General SEcond Order Circuit: 2 Res. - 2 Inductors - Current Source

In summary, the given circuit can be analyzed by using state variables and a matrix equation to solve for the currents I_1 and I_2. The equations for I_1 and I_2 can be obtained by eliminating I_3 and V_1 using the equations for the voltage drop across resistor R_1 and the inductor L_1. This results in two coupled first-order differential equations in terms of the state variables I_1 and I_2.
  • #1
VinnyCee
489
0

Homework Statement



Obtain [itex]i_1[/itex] and [itex]i_2[/itex] for t > 0 in the circuit below.

http://img258.imageshack.us/img258/7765/problem60as1.jpg [Broken]



Homework Equations



[tex]V_L\,=\,\frac{di_L}{dt}[/tex]



The Attempt at a Solution



To get initial conditions, I made a second circuit diagram for t < 0.

http://img219.imageshack.us/img219/8606/problem60part2ch2.jpg [Broken]

Since there is no current before t = 0, both initial currents and voltages are zero.

[tex]i_1(0^-)\,=\,i_1(0^+)\,=\,0\,A[/tex]

[tex]i_2(0^-)\,=\,i_2(0^+)\,=\,0\,A[/tex]

[tex]V_1(0^-)\,=\,V_2(0^-)\,=\,0\,V[/tex]


I also made a third circuit diagram for t > 0.

http://img254.imageshack.us/img254/2797/problem60part3ii5.jpg [Broken]

[tex]i_1\,=\,\frac{di_1}{dt}[/tex]

[tex]i_2\,=\,\frac{V_1\,-\,V_2}{3\Omega}[/tex]

[tex]i_3\,=\,\frac{0\,-\,V_1}{2\Omega}[/tex]


KCL @ [itex]V_1[/itex]:

[tex]4\,A\,+\,i_3\,=\,i_1\,+\,i_2[/tex]

[tex]4\,-\,\frac{V_1}{2}\,=\,\frac{di_1}{dt}\,+\,\frac{V_1\,-\,V_2}{3}[/tex]


KCL @ [itex]V_2[/itex]:

[tex]\frac{V_1\,-\,V_2}{3}\,=\,\frac{di_2}{dt}[/tex]



Here I am stuck, I know that I need to produce a second order differential equation before I can even think about solving this circuit, but I am having trouble finding what equations to use to get such an O.D.E. Please help!
 
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  • #2
Hmm, you have some errors in your analysis. First off, you'd only get a second order equation if you also had capacitance. This circuit will have only first oder DE's. Second, you need to re-look at your equations. For instance, i1=di1/dt makes no sense (I assume it's a typo?).

Finally a heads up that, you drew the arrow for i3 backwards to the current that will flow through that resistor. It won't matter, just expect a minus sign for i3.
 
  • #3
Lemme try again!

http://img164.imageshack.us/img164/6674/problem60part3kw4.jpg [Broken]

[tex]V_1\,=\,\frac{di_1}{dt}[/tex]

[tex]V_2\,=\,\frac{di_2}{dt}[/tex]

[tex]i_2\,=\,\frac{V_1\,-\,V_2}{3\Omega}[/tex]

[tex]i_3\,=\,\frac{V_1\,-\,0}{2\Omega}[/tex]KCL @ [itex]V_1[/itex]:

[tex]4\,A\,=\,i_1\,+\,i_2+\,i_3[/tex]

[tex]4\,=\,i_1\,+\,\frac{V_1\,-\,V_2}{3\Omega}\,+\,\frac{V_1}{2\Omega}[/tex]

[tex]24\,=\,6\,i_1\,+\,2\,V_1\,-\,2\,V_2\,+\,3\,V_1[/tex]

[tex]24\,=\,6\,i_1\,+\,5\,V_1\,-\,2\,V_2[/tex]

Substituting for [itex]V_1[/itex] and [itex]V_2[/itex]:

[tex]24\,=\,6\,i_1\,+\,5\,\frac{di_1}{dt}\,-\,2\,\frac{di_2}{dt}[/tex]

Now what? I have that pesky [itex]i_1[/itex] that I don't know what to do with!
 
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  • #4
The following may be a good starter.

Regards,

Nacer.

http://islam.moved.in/tmp/dd.jpg [Broken]
 
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  • #5
How do you get the last part of your last equation?

