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17 Jan 2013: Experimental Consequences of the S-wave cos(k. x. ) cos(k. y. ) Superconductivity in the Iron-Pnictides. ... B 78, 144514 (2008). Stability and pairing in quasi-one-dimensional Bose-Fermi mixtures.

17 Dec 2013: Analogue ModulationFrequency Modulation. Bandwidth of FM SignalsUsing cos(A B) = cos A cos B sin A sin B we can write. ... Analogue ModulationFrequency Modulation. Bandwidth of FM Signals. Using the relationships sin A sin B = 12 [cos(A B) cos(A B)]

4 Mar 2013: sin αsinh a. =sin βsinh b. =sin γsinh c. cosh a = cosh b cosh c cos α sinh b sinh c. ... cos α = cos β cos γ cosh a sinh β sinh γJ.Rasmussen@dpmms.cam.ac.uk.

4 Oct 2013: cos A cos B =1. 2[cos(AB) cos(A B)]. sin A sin B =1. ... 2[cos(AB) cos(A B)]. sin A cos B =1. 2[sin(A B) sin(AB)].

16 Dec 2013: Analogue ModulationFrequency Modulation. Bandwidth of FM SignalsUsing cos(A B) = cos A cos B sin A sin B we can write. ... Analogue ModulationFrequency Modulation. Bandwidth of FM Signals. Using the relationships sin A sin B = 12 [cos(A B) cos(A B)]

11 Feb 2013: sin αsin a. =sin βsin b. =sin γsin c. cos a = cos b cos c sin α sin b sin c. ... cos α = cos β cos γ sin a sin β sin γ.

26 Jul 2013: evaluated at =. In Fourier notation and given that t= cos t, they become. ... 0. = 0. 2. Pressure and stress field. We have. ̃20 =. 2De1 De21 iDe1. B. 2 cos 2 sin 22, 43. as well as zero pressure p2 = 0, so that the average

4 Mar 2013: Sketch the imageof σ in R3. (a) U = (0, 2π) R, σ(θ, z) = (f (z) cos θ, f (z) sin θ, z), where f (z) > 0.(b) U = R2, ... r, z) = (r cos z, r sin z, z).(c) U = (0, 2π)(0, 2π), σ(θ, φ) = ((a b cos φ) cos θ, (a b cos φ) sin θ, b

17 Dec 2013: a) Convolution: f(t) g(t) F(ω)G(ω)(b) Modulation: f(t) cos(ω0t) 12 [F(ω ω0) F(ω ω0)](c) Parseval’s Theorem:. ... where a1(t) = b cos(2πfxt) is a sinusoidal modulating signal and a2(t) = cos(2πfct)is a unit amplitude carrier signal.

26 Jul 2013: us(ξ,η)= u0s. 2(1 cos ξ)( cos ξ)(1 e2) (1 e2) cos(2ξ) s(ξ,η). ... Here hwis the minimal distance of the centroid to the wall without penetration, hw =(1 e2) (1 e2) cos(2θ)/2.

5 Aug 2013: b). X(t) = cos(2πf0t Φ). f0 is fixed; Φ is uniformly distributed between 0 and ϕmax. ... b) X(t) = cos(2πf0t Φ):. E[X(t)] =. X(t, ϕ) fΦ(ϕ) dϕ = ϕmax0.

16 Dec 2013: sinAsinB=0.5[cos(A-B)-cos(AB)]. fixed constants. cos(AB)=cosAcosB-sinAsinB, where B = 2φ and A fixed constant simplify and show as before. ... E[cos(2φ)] = E[sin(2φ)]. = 0. Condition 2 Ok. Example: random phase sine-wave.

19 Feb 2013: Sketch the imageof σ in R3. (a) U = (0, 2π) R, σ(θ,z) = (f(z) cos θ,f(z) sin θ,z), where f(z) > 0.(b) U = ... r,z) = (r cos z,r sin z,z).(c) U = (0, 2π)(0, 2π),σ(θ,φ) = ((a b cos φ) cos θ, (a b cos φ) sin θ,b

11 Sep 2013: T(t) = A exp{kt} B exp{. kt}, k > 0. T(t) = A cos(|k|t) B sin(. ... Thus k = ω2. Furthermore, we need A = 0 to get ridof the cos term.

19 Jan 2013: Find the time-averaged energy flux vector 〈I〉 for (i) a plane harmonic S-wave with u =B cos(kx ωt), (ii) a plane harmonic P-wave with u = Ak̂ ... An SV-wave with displacement. u = B(cos θ, 0,sin θ)eik(x sin θz cos θ)iωt.

