1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Low-Pass Amplifier Bode Plot

  1. Apr 7, 2014 #1
    1. The problem statement, all variables and given/known data

    I need some help with the following problem (it is from an old test paper).

    Below is the Bode plot of the open loop gain of an amplifier:


    A constant fraction β=0.1 of the output is fed back to the input. This feedback does not load the amplifier.

    Using the Bode plot determine the frequency for which the magnitude of the loop gain of this circuit is equal to unity.

    3. The attempt at a solution

    Clearly looking at the first graph ##|A|_{dB} = 1## at around ##1500 \ Hz##. I've marked this with a pen on the graph. But my answer was marked as incorrect. Why is that?

    Clearly the graph touches 1 dB roughly at ##f = 1500 \ Hz##. So why is my answer wrong? :confused:

    Any help is greatly appreciated.

    P.S. The value of the amplifier's closed loop gain at DC was calculated to be -9.1.

    Attached Files:

  2. jcsd
  3. Apr 7, 2014 #2


    User Avatar

    Staff: Mentor

    If loop gain is defined as A.β
    and you are told β=0.1, then you are looking for f where A=10

    (I like to write this as 10 volts/volt, whatever, as a reminder it is not 10 dB)

    So, how many dB gain are you looking for?

    NOTE: 'loop gain' is quite different to 'closed loop gain'. Can you distinguish the difference?
  4. Apr 7, 2014 #3
    Thank you so much for your response.

    So, in decibel it is ##|A|_{dB}=20 \ log_{10} 10 = 20 \ dB##, therefore the frequency would be @ f=500 Hz. Is that right??

    Yes, I distinguish the difference, loop gain is Aβ whereas closed loop gain is ##A(s)/1-A(s) \beta (s)##.
  5. Apr 7, 2014 #4


    User Avatar

    Staff: Mentor

    That should be right.

    I would mark that last expression wrong, because it is missing an essential set of parentheses. :frown:
  6. Apr 8, 2014 #5
    I meant ##A(s)/[1-A(s) \beta (s)]##. Thank you very much.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted