Razavi Electronics2 Lec43: Intro. To Instability in Feedback Systems

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  • เผยแพร่เมื่อ 2 พ.ย. 2024

ความคิดเห็น • 19

  • @coolwinder
    @coolwinder 3 ปีที่แล้ว +8

    01:25 - Intro and Review
    04:05 - Observation: Delay in the Loop
    16:45 - Analysis for Instability
    33:40 - Examination of a Close-Loop System in Time-Domain
    40:40 - Review of Bode's Rules

  • @devanshrajoria5418
    @devanshrajoria5418 4 ปีที่แล้ว +2

    Thank you so much for this awesome video sir. I can now relate it to what i have learnt from college days.

  • @zinhaboussi
    @zinhaboussi 6 หลายเดือนก่อน

    01:16 Negative feedback systems can become unstable, leading to oscillations.
    04:33 Explanation of loop gain calculation and its impact on feedback systems
    12:47 Long delay in feedback system can change negative feedback to positive feedback
    16:05 Feedback system instability due to phase shift and delay
    21:50 When the loop gain is equal to minus one at a particular frequency, the system becomes unstable.
    24:34 Instability in loop systems requires avoiding specific conditions.
    31:05 Positive feedback causes system instability
    34:27 Exploring the waveforms in a feedback system with a sinusoidal input.
    39:40 Understanding Bode's rules for constructing frequency response
    42:26 Understanding phase contributions in transfer functions
    Crafted by Merlin AI.

  • @dhananjaykumar5114
    @dhananjaykumar5114 3 ปีที่แล้ว +1

    Very Good Lecture

  • @saicharanmarrivada5077
    @saicharanmarrivada5077 4 ปีที่แล้ว

    Excellent!!

  • @jojoesekyisagoe
    @jojoesekyisagoe 6 ปีที่แล้ว +1

    Thank you Sir

  • @NaaJeevitham500
    @NaaJeevitham500 5 ปีที่แล้ว +1

    is instability important only for negative feedback system? not for CS amplifier?

    • @ssayank
      @ssayank 5 ปีที่แล้ว +1

      Yes, most of the amplifier is employed in negative feedback, because of of several advantages. For the CS stage, without negative feedback, the amplifier has only one pole, so it is inherently stable.
      The problem generally comes when we have negative feedback. Even if we have a stable open loop, the closed-loop can become unstable. That's why we generally keep sufficient phase-margin.

  • @sonnyasu
    @sonnyasu 3 ปีที่แล้ว +1

    Shouldn’t we consider the sign of zero? Left hand plane acts a phase up but right hand acts like a pole

  • @muratt6894
    @muratt6894 6 ปีที่แล้ว +1

    Thank You

  • @yihou6433
    @yihou6433 2 ปีที่แล้ว

    It's clear to see instability from closed-loop gain transfer function as the denominator goes to zero. But the explanation from loop gain test 29:00 is reluctant. To make Vf enhances Vtest, we just need phase shift to be 180 degree without the requirement of unity gain. So how can we explain the unity amplitude is required for instability from "loop gain test"?

    • @mohammedabdulhakabdullaabd1121
      @mohammedabdulhakabdullaabd1121 ปีที่แล้ว

      Unity LOOP GAIN KA1 (not unity Gain A1) is required to be having a magnitude of 1 and a phase of 180 degrees shift, which is the same as having the dominator of the transfer function (1+KA1) =0 .

    • @golluiz
      @golluiz 8 หลายเดือนก่อน

      I have the same question. Using the same snapshots excercise (from 34:00 to 40:00) that prof. used. Let's say, for instance, that KH(jw0) is 0.5 (So less than 1) with 180 phase shift. Applying a sinewave with amplitude 1 in X, Vf will present a 180 degree phase shift sinewave with 0.5 amplitude. The subtractor will generate a sinewave that is 1 - (-0.5) = 1.5. That will generate a 0.75 sinewave at VF and so on. So any value of KH(jw0) that is bigger than 0 will add up the signals and become unstable. Could tell me where is the flaw in my reasoning?

    • @hardikjain-brb
      @hardikjain-brb 7 หลายเดือนก่อน

      You will reach input*2 if you keep doing it indefinately @@golluiz
      Even if you take -0.8 you will reach @ 4*input

    • @hardikjain-brb
      @hardikjain-brb 7 หลายเดือนก่อน

      It will converge to 1/1+KH for KH=-0.5 itll go to 2*input
      Because the final output is H/1+KH and the final converged loop output is the input of the block H

  • @ssayank
    @ssayank 5 ปีที่แล้ว +1

    Now. I get the confusion, during the break of this loop, we are not considering the loading effect. Is there any reason? Won't we consider the loading during frequency analysis of if feedback amplifier?
    I get the same confusion when I do the stability analysis of common-mode feedback.

    • @erwinshad9143
      @erwinshad9143 4 ปีที่แล้ว +1

      I think you can assum that the loading effect is considered inside the H(s)

    • @yadukrishnanm6280
      @yadukrishnanm6280 3 ปีที่แล้ว

      I think it's because as Professor mentioned...that the feedback is mostly frequency independent..hence the reason