothing else happens. If they are mismatched however, then we see the second time constant start to take effect.
    If the ratio of the two resistors is such that the junction without caps would force a higher voltage than with the caps, then the voltage will rise. If the ratio is such that a lower voltage would be pressent, then the voltage falls.

    For example, if R1=4k and R2=1k and C1=1uf and C2=1uf, the voltage would rise exponentially due to Rs*C and reach the max level of approximately Vin/2, then start to fall slowly due to exponential R1 R2 and C1 C2 time constants until it reached Vin/5. So with 1v input, it would rise very quickly (but not instantaneously) to 0.5v, then slowly back down to 0.2 volts.

    However, with R1=4k, R2=1k, and C1=1uf and C2=4uf, we'd see a fast rise to 0.2 volts and it would stay there. This is how a "fast" bias network is built up where we need a reference voltage that is filtered but it has to be able to rise quickly for starting the circuit. The capacitor values have to be oppositely matched to the resistor values.

    With the inclusion of Rs, the question of the time constants starts to play a real role in the circuit behavior. When we compare time constants that tells us a little about the basic operation of the circuit already. For example, with very very small Rs we have a 'stiff' differential equation that could be more difficult to solve numerically but with Rs more comparable to the other resistors we have a much more relaxed situation. We like to know the comparative difference sometimes and that comes from knowing the time constants:
    Rs*C
    R1*C1
    R2*C2
     
    Last edited: Sep 28, 2013
  4. dave

    Dave New Member

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  5. steveB

    steveB Well-Known Member Most Helpful Member

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    I worked out the state space system, which provides some interesting information.

    Let the state vector be [V1; V2] and the input vector be [Vi]. The A and B matrix elements are as follows.

    A11=-1/C1/(Rs||R1)
    A12=-1/C1/Rs
    A21=-1/C2/Rs
    A22=-1/C2/(Rs||R2)

    B11=1/Rs/C1
    B21=1/Rs/C2

    Finding the eigenvalues of the A matrix will give the poles, but doing this algebraically is too complex. However, if we allow the capacitor values to be equal and the resistor values to be equal, we can simplify the math and we get.

    A11=-1/C/(Rs||R)
    A12=-1/C/Rs
    A21=-1/C/Rs
    A22=-1/C/(Rs||R)

    which has eigenvalues of -1/R/C and -(2R+Rs)/R/C/Rs

    Now, to find out what happens when the source becomes ideal, take the limit as Rs goes to zero. The eigenvalues then become -1/R/C and -infinity. This translates to time constants of RC and zero. This corresponds to what MrAl was saying early on. The ideal voltage source creates a time constant of zero effectively. This is why I don't think it is good to say that there are two equal time constants in the circuit when the voltage source is ideal. The impractical ideal voltage source constrains the circuit and kills one of the degrees of freedom. This is very different from a circuit that truly had two equal time constants. Such a system would be a critically damped second order system, but that's not what we have here.

    So before when I said that I can't argue against the idea of two equal time constants, I was speaking to soon. I now can argue against it. The argument may not be accepted, but I do think it makes a reasonable argument to put forward. Time constants should be linked to system poles, which translates back to degrees of freedom in the system. Here, we need to say that there is one degree of freedom and hence one time constant, or if we want to allow the second time constant, it should be zero and not equal to the main system time constant.
     
    Last edited: Oct 1, 2013
  6. Ratchit

    Ratchit Well-Known Member

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    steveB,

    What does a zero time constant mean? It means either the cap is zero, or the resistor it interacts with is zero, or both are zero. And yet the circuit shows two physical caps and two resistors all unshorted. I did an analysis the the voltage across one capacitor assuming a 1 volt source with no resistance. It showed two exponential terms with equal time constant values. The TC's of a linear circuit should be the same during a de-energizing cycle because the voltage source does not determine the TC's. The cap value and the resistance it interacts with determines the TC. So I don't see the TC values changing no matter what the voltage is.

