find V(s) and V(t)

Discussion in 'Homework Help' started by yoamocuy, Feb 23, 2010.

1. yoamocuy Thread Starter Active Member

Oct 7, 2009
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1. The problem statement, all variables and given/known data
everything's in the picture.

2. Relevant equations

V=I*R

Ic=Iin*R/(R+C)

3. The attempt at a solution
Ok, I think I got the first part alright. I found the current in terms of Vin to be: Vin/(25+(22.5*225/s)/(22.5+225/s)) which simplifies to Vin*(s+10)/(225s+475). Now that I have current I can use the current divider rule to find Ic and then multiply Ic by C to get Vc. By doing that I get the equation [Vin*(s+10)/(225s+475)]*[22.5s/(225/s+22.5)]*225/s. After much simplification I get Vc=9*Vin/(s+19). I divide that by Vin and get 9/(s+19). Setting the bottom equal to zero I get a pole at s=-19. Setting the entire thing equal to zero I get a zero at s=+or- infinity.

Once I get to the 2nd part I'm a bit confused... if Vin=45 at an angle of 0o, is my Vc(t) just simply equal to 45*9/(0+19) since no s is given?

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2. t_n_k AAC Fanatic!

Mar 6, 2009
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Unfortunately you have made an error at the very start.

The current would be

Vin/(25s+ etc......)

You forgot the 's' operator for the inductor.

Another approach might be to think of the circuit as a voltage divider in which the the upper element is the inductor and the lower element is the parallel RC combination.

Also it sometimes helps to write out an expression in terms of the general circuit elements comprising R's, L's & C's and make the actual value substitutions once you have the expression in its 'simplest' form. This also provides some insight into the characteristic equation forms one encounters in typical solutions for circuits with specific configurations. For instance, in this case you would (at the very least) expect to see a second order characteristic equation with 'reasonable' coefficients - i.e. terms like 1/LC or 1/RC and so on.

3. yoamocuy Thread Starter Active Member

Oct 7, 2009
84
0
ug your right, i thought my inductor was a resistor...

Ok I ended up with the equation Vc/Vin=9/(s^2+10s+9). That gives me poles of -1 and -9 and a zero at infinity. I'm still confused with the next part though Could you point me in the right direction?

4. t_n_k AAC Fanatic!

Mar 6, 2009
5,448
790
Yes - the second part is confusing since the 'implication' is that the source voltage is a sinusoid of magnitude 45V and zero degrees phase angle. However you are not given the source frequency and a specific solution cannot be found.

You should have no difficulty with the 3rd and 4th parts of the question.

5. yoamocuy Thread Starter Active Member

Oct 7, 2009
84
0
so for the 2nd part would Vc(t) just be 45*9/(0^2+10*0+9)? and then the 3rd part I'd have 45*9/(j3^2+10*j3 +9)?

6. t_n_k AAC Fanatic!

Mar 6, 2009
5,448
790
As I suggested the second part cannot be resolved to a specific solution without knowing the sinusoid frequency.

If the second part stated the input voltage was a simple DC value of 45V then you would be able to provide a solution (see my comment below). But since it appears it is a sinusoid you can't find the effective impedances in the absence of a specific frequency term.

Comment:
The question begins by asking for the transfer function in the 's' domain form which suggests the requirement for you to provide a transient + steady-state solution. Are you familiar with the 'difference' between transient and steady-state solutions? And do you think this was anticipated in the question?