Hefei-Lectures 2015 First Lesson: Optical Resonators. Claus Zimmermann, Eberhard-Karls-Universität Tübingen, Germany

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1 Hefei-Lectures 05 First Lesson: Optical Resonators Claus Zimmermann, Eberhard-Karls-Universität Tübingen, Germany October, 05

2 . Power and Field in an Optical Resonator (steady state solutions) The left mirror is the input coupler with a given eld re ectivity r, the right mirror is the output coupler with a re ectivity r. The third mirror re ects the light back to the left mirror and should be as ideal and lossless as technically possible (r 3 = ). The resonator may contain a nonlinear crystal, a cloud of atoms, or something else. ² electric eld in the resonator If the resonator is in a steady state, the eld inside the cavity, right behind the input coupling mirror, E c, should be the same after one round trip E n+ = E n During a round trip of length l the eld accumulates a phase ' = kl; with k being the wave number. Furthermore the eld is reduced by a factor r m := r p L; with the eld-re ectivity of the input coupler r and the power losses L of one round trip including the losses at the output coupler and the third mirror (but without the losses due to transmission back through the input coupler). Finally there is eld coupled in through the input coupler q t E i = re i : In total one obtains or E n+ = r m e i' E n + t E i = E n := E c ; E c E i = t r m e i'

3 ² intensity circulating in the resonator The power time averaged over one oscillation of the light eld is obtained from the modulus of the eld: There is a maximum for P c = je cj P i je i j = t ( r m e i' ) ( r m e i' ) = t r m cos ' + rm ' = q ¼ with q being a natural number. One obtains a series of resonances ' = q ¼ = kl k = q ¼ l º =! ¼ = ck ¼ = c q ¼ ¼ l º F SR : = c l = q c l = q º F SR Close to a resonance one can expand cos(x) ' x = and obtains P c P i ' This is a Lorentzian line shape function in '. ² impedance matching On resonance, ' = 0, one obtains P c P i = t =r m ( r m ) =r m + ' : t ( r m ) = r r p L With the de nition of the power transmission T and the power re ectivity R R : = r T : = t = R R m : = rm = R ( L) we obtain P c P i = R p Rm :

4 Let s calculate the optimal input coupler i.e. the R for which the power in the cavity is largest: p d R R (L ) + L ³ dr p = ³ R ( L) + p p = 0 R (L ) 3 R (L ) p R (L ) = L R ( L) = ( L) R = L This is the impedance matched case. Lets calculate the enhancement A : A := P max P i = R = L : With % losses and perfect impedance matching the power in the cavity is 00 times the power incident on the input coupler. ² re ected intensity The re ected eld consists of a part which is directly at the input coupler and a part which comes from inside the cavity: p E r = r E i + E c Le ikl t t p = r E i + E i p r Le i' Lt e i' The negative sign of the rst term takes care of the time reversal symmetry of a beam splitter: Any energy conserving surface re ects the eld on one side with a ¼ phase shift relative to the re ection from the opposite side. On resonance on gets with the abbreviation E r E i = r + t r l r r l = r + ( r ) r l r r l = ( r ) r l r ( r r l ) r r l = r l r r r l r l := p L The power may be calculated form the eld modulus as above. For perfect impedance matching,i.e. for r = r l, the re ection vanishes and all the incident light is coupled into the resonator. 3

5 ² Question to think about Assume that the only round trip losses are given by the transmission of the output coupler: L = R = T In the impedance matched case one obtains for the intensity behind the output coupler P trans = P c T = P in T T = P in : This holds independently of the value of T, i.e. also for T = 0. Impedance matching then requires that also T = 0. In other words: If you build a resonator with two lossless mirrors with 00% re ectivity the transmission through the cavity on resonance will be 00%. How can that be?. Equation of motion ² di erential equation for the eld We require that the eld after one round trip is the eld at the beginning of the round trip corrected for the round trip losses and the round trip phase shift. In addition, one has to add the eld which is coupled in during on round trip E(t + ) = E(t)r m e i' + t E i For very short round trip times (what is the typical round trip time?) one can expand the electric eld as E (t + ) ' E (t) + de dt 4