    Ratch
     
  7. MrAl

    MrAl Well-Known Member Most Helpful Member

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    Hi,

    Well the problem isnt so much what the VOLTAGE would be across one of the caps, it's what the CURRENT will be through the caps. We must look at both V and I to see what is wrong when we dont have a series resistor for the voltage source. With zero Ohms for the voltage source series resistance we can say superficially that we have two parallel RC circuits, but the circuit is not sound so we cant be sure if we have the right answer or not yet.

    The voltage across the bottom cap for example comes from:
    vC2(s)=((s*C1*R1+1)*R2)/(s*C2*R1*R2+s*C1*R1*R2+R2+R1)

    and with a step change on the input (a DC voltage source) we have:
    vC2(t)=R2/(R1+R2)-((C2*R2-C1*R1))/((C1+C2)*(R1+R2))*e^(-at)
    where a=(R1+R2)/((C2+C1)*R1*R2)

    and with an input voltage of 1 volt and R1=R2 and C1=C2, vC2(t)=0.5 volts which is a constant.

    But when we look next at the current we see:
    I(s)=((s*C1*R1+1)*(s*C2*R2+1))/(s*C2*R1*R2+s*C1*R1*R2+R1+R2)

    and for a step change input if we solve for the initial value I(0) we get:
    I(t)=infinity

    so we need not go any farther :)

    If the input voltage was specified as a ramp or sine or something other than DC then maybe we can go farther without question because those type of sources will not cause an infinite current unless they start out non zero at t=0. So there we could do it if we restrict the input to sources which start out at zero volts and rise slowly rather than as a step change. For a ramp which starts out at zero at t=0 for example we'd get a current I(t) response similar to A+B*e^-at+C*t or something like that which is not infinite at t=0 although it may be at t=infinity.
     
    Last edited: Sep 28, 2013
  8. steveB

    steveB Well-Known Member Most Helpful Member

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    That's a good question, but I don't agree with your answer. My preference is to assign time constants to poles of the system, not to energy storage components. For me, this is more meaningful and useful.

    But, what does the zero value mean? To me, it means that the pole has moved to infinity, and the circuit has degenerated to having one less degree of freedom. The zero time constant will not reveal itself in the analysis when only one differential equation is specified. The A matrix will be a 1X1 matrix with only one eigenvalue. As long as Rs is not zero, we have two different time constants, but when Rs is zero, the system has one time constant.

    At least that's my interpretation of what the circuit analysis says.

    So, how do you interpret what the pole values are saying? The analysis says that one time constant is the value you would expect, and the other one depends on Rs. As Rs gets smaller and smaller, the other time constant also gets smaller and smaller. At this point, there is no question of what the small TC value means. It is just related to a fast pole in the system. But, by your point of view, that TC, which is approaching zero as Rs approaches zero, will suddenly jump to equal the value of the other TC once Rs gets to zero.

    I'm confused by what you say here. How can one capacitor decay with two exponential terms with equal time constant values? If you looked on a Oscope, what would you see that is different than a decay with one exponential term with one time constant value?

    I agree. That's why it is called a time "constant". Constant means that it does not change.
     
  9. Heidi

    Heidi Member

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    I have a few stupid questions.

    Does the 's' in the impedance formula represent a complex number?

    I only know that impedance is the ratio of voltage phasor and current phasor, and a phasor is only related to sinusoidal functions. If a circuit consisting of resistors and capacitors is driven by a DC voltage, does the DC source see an impedance?

    It seems to be a seperate problem which can be found in example 4 at http://faculty.kfupm.edu.sa/EE/husseina/081/STC_Bode_Plot.pdf
     
  10. Ratchit

    Ratchit Well-Known Member

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    steveB,

    Who cares about the poles of the circuit? We are looking for the TC of each storage element, not the circuit response. Anyway, I calculated the voltage across the caps with the voltage source value set to zero. The gamma term represents the initial voltage across each cap. Notice that when the product of the voltage and capacitance of each cap are equal, the TC's cancel out and the voltage is zero. Notice also a TC attached to each of the voltage-capacitance product terms.

