RLC parallel Circuit excited by 2 dc voltage sources

[MrAl]

Hi again,

Lets look at the series of L and RL first, we'll call this zL:
Since L becomes s*L and it is in series with RL we just add these two:
zL=RL+s*L

Also, C becomes 1/(s*C) and we'll call that zC:
zC=1/(s*C)

Now we look at Zx, which is the parallel combination of zL, Rc, and zC:
Zx=1/(1/zL+1/Rc+1/zC)

Now replacing zL and zC with their actual values and simplifying a little we get:
Zx=1/(1/(s*L)+1/Rc+s*C)

To simplify this we multiply top and bottom by s*L and by Rc and we get:
Zx=(Rc*s*L)/(Rc*s^2*C*L+s*L+Rc)

So now we have the equivalent impedance of the lower part of the divider network.

Note that it is more clear to write s*L rather then SL, and that when we have a numerator of say 1 and denominator a^2+b+c we can not write it like this:
1/a^2+b+c

but must use parens in the bottom like this:
1/(a^2+b+c)

otherwise the reader wont know exactly what we mean.

Ok, so back to the circuit...

Now that we have Zx, we can do the voltage division to get Vout:
Vout=Vs*Zx/(Rs+Zx)

Care to try to get the expression for Vout again?

This is quite important because many networks can be solved in a similar manner.

 

[anampiracha]

Vo=Vs*(Rc*s*L)/(Rs*(Rc*s^2*C*L+s*L+Rc)+Rc*s*L)

I hope this time its correct

 

[MrAl]

Hi again,

Yes, it looks like you got it right. I knew you would get it!

Now, care to try to get the current you need, iL ?

 

[anampiracha]

after getting IL what is next?

 

[MrAl]

Hi,

After you find iL that will be your required H(s).
To get h(t), you then need to find the Inverse Laplace Transform of H(s). That's your impulse response.

 

[anampiracha]

hi,

We know that H(s) = Y(s)/X(s) , where Y(s) is out put. So should I treat IL as H(s)?

 

[anampiracha]

what is the use of unit impulse then?do i need to * it with Vs for h(t)?

 

[MrAl]

Hi,


Yes the expression for iL is the H(s) for this problem.

The impulse response characterizes the input/output relationship similar to how the transfer function does, except the impulse response is in the time domain while the transfer function is in the frequency domain. So if you want to find the output of a circuit in the time domain using the convolution integral you need the impulse response so that you can convolve the input with the impulse response (using the convolution integral).

 

[anampiracha]

can you please elaborate the steps for h(t) ? This exercise is worth 1% in my course. The use of it is to develop the connection between 2 subjects and forcing them to try different approaches . None of my class mates have attempted this exercise as it is 1% that's why I am having issues which I have to tackle by my self and I found you very supportive and helpful in field of circuits, I am really thankful to you . I am trying to solve it because for me learning matters instead of grades.

 

[MrAl]

Hi,

Do you understand how to use partial fractions ie a partial fraction decomposition?

 

[anampiracha]

yeah I know that very well .

 

[MrAl]

Hi,

Oh that's great. Then you can try to factor the H(s) and use the Inverse Laplace Transform on each term. That should give you h(t).
Make sense?

Hey, they give the differential form also, so if you want you can just use the Laplace equivalents for the second and first derivatives. Have you ever done it this way before?

 

[anampiracha]

No i never did such a complicated integration/ differentiation

 

[MrAl]

Hi again,


Let me see if i can get you going with this using the Laplace equivalent expressions for the derivatives. Im sure you will find this way too easy

Here is a little table that will show you what to substitute:

FORM    ALT FORM   FORMAL s FORM                     INFORMAL s FORM
f(t)    f(t)       F(s)                               F
f'(t)   df/dt      s*F(s)-f(0)                        s*F-f0
f''(t)  d^2f/dt^2  s^2*F(s)-s*f(0)-f'(0)              s^2*F-s*f0-f'0 or s^2*F-s*f0-df0
f'''(t) d^3f/dt^3  s^3*F(s)-s^2*f(0)-s*f'(0)-f''(0)   s^3*F-s^2*f0-s*df0-ddf0

In the above table, f(t) is the time expression and F(s) is the frequency expression. The values to be substituted are shown as either F or more formally F(s). f(0) is the value of f at t=0, and f'(0) is the value of the first derivative at t=0, etc.

