1. Not finding help here? Sign up for a free 30min 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!

Will an infinite impedance cause no phase shift?

  1. Jul 22, 2016 #1
    1. The problem statement, all variables and given/known data

    0Zc8nQe.png

    imgur link: http://i.imgur.com/0Zc8nQe.png

    2. Relevant equations

    Y-Delta transformations

    Star%2Bto%2BDelta%2Band%2BDelta%2Bto%2BStar%2BTransformation.png

    3. The attempt at a solution

    Since it's a proof, I can't check the answer in the back.

    What I did: I transformed the three impedances in their delta config to a Y config, and my TI89 told me that the only impedance in the Y config that caused a phase shift (by virtue of the entire impedance value being imaginary), was that single impedance connecting down to common ground from in between the other two.

    The calcs wound up as

    [tex]Z_1 = \frac{CL_1L_2\omega^3j}{C(L_1+L_2)\omega^2-1}[/tex]
    [tex]Z_2 = \frac{CL_2L_1^2\omega^4}{(C(L_1+L_2)\omega^2-1)^2}[/tex]
    [tex]Z_3 = \frac{CL_2^2L_1\omega^4}{(C(L_1+L_2)\omega^2-1)^2}[/tex]

    Z1 being the one that connects to ground from in between 2 and 3 which are in the op-amp loop.

    Making Z1 go to infinity by making it's denominator go to zero when the frequency is the fundamental frequency, gets us the equation we need to prove.

    I'm fine with the mathematical reasoning.

    I just don't feel comfortable with the idea that this theoretically infinite impedance is causing no phase shift? Why does current have to flow through a component in order to be phase shifted?
     
  2. jcsd
  3. Jul 22, 2016 #2

    Simon Bridge

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    What is the physical meaning of infinite impedence in this case? What has to happen for the impedence to get bigger?
     
  4. Jul 23, 2016 #3

    NascentOxygen

    User Avatar

    Staff: Mentor

    In the Hartley there is some coupling between L1 and L2.
     
  5. Jul 23, 2016 #4

    NascentOxygen

    User Avatar

    Staff: Mentor

  6. Jul 25, 2016 #5

    LvW

    User Avatar

    kostoglotov - in case you are still interested in finding the oscillation frequency, here is what you must do:

    1.) Insert an additional resistor R0 between the opamp output and the rest of the feedback network. This is necessary - otherwise the inductor L1 has no meanning and no influence - and the circuit does not oscillate.

    2.) Determine the feedback function HF(s) including the resistor Ri (which can be considered to be grounded - due to the virtual ground principle for the inverting opamp)
    The result will be:
    HF(s)=numerator/denomionator=N(s)/D(s) with
    N(s)=(C L1 L2 Ri) s^3
    and
    D(s)=R0 Ri + ( L2 R0 + L1 Ri) s + (C L2 R0 Ri + C L1 R0 Ri + L1 L2)s^2 + (C L1 L2 Ri + C L1 L2 R0) s^3.

    3.) After setting s=jω we have a complex function - and we need the frequency where the phase shift is -180deg (because the inverter will contribute another 180 deg).
    That means: The imag. part of the transfer function HF must be ZERO (the result must ne negative-real).
    Because the numerator N(jw) always is imaginary, we require that the denominator also must be imaginary at the desired frequency.
    Hence, we are setting the real part of the denominator equal to zero: Re[D(jw)]=0.
    Solving this equation, we get the oscillation frequency.

    4.) Proof: If the resistor Ri is very large (infinite), the resulting oscillation frequency will be as given in the task description (post#1).
    Note that this expression for the oscillation frequency is an approximation only because it may be applied for Ri⇒infinite only!
     
    Last edited: Jul 25, 2016
  7. Jul 25, 2016 #6

    NascentOxygen

    User Avatar

    Staff: Mentor

    There seems to be some misunderstanding on how the Hartley works. It needs some selectivity in the loop so that the frequency is predictable, but phasing is the responsibility of the transformer. With the centre-tap earthed, the second winding on the autotransformer provides the required 180° phase shift. The turns ratio can make up for losses and restore loop gain to a little over unity.
     
  8. Jul 29, 2016 #7

    NascentOxygen

    User Avatar

    Staff: Mentor

    The homework help at Physics Forums usually works well, but sadly this thread has not. Despite a storm of technical talk, the responses have contributed nothing that addresses the OP's question on his level. Here's a reminder of what is being asked:
    From the formula given for the oscillation frequency it is clear that only the general principle of operation is under discussion, nothing more complicated than resonance in one of the feedback loops. A treatise on oscillator minutiae is not called for, nor do we need drawn-out debate on what this configuration should or should not be named; all this is off-topic and off-putting, and doing nothing to assist the struggling student with his question. In fact, quite the contrary: when a simple question starts a bunfight the first one to flee is often he who asked the question.

    Ralph Hartley was around for over 50 years while the Hartley Oscillator with its tapped coil was popular among radio hobbyists, so he had ample opportunity to disassociate himself and his name from the oscillator in that form if he objected to the association, and he clearly did not. So decades after his death he doesn't need anyone to go into bat on his behalf now.

    Regrettably, it looks like OP has long ago departed the discussion, and none the wiser for his asking.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Will an infinite impedance cause no phase shift?
  1. Phase shifting ckt (Replies: 26)

  2. Filter Phase Shift (Replies: 10)

  3. Phase shift oscillator (Replies: 26)

Loading...