4 Oct 2013: TA sin θATA cos θA. µgδxT. (3). Now. TB sin θBTB cos θB. ... T(t) = cos jmnct and sin jmnct. Putting all this together i.e.

26 Jul 2013: cosh cos ,. it is found that. θ. (a) (b) (c). ... cos cos cosh cos. 4 sin2. cosh cos 2 sin sin.

26 Jul 2013: b) Two dimers. We now consider the case of two dimersa and b. ... 482 2d3, 3a. V = ̄ab2 b1b1b2. ̄ba2 a1. a1a2 f a f b.

28 Nov 2013: X = 0 is given byF′ G = c0, (cos X σ0) G sin XF = 0, (4.4a,b). ... Θ = Ae(1κ2)τ cos κX Z, W = Ae(1κ2)τ cos κX, (5.1a,b)for any amplitude A.

26 Jul 2013: With this in mind we let g2 = g1(z φ) with g1(z) = cos(z), andwe set equal for both sheets the bending stiffness B and the forcing amplitude A. ... B(576 B2. ) C, (3.7a)η2(z) = A. 12B [cos z cos(z φ)] [288 sin z (288 B2) sin(z φ)]B(576 B2).

30 Jan 2013: φ(Ai) = A′i. Trigonometry. If 4ABC has sides a, b, c and opposite angles α, β, γ, then. ... sin αa. =sin β. b=. sin γc. c2 = a2 b2 2ab cos γ.J.Rasmussen@dpmms.cam.ac.uk.

26 Jul 2013: 1x, = U1ey , D4a. 1x,0 = b cos x tex a sinx t ey , D4b. ... UN =12 b. 2 2ab cos a2. D17. Allowing tangential motion does therefore reduce the swim-ming speed by the same amount, and the final results of thepapers, Eq.

16 Dec 2013: 1. Find the General Solution with f (t) = 1. Complementary function is y = A cos(3t) B sin(3t). ... Particular integral is y = 19General solution is y = 19 A cos(3t) B sin(3t).

6 Aug 2013: b) The canonical momentum conjugate to x is. px =L. ẋ= (m1 m2)ẋ m2lφ̇ cos φ. ... circular motion around the z axis with constantangular velocity. In terms of the z coordinate, the particle undergoes simpleharmonic motion, z(t) = a sin κt b cos κt

26 Jul 2013: xm(x, t) =x 'a cos(x Re ω(t)t φ)and ym(x, t) = 'b sin(x Re ω(t)t),. ... xy|y=0 b sinx. 2ψ1. y2|y=0, (3.15). ψ2,x |(x,0) = a cos(x φ)2ψ1.

15 Aug 2013: b) xn = A cos(n0) B sin(n0) where A and B are uncorrelated random vari-ables with zero mean and identical variances. ... rXX[k] = 0.5A2 cos(k0). ]. 5. A stationary random process has autocorrelation function:.

19 Oct 2013: expected. (ii) all A’s, B’s zero except A1:. φ = A1r cos θ = A1z , u = φ = A1ez. ... z. (iii) all A’s, B’s zero except B1:. φ =B1 cos θ.

31 Jan 2013: The polar triangle A′B′C′ is defined by theunit vectors in the directions B C, C A, and B A. ... Deduce that. sin α sin β cos c = cos γ cos α cos β.

26 Jul 2013: 4 m1=1. m2=1. 1. m12 m221 m12. 2m12 m2. 2cos 2 b m1x cos 2 b m2y , 18from which we note that large contributions to Eqs. ... cos m2w W. m1=1. 1m Cm 12m1 12 ln S sdwds , 20b.

26 Jul 2013: uy=y1 = 1 ṠR cos y=y1 , 11a. vy=y1 = ṠR sin y=y1 , 11b. ... uy=y2 = 1 ṠR cos y=y2 U , 11c. vy=y2 = ṠR sin y=y2.

26 Jul 2013: mal a or b at given Pe). Starting from a random initial condi-. ... For t 1, the swimming stroke,. b cos t b1 þ sin t bk; (44).

21 May 2013: For M, N N, we use the abbreviatednotationB(M, N) = B(M, M; 0, N). ... A horizontal crossingof B = B(M1, M2; N1, N2)is an open path ofB linking some vertexvM1,n1 to some vertexvM2,n2; a verticalcrossinglinks somevm1,N1 to somevm2,N2.

16 Dec 2013: b) -1, 2, -1. (c) -1, 0, 2, 0, -1. (d) 1, 2, 2, 1. ... b) Explain how you would use the Hamming window. w (n) = 0.54 0.46 cos. (.