[tex]\frac{di_1}{dt}\,=\,3\,\left(4\,-\,i_1\,-\,\frac{1}{2}\,\frac{di_1}{dt}\right)\,+\,\left(-\frac{di_1}{dt}\,-\,\frac{1}{2}\,\frac{d^2\,i_1}{dt^2}\right)[/tex]

How do you get the last squared term?

[tex]\frac{di_2}{dt}\,=\,-\frac{di_1}{dt}\,-\,\frac{1}{2}\,\frac{d^2\,i_1}{dt^2}[/tex]
 
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  • #6
For such a simple circuit, the math gets pretty messy! Here's a solution using state variables. The first eqn is
[tex]I_0=I_1+I_2+I_3[/tex]
where I_0(t) is the source current (0 for t<=0, i0 later). Since we’re interested in I1 and I2 we'll eliminate I_3 and V_1 from the equations using

[tex]I_3=I_0-I_1-I_2,[/tex]

[tex]V_1=I_3 R_3=L_2 \dot{I_2}=I_1 R_1+L_1 \dot{I_1} .[/tex]

I've used symbols instead of actual values for the R's and L's in the branches so you can follow what's what.
We get

[tex]\dot{I_1}=\frac{R_3 I_0-R_3 I_2 -(R_1+R_3) I_1}{L_1},[/tex]

[tex]\dot{I_2}=\frac{R_3(I_0-I_1-I_2)}{L_2} .[/tex]

These are two coupled first order DE's in the 4 state variables (Idot and I for subscripts 1 and 2). We know the solutions will be of the form
[tex]I=a e^{-t/\tau}+b[/tex]
for t>0 so we can substitute this form and solve the resulting 2 simultaneous algebraic equations.

To keep track of the multitude of coefficients it's easiest to put this into matrix form. It's no different then the above, but it’s easier to write the matrix “A” instead of writing a bunch of R's and L's on each line. Define the matrix

[tex]A=-\left(\begin{array}{cc}(R_1+R_3)/L_1 & R_3/L_1\\R_3/L_2 & R_3/L_2\end{array}\right),[/tex]

whose terms have dimensions 1/second, and column vectors

[tex]b=\left(\begin{array}{cc}R_3/L_1 \\ R_3/L_2\end{array}\right)[/tex]

and
[tex]x=\left(\begin{array}{cc}I_1 \\ I_2\end{array}\right),[/tex]

then
[tex]\dot{x}=Ax+bI_0 .[/tex]

Try the solution
[tex]x=e^{At}z[/tex]
where the meaning of the exponential is a matrix where each term in A is multiplied by t and the product is exponentiated. This gives

[tex]\dot{z}=[e^{At}]^{-1} bI_0= e^{-At}bI_0 .[/tex]

I won’t fill in all the steps, it’s straightforward to integrate this, use the initial condition x(0)=(0;0), and get for currents I_1 and I_2

[tex]x=-i_0 A^{-1} (1-e^{At}) b.[/tex]

Terms in A are negative and have the form of 1/tau where tau is a classic L/R time constant, so the currents start out at zero and grow exponentially, reaching a steady state DC value after a time long compared to the time constants.

EDIT: clean up the equations (as usual...)
EDIT 2: remove erroneous dot from x=exp(At)z
EDIT 3: N.B.: I defined I_3 in same direction as other currents, according to second diagram...
 
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  • #7
I still don't understand, how do you get the last term in this EQ:

[tex]V_1=I_3 R_3=L_2 \dot{I_2}=I_1 R_1+L_1 \dot{I_1}[/tex]

And what were your calculationis to get [itex]\dot{I_1}[/itex] and [itex]\dot{I_2}[/itex] after that?
 
  • #8
VinnyCee said:
I still don't understand, how do you get the last term in this EQ:

[tex]V_1=I_3 R_3=L_2 \dot{I_2}=I_1 R_1+L_1 \dot{I_1}[/tex]

And what were your calculationis to get [itex]\dot{I_1}[/itex] and [itex]\dot{I_2}[/itex] after that?
First I see I swapped currents--my equations correspond if you swap I1 and I2 in your diagram. Sorry for that.