6 and obtains E (t) + de dt = E(t)r me i' + t E i de dt = E(t) r m e i' + t E i Near resonance, ', we expand the exponential and obtain ² resonators with low losses e i' ' + i' de dt = E (r mi' ( r m )) + t E i: We rst write down the round trip transmission then solve for For small losses r m ' : with this one obtains t m := p r m = p + r m p rm r m = t m + r m : t m ' t m + r m de = (i ) E + dt ' : = r m ' ', cavity detuning : = r m : = t E i, eld pump rate ' t m = L, eld decay rate We look for the solution on resonance of a lled cavity after the pump eld has turned of: de dt = E E(t) = E 0 e t : measuring the decay rate is one way to determine the losses of the resonator (works only for very good cavities and fast detectors). Which decay rates do you expect for typical resonators? 5

7 ² general solution is E (t) = i + Ce(i )t C is given by the initial conditions. After several decay times the second term decays and one obtains (compare to the steady state solution above): E c = i = i + ( i ) ( + i ) = i The eld oscillates with a dispersive and an absorptive amplitude (plot the two terms, does it remind you to something?). In fact, the cavity behaves like a driven, damped mechanical harmonic oscillator only that the phase µ between the driver and the oscillator is di erent. If the phase of the driving eld is taken as reference, the relative phase is given by the phase of the eld inside the cavity: µ = arctan Im (E c) Re (E c ) + + = arctan i = arctan : For À the phase shift is ¼ relative to the limiting case of. This is the same for a mechanical resonator. What di ers is the phase at resonance which is 90 ± for the mechanical oscillator and 0 ± for the optical resonator..3 Line width, nesse, decay rate ² line width is de ned as full width at half maximum (FWHM) of the power resonance function. P c P i = t r m cos ' + r m close to resonance (' ) the expansion yields P c P i ' half maximum is obtained for t r m ' + r m = p rm ' = = r m t ( r m ) + r m ' 6

8 which corresponds to a detuning ± = º 0 ¼ = ± = ' = so ± F W HM = ± = = º 0 ¼ ' = = º 0 r m p : ¼ rm ² nesse is de ned as the ratio with p rm F = º 0 = ¼ ± = r m r m = r p L = p ( T ) ( L) and in the impedance matched case r = p L; one obtains p p p r L L F = ¼ p = ¼ : r L L and for small losses, L, one nally gets F ' ¼ L : ² decay rate the power decay rate P is twice the eld decay rate P := = r m : 7

9 with this, one obtains ± F W HM = º 0 ¼ r m = ¼ ' r m ¼ = P ¼ = ¼ T res ; where T res is the =e-lifetime of the power in the resonator. The nesse and the life time is connected according to F = º 0 ± F W HM = ¼º 0 T res =! 0 T res :.4 Optical Beams (Kogelnik, Li, Applied Optics 5, 550, (966)) To understand laser beams in resonators we rst need to talk about geometric optics. ² beam vector we introduce a vector µ x ~x := x 0 = µ distance to optical axis slope of the beam ² ABCD-matrices optical elements may be described by matrices ) path of length d ) lens M d = M f = (convex lenses f > 0, concave lenses f < 0) µ d 0 µ 0 f 3) path within a material with index of refraction µ d=n M _ = d 0 8

10 In general, the matrix µ A B M = C D is called ABCD-matrix. It transforms an optical beam into a new one ~x = M ~x ² example : propagation along a distance d µ µ µ x d x = 0 x 0 x 0 µ x + d x = 0 x changes distance to optical axis but keeps the slope constant ² lens with focal length f: µ 0 f µ x x p = µ x x 0 x f = µ x x 0 changes slope but keeps the distance µ 0 µ x f 0 = µ x x f The new beam intersects the optical axis at a distance f from the lens. ² optical systems composite systems are described by the product of the corresponding matrices. Example: path d, lens f, path d µ µ µ µ d 0 d A B M = 0 = f 0 C D 9

11 ² lens conductor The possible trajectories between two mirrors with radius of curvature r corresponds to a periodically repeating series of lenses with focal length f = r One round trip is described by the Matrix M = M d M f M d M f 0

12 A number of n round trips are described by M n = (M d M f M d M f ) n ² Sylvester Theorem µ n A B = µ A sin(n µ) sin ((n ) µ) B (sin n µ) C D sin µ C sin (n µ) D sin (n µ) sin ((n ) µ) with cos µ := (A + D) periodic focusing and defocussing with the frequency µ. ² stability unstable solutions are obtained if µ is unde ned i.e. (A + D) >. Stability range is thus < (A + D) < standing wave resonator: ² stability diagram 0 < µ µ dr dr < {z } {z } :=g :=g 0 < g g < Since g g > 0 the point (g g ) is in the rst and third quadrant. The stability boundaries ful ll g g =, i.e. they form the functions g = g :

