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Homework Statement
https://imageshack.us/scaled/large/708/capturehdx.png
Homework Equations
https://imageshack.us/scaled/large/845/capturevtq.png
V = IR
https://imageshack.us/scaled/large/18/captureid.png
[itex]I_{C}(t) = C\frac{dV_{x}(t)}{dt}[/itex]
https://imageshack.us/scaled/large/818/captureplr.png
[itex]V_{x}(t) = L\frac{di_{L}}{dt}[/itex]
The Attempt at a Solution
I'll start by solving for the equation of the node voltage [itex]V_{x}(t)[/itex]
[itex]\frac{V_{s} - V_{x}(t)}{R} = C_{1}\frac{d[V_{x}(t) - V_{s}]}{dt} + C_{2}\frac{d[V_{x} - 0]}{dt} + \frac{1}{L}∫_{0}^{t}[V_{x}(t) - 0]dt + I_{L}(t=0^{+}) [/itex]
I use the fact that [itex]I_{L}(t=0^{-}) = 0 A[/itex] and that because the current through a inductor can't change instantaneously so [itex]I_{L}(t=0^{-}) = I_{L}(t=0^{+}) = 0 A[/itex]. I also simplify some mathematically.
[itex]-\frac{V_{x}(t)}{R} = C_{1}\frac{dV_{x}(t)}{dt} + C_{2}\frac{dV_{x}}{dt} + \frac{1}{L}∫_{0}^{t}V_{x}(t)dt[/itex]
Simplify some more
[itex]-\frac{V_{x}(t)}{R} = (C_{1} + C_{2})\frac{dV_{x}(t)}{dt} + \frac{1}{L}∫_{0}^{t}V_{x}(t)dt[/itex]
Take the derivative with respect to time of both sides
[itex]-\frac{d}{dt}\frac{V_{x}(t)}{R} = \frac{d[(C_{1} + C_{2})\frac{dV_{x}(t)}{dt} + \frac{1}{L}∫_{0}^{t}V_{x}(t)dt]}{dt}[/itex]
Simplify
[itex]-\frac{1}{R}\frac{V_{x}(t)}{dt} = (C_{1} + C_{2})\frac{d^{2}V_{x}(t)}{dt^{2}} + \frac{1}{L}V_{x}(t)[/itex]
Set equal to zero
[itex]0 = (C_{1} + C_{2})\frac{d^{2}V_{x}(t)}{dt^{2}} + \frac{1}{R}\frac{V_{x}(t)}{dt} + \frac{1}{L}V_{x}(t)[/itex]
Replace the constants with constants A, B, and C
[itex]A = C_{1} + C_{2}[/itex]
[itex]B = \frac{1}{R}[/itex]
[itex]C = \frac{1}{L}[/itex]
[itex]0 = A\frac{d^{2}V_{x}(t)}{dt^{2}} + B\frac{V_{x}(t)}{dt} + CV_{x}(t)[/itex]
This is a second order differential equation so it is known that
[itex]V_{x}(t) = Ke^{st}, \frac{dV_{x}(t)}{dt} = Kse^{st}, \frac{d^{2}V_{x}(t)}{dt^{2}} = Ks^{2}e^{st}[/itex]
So plugging this into the equation
[itex]0 = AKs^{2}e^{st} + BKse^{st} + CKe^{st}[/itex]
Divide through by [itex]ke^{st}[/itex]
[itex] 0 = As^{2} + Bs + C[/itex]
This is a quadratic equation so it's known that
[itex]s = \frac{-B ± \sqrt{B^2 - 4AC}}{2A}[/itex]
It can be seen from the equation for s that there are two values for s, I'll call them [itex]s_{1}[/itex] and [itex]s_{2}[/itex]. So my voltage equation becomes
[itex]V_{x}(t) = A_{1}e^{s_{1}t} + A_{2}e^{s_{2}t}[/itex]
In order to solve for [itex]A_{1}[/itex] and [itex]A_{2}[/itex] I need to know the initial condition of the node voltage.
At time [itex]0^{-}[/itex] the circuit looks like the one below
https://imageshack.us/scaled/large/833/capturetiru.png
[itex]\sum Q = 0 = 4 uF(V_{x}(t = 0^{-}) - 10) + 1 uF(V_{x}(t = 0^{-}) - 0)[/itex]
Divide through by uF
[itex]0 = 4V_{x}(t = 0^{-}) - 4(10) + V_{x}(t = 0^{-})[/itex]
simplify
[itex]40 = 5V_{x}(t = 0^{-})[/itex]
so
[itex]V_{x}(t = 0^{-}) = 8 V[/itex]
I use the fact that the voltage across a capacitor can't change instantaneously so [itex]V_{x}(t = 0^{-}) = V_{x}(t = 0^{+})[/itex].