For example, say we have this to transform:
y"+y'+y=0

with all initial conditions equal to zero.

We need the expressions for y" and y' and y:
Y" => s^2*F(s)-s*f(0)-f'(0)
y' => s*F(s)-f(0)
y => F(s)

So all we do is substitute those into the original equation and set all the f^n(0) to zero.

Make sense?

 

[anampiracha]

hi,

I Tried to use this Laplace method , but by using it, the equation formed is " IL*(s^2+A*s+B) =D"... is it? Now from here I will get IL... which we found early .... and we call it as H(s) .. Isn't?

 

[MrAl]

Hi,

Yes, isnt that cool?

Now there are a couple methods for finding the impulse response itself. Here's one which can be found on the web which seems to be a good method because we already have the
differential equation and it is defined in terms of A, B, and D as required, so here it is...

Find the step response, then take the first derivative. In other words:

h(t)=d(stepresponse)/dt

Do you know how to find the step response?

 

[anampiracha]

If I were given the values for all the active passive elements of the circuit I could solve the whole exercise in seconds .. But these derivations are making me crazy. I almost spent whole day on it and i didn't get anything valuable

s^2+AS+B , I couldn't factorize it to get the roots, to find IL by this way so exhausted

 

[MrAl]

Hi,

Oh sorry to hear that, lets see if we can make this a little easier. Get some rest first though, that's important.

s^2+A*s+B

can be factored, but this isnt a straight forward as when you can do it perfectly and definitely like (s+1)*(s+1). But it's almost the same. The only difference is we use the quadratic formula that is for:
a*s^2+b*s+c=0

we have:
s=(-b(+/-)sqrt(b^2-4*a*c))/2a

which gives two solutions:
with D=sqrt(b^2-4*a*c)
r1=(-b+D)/2a
r2=(-b-D)/2a

where i called the solutions r1 and r2 because they are roots. The original equation is then factored as:
a*s^2+b*s+c=(s-r1)*(s-r2)

You can apply that to your equation setting a=1 and b=A and c=B and see what you get.
The interesting part comes up when D above is imaginary, and also divides the solutions into different kinds of responses.

Another approach that i was telling you about is to find the step response directly from the differential equation, then take the derivative.

To find the step response we set the differential equation equal to 0 then find the complementary solution, then the particular integral, then the general solution, then set the boundary conditions f(0)=0, and f'(0)=0, that gives the step response. We then take the derivative to get the impulse response.

In these second order systems we always have to be aware that when the values of the components in the circuit are not yet specified (specified as variables) that means we can end up with either a response with only an exponential, or a response with an exponential and sine and cosine terms. This is alternately expressed as complex exponentials, but it does get a little tricky so make sure you get plenty of rest before attempting any of this

Here's an example i actually found on the web...

d''+13y'+12y=f(t)

Find the solution with f(t)=0, which is y=A*e^-12t+B*e^-t.
Particular integral is y=1/12
General solution then is y=1/12+A*e^-12t+B*e^-t.
Set boundary conditions y(0)=0 and y'(0)=0, then we get two equations:
1/12+A+B=0
-12*A-B=0
Solving, we get A=1/132 and B=-1/11.
Thus the step response is:
y=1/12+(1/132)*e^-12t-(1/11)*e^-t
Then take the first derivative of that we get:
h(t)=e^(-t)/11-e^(-12*t)/11

 

[anampiracha]

hi,

I just read the demands of exercise now carefully, They are asking to express the answer in form of A,B,D and P.

I attempted part a and b

In part a , am stuck after , can you please check and let me know if i am doing right?

 

[MrAl]

Hi,

That looks right so far, but it is kinda hard to read.