13 ² boundary at g g = for resonators with equal mirrors for r = r = r; i.e. g g = µ d µ d = r r d r d r = = 0 or d r = ² plane resonator d r = 0 r is in nite and the resonator consists of two plane mirrors. The resonator is unstable against an in nitesimal tilt of the beam. ² concentric resonator d r = The radius is half the distance and the mirror planes lie on a sphere. The resonator is unstable against an in nitesimal transversal shift of the beam. ² confocal resonator, boundary at g g = 0 for equal mirrors one obtains d r = 0 d = r: The mirrors have their focal point at the same position. The resonator is stable. However an in nitesimal di erence in the radius makes the resonator unstable. ² Questions to think about After how many round trips does the beam comes back to the starting point in a confocal resonator if the beam is injected not along the optical axis? What is the free spectral range in this case? If the cavity has two equal mirrors and no losses, is the cavity still impedance matched?

14 .5 Gaussian beams (Kogelnik, Li, Applied Optics 5, 550, (966)) ² paraxial wave equation In vacuum, Maxwells equations can be used to derive the Helmholtz equation. r u(~r) + k u(~r) = 0 with k = ¼ : Ansatz for the solution. u(~r) = ª(x; y; z) e ikz ª is slowly varying in space. The Helmholtz equation then becomes the paraxial = 0: The ª has been neglected. It would by responsible for a fast spatial modulation which is already taken care of in the Ansatz. The remaining changes of ª in z z-direction should stay linear. This is the paraxial approximation. ² solution The most simple solution (fundamental mode) has the form ª(r; z) = e i(p (z)+ k q(z) r ) with the functions q(z) and P (z) obeying the di erential equations and also From an gets with q 0 can be complex. dq(z) dz dp (z) dz = = i q(z) r := x + y : dq dz = q = q 0 + z 3

15 We choose the origin such that the real part of q 0 vanishes. q 0 is then complex and we can write q 0 = iz 0 with the real valued Rayleigh-length z 0. This is the most important parameter of a laser beam ² beam waist, Rayleigh length, confocal parameter with q = iz 0 + z the solution ª writes ª = e ip (z) k q i e q q r ª = e ip (z) e i k z+iz 0 z +z 0 r ª = e ip (z) i k r z z +z 0 e r w with w := k z + z 0 z 0 = z 0 k à µ! z + : z 0 The eld amplitude has an envelope according to a Gaussian e r w radius w of s µ z w = w 0 + : The beam waist w 0 is the radius at z = 0: z 0 with a e -beam w 0 := z 0 k = b k The confocal parameter is de ned as twice the Rayleigh length b := z 0 : 4

16 ² far eld angle in the limit jzj À jz 0 j one gets µ is the far eld angle ² wave fronts The solution of is w ' w 0 z z 0 w z = w 0 z 0 = tan µ tan µ = w r r 0 = = = z 0 kz 0 ¼ z 0 dp dz = i q = i z + iz 0 ip (z) = ln( i z z 0 ); ¼ w 0 (check it by taking the derivative). Decomposition it in real and imaginary part yields Re (ip (z)) = µ ln( i z ) + ln( + i z ) z 0 z 0 = µ( ln i z )( + i z ) z 0 z 0 = µ ln + ( z r ) = ln + ( z ) z 0 z 0 and Im(iP (z)) = µ ln( i z ) ln( + i z ) i z 0 z Ã! 0 = i z i ln z 0 + i z z 0 Ãs! i z z = i ln 0 + i z z 0 q (note that arctan x = i ln ix +ix ip (z) = ln = arctan z z 0 ). As result one obtains r + ( z z 0 ) i arctan( z z 0 ): 5

