step response of series RLC circuit derivation

[PG1995]

Hi

Please have a look on the following attachments to see my questions. Please help me with them.

The following two attachments have the queries:
1: https://www.electro-tech-online.com/attachments/rlc-der1-jpg.58875/
2: https://www.electro-tech-online.com/attachments/rlc-der2-jpg.58876/

Eq. 8.4 #1: https://www.electro-tech-online.com/attachments/8-4b-jpg.58878/
Eq. 8.4 #2: https://www.electro-tech-online.com/attachments/8-4a-jpg.58877/

Regards
PG

 

[MrAl]

Hi

Please have a look on the following attachments to see my questions. Please help me with them.

The following two attachments have the queries:
1: https://www.electro-tech-online.com/attachments/rlc-der1-jpg.58875/
2: https://www.electro-tech-online.com/attachments/rlc-der2-jpg.58876/

Eq. 8.4 #1: https://www.electro-tech-online.com/attachments/8-4a-jpg.58877/
Eq. 8.4 #2: https://www.electro-tech-online.com/attachments/8-4b-jpg.58878/

Regards
PG

Hi PG,


Q1:
The associated quadratic is the denominator, not really just the differential equation as is, but
the left side only and the right side is the numerator. Thus if the right side is 0 or 1 or any
other constant (like Vs/LC) that doesnt change anything.
The associated equation is really RightSide/LeftSide, not simply the entire equation taken as
an algebraic equation. That means to find the roots we only find them for the LeftSide.

Q2:
If you are not given the initial conditions you want, then you have to try to solve for them.
This is often different for every circuit.
For a simple example consider an RL circuit excited by a unit step (like a battery). If you
are given iL(0) and you want vL(0), then you have to multiply iL(0) times R and that gives
you a voltage drop for the resistor which can then be subtracted from the battery voltage.
Since this is for t=0, you end up with vL(0).

Q3:
They are simply stating that the total response is the sum of the natural unforced response
plus the forced response.

Qx:
Why didnt you reply to your other post again? You were going to find two expressions to
define that function piecewise right?

 

[PG1995]

Hi PG,
Q1:
The associated quadratic is the denominator, not really just the differential equation as is, but
the left side only and the right side is the numerator. Thus if the right side is 0 or 1 or any
other constant (like Vs/LC) that doesnt change anything.
The associated equation is really RightSide/LeftSide, not simply the entire equation taken as
an algebraic equation. That means to find the roots we only find them for the LeftSide.

Hi MrAl

Thank you for the reply. But I'm still very confused about Q1. Here, this is the standard linear homogeneous differential equation we are dealing dealing with along with its solution. Please also have a look here.

Below I have included complete treatment of Eq. 8.4 I was talking about in my previous post. The link "Eq. 8.4 #2" has important parts highlighted which you might mind useful. Please help me.

Eq. 8.4 #1: https://www.electro-tech-online.com/attachments/8-4-1-jpg.58887/
Eq. 8.4 #2: https://www.electro-tech-online.com/attachments/8-4-2-jpg.58888/
Eq. 8.4 #3: https://www.electro-tech-online.com/attachments/8-4-3-jpg.58889/

Qx:
Why didnt you reply to your other post again? You were going to find two expressions to
define that function piecewise right?

Seriously, I'm occupied with many different things for last some days. I will give it a shot as soon as I find some time. Thanks.

Best regards
PG

 

[MrAl]

Hello again,


Here's the thing, when you have an RLC circuit not excited by any forcing function you get a certain characteristic equation. When you apply a forcing function that is a step you get the same characteristic equation. Same characteristic equation, same roots.

With no input, we get (s*R)/L+1/(C*L)+s^2 as the denominator, and with a step input we get the same denominator but the numerator changes.
Same denominator, same roots.

Here's a full time domain solution for both cases (exponential solution cases):

No step:
-(C*R*e^((t*sqrt(C^2*R^2-4*C*L))/(2*C*L)-(t*R)/(2*L)))/(2*L*sqrt(C^2*R^2-4*C*L))+e^((t*sqrt(C^2*R^2-4*C*L))/(2*C*L)-(t*R)/(2*L))/(2*L)+(C*R*e^(-(t*sqrt(C^2*R^2-4*C*L))/(2*C*L)-(t*R)/(2*L)))/(2*L*sqrt(C^2*R^2-4*C*L))+e^(-(t*sqrt(C^2*R^2-4*C*L))/(2*C*L)-(t*R)/(2*L))/(2*L)

With a step:
(C*E*e^((t*sqrt(C^2*R^2-4*C*L))/(2*C*L)-(t*R)/(2*L)))/sqrt(C^2*R^2-4*C*L)-(C*E*e^(-(t*sqrt(C^2*R^2-4*C*L))/(2*C*L)-(t*R)/(2*L)))/sqrt(C^2*R^2-4*C*L)

