Transfer Function at Small Complex Frequency

Discussion in 'Homework Help' started by tquiva, Feb 24, 2011.

  1. tquiva

    Thread Starter Member

    Oct 19, 2010
    176
    1
    Could someone please help me with this problem:

    [​IMG]

    (a) Use physical arguments to determine the value of the transfer function at very small values of the complex frequency, and at very large values of the complex frequency.
    (b) Let the input to the above circuit be unit for negative times and zero for positive times. If possible, determine the value of the circuit output at negative times, just after the input switched to zero, and at very large times.

    I know that when the frequency is low, then the capacitor becomes an open circuit. When the frequency is high, the capacitor becomes a short.

    I redrew the circuit for a low frequency with opens at both capacitors. I see the the input voltage becomes disconnect from the positive terminal of the op-amp, but is still connected to the negative terminal. Does this mean that the output voltage at low frequency is a 0?

    At a high frequency, I redrew the circuit again, and noticed that there are shorts connected to both the 5R and 3R resistors. Does this mean I remove those resistors from the circuit as well?

    For part (b), could someone please point me in the right direction to get started?
     
    Last edited: Feb 24, 2011
  2. Georacer

    Moderator

    Nov 25, 2009
    5,142
    1,266
    At low frequencies, the capacitor is open indeed, but be careful. Pin- is still connected to the input, and pin+ is still at the same voltage as pin-. This is enough to produce a legit output.
    At high frequencies you may disregard 3R and 5R, since the output is the same as pin+, which is the same as pin-.

    For the second part, unless you have some data about the OpAmp itself, I guess you have to recall something you have said in class.
     
    tquiva likes this.
  3. tquiva

    Thread Starter Member

    Oct 19, 2010
    176
    1
    For the second part:
    At t<0, the switch is closed for a long time. Therefore, the circuit is in steady state and the capacitors become open.

    Is this correct?
     
  4. Georacer

    Moderator

    Nov 25, 2009
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    Yes, I thought we had established that. The same goes for t>>0.
     
  5. tquiva

    Thread Starter Member

    Oct 19, 2010
    176
    1
    the capacitors are also open for t > 0 ?
     
  6. Georacer

    Moderator

    Nov 25, 2009
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    Yes, but long after 0, not directly after it. That is the meaning of "The same goes for T>>0". The voltage is stable (DC) at that time, isn't it?
     
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