23 Oct 2013: 4. Let (fn) be a sequence of real-valued continuous functions on a closed, bounded interval [a,b], and. ... m=1 mam cos mx. (b) Show that f(x) =. m=1 am sin mx defines a continuous function on [π,π], but that the seriesm=1 mam cos mx need

26 Jul 2013: ̃1 =ir2. 4!cos 2" 1" ,! 16a". ũ1r = ir. 2sin 2" ,! 16b". ũ1" =ir. 2!1 cos 2""! 16c". B. Second-order solution. 1. ... 4-̃4!0" = 0. A4 sin 2" B4 cos 2" C4 sin 4"r2. ,!

15 Aug 2013: b) Explain how you would use the Hamming window. wn = 0.54 0.46 cos. ... 10. A digital filter has the transfer function (1bz1)/(1az1), where a > 0,b > 0, b < a.

6 Aug 2013: E m2gl cos φ. [9]. 2 A harmonic oscillator is weakly perturbed by a cubic potential λx3 so thatits Hamiltonian has the form. ... 19]. (b) Find the temperature T (x, t) of the system if s(x, t) = cos(px)δ(t t0)and T (x, t < t0) = 0, for the two cases

21 May 2013: n. i ii. P P x a x b=. =. ( ) ( )data;model data, ;modelb. ... and D from the two distributions. ( ) ( ) ( ) ( ) ( )[ ]exp cos sin cos 2 sin 2P A B C Dα α α α α.

29 Nov 2013: Dfm|x(h)).(b) Define f: R3 R2 by f(x,y,z) = (exyz, cos x2y). ... 10. Use the contraction mapping theorem to show that the equation cos x = x has a unique.

4 Oct 2013: B(cos(1 ξ)) A cos ξ = 1. Solving these two equations for A and B gives the Green’s function to be:. ... Then. I. λ=. 0. cos(λx)eαxdx = <(. 0. e(αλi)xdx. ),. = <[e(αλi)x.

26 Jul 2013: 1(a). Our new result is illustrated in Fig. 1(c), where we showthe dynamics of swimmers undergoing reciprocal motionwith velocity Uswim UðtÞe and UðtÞ U cos!t withU 5 ... a) No swimming, (b) steady swimming at speed U 5 m=s, (c) reciprocalswimming at

26 Jul 2013: 3) leads to 2p = 0. A solutionof the harmonic equation for the pressure is, at the first order inkb, p = p1(r,θ) (A cos s B sin s) with s = ... xi = x0 b cos β sin (s φ) , (34)yi = y0 b sin β sin (s φ).

5 Feb 2013: The polar triangle A′B′C′ is defined by theunit vectors in the directions B C, C A, and B A. ... Deduce that. sin α sin β cos c = cos γ cos α cos β.

22 May 2013: i/2)n = (1/2)neinπ/2 = (1/2)n(. cos(nπ/2) i sin(nπ/2)). Thus for some B′ and C′,. ... p(n)11 = A (1/2). n(. B′ cos(nπ/2) C′ sin(nπ/2)). We then use facts that p(0)11 = 1, p.

16 Dec 2013: 1. Find the General Solution with f (t) = 1. Complementary function is y = A cos(3t) B sin(3t). ... Particular integral is y = 19General solution is y = 19 A cos(3t) B sin(3t).

26 Jul 2013: a) The no-shear regions are transverse to the flow direction. (b) Theno-shear regions are parallel to the flow direction. ... Thesolution to this problem was derived by conformal mapping in Philip (1972a) and isrecalled in Appendix B.

19 Oct 2013: be similar to that of the spherical harmonicP4(cos ), and so we adopt the estimate L 4:5. ... Then equations(1) and (2) imply T=T r2c 2i =gL and juj 2g T=2N2T , where we have takensin cos 21=2.

13 Aug 2013: σ1 = sin θ sin ψdφ cos ψdθ,σ2 = sin θ cos ψdφ sin ψdθ,σ3 = cos θdφ dψ. ... 1 a. 4. ρ4. ). 4(dψ cos θdφ)2+. ρ2. 4(dθ2 sin2 θdφ2. ).

10 Nov 2013: 2t cos ka)c†kσckσ. /a. (k). /a k-. B.Z. As expected, as k 0, spectrum becomes free electron-like:k 2t t(ka)2 (with m = 2/2a2t). ... k. cos k a†kak. Ĥ = JNS2 B.Z.k. ωka†kak O(S.

21 May 2013: n. i ii. P P x a x b=. =. ( ) ( )data;model data, ;modelb. ... and D from the two distributions. ( ) ( ) ( ) ( ) ( )[ ]exp cos sin cos 2 sin 2P A B C Dα α α α α.