Notice we get the same value for the exponents in both cases (positive or negative).
t*sqrt(C^2*R^2-4*C*L))/(2*C*L)-(t*R)/(2*L)
t*sqrt(C^2*R^2-4*C*L))/(2*C*L)-(t*R)/(2*L)


One more easier to read example:

y''+A*y'+B*y=0
we get:
y=e^(-(x*A)/2)*(K1*sin((x*sqrt(4*B-A^2))/2)+K2*cos((x*sqrt(4*B-A^2))/2))

and for:
y''+A*y'+B*y=1
we get:
y=e^(-(x*A)/2)*(K1*sin((x*sqrt(4*B-A^2))/2)+K2*cos((x*sqrt(4*B-A^2))/2))+1/B

What's the difference? The second one has an added 1/B. That 1/B doesnt change the frequency in any way.



Unless i dont understand your question.

 

[PG1995]

Hello again,


Here's the thing, when you have an RLC circuit not excited by any forcing function you get a certain characteristic equation. When you apply a forcing function that is a step you get the same characteristic equation. Same characteristic equation, same roots.

With no input, we get (s*R)/L+1/(C*L)+s^2 as the denominator, and with a step input we get the same denominator but the numerator changes.
Same denominator, same roots.

Here's a full time domain solution for both cases (exponential solution cases):

No step:
-(C*R*e^((t*sqrt(C^2*R^2-4*C*L))/(2*C*L)-(t*R)/(2*L)))/(2*L*sqrt(C^2*R^2-4*C*L))+e^((t*sqrt(C^2*R^2-4*C*L))/(2*C*L)-(t*R)/(2*L))/(2*L)+(C*R*e^(-(t*sqrt(C^2*R^2-4*C*L))/(2*C*L)-(t*R)/(2*L)))/(2*L*sqrt(C^2*R^2-4*C*L))+e^(-(t*sqrt(C^2*R^2-4*C*L))/(2*C*L)-(t*R)/(2*L))/(2*L)

With a step:
(C*E*e^((t*sqrt(C^2*R^2-4*C*L))/(2*C*L)-(t*R)/(2*L)))/sqrt(C^2*R^2-4*C*L)-(C*E*e^(-(t*sqrt(C^2*R^2-4*C*L))/(2*C*L)-(t*R)/(2*L)))/sqrt(C^2*R^2-4*C*L)

Notice we get the same value for the exponents in both cases (positive or negative).
t*sqrt(C^2*R^2-4*C*L))/(2*C*L)-(t*R)/(2*L)
t*sqrt(C^2*R^2-4*C*L))/(2*C*L)-(t*R)/(2*L)


One more easier to read example:

y''+A*y'+B*y=0
we get:
y=e^(-(x*A)/2)*(K1*sin((x*sqrt(4*B-A^2))/2)+K2*cos((x*sqrt(4*B-A^2))/2))

and for:
y''+A*y'+B*y=1
we get:
y=e^(-(x*A)/2)*(K1*sin((x*sqrt(4*B-A^2))/2)+K2*cos((x*sqrt(4*B-A^2))/2))+1/B

What's the difference? The second one has an added 1/B. That 1/B doesnt change the frequency in any way.



Unless i dont understand your question.

Thanks a lot, MrAl.

I'm viewing your post on my cell phone. I don't get that denominator and numerator talk in the part set in red. Could you please clearify it? I hope I'm not missing something obvious. Many thanks.

Best wishes
PG

 

[MrAl]

Hi,

Well you know that if two equations have the same denominator then they have the same roots right?

Well BOTH of the equations you were talking about have the same denominator. In other words, we dont solve the equation on the left AND right to get the roots, we only solve the equation on the left alone to get the roots, given a constant on the right (which is a step).

Let me try to illustrate this...

y''+A*y'+B*y=1

We wont be doing this:
y''+A*y'+B*y-1=0

We'll be doing this:
y''+A*y'+B*y=0

even though there is a '1' on the right hand side in the original equation.

If you look at the original differential equation you'll see 1/LC on the right. If you look at the equation in the frequency domain, you'll see a denominator
and a numerator with the numerator being s/L after dividing the denominator by LC to get the s^2 term alone without any coefficient. The denominator is then s^2+A*s+B for example, and for the step response the numerator changes to 1/L but the denominator stays the same. Thus either way the denominator is always the same, so the roots can not change.

Does that make sense?

Lets do it...

In the original RLC circuit, we have:
Z=R+s*L+1/(s*C)
Now inverting that as 1/Z we get:
(s*C)/(s^2*C*L+s*C*R+1)

Divide top and bottom by LC we get:
(s/L)/(s^2+(s*R)/L+1/(C*L))

So we have that denominator as shown. The step response is:
(1/L)/(s^2+(s*R)/L+1/(C*L))

and it's clear that we didnt have to change the denominator. With the same coefficients, we get the same roots. That's all they were saying.