    Heidi1.JPG

    Ratch
     
  11. Ratchit

    Ratchit Well-Known Member

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    s = sigma + j*omega, where sigma and omega are real numbers

    If you see the impedance of a circuit or component represented by a expression containing "s" terms, you can get its sinusoidal impedance by simply substituting "j*omega" for each "s" in the expression.

    The DC voltage sees everything that an AC voltage sees if attached to the same point of the circuit. The components of the circuit do not change just because the voltage source does. Impedance is only defined with respect to sinusoidal stimulation.

    Ratch
     
  12. steveB

    steveB Well-Known Member Most Helpful Member

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    I care about poles in a circuit. Poles tell us so much and my point is that time constants are circuit related values, not component related. When you specify a time constant of a simple RC or RL circuit, what are you really talking about? It's the inverse of the pole frequency. A time constant (or pole in more general cases) is associated with a particular degree of freedom in the system (EDIT: well kind of, - strictly, poles can mix among the variables and you get into partitioning analysis which is far too advanced for this discussion).

    All you are doing is taking the one system pole, which relates to a time constant, and assigning it to both capacitors. As long as one understands that's what you are doing, then there is no harm done in using that viewpoint. But, as I pointed out mathematically, if you introduce a nonzero Rs, you will get another time constant in the system. It will be a much faster TC for a good voltage source, but it is a real time constant that relates to the responses of system variables (or degrees of freedom). So what do you say in that case? Are there now 3 time constants? Or, do you just ignore the fast TC and say it is not relevant for the two capacitors? I think you create a lot of confusion and inconsistency with these viewpoints.

    It is interesting to note that if you try to assign two equal time constants to one variable, it doesn't work. Consider v(t)=A exp(-t/T) +B exp(-t/T). This is equal to v(t)=(A+B) exp(-t/T). So, your equation above does not show anything real or distinct from the viewpoint of having one time constant.

    In general, this TC approach does not help because once system poles come off the real axis, you have complex or imaginary poles and the dynamics is more complicated than what a time constant can tell you. You will have oscillatory effects, not just damping effects. This is why people, in addition to myself, also care about poles.
     
    Last edited: Sep 29, 2013
  13. Ratchit

    Ratchit Well-Known Member

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    steveB,

    "I care about poles in a circuit. Poles tell us so much and my point is that time constants are circuit related values, not component related."

    Circuits and TC's have a relationship, but I don't believe the whole circuit has a time constant. Each individual storage component does.

    "When you specify a time constant of a simple RC or RL circuit, what are you really talking about? It's the inverse of the pole frequency."

    Perhaps for a simple RC/Rl circuit.,

    "All you are doing is taking the one system pole, which relates to a time constant, and assigning it to both capacitors. As long as one understands that's what you are doing, then there is no harm done in using that viewpoint. But, as I pointed out mathematically, if you introduce a nonzero Rs, you will get another time constant in the system. It will be a much faster TC for a good voltage source, but it is a real time constant that relates to the responses of system variables (or degrees of freedom). So what do you say in that case? Are there now 3 time constants? Or, do you just ignore the fast TC and say it is not relevant for the two capacitors? I think you create a lot of confusion and inconsistency with these viewpoints."

    No, it is one time constant for each storage element that happens to have the same value for both. Since the circuit has a symetrical topological layout, introducing a third voltage source resistance will change the value of the TC's for both, but they should still have the same values. Why would there be three TC's when there are only two storage elements?

    "It is interesting to note that if you try to assign two equal time constants to one variable, it doesn't work. Consider v(t)=A exp(-t/T) +B exp(-t/T). This is equal to v(t)=(A+B) exp(-t/T). So, your equation above does not show anything real or distinct from the viewpoint of having one time constant."

