I think that if the video quality was improved, these lectures would gain more traction. Big thanks to the uploader and Dan Gelbart for this awesome content.
1:06:10 - I take issue with the statement "the phase shift can be much more than 180 degree" --> How? In a 2nd order resonant system, it shall be _exactly_ 180 degree at its worst. A sharp resonance-peak makes the transition from 0 deg to 180deg _much_ faster, when sweeping frequency, but it will never exceed 180 deg, no matter how high the system-Q. @OP - do you perhaps have an explanation to the meaning of this statement?
@@theA731N Okay, for the sake of argument, plot x/F [m/N] i.e. 'mechanical gain' on the y-axis and frequency [rad/s] on the x-axis. For the 2nd order system, let the mass be *m* and stiffness of your cantilever be *k* . The resonance-peak shall appear at *sqrt(k/m)*. Far before resonance, the x and the F shall be in phase, i.e. the so called "stiffness line" following F = k.x , i.e. the phaseshift shall be zero. Far right of the resonance peak, the mass shall dominate on the co called "mass line", primarily following F = 1/(m.s^2). The s^2 in the denominator already tells you that the phaseshift shall be exactly -180 degrees. between the left of the resonance and the right of the resonance, lies the resonance itself. Here, the system-damping shall dominate the response, and the phaseshift between F and dx/dt shall be zero, in other words, the phaseshift of x/F shall be -90 degree (aka -pi/2 rad). I agree that the rate of change of phaseshift shall be enormous, for a system with low damping, i.e. a high q-factor, as the value rapidly changes from 0 to -180deg ; however, the phaseshift shall merely be pi/2 near the resonance-frequency.
The phase shift can be much more than 180 degrees. One example is if you have a delay in the loop. Imagine a delay line of T in the loop followed by a resonant system resonating at frequency f. If 1/f
@@dgelbart Thanks for the clarification! I was thinking of the "second order system" you mentioned at 1:05:50. An ideal delay-line could indeed introduce phaseshift without increasing system-order. Still one thing bugging me, though - you mention about "the slope of the plot" 1:06:35 corresponding with the phaseshift. I completely agree on the "-6dB/octave" part. However, near the resonance-peak itself, (in the absence of delay lines, but for a high Q-factor) there can still be a large slope, without it being indicative of a large phaseshift - what am I missing?
To clarify further: what I said in the lectures needs correction. It is true that in a multi-pole system each pole adds 90 deg phase shift and 6dB/Octave slope to the Bode plot. It is also true that a second order (2 pole) resonant system can only add 180 phase shift, as you pointed out. However, every amplifier or "plant" has a built in delay and the delay adds a phase shift that can be arbitrarily high. Actually, without this delay a second order negative feedback system will never oscillate as the phase shift never reaches 180 degree, it just approaches it assymptotically. Thanks for pointing this out! Dan
There's an interesting phenomenon with audiophiles swapping normal opamps for high bandwidth ones and those new opamps oscillating at like 10mhz and dissipating a watt....
~20min mark talking about distortion is a bit unclear, the amplifier doesn't naturally cancel or fight any distortion, it is just the characteristics of a typical real world amplifier are such that, when used at low gain, they will not introduce a lot of distortion, that is a byproduct of the technology at the time (and to some extent even now), not a law related to amplifiers in general that is why it is quite common in electronics to see couple opamps in series to do work which one could do, and a lot of times extra opamp is "free" anyway, since there usually is a number of them in a package
An open loop amplifier can only add distortion, not reduce it. An amplifier with negative feedback, like an op-amp, actively reduces distortion, theoretically to zero if the gain is high enough. This distortion reduction is true even if the amplifier itself causes distortion.
Thank you for posting Dan Gelbart lectures! Great practical information.
I think that if the video quality was improved, these lectures would gain more traction.
Big thanks to the uploader and Dan Gelbart for this awesome content.
I am so hype to watch all these. Thank you so much
@cyclosgarage Hello, I watched also all of your videos! It is great so see whats possible. Good luke! : ) greetings from germany
It would be more helpful for students if you could upload a higher quality video. Hard to read the board.