17 With µ i P (z) + k r z z + z 0 r = ln + ( z ) z 0 µ +i arctan( z ) k z 0 r z z + z 0 : the total eld can now be written u(~r) = ª(x; y; z) e ikz u(~r) = ª(x; y; z) e ikz = ª 0 q + ( z z 0 ) {z } conservation of total power of u e ( r w(z) ) {z } envelope e i ³ kz arctan( z ) z 0 {z } phase factor e i k r R(z) {z } curved wave fronts With We now discuss the di erent terms: ² wave front curvature R(z) := z ³ + ( z 0 z ) R(z) is the radius of the wave fronts which can be seen by looking at the position of constant phase k r R(z) + kz + arctan( z ) = const: z 0 arctan( z z 0 ) depends only slowly on z and can be neglected. One gets z(r) = const k r R(z) : The wave front runs along a parabola z(r)» r with a curvature Plotting =R (z) yields: d dr z(r) = R(z) : 6

18 Strongest curvature is at z 0. ² Gouy phase nally the phase arctan( z z 0 ) is called Gouy phase. While going through the focus there is an overall phase shift of ¼. This phase is fundamentally responsible for the absorption of light by a point like scatterer as for instance an atom. (see also optical theorem) ² transverse modes Other solutions are obtained with the Ansatz ª = g( x w ) h( y k (z)+ )e i(p q(z) (x +y )) w g and h are real functions with real variables. The Helmholtz equation is solved if g is one of the functions ³p x N m H m w 7

19 with and The same holds for h accordingly. d dx H m x d dx H m + mh m = 0: N m = r q w ¼ m m! Thus ³p x g m h n = N m H m w The rst few Hermitian polynomials are The Gouy phase now reads: H 0 (x) = H (x) = x H (x) = 4x H 3 (x) = 8x 3 x ³p y N n H n : w Á(m; n; z) = (m + n + ) arctan( z z 0 ): Theses solutions are called transverse electrical modes (TEM nm ). ² intensity pro les 8

20 ² transformation of a Gaussian beam and ABCD-law The q-parameters of a Gaussian beam before (q ) and after (q ) an optical system is connected by the ABCD-law q = Aq + B Cq + D A, B, C, D are the Elements of the above ray matrix M of geometrical optics. A proof is found in A. Siegman: Laser, University Science Books, 986, chapter 0..6 Modes of a standing wave resonator ² self-consistency of q. After one round trip q should reproduce: with q = Aq + B Cq + D µ A B M = D C describing the optical system of one round trip. In general q is a complex number q = z + iz 0 The real part gives the position of the beam waist, the imaginary part is the Rayleigh length of the beam. z 0 = b= = z 0 = kw 0 9

21 The condition for self-consistency yield z = A D s C z 0 = z 0 = r µ z B C A D B C C with A, B, C, D containing the geometrical variables such as r = f and d. ² Standing wave resonator with two curved mirrors The matrix is µ A B M = = C D µ d 0 µ 0 =f µ d 0 µ 0 =f µ d 0 From this one can calculate the matrix elements which are plucked in the general solution above. For two equal mirrors (f = f ) one obtains: ² stability range with the de nition one may write or z 0 = p d(r d) Rayleigh-length z = 0 position of the waist g = g := d r µ d r = + = r rd d = (z 0) r µ g z0 + = r µ d d r r g the Rayleigh-length z 0 normalized to the only length scale of the system, f = r=, form half a circle. : 0

22 ² spectrum The resonance frequencies are given by the phase that is collected by propagating from one mirror to the other including the Gouy phase ' = k d (m + n + ) arctan( d= z 0 ) {z } Gouy-phase The phase of a complete round trip, ', must be a multiple of ¼. ' = k d 4(m + n + ) arctan( d= p ) = ¼q d(r d) k = ¼ d q + d (m + n + ) arctan( d= p ) d(r d) The corresponding frequency in units of the free spectral range º 0 = c d º(q; m; n) = q + º 0 ¼ (m + n + ) arctan( q ): r d then reads and since one nally gets: arccos( x) = arctan( q ) x º(q; m; n) º 0 = (q + ) + ¼ (m + n + ) arccos( d r ): The natural number q counts the longitudinal modes. The numbers m and n count the transverse modes. The additional in the brackets, (q + ), is a matter of convention: The mode with the lowest frequency is here counted as the zeroth longitudinal mode.

23 By introducing the transverse oscillation frequency the spectrum has the very simple form ² plane cavity º t := ¼ arccos( d r ) º (q; m; n) = (q + ) º 0 + (m + n + )º t : d r = 0 arccos() = 0 º(q; m; n) º 0 = q + º t = 0 The mode consists only of the near eld part of the Gaussian beam. The Gouy phase can be neglected. The transverse frequency is zero. All transverse modes are degenerate. ² concentric resonator d r = arccos( ) = ¼ º(q; m; n) = (q + ) + (m + n + ) º 0 º t = º 0 Each transverse mode is degenerate with a longitudinal mode. As in the plane cavity, all modes are degenerate.

24 ² confocal resonator d r = arccos(0) = ¼ º(q; m; n) º 0 = (q + ) + (m + n + ) º t = º 0 Between two longitudinal modes there is always a transverse mode. In addition, half of the transverse modes are degenerate with a longitudinal mode. ² Interpretation For a xed q, the transverse modes for the spectrum of a harmonic oscillator of frequency º t. The light swings around the optical axis with a frequency that depends on 3

25 the curvature of the mirrors. Small curvature results in a low frequency, no curvature in a vanishing frequency of the degenerate plane cavity. In the confocal cavity the light bounces twice, i.e. it needs two round trips before it comes back to the initial position: the transverse frequency is twice as large as the free spectral range which gives raise to a semi degenerate spectrum. The concentric resonator both frequencies are the same and the spectrum is fully degenerate. The mode pro le is given by Hermitian polynomials since this is the solution of the quantum harmonic oscillator. For the photons in the resonator Helmholtz equation plays the same role as Schrödinger s equation for a massive particle in a harmonic potential. The resonator is a harmonic trap for the transverse motion of the photon..7 Mode Matching ² mode matching How can the output mode of a laser (or other source) be transformed such that is matches with the mode of the resonator (or other device)? ² Lens Transformation of a Gaussian beam by a lens. The matrix of the lens µ A B M = C D the beam parameter of the input beam = µ 0 f q = z + iz 0 = z + i b 4

26 and that of the output beam q = z + iz 0; = z + i b are connected by the ABCD-law q = z + i b = Aq + B Cq + D = q + 0 q f + = z + iz 0 z iz0 + f f = (z + iz 0)( z f + i z 0 f ) ( z f ) + ( z 0 f ) = z( z ) z 0 f f ( z f ) + ( z 0 f ) + i z z 0( ) + z 0z f f ( z f ) + ( z 0 f ) : This yields an equation for the real and the imaginary part. ² position of the new waist We use dimensionless quantities. z = z( z f ) z 0 f ( z f ) + ( z 0 f ) : b = z 0( z f ) + z 0z f ( z f ) + ( z 0 f ) : ~z : = z f ; ~q : = q f ; and obtain ~z 0 : = z 0 f ~d : = d f ~d : = d f ~z = ~z( ~z) ~z ~z ~z ( ~z) + ez 0 = ( ~z) ~z 0 + ~z ( ~z) + ~z 0 ~z = + : ( ~z) + ~z 0 5

27 ² diagram As seen from the input beam the lens is at a distance d from the waist. z = d As seen from the output beam the lens is at a distance d from the waist. z = d This yields Introducing ~ d = ~ d ( ~ d ) + ~z 0 : p : = d ~ p : = d ~ one obtains a dispersive Lorentzian curve with a width ~ b p = p : p + ~z 0 At the origin of the diagram p = 0 and p = 0, i.e. d = f and d = f: a waist of the input beam at a distance f from a lens is mapped to a waist of the output beam at the same distance f from the lens 6

28 In contrary, geometric optics images an object at f to an image at in nite distance (dashed line)! The largest possible distance from the output waist to the lens is e b. For a collimated input beam (p = ) the output is p = 0 i.e. d = f. This agrees with geometric optics. For small waists the transformation approaches the classical limit. ² waist of the output beam As above one obtains That is an absorptive Lorentzian: ~ b = ~ b p + ~z The largest b (collimated output)) is obtained for p = 0 i.e. d = f: The value is b ;max = 4 e b..8 Ring resonators ² Bowtie resonator Bowtie-resonators are often used for single mode lasers and frequency doublers 7

29 With the de nitions g := d g := d r r the solution may be written as (calculation is a very good exercise and not very di cult) µ b = g + g r g µ b = g + g : r g Again one obtains a semi circle with a radius b max = 8 r d r :

30 A strongly focused beam is obtained for d À d. This is useful for applications where a high intensity is needed (laser, nonlinear optics ). Similarly focussed beams may be obtained with standing wave cavities only if they are very small, which is often unpractical. ² astigmatism in a ring resonator Hitting the mirror under an angle leads to di erent focal length in x and y direction. f x = r x = r cos(µ) : f y = r y = r cos(µ); with the angle of incidence µ. One may write a Gaussian beam as the product of two Gaussian functions I = I 0 exp( x =w x) exp( y =w y); which describe the pro le for the two transverse directions. The waists w x and w y may be calculated independently as above taking into account the two di erent e ective radii (see also Jenkins and White, Fundamentals of Optics, McGraw-Hill, New York, 957, page 95). ² Spectrum By properly taking into account the Gouy phase one obtains (after some lengthy but simple calculation) µ µ º(q; m; n) = qº m º tx + + n º ty ³ ³ d= z x + arctan d= z x with the free spectral range arctan º tx : = º 0 º ty : = º 0 º 0 : = c L ³ 4¼ arctan d= z y + arctan º 0 := c L ² geometric phase for odd number of mirrors: transverse modes 4¼ ³ d= z y the above spectrum only hold for resonators with a even number of mirrors. Each mirror creates the mirror image of the function which describes the pro le in the plane of incidence H n (x)! H n ( x): 9

31 Symmetric transverse mode functions have an even index. They are left unchanged by the mirror. H n (x)! H n ( x) = H n (x): Mode functions with odd index are antisymmetric and thus change sign. H n+ (x)! H n+ ( x) = H n+ (x): The sign change corresponds to an additional phase of ¼ collected during one round trip and thus the spectrum of the odd modes is shifted by half a free spectral range. ² geometric phase for odd number of mirrors: polarization Similarly, light which is linearly polarized in the plane of the resonator su ers a sign change at each re ection. For cavities with an odd number of mirrors the spectrum for this polarization is shifted by half a free spectral range. Such cavities are very good polarization lters..9 Locking a laser to a cavity ² Generating an error signal, setup The method of Pound, Drever and Hall (983) is based on detecting the phase shift of the light re ected from the resonator. A radio frequency generator drives an electro-optical modulator (EOM) which modulates the phase of the laser light with the frequency (0 00MHz). The re ected light is recorded with a photodiode. The electric signal is sent to one of the inputs of a mixer. The output of mixer is proportional to the product of the two input signals. After time averaging with a low pass lter one obtains the desired error signal. 30

32 ² Generating an error signal, qualitative explanation Qualitative explanation: the EOM generates two sidebands frequency shifted from the carrier frequency! by. If the resonator is replaced by a mirror one cannot observe the modulation with the photo diode. The beat signal between the carrier and the left side band would be 80 ± out of phase with the beat signal between the carrier and the right side band and both beat signals exactly cancel. If we replace the resonator and tune the carrier close to resonance the phase of the carrier is shifted by the resonator. The two side bands are far detuned from resonance and their phases stay unchanged. Now the two beat signals do not cancel any more and one obtains an electric signal. The phase of the signal is analyzed by rst mix it with the driving oscillator and than lter out harmonics. For a quantitative analysis we rst have to understand the spectrum of the light behind the EOM. ² spectrum of a phase modulated light eld The modulated eld is given in complex writing by i(!t+m cos t) E(t) = E 0 e with the modulation index M. We use the so called Jacobi Anger identity: e i(!t+m cos t) = J 0 (M) e i!t + +X n= i n J n (M) cos (n t) e i!t : with the Bessel- functions of rst kind J n (M) ( J(M) M -0.4 The physical electric eld is the real part: Ã Re (E (t)) = E 0 J 0 (M) cos!t + 3 +X n= J n (M) Re e i!t+in ¼ cos(n t)!

33 With Re e i!t+in ¼ cos(n t) = Re µe i!t+in ¼ e in t + e in t = Re e i(!+n )t+n ¼ + e i(! n )t+n ¼ = ³ cos (! + n ) t + n ¼ + ³ cos (! n ) t + n ¼ on obtains à E 0 J 0 (M) cos!t + X n= ³ ³ J n (M) cos (! + n ) t + n ¼ ³! + cos (! n ) t + n ¼ The rst term is the carrier. It has the intensity J 0 (M). The sum contains pairs of side bands of equal intensity J n(m). Their frequency is shifted relative to the carrier by n. The n-th side bands oscillate in phase but shifted relative to the carrier by n 90 ±. ² generating an error signal, quantitative analysis We look at the beat signal between the carrier and the left side band. The phase of the carrier is shifted due to the resonator by '. The beat signal is proportional to cos(!t + ') cos(! + ¼=)t = cos(!t + ') sin(! )t = sin (!t + ' t ) + sin (' + t ) {z } vanishes by averaging over one optical cycle ' sin( t + '): The beat signal with the right side band is obtained by replacing! : cos(!t + ') sin(! + )t ' sin( t + ') 3