    Yes it does. "A" represents the contribution of one storage element, and "B" represents the other storage element. Algebraically it shows that a TC is assigned to each storage element. Only when you assign values to A and B can they be put together as one term.

    "In general, this TC approach does not help because once system poles come off the real axis, you have complex or imaginary poles and the dynamics is more complicated than what a time constant can tell you. You will have oscillatory effects, not just damping effects. This is why people, in addition to myself, also care about poles."

    As I said before, we are not trying to analyze the circuit's response. We are trying to find the TC of the storage elements. Since there is no inductance or feedback, there will be no oscillations in this circuit. I don't think poles are appropriate for what we are trying to do here.

    Ratch
     
  14. MrAl

    MrAl Well-Known Member Most Helpful Member

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    Hi again,

    A time constant has dimensions t and appears in the exponential as in 1/t. So really anything we can express that way we might be able to call a time constant. Once we multiply that 1/t by t we get a constant.

    Looking over Heidi's linked paper i see that they make a habit of making the assumption where we are to ignore the infinite current during startup with a DC source, and just approach the problem from a more simplistic viewpoint where we just short out the DC source and go from there. See problem #9 for another such example.

    But as to the time constant(s), sometimes we view a circuit as having two time constants even when it only has one true exponential. That's because sometimes we want to make a rough comparison of the "two" time constants as for example when we want to try to figure out which is the "dominant" time constant. In that case we would be seeking to try to eliminate one of the "two" time constants so that we could simplify the analysis. So we might have an exponential part that looks like this:
    e^(-t*(1/T1+1/T2))

    where T1=R1*C1 and T2=R2*C2.

    Clearly we can combine the "two" time constants and obtain:
    e^(-t*(1/T3))

    where T3 is equal to T1*T2/(T1+T2).

    So we can call this either one time constant or two time constants, depending on what we are seeking to understand about the circuit. It's also obvious that if T2 was much larger than T1 then T1 would dominate and we would end up with:
    e^(-t/T1)

    which would make an approximate analysis much simpler.

    In the strictest sense of the definition there's only one time constant but that view is often too restrictive when there is an obvious significance to the view with more than one time constant. For this circuit for example we might want to understand what we have when R1C1>>R2C2, and although the response itself only has one exponential it will certainly be of significance to think of having two time constants, especially when they are very much different from each other numerically. Couldnt we also for some reason want to look at the time constants R2C1 and R1C2? If we find significance in this calculation (and i think we can) then we will want to think of them as time constants as well.

    So it all depends on the viewpoint. For the original problem however it appears from the linked text that they want a simple combination that results in only one time constant, and that comes from shorting the power supply source.
     
    Last edited: Sep 29, 2013
  15. steveB

    steveB Well-Known Member Most Helpful Member

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    OK, quite right that there should not be three time constants. I'm glad we agree on that point. So, for the case of nonzero source resistance Rs, we both agree that there are two time constants.

    Now, you are saying that the two time constants are equal and one is assigned to one capacitor and the other is assigned to the other capacitor. My position is that this does not hold up in analysis. Above, I showed that there are two different values for the time constants, as follows, where I'm assuming the R1=R2=R and the C1=C2=C, with source resistance Rs.

    T1=R C and T2=R C Rs /(2R+Rs)

    Again, these come from the pole values. Now, why are these two values important? It's because the time response for both V1(t) and V2(t), which are the capacitor voltages, will have responses with these two time constants for exponential decay terms. Let's not even look at the forced response of the system, and just look at the free response from initial voltages on the caps V1(0) and V2(0).

    V1(t)=(V1(0)+V2(0)) exp(-t/T2) / 2 + (V1(0)-V2(0)) exp(-t/T1) / 2
    V2(t)=(V1(0)+V2(0)) exp(-t/T2) / 2 - (V1(0)-V2(0)) exp(-t/T1) / 2

    So, again I stress that time constants here are associated with the circuit responses to the state variables, which I've chosen to be V1 and V2, although you could define different state variable is you wanted. If you want to assign time constants to the capacitors, presumably because these voltages are the capacitor voltages, then you need to assign both time constants to both capacitors, and the two time constants are different values, not the same values as you stated. So by your logic, it's not 2 or 3 time constants, but now it's 4 because we assign 2 values to 2 capacitors. Sorry, but it just doesn't work the way you explained it.
     
  16. steveB

    steveB Well-Known Member Most Helpful Member

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    Heidi,

    I hope you will have noticed that the title of this document is "Single Time Constant Circuits". So Ratch's claim that the circuit has two time constants (one for each capacitor) is in stark contrast to what the document is saying. I find this document you referenced sound, and do not agree with Ratch's perspective on this.
     
  17. MikeMl

    MikeMl Well-Known Member Most Helpful Member

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    TC.jpg
     
  18. Ratchit

    Ratchit Well-Known Member

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    steveB,

    "So, again I stress that time constants here are associated with the circuit responses to the state variables, which I've chosen to be V1 and V2, although you could define different state variable is you wanted. If you want to assign time constants to the capacitors, presumably because these voltages are the capacitor voltages, then you need to assign both time constants to both capacitors, and the two time constants are different values, not the same values as you stated. So by your logic, it's not 2 or 3 time constants, but now it's 4 because we assign 2 values to 2 capacitors. Sorry, but it just doesn't work the way you explained it."

    Well, I calculated the voltages across each capacitor using Laplace transforms. It came out with each storage component having the same TC. Not surprising since it is a symetrical topology. One single TC for each cap. I see two and only two TC's . I don't assign TC's to capacitors. I just calculate what they are from the voltage response to a step input or an immediate cutoff.

    Ratch
     
  19. Ratchit

    Ratchit Well-Known Member

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    MikeML,

    The time constant for each cap is 40us. Your Spice shows only a 10ns pulse. I suggest a 200us pulse to see an exponential effect.

    Ratch
     
    Last edited: Sep 29, 2013
  20. MikeMl

    MikeMl Well-Known Member Most Helpful Member

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    How about 200us? (5 "time-constants") :D

    TCa.jpg
     
    Last edited: Sep 29, 2013
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  21. steveB

    steveB Well-Known Member Most Helpful Member

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    You calculated the TC for the OPs given circuit and you obtained one value for a time constant. You found this value in two different ways. That is fine. However, you then decided on your own to say this means there is two values, one for each capacitor. This decision is not based on any standard practice or established convention and does not make sense in a mathematical analysis, as I showed.

    In the simpler case with Rs=0, There is only one state variable. It is a property of state space systems that all variables can be related back to state variables and input variables. Any variable that depends on the state variable will show the time constant value in its response. If you choose V1 as the state variable, then V2 will just be another variable that depends on V1 and it will show the same time constant. And vice versa, if you choose V2 as the state variable. You could also define a new state that is a linear combination of V1 and V2.

    I showed the more general case with nonzero Rs. Here V1 and V2 respond with 2 time constants. Both state variables show both time constants because I didn't choose the natural state variables that would diagonalize the A matrix. When you do diagonalize the A matrix, each state variable would only show one time constant, and any other system variable that depends on both state variables would show both time constants in the response. In this example V1 and V2 will always be mixed modes that show both time constants.

    Time constants are always traceable back to poles (assuming the poles are on the real axis). Knowing the number of poles is important because it shows the order of a system. It's important to not say that a first order system (such as the OPs example) has 2 time constants because this will trick people into thinking it's a second order system. The more general circuit is a second order system, so there we do want to say there are two time constants, but those are not equal values, even in the case of perfect symmetry with both capacitors equal and both resistors equal.
     
    Last edited: Sep 29, 2013

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