As for your questions:
1. V1 is the sum of the voltage drop across resistor R1 (I1*R1) plus that across the inductor ([tex]L_1 dI_1/dt[/tex], which I labelled [tex]L_1\dot{I_1}).[/tex]

2. From the V1 equation,
[tex]\dot{I_1}=\frac{I_3 R_3-I_1 R_1}{L_1}.[/tex]
Rearrange
[tex]I_3=I_0-I_1-I_2[/tex] Eq. (1)
to give
[tex]I_3=I_0-I_1-I_2[/tex] Eq. (2)
and substitute into the equation above to eliminate I3. This gives [tex]\dot{I_1}[/tex] in terms of [tex]I_1[/tex] and [tex]I_2[/tex] as in the post.

Now do the same for I2 from the V1 equation. Get
[tex]\dot{I_2}=\frac{I_3 R_3}{L_2}[/tex]
and again substitute Eq. (2) to eliminate I3.

Choosing which state variables to work with are a key to making the problem easier. You could work with three coupled equations (I1, I2 and I3) but they contain redundant information as expressed by Eq. (1) so something will eventually drop out. I chose to express [tex]\dot{I_1}[/tex] and [tex]\dot{I_2}[/tex] in terms of [tex]I_3[/tex], which is then eliminated early, leaving two coupled equations expressing [tex]\dot{I_1}[/tex] and [tex]\dot{I_2}[/tex] in terms of [tex]I_1[/tex] and [tex]I_2[/tex].

EDIT: equation cleanup
 
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  • #9
marcusl said:
Hmm, you have some errors in your analysis. First off, you'd only get a second order equation if you also had capacitance. This circuit will have only first oder DE's. .

No, if you have two energy storing elements that cannot be associated in series or in parallel, you will have a second order system.
 
  • #10
Oh! Well, I'm always interested in learning. Can't you consider these two elements to be in parallel? Is my solution incorrect?
 
  • #11
marcusl said:
Oh! Well, I'm always interested in learning. Can't you consider these two elements to be in parallel? Is my solution incorrect?
They are not in parallel since there is a 3 ohm resistor between them.
 
  • #12
Ok, if I do this with Laplace transforms I get a different answer. The current source [tex]I(s)= i_0/s[/tex] looks into an impedance Z(s)

[tex]\frac{1}{Z(s)}=\frac{1}{R_3}+\frac{1}{sL_2}+\frac{1}{R_1+sL_1}[/tex] or

[tex]Z(s)=\frac{sR_3 L_2(R_1+sL_1)}{sL_2(R_1+sL_1)+R_3(R_1+sL_1)+sR_3 L_2}.[/tex]

The voltage is [tex]V_1 (s)=Z(s)I(s)[/tex] which works out to

[tex]V_1 (s)=\frac{i_0 R_3 L_2(R_1+sL_1)}{s^2 L_1 L_2+s(R_1 L_2+R_3 L_1+R_3 L_2)+R_1 R_3}.[/tex]

This could be a damped oscillation, having both sine and cosine terms, for certain choices of components. With the present values it still looks non-oscillatory.

So the question is: which derivation is correct, and why?
 
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  • #13
marcusl said:
Ok, if I do this with Laplace transforms I get a different answer. The current source [tex]I(s)= i_0/s[/tex] looks into an impedance Z(s)

[tex]\frac{1}{Z(s)}=\frac{1}{R_3}+\frac{1}{sL_2}+\frac{1}{R_1+sL_1}[/tex] or

[tex]Z(s)=\frac{sR_3 L_2(R_1+sL_1)}{sL_2(R_1+sL_1)+R_3(R_1+sL_1)+sR_3 L_2}.[/tex]

The voltage is [tex]V_1 (s)=Z(s)I(s)[/tex] which works out to

[tex]V_1 (s)=\frac{i_0 R_3 L_2(R_1+sL_1)}{s^2 L_1 L_2+s(R_1 L_2+R_3 L_1+R_3 L_2)+R_1 R_3}.[/tex]

This could be a damped oscillation, having both sine and cosine terms, for certain choices of components. With the present values it still looks non-oscillatory.

So the question is: which derivation is correct, and why?

The signs of the elements of your matrix A in the state space are wrong. If you correct them, the eigenvalues of the matrix will be identical to the poles of your function in the s plane.
Both solutions are correct, giving real natural frequencies. This means that your circuit is overdamped.
 
