Hi,

I'm a bit rust on my s-domain circuit analysis but this one looks so simple and I'm having a difficult time figuring out where I'm going wrong. If anyone can help I would greatly appreciate it.

Here's the problem:

David

- posted
15 years ago

Hi,

I'm a bit rust on my s-domain circuit analysis but this one looks so simple and I'm having a difficult time figuring out where I'm going wrong. If anyone can help I would greatly appreciate it.

Here's the problem:

formatting link

Thanks.
David

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- Vote on answer
- posted
15 years ago

---------------------- You have carried all the R1+... and CR terms along too far and this leads to confusion. You can look at the terms in the start of the 3rd line on the first page and put them into the form A (1+sT1)/(1+sT2) where T1 and T2 have the form CRequivalent This gives the transfer function.

The time domain V0/V1 depends on the input signal which you haven't specified.

- Vote on answer
- posted
15 years ago

I am not working through the complete thing, but you can make life easier for yourself by realising that since R2 & R3 are pure resistors if you know the voltage across R2 + R3 the output voltage that you are looking for is simply

Vout= R3/(R2+R3) * (Vacross R2 & R3)

So I suggest that you let RL = R2 + R3

and then work out the voltage across RL then multiply this answer by R3/RL to get the voltage that you need.

John

- Vote on answer
- posted
15 years ago

Hi John,

Thanks for the response.

I believe I've solve it with phasors (checking with pspise). I am only required to find magnitude & phase response.

I wanted to find s-domain solution and transient for understanding reasons.

I see this wasn't the easiest way to approach the solution. At this point I'm convinced if there isn't something incorrect in my initial setup that everything else is correct.

I used Mathcad to check my math through the problem and the initial setup and final s-domain equations evaluate to the same value. I also checked the inverse Laplace transform and it is correct as well.

I've been thinking though- with no input applied the inverse transform isn't very useful. If the transfer function was multiplied by an input signal then the dirac function would no longer be present.

So perhaps seeing the dirac in the transfer function isn't such an odd thing?

David

- Vote on answer
- posted
15 years ago

n message

Thanks Don. I've been trying to find a way to reach the form you suggested, the best I can do is

The "R1 +" term in the denominator is confusing me when trying to reach the form you suggested. I can reach the form A(1+sR1C1) / [ B

+(1+sR1C1)] but no idea how to remove that B.- Vote on answer
- posted
15 years ago

Thanks Don. I've been trying to find a way to reach the form you suggested, the best I can do is

The "R1 +" term in the denominator is confusing me when trying to reach the form you suggested. I can reach the form A(1+sR1C1) / [ B

+(1+sR1C1)] but no idea how to remove that B.--------------------------- You then have in the denominator B+1 +sR1C1 =(B+1)(1+sR1C1/(B+1))

That is R3(1+sC1R1)/[R1 +(R2+R3)(1+sR1C1)] from your 3rd line

=R3(1+sC1R1)/[R1+R2 +R3 +(R2+R3)sC1R1]

=R3(1+sC1R1)/(R1+R2+R3)(1+[(R2+R3)R1C1/(R1+R2+R3)])

=A(1+sC1R1)/(1+sC1Re)

where A=R3/(R1+R2+R3) and Re =(R2+R3)R1/(R1+R2+R3)

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