34 Adding both signals results in (sin( t + ') + sin( t + ')) = cos( t) sin ' The electric signal from the diode is mixed (i.e. multiplied with cos ( t)) and averages by the low pass lter: U ' T TZ 0 cos( t) cos( t) sin ' = sin ' ' ' Close to the resonance ' is small and the sin can be approximated by its argument. The signal is proportional to the phase shift of the carried which is antisymmetric with detuning and may thus serve as an error signal. ² servo loops A actual time depending signal s (t) is compared to a reference value S and then fed into a an electronic servo controller. Its output acts on the system such that the actual signal equals the reference value. The actual value of the system is measured and gives a new signal T (s (t) S). In general, the transfer function T can be an operator that acts on the function s (t) and describes the properties of the servo and the system. In the steady state one requires that ² proportional controller T (s(t) S) = s(t): The minimum requirement is a negative feed back. The transfer-oparator thus inverts and ampli es the signal, i.e. multiplies it with a factor G: T s (t) = G s (t) 33

35 The steady state condition now reads: G (s(t) S) = s(t) or s(t) = G + G S: The actual value s equals the reference value S only for in nite gain: lim G! G + G = : ² Integral controller The transfer operator is now an integrator. It transforms then signal according to: with a gain factor G. The steady state condition yields Z t T s (t) = G 0 s (t 0 ) dt 0 s(t) = T (s(t) S) = G Z t (s(t 0 ) S)dt 0 0 or _s = G (s S): If the signal is constant in time, _s = 0, it matches the reference value s = S also for nite gain G. ² noise If we add noise R to the signal one obtains Z t s(t) = G 0 (s(t 0 ) S)dt 0 + R and In steady state one obtains _s = G(s(t) S) + _R s = _ R G + S: 34

36 We look at noise with a given frequency! : inserting into the steady state solution R(t) = R(!) e i!t _R = i! R(!) e i!t ; s(!) = i! R(!) + S; G i.e. small frequencies are more suppressed than larger frequencies. (/f-noise). ² complex transfer-function We now expand the signal in the base of harmonic oscillations e i!t Z s (t) = s (!) e i!t d!: We also require that the transfer-operator T is orthogonal in this base such that it obeys an eigenequation: T e i!t = T (!) e i!t : The complex function T (!) is called the spectrum of the operator. One can now describe the action of the operator T on the function s (t) by an integral: Z T s(t) = T s (!) e i!t d! Z = s (!) T e i!t d! Z = s (!) T (!) e i!t d! A linear servo-loop is completely described by its spectrum, i.e. its transfer-function. ² question to think about What is the spectrum of an integrator, a proportional controller, or a di erentiator? Plot them in a log-log-plot. Determine the slopes. In an ideal servo loop perturbations at all frequencies are compensated with the maximum gain What would be the best controller? ² stability A servo controller cannot follow arbitrarily fast perturbations and allways reacts only with some time delay. Each controller thus can be characterized by its critical frequency! c where the phase delay exceeds 90 ± : ' (! c ) = 90 ±. The sign of the real part of the transfer function changes from negative to positive. A periodic perturbation with a frequency above! c cannot be compensated any more but will be ampli ed. This can be avoided if the gain is smaller than for! >! c : jt (!)j <. The unity gain -frequency must be smaller than the critical frequency. 35

37 ² Nyquist-criterium One can show from complex integral analysis that oscillations can be avoided if the slope at unity gain is smaller than 6 db Oct. The transfer function T (!) can be represented by a curve in the complex plain parameterize by the frequency!. The servo loop is stable if the the point lies outside this curve. 36

38 ² Question to think about A race car is kept on track by the driver. What is the controller and what is the system? What is the actual signal and what is the reference value? Which part of the loop determins the critical frequency? Start thinking about the problem by assuming a race track that resembles a sin curve. Think about strategies to improve the servo loop. 37