  • #14
A new circuit diagram using the s-domain:

http://img338.imageshack.us/img338/8682/problem60part4vk9.jpg [Broken]

KCL @ [itex]V_1[/itex]:

[tex]4\,=\,i_1\,+\,i_2\,+\,i_3[/tex]

[tex]4\,=\,\frac{V_1}{2}\,+\,\frac{V_1}{j\omega}\,+\,\frac{V_1}{j\omega\,+\,3}[/tex]

[tex]8j\omega(j\omega\,+\,3)\,=\,V_1j\omega(j\omega\,+\,3)\,+\,2V_1(j\omega\,+\,3)\,+\,2V_1j\omega[/tex]

[tex]V_1\,=\,\frac{8j\omega(j\omega\,+\,3)}{j\omega(j\omega\,+\,3)\,+\,2(j\omega\,+\,3)\,+\,2j\omega}[/tex]

Now what?
 
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  • #15
[tex]i_2\,=\,\frac{8\,V}{j\omega}[/tex] <----- Is that right?
 
  • #16
SGT said:
The signs of the elements of your matrix A in the state space are wrong. If you correct them, the eigenvalues of the matrix will be identical to the poles of your function in the s plane.
Both solutions are correct, giving real natural frequencies. This means that your circuit is overdamped.
Hmm, I don't see it. The sign seems right to me, and the solution oscillates only if A is complex.
VinnyCee said:
A new circuit diagram using the s-domain:

http://img338.imageshack.us/img338/8682/problem60part4vk9.jpg [Broken]

KCL @ [itex]V_1[/itex]:

[tex]4\,=\,i_1\,+\,i_2\,+\,i_3[/tex]

[tex]4\,=\,\frac{V_1}{2}\,+\,\frac{V_1}{j\omega}\,+\,\frac{V_1}{j\omega\,+\,3}[/tex]

[tex]8j\omega(j\omega\,+\,3)\,=\,V_1j\omega(j\omega\,+\,3)\,+\,2V_1(j\omega\,+\,3)\,+\,2V_1j\omega[/tex]

[tex]V_1\,=\,\frac{8j\omega(j\omega\,+\,3)}{j\omega(j\omega\,+\,3)\,+\,2(j\omega\,+\,3)\,+\,2j\omega}[/tex]

Now what?
It's more conventional to replace jomega with s. You need to also transform your source. For a step function I0(s)=4/s, whcih will cancel the leading factor of s in your expression for V1(s).

Now it's just a matter taking the reverse transform of your expression using the Laplace transform table of your choice, for instance, entry 2.27 from
http://www.vibrationdata.com/Laplace.htm" [Broken]
 
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  • #17
marcusl said:
Hmm, I don't see it. The sign seems right to me, and the solution oscillates only if A is complex.

I didn´t see the minus sign before the matrix. To find the eigenvalues, that are the natural frequencies of the circuit you must calculate
[tex]|\lambda I - A| = 0[/tex]
In your case:

[tex]\left|\begin{array}{cc}|lambda + 5 & 3\\3 & \lambda + 3\end{array}\right|, =, 0[/tex]

This gives [tex]\lambda^2 + 8\lambda + 6 = 0[/tex] which has real roots. So, no oscillation.
 
  • #18
Of course, I should have calculated the eigenvalues of A sooner. The state variable and s-domain methods give the same result as expected.

Thx SGT.
 

What is a second order circuit?

A second order circuit is an electrical circuit that contains two energy storage elements, such as capacitors or inductors. These elements allow the circuit to exhibit more complex behavior and response compared to a first order circuit, which only contains one energy storage element.

What is the difference between a resistor and an inductor?

A resistor is a passive component that resists the flow of electric current, while an inductor is an active component that stores energy in the form of a magnetic field. This means that a resistor dissipates energy, while an inductor can store and release energy.

How do you calculate the total resistance of a circuit?

The total resistance of a circuit can be calculated by adding the individual resistances of each component in series, or by using the parallel resistance formula for components in parallel. In a second order circuit with two resistors, the total resistance would be the sum of the two resistances.

What is the role of a current source in a circuit?

A current source is a component that maintains a constant current through a circuit, regardless of the voltage across it. In a second order circuit, a current source can provide a constant current to the inductors, which can then store and release energy as needed.

How does a second order circuit respond to changes in input?

A second order circuit can exhibit more complex behavior compared to a first order circuit, meaning it can respond to changes in input in a variety of ways. This can include oscillations, resonance, and other dynamic responses, depending on the specific circuit components and their values.

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