# Linear Oscillator Frequency Problem

I was doing one of the problem sets when I stumbled upon this gem.

I know that for oscillation to occur, the gain has to be greater than 1.

The loop gain I computed is

Could anyone suggest any approach towards the problem? Thanks.

• Might be worth working out the oscillation frequency first, as the question suggests. Jun 7, 2015 at 4:00

Some of your gain calculations are wrong: should contain terms: R/(1+RCs), 2R/(1+2RCs), 1/2RCs and, of course, gm

Derive the closed loop transfer function, which is 3rd order. For steady-state oscillation the denominator of the CLTF must factorise to the form: (s^2 + wn^2)(s + a), where wn is the resonant frequency in rad/sec. The transient, e^-at, decays to zero, leaving the steady-state sinusoid at wn rad/sec. Compare TF denominator coefficients to find the values of wn and gm

Rough calculation gives: wn = 1/RCsqrt(2); gm = sqrt(3/2)/R, but these need checking.

The above is based on the observation that the denominator of the closed loop TF of a 3rd order system can be written: (As^3 + Bs^2 + Cs + D), which can be simplified by dividing throughout by A, giving the form (s^3 + Ps^2 + Qs + R). Note that the denominator of the CLTF represents the characteristic equation; the numerator only serves to augment the natural behaviour.

The relative stability of a 3rd order CLTF can be determined by comparing the products of the inner two coefficients and the outer two coefficients. Thus, BC>AD = stable; BC=AD = critically stable (i.e. sinusoidal); BC < AD = unstable. This is a pretty useful design ROT for 3rd order systems generally.

So, taking the simplified TF and setting the sinusoidal condition, PQ=R, gives the denominator TF: (s^3 + Ps^2 + Qs + PQ) and this factorises to (s^2 + Q)(s + P), which immediately gives the resonant frequency as wn=sqrt(Q) rad/sec

• Question to Chu: How can I derive a closed-loop transfer function without any input?
– LvW
Jun 7, 2015 at 16:37
• @LvW, that's a very good question. I view an oscillator as a feedback system with a set-point (input demand) of zero. Since the configuration is critically stable, the smallest disturbance, however it may arise, will invoke the complementary function solution (or 'homogeneous solution') of the differential equation.
– Chu
Jun 7, 2015 at 20:45
• I know what you mean - and, yes, the denominator will be independent on the node at which such a disturbance is injected. Nevertheless, mentioning a closed-loop tzransfer function with out definition of input and output can create some confusion. Don`t you agree that analyzing the loop gain is much more efficient and easy to understand?
– LvW
Jun 7, 2015 at 21:26
• @LvW, Possibly. Perhaps I've been a control engineer too long! But it's good to have a different perspective on problems. I wasn't sure that the gain/phase method would work properly, given that the system is 3rd order. What answers do you get?
– Chu
Jun 7, 2015 at 22:34
• @LvW, just checked - same results for wn and gm as TF method, so two answers for the price of one today!
– Chu
Jun 7, 2015 at 22:48

Here comes my approach: It is a straighforward procedure based on the oscillation condition.

1.) Remember the oscillation condition: Loop gain real and (slightly) larger than unity (real means: Loop phase=0).

2.) Determine the loop gain A(loop) correctly. For each FET stage: Gain A=-gm*Z (Z:parallel combination). For convinience: Set R||(1/sC)=1/(sC+1/R)

3.) It turns out that the numerator of the 3rd-order loop gain function will be real and negative. Hence, for a real loop gain function the denominator also must be real and negative.

4.) This leads to the condition: Im(denominator)=0 (after setting s=jw). This equation gives you the oscillation frequency. The result will be a very simple expression.

5.) With this frequency you can determine the value of gm which makes the remaining real part equal to (or slightly larger than) unity.