Transfer Function of a Circuit

Sounds like a homework question.

Use the theories and methods that were taught in class lecture and are found in the text.

Sounds like a homework question.

Indeed it does. This is from the CR4 FAQ:

Do your own homework. CR4 is not a homework cheat site. While some here might relish the opportunity to sharpen up old rusty skills by working the homework problem, consider this and consider it well. If you cheat on your homework by using someone else's answers, you are only cheating yourself, because the purpose of any homework or other college assignments is to help you learn, by practice, repetition, and self-discovery.

Sorry, dilipkumar, I expect the CR4 admin will soon disable this thread.

regards,

Vulcan

Arrogant fools!

Think back to your own days in school. What would have been the result if every instructor would have told you to "do your own homework"?

Offering a bit of help on a difficult problem DOES NOT constitute cheating!

Knowledge is meant to be shared.

(He did just ask how to proceed... maybe he just wants a hint or some ideas / help?)

Consider the voltage sources: the +5V and the Vin. Your resistors do not appear to be dependant - they look to be constant.

You are looking to find the Vout at the 'long line' in the picture, as I understand.

Thus; how does the 5V affect that point? What about Vin?

What about both the 5V and Vin at the same time?

Ohm's law?

KVL?

What would help to figure out what each voltage is doing to that point?

Yeah! What he said.

Sometimes it helps to find "equivalents" so as to minimize the complexity of circuits.

Nice homework...

If it had been a homework question, he wouldn't have had that misleading formula to start off with: so I'm going to help if I can.

First put all the inputs and variables in a spread sheet.

Now imagine that the junction of R1/R2 is not connected to R5 and calculate the voltage (V₁₂) at that junction.

Now calculate the parallel value of R1 and R2 (RP₁₂).

As far as R5 is concerned looking into R1/R2 it looks like RP₁₂ to V₁₂

Now the R1 R2 R5 combination just looks like (R5 + RP₁₂) to V₁₂

Do the same for R3/R4 as for R1/R2: so that you have RP₃₄ to V₃₄

Now you can calculate the voltage at the junction of R3, R4 and R5. It's just a potential divider with (R5 + RP₁₂) to V₁₂ on one side and RP₃₄ to V₃₄ on the other side.

That's the voltage; it has a source impedance of (R5 + RP₁₂) in parallel with RP₃₄.

Look up Thevenin's theorem on Wikipedia (or anywhere else) for more information.

Hello,

I feel that my question is misunderstood. We can get the equation easily with electrical network theory and the equation that i got is

Vout={Vx*(R3 parallel R4)}/{Ry+(R3 parallel R4)}+{Vin*(Ry parallel R4)}/{R3+(Ry parallel R4)}

Where, Ry= R5+(R1 parallel R2) and Vx=(5V/R1)*(R1 parallel R2).

but my point of discussion is that can we get the same expression in terms of co-efficients of variables(below is the format i am looking the expression to be). because that would be a better expression for a designer who just wants to change the value of resistors to get diffrent level of Vout.(here vout= .524 to 2.602V with values as shown)

Vout = C + (x)Vin + (a)R1 + (b)R2 + (c)R3+ (d)R4 + (e)R5

is there any simulation tool which simulates and gives the expression at selected node?

Please indicate the location of Vout in your circuit

Vout is the junction of R3, R4 and R5 as shown with a long line.

Frankly Dilip, I think the answer is no: all the component effects are interdependent. For example if you make R5 huge compared to the other components then R1 and R2 become irrelevant; if you make R3 very small compared to the others then V_out just equals V_in.

What sort of terms were you expecting in positions C, (x), (a), (b), (c), (d) and (e) ?

Any free Spice simulator will do what you want, but, why not just use a spread sheet?

okay Randall. Here C, (x), (a), (b), (c), (d) and (e) are the co-efficients. Actually long back i tried regression analysis on other network analysis. For Example,for a voltage divider, Vout(across R2)=-0.5R1+0.5R2+0.5vin. Definately the Excel sheet solves the needs.

Why not redefine the values of all resistors as "1" since you want to assess ratios vs. Vout. Using a "unit value" viewpoint can be instructive.

Can you please elaborate your thought