So when you plug in 0 into the voltage equation you get the following equation
[itex]V_{x}(t = 0^{+}) = A_{1} + A_{2}[/itex]
Being that I have two unknowns I need another equation. I use the fact that the voltage through a capacitor can't change instantaneously
So it can be seen that [itex]\frac{dV_{x}(t)}{dt}[/itex] represents the change in voltage across [itex]C_{2}[/itex] which is zero. So I have the following second equation
[itex]0 = A_{1}s_{1} + A_{2}s_{2}[/itex]
So I choose to solve for [itex]A_{1}[/itex] and [itex]A_{2}[/itex] using matrix and putting them in reduce row echelon form
1, 1, [itex]V_{x}(t = 0^{+})[/itex]
[itex]s_{1}, s_{2}, 0[/itex]
I divide row 2 by [itex]s_{1}[/itex]
1, 1, [itex]V_{x}(t = 0^{+})[/itex]
[itex]1, \frac{s_{2}}{s_{2}}, 0[/itex]
I multiply row 1 by negative one and add it to row two to get a new row 2
1, 1, [itex]V_{x}(t = 0^{+})[/itex]
[itex]0, \frac{s_{2}}{s_{2}}-1, -V_{x}(t = 0^{+})[/itex]
I divide row two through by [itex]\frac{1}{\frac{s_{2}}{s_{2}}-1}[/itex]
1, 1, [itex]V_{x}(t = 0^{+})[/itex]
[itex]0, 1, -\frac{V_{x}(t = 0^{+})}{\frac{s_{2}}{s_{2}}-1}[/itex]
I multiply row two by negative one and add to row one to get a new row one
1, 0, [itex]V_{x}(t = 0^{+}) + \frac{V_{x}(t = 0^{+})}{\frac{s_{2}}{s_{2}}-1}[/itex]
[itex]0, 1, -\frac{V_{x}(t = 0^{+})}{\frac{s_{2}}{s_{2}}-1}[/itex]
So I get the following equations, I factored the first one
[itex]A_{1} = V_{x}(t = 0^{+})(1 + \frac{1}{\frac{s_{2}}{s_{2}}-1})[/itex]
[itex]A_{2} = -\frac{V_{x}(t = 0^{+})}{\frac{s_{2}}{s_{2}}-1}[/itex]
The quantity [itex]\frac{s_{1}}{s_{2}}[/itex] seems to be an interesting quantity.
[itex]\frac{s_{1}}{s_{2}} = \frac{-B - \sqrt{B^{2} - 4AC}}{-B + \sqrt{B^2 - 4AC}}[/itex]
I multiply by a form of one [itex]\frac{-B - \sqrt{B^{2} - 4 AC}}{-B - \sqrt{B^{2} - 4 AC}}[/itex] and get the following.
[itex]\frac{s_{1}}{s_{2}} = \frac{(B + \sqrt{B^{2} - 4AC})^{2}}{B^{2} - B^{2} + 4AC} = \frac{B^{2} + 2B\sqrt{B^{2} - 4AC} + B^{2} - 4AC}{4AC} = \frac{2B^{2} + 2B\sqrt{B^{2} - 4AC} - 4AC}{4AC} = \frac{B^{2}}{2AC} + \frac{B}{2AC}\sqrt{B^{2} - 4AC} - 1[/itex]
So then
[itex]\frac{s_{2}}{s_{1}} - 1 = \frac{B}{2AC}\sqrt{B^{2} - 4AC} + \frac{B^{2}}{2AC} - 2[/itex]
Now in my formulas I have the following relation [itex]\frac{1}{\frac{s_{2}}{s_{1}}-1}[/itex] So I solve for this relation.
