Converting discrete circuits into block diagrams

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TecHNooB

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Sep 10, 2005
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If you are given a circuit diagram, how do you turn it into a very basic block diagram (nonreduced)? I can do it given the transfer function which I can easily find but I don't think that's the point of this problem I'm working on. Here's a picture: Circuit

And.. unless I screwed up, here's the transfer function.

H(s) = R2 / [ s^2 ( R2LC + R1LC ) + s R1R2C + R2 + R1 ]
 

esun

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Nov 12, 2001
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What exactly do you mean by a block diagram? Sometimes circuits are represented as blocks with transfer functions, but in that case you'd just find the transfer function. You could reduce R1, R2, and Vi using a Thevenin equivalent and make that into a "block" (and reducing the circuit to a basic series RLC) but I don't think that's what you mean either.
 

TecHNooB

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Sep 10, 2005
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Apparently there's a 'block diagram method' for finding the transfer function. Typically, you're either given the transfer function and asked to create a block diagram or you are given the block diagram and asked to find the transfer function. In this case, you're given the actual circuit and asked to find the transfer function via block reduction method (I'm assuming this means without using laplace transforms). The problem is, I don't know how to turn the circuit into a block diagram without finding the transfer function first via laplace transforms. But that kind of defeats the purpose. I'll ask the prof tmrw I guess.
 

TecHNooB

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Sep 10, 2005
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Okay, turns out this was actually really simple. Basically, you write some KVL or KCL equations to relate the input and output. Then you use the expression with V_in and add whatever you want to it to get values you need to get to the output. Here's one possible solution in case anyone's interested.

http://img35.imageshack.us/img35/6250/solutionx.jpg
 

blahblah99

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Oct 10, 2000
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Well, in the example above you have three networks in series. You'd have to find the transfer function of each network (the resistor divider, the inductor, and the capacitor) and multiply the three to arrive at the answer. You can do this in the laplace domain or using traditional kvl/kcl. Either way, you will have to convert the resulting equation into a transfer function.
 
Dec 30, 2004
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V.o= V.i((R.1R.2C+(R.1CL+R.1)s)(s))/((R.2CL+1)(CL+1)s^2)

inductor time constant is L*s, capacitor time constant is s/C, so from there it's just voltage division to get the node to the right of R.1, and then voltage division again to get the node to the right of L.

Is this what you were looking for?
 
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