Simple transfer function - algebra giving me problems

AI Thread Summary
The discussion revolves around finding the transfer function for a circuit involving impedances Z_1(s) and Z_2(s). The initial attempt at deriving the function led to confusion, as the user misread the problem and incorrectly applied the formula for a non-inverting amplifier. After realizing the mistake, they corrected their approach to use the appropriate transfer function form, which is Z_1(s) + Z_2(s) over Z_1(s). The correct final solution involves a more complex polynomial expression in terms of s. Overall, the thread highlights the importance of accurately interpreting circuit configurations in transfer function calculations.
trickae
Messages
82
Reaction score
0

Homework Statement



Find a transfer function: \frac{V_o(s)}{V_i(s)} = \frac{Z_2 (s)}{-Z_1(s)}

Homework Equations



Z_1(s) = R_1 + \frac{1}{C_1s}
Z_2(s) = \frac {\frac{R_2}{C_2s}}{R_2 + \frac{1}{C_2s}}final solution should be:
G(s) = \frac{V_o(s)}{V_i(s)} = \frac{C_1C_2R_1R_2s^2 + (C_2R_2 + C_1R_2 + C_1R_1)s + 1}{C_1C_2R_1R_2s^2 + (C_1R_1 + C_2R_2)s + 1}

The Attempt at a Solution



- Give me a second I'm still typing up the latex commands

G(s) = \frac{V_o(s)}{V_i(s)}= \frac{-\frac {\frac{R_2}{C_2s}}{R_2 + \frac{1}{C_2s}}}{R_1 + \frac{1}{C_1s}}

= -\frac {\frac{R_2}{C_2s}}{(R_2 + \frac{1}{C_2s})(R_1 + \frac{1}{C_1s}) }

= \frac{-R_2}{(C_2s)(R_2 + \frac{1}{C_2s})(R_1 + \frac{1}{C_1s})}

= \frac{-R_2(C_1C_2s)}{(C_2s)(C_1C_2R_1R_2s^2 + (C_1R_1 + C_2R_2)s + 1)}

=\frac{-R_2(C_1)}{(C_1C_2R_1R_2s^2 + (C_1R_1 + C_2R_2)s + 1)}
which is no where near the solution.
 
Last edited:
Physics news on Phys.org
I misread the problem - apologies - the transfer function for a Non inverting amplifer is in the form

\frac{Z_1(s) + Z_2(s)}{Z_1(s)} - now i get the right answer
 
trickae said:
I misread the problem - apologies - the transfer function for a Non inverting amplifer is in the form

\frac{Z_1(s) + Z_2(s)}{Z_1(s)} - now i get the right answer

hehe good. Cause I quickly did it, and definitely did not get the "answer".

I feel bad for you typing all of that up in Latex. Probably took a few ;)
 
Thread 'Have I solved this structural engineering equation correctly?'
Hi all, I have a structural engineering book from 1979. I am trying to follow it as best as I can. I have come to a formula that calculates the rotations in radians at the rigid joint that requires an iterative procedure. This equation comes in the form of: $$ x_i = \frac {Q_ih_i + Q_{i+1}h_{i+1}}{4K} + \frac {C}{K}x_{i-1} + \frac {C}{K}x_{i+1} $$ Where: ## Q ## is the horizontal storey shear ## h ## is the storey height ## K = (6G_i + C_i + C_{i+1}) ## ## G = \frac {I_g}{h} ## ## C...

Similar threads

Back
Top