+ Post New Thread

Results 1 to 5 of 5

- 16th July 2005, 22:57 #1

- Join Date
- Mar 2005
- Posts
- 14
- Helped
- 3 / 3
- Points
- 1,468
- Level
- 8

## What is the basic condition for oscillation ?

If the input reflection coefficient of a 2 port network is greater than 1, is that condition enough for me to say that the network is oscillatory or i have to check another condition like barkhausen condition to be sure that oscillations occurs ?!

- 16th July 2005, 22:57

- 17th July 2005, 00:22 #2

- Join Date
- Jul 2002
- Location
- Middle Earth
- Posts
- 4,630
- Helped
- 481 / 481
- Points
- 37,801
- Level
- 47

## Re: What is the basic condition for oscillation ?

The very old Barkhausen criteria comes from control systems. It states that if and only if there are poles in the right half plane, the system will oscillate at the frequency where the nonlinearities reduce the loop gain to unity and also the loop phase is a multiple of 360 degrees. This is based on complex number theory.

For some reason everyone reverses this and says that the gain and phase meeting certain criteria will cause poles to exist in the right half plane.

The reason they can get away with this is that they only apply it to well known circuits that are intentionally designed to have right hand poles. It is easy to have a system with gain greater than one at 360 degrees. One standard way to test for right hand poles is to do a contour integration from -infinite frequency to +infinite frequency. The number of times the vector from the point -1 +j0 to the gain curve rotates counterclockwise is the number of poles minus the number of zeros in the right half plane. There are many gain curves that intersect the 360 degree mark beyond the gain =1 point but curve back and do not encircle the point.

Anyway, for your one port oscillator, you need the losses in the external network to be less than the reflection gain of the one port. In years past it was a standard method to analyze oscillator circuits for the presence of a negative real part of impedance presented to the tuned circuit. If this resistance was greater in magnitude than the equivalent series loss resistance, the circuit would oscillate.

- 17th July 2005, 00:22

- 17th July 2005, 09:36 #3

- Join Date
- Apr 2005
- Posts
- 10
- Helped
- 3 / 3
- Points
- 1,500
- Level
- 8

## Re: What is the basic condition for oscillation ?

as flatlutenet has mentioned ,with refrence to the NYquist stability criterion shows weather the system is oscillatory or not ,this is basiclly determined by studying the open loop transfer function of the desired system when the O.l TF HAS NO poles in the right hand planes then the encirclements of 1+j0 in a cw direction determines the stability of the system ,for a certain system if you are checking if it's the potential to oscillate or not ,both the loop gain >1 and loop phase =2π n should be checked at the suspected frequency ,where any of them to be violated the system willnot oscillate .for example in the analysis of a fabry perot cavity the distance between the 2 reflectors was chosen such that a complete round trip phase = multiples of 2 pi. elseif , though the active medium supplies a round trip gain > 1 , the distance was adjusted for distructive interference( RTPS of pi) the circuit will not oscillate ,so both conditions must be satisfied simultaniously and i/p reflection >1 alone is not satisfactory.

h0op

- 17th July 2005, 09:36

- 25th July 2005, 13:34 #4

- Join Date
- May 2005
- Posts
- 288
- Helped
- 5 / 5
- Points
- 3,090
- Level
- 13

## Re: What is the basic condition for oscillation ?

Hello;

Simply saying, a circuit with positive feedback will oscillate.

- 25th July 2005, 17:45 #5

- Join Date
- Dec 2004
- Location
- New England, USA
- Posts
- 4,249
- Helped
- 1180 / 1180
- Points
- 30,463
- Level
- 42

## Re: What is the basic condition for oscillation ?

I usually find these 2 basic conditions will insure a steady oscillation:

1) attempt to design an amplifier

2) measure it

+ Post New Thread

Please login