[itex]\frac{1}{\frac{s_{2}}{s_{1}}-1} = \frac{1}{\frac{B}{2AC}\sqrt{B^{2} - 4AC} +( \frac{B^{2}}{2AC} - 2)}[/itex]
I multiply by a form of one [itex]\frac{-\frac{B}{2AC}\sqrt{B^{2}-4AC} + \frac{B^2}{2AC} - 2}{-\frac{B}{2AC}\sqrt{B^{2}-4AC} + \frac{B^2}{2AC} - 2}[/itex]
[itex]\frac{1}{\frac{s_{2}}{s_{1}}-1} = \frac{-\frac{B}{2AC}\sqrt{B^{2}-4AC} + \frac{B^2}{2AC} - 2}{-\frac{B^{2}}{4A^{2}C^{2}}(B^{2} - 4AC) + (\frac{B^{2}}{2AC} - 2)^{2}} = \frac{-\frac{B}{2AC}\sqrt{B^{2}-4AC} + \frac{B^2}{2AC} - 2}{-\frac{B^{4}}{4A^{2}C^{2}} + \frac{B^{2}}{AC} + \frac{B^{4}}{4A^{2}C^{2}} - \frac{4B^{2}}{2AC} + 4} = \frac{-\frac{B}{2AC}\sqrt{B^{2}-4AC} + \frac{B^{2}}{2AC} - 2}{-\frac{B^2}{AC} + 4}[/itex]
At this point I multiply through by a form of one [itex]\frac{AC}{AC}[/itex]
[itex]\frac{-\frac{B}{2}\sqrt{B^{2} - 4AC} + \frac{B^{2}}{2}-2AC}{4AC - B^{2}}[/itex]
I multiply through by another form of one [itex]\frac{2}{2}[/itex]
[itex]\frac{-B\sqrt{B^{2} - 4AC} + B^{2} - 4AC}{2(4AC - B^{2})}[/itex]
Simplify
[itex]\frac{B\sqrt{B^{2}-4AC} + B^{2}}{2(4AC - B^{2})} - \frac{2AC}{4AC - B^{2}}[/itex]
My equations for [itex]A_{1}[/itex] and [itex]A_{2}[/itex] then become
[itex]A_{1} = V_{x}(t = 0^{+})(1 + \frac{B\sqrt{B^{2}-4AC} + B^{2}}{2(4AC - B^{2})} - \frac{2AC}{4AC - B^{2}})[/itex]
[itex]A_{2} = -V_{x}(t = 0^{+})(\frac{B\sqrt{B^{2}-4AC} + B^{2}}{2(4AC - B^{2})} - \frac{2AC}{4AC - B^{2}})[/itex]
My voltage equation then becomes
[itex]V_{x}(t) = V_{x}(t = 0^{+})(1 + \frac{B\sqrt{B^{2}-4AC} + B^{2}}{2(4AC - B^{2})} - \frac{2AC}{4AC - B^{2}})e^{(\frac{-B + \sqrt{B^2 - 4AC}}{2A})t} - V_{x}(t = 0^{+})(\frac{B\sqrt{B^{2}-4AC} + B^{2}}{2(4AC - B^{2})} - \frac{2AC}{4AC - B^{2}})e^{(\frac{-B - \sqrt{B^2 - 4AC}}{2A})t}[/itex]
Plugging back in the values for A, B, and C
[itex]V_{x}(t) = V_{x}(t = 0^{+})(1 + \frac{\frac{1}{R}\sqrt{(\frac{1}{R})^{2}-4(C_{1}+C_{2})\frac{1}{L}} + (\frac{1}{R})^{2}}{2(4(C_{1}+C_{2})\frac{1}{L} - (\frac{1}{R})^{2})} - \frac{2(C_{1}+C_{2})\frac{1}{L}}{4(C_{1}+C_{2})\frac{1}{L} - (\frac{1}{R})^{2}})e^{(\frac{-\frac{1}{R} + \sqrt{(\frac{1}{R})^2 - 4(C_{1}+C_{2})\frac{1}{L}}}{2(C_{1}+C_{2})})t} - V_{x}(t = 0^{+})(\frac{\frac{1}{R}\sqrt{(\frac{1}{R})^{2}-4(C_{1}+C_{2})\frac{1}{L}} + (\frac{1}{R})^{2}}{2(4(C_{1}+C_{2})\frac{1}{L} - (\frac{1}{R})^{2})} - \frac{2(C_{1}+C_{2})\frac{1}{L}}{4(C_{1}+C_{2})\frac{1}{L} - (\frac{1}{R})^{2}})e^{(\frac{-(\frac{1}{R}) - \sqrt{(\frac{1}{R})^2 - 4(C_{1}+C_{2})\frac{1}{L}}}{2(C_{1}+C_{2})})t}[/itex]
That's kind of ugly. But anyways if my equation
[itex]V_{x}(t) = V_{x}(t = 0^{+})(1 + \frac{B\sqrt{B^{2}-4AC} + B^{2}}{2(4AC - B^{2})} - \frac{2AC}{4AC - B^{2}})e^{(\frac{-B + \sqrt{B^2 - 4AC}}{2A})t} - V_{x}(t = 0^{+})(\frac{B\sqrt{B^{2}-4AC} + B^{2}}{2(4AC - B^{2})} - \frac{2AC}{4AC - B^{2}})e^{(\frac{-B - \sqrt{B^2 - 4AC}}{2A})t}[/itex]
is correct then I have developed a generic formula for circuits in which you can find a second order homogeneous equation. Does it look right? Thanks for your help!
When I plug the values pack in I get the following equation
[itex]V_{x}(t) = \frac{4}{49}e^{2000t}((7j - 401)e^{14000jt} - (7j - 499)e^{-14000jt})[/itex]
Which dosen't seem right. Not sure what I did wrong though.
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