MIT Electronic Feedback Systems (1985) View the complete course: ocw.mit.edu/RES6-010S13 Instructor: James K. Roberge License: Creative Commons BY-NC-SA More information at ocw.mit.edu/terms More courses at ocw.mit.edu
the video before this one made more sense to me, he shows waveforms on the scope. I was a tech with AT&T and learned most of the job by hands on experience.
a voltage regulator is a Voltage controller, a Constant Current source is current controller and like that a Phase locked loop is a more or less a Phase controller or frequency controller , every thing is in a loop , we have a reference and also an error amplifying device and some sort of control mechanism , all are one and the same if our view is abstract, that's it and may the Professors Soul Rest in Peace
What a concert of science . Thanks professor ! I just try to watch other video on the same subject and it was senseless mumbling in bad English with lots of math . I had imposition that guy try to convince him self about his own greatness .
An instantaneous frequency is the derivative of phase in the time domain, and reciprocally, the instantaneous phase is the integral of instantaneous frequency.
@Ralph M Look if p is the phase of a signal x(t) = sin(2*pi*f*t) in degrees. Then p = 360*t*f, so I see how integrating a constant frequency value will give us phase. But since the frequency is constant - 8:28 you can't have a domain of it. no df is possible right? Also @panier66 I agree with you, but how does this relate to what the prof is saying - integrate frequency error and get phase error. Any idea??
8:25 I don't see what he means by phase error between two signals of _different_ frequencies. I mean sure you can find the difference in frequencies when you have two signals with different constant frequencies. But Phase difference is *defined as* the difference, expressed in degrees or radians, between two waves having the same frequency and referenced to the same point in time. I really don't see how integrating the frequency error (a constant since he says both are constant and different) gives us the phase error.
Technically you are correct, but the term is used loosely because it's understood that the feedback clock will never be EXACTLY the same frequency as the reference clock (although, that is what the goal is, along with getting the rising edges to be perfectly aligned.)
I think it's done because that's what the circuit does.. Just find what percentage of the waveform of a signal has been traversed. (π/2 , π/4...) Do the same for another signal. And find a phase. Subtract to get the phase difference.
the phase definition to which you are referring is for mathematics. in electrical engineering, phase is the integral of angular frequency or the x in sin(x); with the definition which he describes at 8:13 you can compare the phase of any two signals even with different frequencies.
Then the phase detector will produce an error signal until the feedback clock rising edge is aligned with the refclock rising edge. One thing to note is that you never get "perfect" phase and frequency alignment. The loop is always doing some error correction. Sometimes the error correction is very small, like when the loop is "locked". This is why you always have finite jitter. Jitter is never zero, although that is ideal.
Assuming you start same freq and same phase, and then alter the phase of input sig : the pll sig would at first go up or down in freq, but a moment later would adjust to original freq, but with the new phase.
@@nicholaselliott2484 thanks for saying that. There's a steady state adjustment like with ordinary AC circuits. Following a phase change to the input signal, the XOR (or another phase comparator) would drive the oscillator with a different voltage, making the frequency jump momentarily. But this can't be a steady state, as the feedback would work to bring the oscillator signal back to the input signal's frequency, but with the new phase. (You understood everything quickly, but in case the original comment wasn't enough for somebody viewing this later, added a bit of extra).
