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RLC parallel Circuit excited by 2 dc voltage sources

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anampiracha

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can somebody help me please in solving this application exercise. I am not getting it. I am stucked at the very first step. formation of equation
 

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  • MAT290 Application Exercise 1(1).pdf
    170.5 KB · Views: 495
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Hi there,

I guess this is for a class or something?

Let me see if i can get you started...

Instead of writing di/dt and d^2i/dt^2 and stuff like that that isnt that easy to read in text, i'll use the following much simplified notation:

X1 will be the voltage across the capacitor,
X2 will be the current through the inductor.
(note those are upper case X's).
Since we now have these defined, we can define:
x1*C is the current through the capacitor (x1 is the time derivative of X1 and iC=C*dv/dt which is the same thing here),
x2*L is the voltage across the inductor (x2 is the time derivative of X2, and vL=L*di/dt which is the same thing here).
(note the derivatives are always lower case x's).

Given those four definitions, we can then write two ODE's for this circuit starting with the first voltage source Vs1:
(Vs1-X1)/Rs=x1*C+X1/Rc+X2
(X1-L*x2)/RL=X2

and solving for the derivatives x1 and x2 we get:
x1=-X2/C-X1/(Rs*C)-X1/(Rc*C)+Vs1/(Rs*C)
x2=X1/L-(RL*X2)/L

That's the two ODE's, and they are written in terms of the capacitor voltage and derivative and inductor current and derivative.
To get the equation shown in part (a) of the paper however, we have to convert this into a single differential equation in terms of only the inductor current and two of it's derivatives and no capacitor voltage or derivative.

One way to do this which is quite interesting is to draw a flow graph of the two ODE's using integrators and make X2 the output,
then solve for X2, then equate to the required Vs1/RsLC and then we can interpret powers of s as derivatives and so we get the
same equation as shown in part (a).
There's another way too which you may like but i'll save that for next time.

Would you like to try this first or do you just want to see how to do it assuming you've never done this before?

Since the central part of the circuit is the same for both sources and they both drive through the same resistance Rs, we would
have the same equation for Vs2 also.
 
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besides your flow graph method, what's another way to do this using differentials ? i can't seem to get the same equation they have given
 
Hi,

The other method was simply to use Mason's Gain Formula.

This was a little more work than i thought at first but here is the procedure i talked about...

Refer to the attachment for the figures and equations. This is a step by step
procedure following the diagrams in the attachment from top to bottom.
We start with the two ODE's, draw a flow graph, reduce the flow graph,
then factor into the required form. We end up with a single differential
equation in iL and its derivatives.

We start first by forming two ODE's for the circuit. That's shown
right under the circuit diagram in the attachment.

Fig1:
We draw a flow graph using the two ODE's. The rectangles are gain blocks
and the triangles are integrators. The output X2 is the inductor current
iL and we keep it that way throughout the reduction process. There are
a total of three feedback loops in this flow diagram.

Fig2:
We reduce the two loops with the two integrators in them. The result leaves
us with only one remaining feedback loop.

Fig3:
We reduce the three forward path gains to a single gain. We still have one
feedback loop.

Fig4:
We reduce the remaining feedback loop to a single forward gain.

Fig5:
We include the denominator of the input in the foward gain and leave the
original input as the input signal.

Now from Fig 5 we form the equations...

Equ1:
We equate the flow graph to X2, the output.

Equ2:
We solve to get Vs1 on the right in some form.

Equ3:
We expand to get all terms by themselves.

Equ4:
We divide by Rc to get rid of Rc on the right (it's not supposed to be there in
the final equation).

Equ5:
We divide by Rs*L*C because that is supposed to be on the right side in the
final equation.

Equ6:
We group terms according to their powers of the variable 's'.

Equ7:
We factor in terms of powers of 's' and 'X2'.

Equ8:
We interpret powers of 's' as derivatives, s^2 is second derivative, s is first, no
's' in the term means just the variable itself (no derivatives).

Equ9:
We replace all X2 with iL, so X2 becomes iL, X2' becomes iL', X2'' becomes iL''.

Result:
We factor each term to get into the same form as the desired equation and we're done.
Comparing the result with the equation shown in the paper, we see that they are
the same.
 

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  • Flow-01.gif
    Flow-01.gif
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This will be the circuit for Vs1? and capacitor initially will act as a short circuit and inductor as an open circuit?
 

