Barkhausen's Criteria

Thread Starter

lkgan

Joined Dec 18, 2009
117
Hi everyone,

From the attached picture, why when βH(jω) = -1, the circuit will oscillate at frequency ω? What's the purpose of letting the circuit to oscillate?

The total phase shift around the loop at ω is 360°, what is the reason? I know the negative feedback itself introduces 180° of phase shift, where's the other 180° comes from?

Last but not least, why do we need 360° phase shift for oscillation?

Appreciate if anyone could answer my questions, thank you......
 

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t_n_k

Joined Mar 6, 2009
5,455
I think the shift around the loop is 180° (not 360°) at the condition for instability - or "oscillation".

This is based on the relationship

Y(s)/x(s)=H(s)/(1+βH(s))

The instability point occurs when the denominator (1+βH(s))=0
or
βH(s)= -1 = magnitude of 1 @ 180°

What's the purpose of letting the circuit oscillate? Non whatsoever - unless you are building an oscillator. Electronics designers normally go to some lengths to avoid oscillation and instability in a circuit. This "analysis" hopefully provides us with some insight into where something might go "wrong" and hence what might be done to avoid the occurrence of instability.
 

Thread Starter

lkgan

Joined Dec 18, 2009
117
Well, what you mentioned did make sense. Currently I am studying Design of Analog CMOS Integrated Circuits and the paragraph which I don't understand that explain this section is :

Note that the total phase shift around the loop at ω is 360° because negative feedback itself introduce 180° of phase shift. The 360° phase shift is necessary for oscillation since the feedback signal must add in phase to the original noise to allow oscillation buildup. By the same token, a loop gain of unity (or greater) is also required to enable growth of the oscillation amplitude.

So why did the author state that total phase shift is 360°? And what do he meant about the signal must add in phase to the original noise?
 

t_n_k

Joined Mar 6, 2009
5,455
Well, what you mentioned did make sense. Currently I am studying Design of Analog CMOS Integrated Circuits and the paragraph which I don't understand that explain this section is :

Note that the total phase shift around the loop at ω is 360° because negative feedback itself introduce 180° of phase shift. The 360° phase shift is necessary for oscillation since the feedback signal must add in phase to the original noise to allow oscillation buildup. By the same token, a loop gain of unity (or greater) is also required to enable growth of the oscillation amplitude.

So why did the author state that total phase shift is 360°? And what do he meant about the signal must add in phase to the original noise?
Sure - if you include the -ve sign at the summing junction then the total phase shift will effectively be 360°. That summing junction bit is always 180° - with an ideal signal inverter. I guess the critical issue is the 180° shift through the frequency dependent elements - i.e. the main transfer part H(s) and the feedback part β in tandem. That's where the other 180° shift potentially occurs, which is what you asked about.

To visualize how that unity gain plus 180° phase shift occurs involves solving the equation

βH(s)=-1

for the particular frequency at which the condition is met.

To achieve 180° requires the combined function βH(s) to be of a form that will introduce at least 180° phase shift - a first order lag won't do it for instance. And two conditions must be met at a given frequency - both unity gain and 180° shift.
 

Thread Starter

lkgan

Joined Dec 18, 2009
117
Sorry, I couldn't really understand what summing junction bit is always 180° - with an ideal signal inverter means? :confused:
 

Thread Starter

lkgan

Joined Dec 18, 2009
117
Hi,

This is what I have understood, the negative feedback β already introduce 180° phase shift because the output of the β amplifier is inverted right? Furthermore, the other 180° phase shift will occur at the summing point where there have a negative sign there. So the total phase shift is 360° and wouldn't affect the input to the H(s) as they are same phase. Am I right?
 

t_n_k

Joined Mar 6, 2009
5,455
Hi,

This is what I have understood, the negative feedback β already introduce 180° phase shift because the output of the β amplifier is inverted right? Furthermore, the other 180° phase shift will occur at the summing point where there have a negative sign there. So the total phase shift is 360° and wouldn't affect the input to the H(s) as they are same phase. Am I right?
1. About the negative feedback matter.

Referring to the diagram - a signal component related to the output is fed back to the "summing junction" (the circular object with the + sign inside it). This feedback value is βY(s). How do we know it is negative feedback? Because the feedback value is subtracted from the input X(s) at the "summing junction" - as denoted by the negative sign at the feedback connection point at the summing junction.

In other words the resulting error signal (at the summing junction output terminal) being fed as input to the system transfer function H(s) is X(s)-βY(s). So it's at the summing junction that the negative feedback is implemented. The β feedback multiplier or amplifier has no indication as to whether it introduces a 180° or any other phase shift - it just has a generic value β. The β block may simply amplify or (more likely) attenuate the value of Y(s) being fed back to the summing junction.

For instance, β could have a straight ratio value less than unity - e.g. β=+0.5. It may also include a phase shift as well as a magnitude change - which would be frequency dependent. That depends on the specification of the β block component. How one specifies the feedback element depends primarily on what one is hoping to achieve by introducing feedback in the first place.

2. Where does the 180° frequency dependent phase shift arise?

It arises from the the phase change one would observe from the (error) input at H(s) to the negative feedback point at the summing junction. Add that phase change to the 180° change introduced at the negative feedback input of the summing junction and you get your 360° total change at a particular frequency.

The frequency at which the shift is exactly 180° through the signal path formed by βH(s) depends on the physical system under consideration. You need to know what β and H(s) actually are.
 
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