Simple transfer function - algebra giving me problems

In summary, the transfer function for a non-inverting amplifier is given by G(s) = \frac{-R_2(C_1)}{(C_1C_2R_1R_2s^2 + (C_1R_1 + C_2R_2)s + 1)}.
  • #1
trickae
83
0

Homework Statement



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

Homework Equations



[tex]Z_1(s) = R_1 + \frac{1}{C_1s}[/tex]
[tex]Z_2(s) = \frac {\frac{R_2}{C_2s}}{R_2 + \frac{1}{C_2s}}[/tex]final solution should be:
[tex]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}[/tex]

The Attempt at a Solution



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

[tex]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}}[/tex]

[tex] = -\frac {\frac{R_2}{C_2s}}{(R_2 + \frac{1}{C_2s})(R_1 + \frac{1}{C_1s}) }[/tex]

[tex] = \frac{-R_2}{(C_2s)(R_2 + \frac{1}{C_2s})(R_1 + \frac{1}{C_1s})} [/tex]

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

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

[tex]\frac{Z_1(s) + Z_2(s)}{Z_1(s)}[/tex] - now i get the right answer
 
  • #3
trickae said:
I misread the problem - apologies - the transfer function for a Non inverting amplifer is in the form

[tex]\frac{Z_1(s) + Z_2(s)}{Z_1(s)}[/tex] - 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 ;)
 

FAQ: Simple transfer function - algebra giving me problems

What is a simple transfer function?

A simple transfer function is a mathematical representation of a system's input-output relationship. It describes how the output of a system is affected by its input, and is typically expressed as a ratio of polynomials in the complex variable s.

How is a transfer function different from an algebraic equation?

A transfer function is a type of algebraic equation that specifically relates to a system's input-output relationship. Unlike a traditional algebraic equation, which may have multiple variables and unknowns, a transfer function only has one independent variable (usually s) and represents the output in terms of the input.

What are some common problems encountered when working with transfer functions?

Some common problems when working with transfer functions include solving for unknown coefficients, simplifying complex expressions, and determining the stability of a system based on its transfer function. It is important to have a strong foundation in algebra and complex numbers to effectively work with transfer functions.

How do I solve problems involving transfer functions?

To solve problems involving transfer functions, it is important to first understand the properties and characteristics of transfer functions. This includes knowing how to manipulate them algebraically, understanding the effects of different parameters on the transfer function, and being able to interpret graphical representations of transfer functions.

What are some applications of transfer functions?

Transfer functions are widely used in the fields of engineering, physics, and mathematics to model and analyze systems. They are particularly useful in control systems, signal processing, and circuit analysis. Transfer functions can also be used to design and optimize systems for desired performance and stability.

Similar threads

Replies
7
Views
2K
Replies
4
Views
1K
Replies
15
Views
2K
Replies
36
Views
4K
Replies
1
Views
1K
Replies
16
Views
1K
Replies
5
Views
2K
Back
Top