On Governors is a must read!
To bad the resolution is so bad. I can't read the equations.
it is so great to see all those lectures. Thanks a lot. How did you become the videos? Are you student of Gelbart?
1:06:10 - I take issue with the statement "the phase shift can be much more than 180 degree" --> How? In a 2nd order resonant system, it shall be _exactly_ 180 degree at its worst. A sharp resonance-peak makes the transition from 0 deg to 180deg _much_ faster, when sweeping frequency, but it will never exceed 180 deg, no matter how high the system-Q. @OP - do you perhaps have an explanation to the meaning of this statement?
I think he’s speaking in terms of the feedback control levels.
@@theA731N Okay, for the sake of argument, plot x/F [m/N] i.e. 'mechanical gain' on the y-axis and frequency [rad/s] on the x-axis. For the 2nd order system, let the mass be *m* and stiffness of your cantilever be *k* . The resonance-peak shall appear at *sqrt(k/m)*. Far before resonance, the x and the F shall be in phase, i.e. the so called "stiffness line" following F = k.x , i.e. the phaseshift shall be zero. Far right of the resonance peak, the mass shall dominate on the co called "mass line", primarily following F = 1/(m.s^2). The s^2 in the denominator already tells you that the phaseshift shall be exactly -180 degrees. between the left of the resonance and the right of the resonance, lies the resonance itself. Here, the system-damping shall dominate the response, and the phaseshift between F and dx/dt shall be zero, in other words, the phaseshift of x/F shall be -90 degree (aka -pi/2 rad).
I agree that the rate of change of phaseshift shall be enormous, for a system with low damping, i.e. a high q-factor, as the value rapidly changes from 0 to -180deg ; however, the phaseshift shall merely be pi/2 near the resonance-frequency.
The phase shift can be much more than 180 degrees. One example is if you have a delay in the loop. Imagine a delay line of T in the loop followed by a resonant system resonating at frequency f. If 1/f
@@dgelbart Thanks for the clarification! I was thinking of the "second order system" you mentioned at 1:05:50. An ideal delay-line could indeed introduce phaseshift without increasing system-order.
Still one thing bugging me, though - you mention about "the slope of the plot" 1:06:35 corresponding with the phaseshift. I completely agree on the "-6dB/octave" part. However, near the resonance-peak itself, (in the absence of delay lines, but for a high Q-factor) there can still be a large slope, without it being indicative of a large phaseshift - what am I missing?
To clarify further: what I said in the lectures needs correction. It is true that in a multi-pole system each pole adds 90 deg phase shift and 6dB/Octave slope to the Bode plot. It is also true that a second order (2 pole) resonant system can only add 180 phase shift, as you pointed out. However, every amplifier or "plant" has a built in delay and the delay adds a phase shift that can be arbitrarily high. Actually, without this delay a second order negative feedback system will never oscillate as the phase shift never reaches 180 degree, it just approaches it assymptotically. Thanks for pointing this out!
Dan
very nice to see more Dan Gelbart videos! But the quality is not good...
There's an interesting phenomenon with audiophiles swapping normal opamps for high bandwidth ones and those new opamps oscillating at like 10mhz and dissipating a watt....
~20min mark talking about distortion is a bit unclear, the amplifier doesn't naturally cancel or fight any distortion, it is just the characteristics of a typical real world amplifier are such that, when used at low gain, they will not introduce a lot of distortion, that is a byproduct of the technology at the time (and to some extent even now), not a law related to amplifiers in general
that is why it is quite common in electronics to see couple opamps in series to do work which one could do, and a lot of times extra opamp is "free" anyway, since there usually is a number of them in a package
An open loop amplifier can only add distortion, not reduce it. An amplifier with negative feedback, like an op-amp, actively reduces distortion, theoretically to zero if the gain is high enough. This distortion reduction is true even if the amplifier itself causes distortion.
The quality is fine you must have been born recently if you think otherwise. And what do you want them to do exactly, go back in time?