Quick question regarding some algebra?

  • Thread starter Thread starter papasmurf
  • Start date Start date
  • Tags Tags
    Algebra
Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
1 reply · 1K views
papasmurf
Messages
21
Reaction score
0
The gain of an internally compensated op amp may be expressed by

Equation 1) A(s) = A0 / ( 1 + s/[itex]\omega[/itex]b) where [itex]\omega[/itex]b is the 3dB frequency or "break frequency" and A0 is the dc gain.

Plugging in s=j[itex]\omega[/itex] and [itex]\omega[/itex] = 2*pi*f gives

Equation 2) A= A0 / (1 + j*f / fb) where fb is still the "break frequency"

I understand how all of that works.

What I don't understand is how you can go from Equation 2 to this:

Equation 3) A = A0 / sqrt( 1 + (f/fb)^2).

I hope all of this makes sense and someone can help me. Thank you.
 
Physics news on Phys.org
papasmurf said:
The gain of an internally compensated op amp may be expressed by

Equation 1) A(s) = A0 / ( 1 + s/[itex]\omega[/itex]b) where [itex]\omega[/itex]b is the 3dB frequency or "break frequency" and A0 is the dc gain.

Plugging in s=j[itex]\omega[/itex] and [itex]\omega[/itex] = 2*pi*f gives

Equation 2) A= A0 / (1 + j*f / fb) where fb is still the "break frequency"

I understand how all of that works.

What I don't understand is how you can go from Equation 2 to this:

Equation 3) A = A0 / sqrt( 1 + (f/fb)^2).

I hope all of this makes sense and someone can help me. Thank you.

That's the magnitude of the complex gain... See Equation #9 here for example:

http://cheme.eng.wayne.edu/neuron/auth/ece3310/experiment_12.htm

.
 
Last edited by a moderator: