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somasimple

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somasimple

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Bob S

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- #3

Proton Soup

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what is the second circuit for, DC analysis?

- #4

Phrak

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Both are a bunch or parts connected together. But is either one a circuit?

- #5

somasimple

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Since R

- #6

uart

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In general though it will be a good approximation at sufficiently low frequencies but very much a poor approximation at frequencies starting somewhere about [itex]\frac{1}{2 \pi R_1 C_1}[/itex] and higher.

- #7

somasimple

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So what you were really asking is whether the second circuit is valid as an approximate equivalent of the first. Even that question is a little vague. Equivalent in terms of input impedance or equivalent in terms of some two-port parameter like vout/vin or whatever (though no vout or vin are labelled on the diagram).

Exactly. It is a huge simplification but it may help to compute the resulting (parallel/serial) resistance and capacitor.

- #8

ygolo

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It really depends on what you're trying to do with the circuit. In many ways it is the [tex]R_2[/tex] that is irrelevant (essentially an open in many cases).

It's late, but, it seems like you are modeling some sort of delay or transmission line. The voltage transfer function is just 3 voltage dividers back to back, giving (check my math on this): [tex]\frac{1}{6 \ast \left(s \ast R_1 \ast C_1 +1 +\frac{R_1}{R_2}\right)^3}[/tex] with [tex]R_1<<R_2[/tex], this reduces to [tex]\frac{1}{6 \ast \left(s \ast R_1 \ast C_1 +1 \right)^3}[/tex] giving you 3 poles close to [tex]-\frac{1}{R_1 \ast C_1}[/tex].

From the transfer function, you know its, ac response, and it's impulse and step responses. You can even do the root-locus if you really wanted.

This is assuming the load impedance is essantially an open.

It's late, but, it seems like you are modeling some sort of delay or transmission line. The voltage transfer function is just 3 voltage dividers back to back, giving (check my math on this): [tex]\frac{1}{6 \ast \left(s \ast R_1 \ast C_1 +1 +\frac{R_1}{R_2}\right)^3}[/tex] with [tex]R_1<<R_2[/tex], this reduces to [tex]\frac{1}{6 \ast \left(s \ast R_1 \ast C_1 +1 \right)^3}[/tex] giving you 3 poles close to [tex]-\frac{1}{R_1 \ast C_1}[/tex].

From the transfer function, you know its, ac response, and it's impulse and step responses. You can even do the root-locus if you really wanted.

This is assuming the load impedance is essantially an open.

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- #9

Blenton

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Why don't you calculate both, compare results and post us in :)

- #10

Bob S

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- #11

somasimple

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That is good idea and I'll simulate these circuits.Why don't you calculate both, compare results and post us in :)

I use a demo version of http://www.spectrum-soft.com/index.shtm" [Broken]

I'll take a look at these hits.

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