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  • Untitled.png
    Untitled.png
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Hi again,


Yes that's right, and you're welcome :)

Swoosh:
Here's another way to do it using the ODE's and skipping the flow graph (which is cheating if you ask me ha ha):

Starting with the original set of ODE's again:
x1=-X2/C-X1/(Rs*C)-X1/(Rc*C)+Vs1/(Rs*C)
x2=X1/L-X2*RL/L

we note that xN is just the derivative of XN, so using the differential operator 's' we can write:
x1=s*X1, and
x2=s*X2

This cuts down on the work required but doesnt offer as much insight into what is actually happening.

When we substitute into the original set of ODE's we get:
s*X1=-X2/C-X1/(Rs*C)-X1/(Rc*C)+Vs1/(Rs*C)
s*X2=X1/L-X2*RL/L

and all we have to do now is solve this set of simultaneous equations in the variables X1 and X2, and all we want is the solution for X2, so we end up with:
X2=(Rc*Vs1)/(Rc*Rs*s*C*RL+Rs*RL+Rc*RL+Rc*Rs*s^2*C*L+Rs*s*L+Rc*s*L+Rc*Rs)

which is the same as when using the flow graph, and after rearranging a little we again get:
(RL*X2)/(Rs*C*L)+(RL*X2)/(Rc*C*L)+(s*RL*X2)/L+X2/(C*L)+(s*X2)/(Rs*C)+(s*X2)/(Rc*C)+s^2*X2=Vs1/(Rs*C*L)

which is again the same as before, so after regrouping and replacing all s^n*X2 with the n^th derivatives of iL, we again get the same solution as the paper.
 
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lol... I don't consider it a cheating because your posts are very informative in fact i am making my concepts clear by solving it again and again by different methods.

A question from the same exercise ... Part b) what is the purpose of dummy variable here? how we are going to treat it?
 
Hi,

A dummy variable is used when we dont want to use the actual variable for example in convolution. The dummy variable is tau because we are already using t.
Take a look at the convolution integral for a good example of this.
 
Hi,

Sorry i cant help there as i dont use Matlab. Im sure there is someone else here who can help though.
 
Laplace

How will be do the part 1 for this ??
 

Attachments

  • MAT290 Application Exercise 2.pdf
    190.3 KB · Views: 353
Last edited:
Hi,

I could help you find the impulse response and things like that, but if you need to use Matlab you'll have to read up on that.
 
Hi,

You know how to find the transfer function?
 
Hi,

Do you understand the frequency models of the components, such as 1/(s*C) and s*L ?
Also, did you do any network analysis, at least with resistors?
 
Hi,

Yeah I know about the frequency models and transformation with regards to laplace, and I also did the network analysis In circuit analysis (1) . these days we are studying circuit analysis part 1.
 
Hi,

Ok good, then let me see if i can get you started, and you can take it from there and try it yourself first ok?

To start with, you know that when you have a resistor R1 in series with another resistor R2 you just add the two: RT=R1+R2.
Secondly, you know that when you have a resistor R1 in parallel with another resistor R2 you use a little formula like this:
RT=(R1*R2)/(R1+R2)
or if you have more than two resistors in parallel then you use:
RT=1/(1/R1+1/R2+1/R3+...+1/Rn)

You also know that the transformation for the inductor is:
s*L
and for the capacitor it is:
1/(s*C)

Now those transformations act the same as the resistors do in the math. So for a resistor R1 in series with an inductor L1 we have:
Z=R1+s*L1

So it's just the same as if s*L1 were a resistor.

IN parallel, we would have:
Z=(R1*s*L1)/(R1+s*L1)

Also, a voltage divider circuit with two resistors is:
Vout=Vin*R2/(R1+R2)
where R1 is the upper resistor and R2 is the lower resistor.

I talk about this just to make this analysis quick, because we have a voltage divider once we put C, Rc, and RL+s*L in parallel,
with Rs being the upper resistor and the other components Z making up the lower impedance (like a resistor).

So a quick way to do this is as follows:
Calculate RL in series with s*L,
calculate that in parallel with 1/(s*C) and in parallel with Rc, call that Zx,
then calculate the voltage divider made up of Rs (upper) and Zx (lower).
You'll come out with the transfer function in this form:
Vout/Vs=Zx/(Rs+Zx)

See if you can get at least that far.

Once you get that, you'll then know what Vout is. Knowing Vout, you can easily
calculate iL because iL just equals Vout/Z where Z is the series combination of
RL and s*L.

See if you can get at least Vout, then try to get iL as described.
 
hi

I got the values like

Zx= RL*SL/ (S²RL*LC)+(RL*SL)+1

and

Vo=VS*RS*SL/ (RS*SL)+(S²*RS*RL*LC)+(RS*RL*SL)+RS

is it right??
 
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