19. Phase-locked Loops

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Күн бұрын

Пікірлер: 42

@andrzejpl9897 10 ай бұрын

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 .

@ElectronicsNotes 8 жыл бұрын

Well explained and a very useful introduction to phase locked loop technology. Thanks.

@brig.4398 8 жыл бұрын

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.

@w0313440 Жыл бұрын

RIP Prof Roberge .

@435345dfhgjs 11 жыл бұрын

I didn't know that Nixon knew so much about electronics... BTW great video

@automationsolution 7 жыл бұрын

handsome hunk he is

@glitter-pixiet2335 8 жыл бұрын

sir,thank u so much for this lecture,i completely understood everything about phase locked loop, you are really great sir.

@xephyr417 8 жыл бұрын

he's dead :/

@glitter-pixiet2335 8 жыл бұрын

i know but still

@bhavani1089shankar 10 жыл бұрын

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

@findharsha 10 жыл бұрын

***** newsoffice.mit.edu/2014/james-k-roberge-professor-of-electrical-engineering-dies-at-75

@arnavrawat9864 4 жыл бұрын

Good observation. It would be much easier if people were taught this unifying philosophy and then explained these 3 systems.

@coderhex1675 7 жыл бұрын

he is like an artist

@bayareapianist 21 күн бұрын

Interestingly, nothing has changed in the last 40 years. If it were a computer science lecture, there would be a patch added every week.

@mikedrz 7 жыл бұрын

You can tell this is older, because it's in English. Now a days you can only find these lectures in Hindi.

@ashwin372 2 жыл бұрын

maybe , sometimes its indian guy explaining in english

@yaswanthraparti8641 Жыл бұрын

@LinuxDebugger you are free to not listen to the content you don't like instead of making useless, condescending remarks.

@7yep4336dfgvvh 6 ай бұрын

@yaswanthraparti8641 they're not useless though

@dasilvaleandro21 7 жыл бұрын

That's a very nice lecture. Would be amazing to be in that class, I wasn't even born though :(

@hrh2092 8 ай бұрын

PLL and he used Fc and Fr in block diagram analysis. Was it not better if he used Pc and Pr as PLL actually is a loop controlling phase not frequency?

@koerd85 4 ай бұрын

Awsome technology

@doid3r4s 7 жыл бұрын

What if the signals are out of phase to begin with but their frequencies are the same?

@circuitdesign 5 жыл бұрын

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.

@danielfromca 4 жыл бұрын

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 Жыл бұрын

@@danielfromca Thanks, this was the exact point I was missing!

@danielfromca Жыл бұрын

@@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).

@lexsearch8411 2 ай бұрын

This poor dude stressed out

@AdityaPrasad007 6 жыл бұрын

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.

@circuitdesign 5 жыл бұрын

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.)

@arnavrawat9864 4 жыл бұрын

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.

@ahmedelsayed8570 2 жыл бұрын

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.

@emylrmm 6 жыл бұрын

at 8:10, the phase is just the integral of the frequency, but only in the linearized frequency domain not in the time domain

@panier66 6 жыл бұрын

An instantaneous frequency is the derivative of phase in the time domain, and reciprocally, the instantaneous phase is the integral of instantaneous frequency.

@AdityaPrasad007 6 жыл бұрын

@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??