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induction to linear time invariant circuit

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http://i44.tinypic.com/xfvv28.gif

i know that i am supposed to use a convolution here.

i tried to build Kcl on both nodes
on the capacitor kcl:
the voltage is Vc=L(I')

-Is+CL(I'')+I2=0

on the resistor kcl:
-I2+(L(I'))/R+I=0

(I is the current on the inductor which is given)

this is as far as i could go

??
 

MrAl

Well-Known Member
Most Helpful Member
Hi BrownOut,


Maybe you can walk him though this one.
 

BrownOut

Banned
I can't see the attachment good enough to know what is being asked. I can see the ckt good enough to know it only has 1 independent node.

Is-Ic-Ir-Il=0

Besides, I'd probably make a mistake somewhere.
 

MrAl

Well-Known Member
Most Helpful Member
Hi again,


Oh, I didnt realize that some people couldnt read some drawings. I wonder
why. In any case, here is a new drawing for anyone who cant read the original drawing. This might help a little.
 

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BrownOut

Banned
How much do you need this? I might be able to help, but you would have to wait till I can review 2nd order non-homogeneous equations. Refreshe me on a couple things, is the curly-delta the impulse function? If so, do you know the general solution to the impulse? Also, what is t'? The notation looks familiar, but remember I did this class almost 20 years ago.

If you do a KVL, you get, after some manipulation:

d^V/dT^2 + 1/RCdV/dT + 1/LCV = 1/C dIS/dT

The slolutions to the homogeneous poly are

S1 = -1/2RC + SQRT((1/RC)^2 - 4/LC); S2 = -1/2RC - SQRT((1/RC)^2 - 4/LC)

(please check this for correctness)

From here, you normally evaluate the discriminant to see if it's real, imaginary or zero.

Can you take it from here? Or do I have to do some more review?
 
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MrAl

Well-Known Member
Most Helpful Member
How much do you need this? I might be able to help, but you would have to wait till I can review 2nd order non-homogeneous equations. Refreshe me on a couple things, is the curly-delta the impulse function? If so, do you know the general solution to the impulse? Also, what is t'? The notation looks familiar, but remember I did this class almost 20 years ago.

If you do a KVL, you get, after some manipulation:

d^V/dT^2 + 1/RCdV/dT + 1/LCV = 1/C dIS/dT

The slolutions to the homogeneous poly are

S1 = -1/2RC + SQRT((1/RC)^2 - 4/LC); S2 = -1/2RC - SQRT((1/RC)^2 - 4/LC)

(please check this for correctness)

From here, you normally evaluate the discriminant to see if it's real, imaginary or zero.

Can you take it from here? Or do I have to do some more review?

Hi again,


I would do this in latex too but gee it takes so long to form the equations
even using the editor. It sure does make a nice looking set of equations
though.


Following BrownOut's kick start...

The circuit contains two nodes, and with nodal analysis we need to look at
N-1 voltages and N-1 equations, but unfortunately, we dont need to analyze
the voltage so forming an equation that contains voltages wont be as easy.
Instead, since we need to look at the current in the inductor anyway we start
looking at the currents with KCL:

vC/R+C*d(vC)/dt+iL=Is (vC is cap voltage, iL is inductor current)

which is of course the sum of currents equated to the source.

This equation still depends on voltage however, and we wish to eliminate that
so that we can end up with an equation that depends entirely on current, so
taking into account the following relationship between current and voltage in
the inductor:

v=L*di/dt

where i is the current in the inductor and v is the voltage across the cap, so

vC=L*d(iL)/dt

Substituting this into the first equation above we get:

(L*d(iL)/dt)/R+C*d(L*d(iL)/dt)/dt+iL=Is

Doing the math we get:

L/R*d(iL)/dt+C*L*d^2(iL)/dt^2+iL=Is

and rearranging a little and replacing iL with i we get:

C*L*d^2i/dt^2+L/R*di/dt+i=Is

which is a second order differential equation that depends only on the inductor
current i=iL.

[Note: a little more accurate representation of this equation would be:
C*L*d^2 i(t)/dt^2+L/R*d i(t)/dt+i(t)=Is
but we are using 'i' as shorthand]

Now since the first step is to make Is equal to an impulse, the response will
be the impulse response and from the interpretation of the impulse response
for a network we get:

C*L*d^2i/dt^2+L/R*di/dt+i=0

That is, the response to an impulse (curly delta) is simply the unforced response
of the network.

The next step would be to solve the differential equation and find the initial
conditions corresponding to t=0+ .


BrownOut:
I had to refresh a little too as i had learned other techniques that are much
quicker than the approach that problem apparently wanted us to use.
I also used Laplace to find the initial conditions, perhaps you can remember
another simpler way to find these and refresh us both :)
 
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BrownOut

Banned
I would have used Laplace too, but generally when a student is solving equations using this method, they don't have Laplace yet. That's why I had to review this method.

Or maybe I made an assumption about what you had to use. Is this a class assignment? For some reason, I made that assumption.
 

MrAl

Well-Known Member
Most Helpful Member
I would have used Laplace too, but generally when a student is solving equations using this method, they don't have Laplace yet. That's why I had to review this method.

Or maybe I made an assumption about what you had to use. Is this a class assignment? For some reason, I made that assumption.
Hi,


Yeah, i was thinking that too that's why i was thinking about another
way to find the initial conditions at t=0+ . In particular, di/dt (0+)
where 'i' is again the inductor current.
 
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