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Amplitude and Frequency/Phase Modulation

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Amplitude and Frequency/Phase Modulation Amplitude and Frequency/Phase Modulation I always had difficulties in understanding frequency modulation (FM) and its frequency spectrum. Even for the pure tone modulation, the FM spectrum consists of an infinite number of sidebands whose signs change from one sideband to the next one and whose amplitudes are given by the horrible Bessel functions. I work for Zurich Instruments, a Swiss maker of lock-in amplifiers and it has become a professional inclination to always think in terms of demodulation. The pictorial representation of AM/FM that I like comes with a (pictorial) digression on lock-in demodulation. Reader not interested in demodulation should read only Sect. 1.1.3 and Sect. 1.21 1.1 Lock-in Demodulation as a Change to a Rotating Frame of Reference A lock-in amplifier can find the amplitude AS and phase 'S of an input signal of the type VS(t) = AS cos(!St + 'S): (1.1) by a process called demodulation. The existing literature explains demodulation with the multiplication of the input signal by sine and cosine of the reference phase, also referred to as the in-phase and quadrature of the reference: this reflects the low level implementation of the scheme (it is a real multiplication) but I could never really understand it. I prefer more a different explanation using complex numbers: first I will present the math, then an equivalent pictorial representation. 1.1.1 Demodulation of the Signal: a Mathematical Approach The signal of eq. (1.1) can also be written in the complex plane as the sum of two vectors (phasors), AS each one of length 2 rotating at the same speed !S, in opposite directions: AS h i V (t) = ej(!S t+'S ) + e−j(!S t+'S ) : S 2 Multiplying by 2e−j!S t stops the counter-clockwise rotating phasors [described by ej(!S t+'S )] and makes the clockwise one precesses at −2!S: −j!S t j'S −j[2!S t+'S ] VS(t)2e = ASe + ASe : Let us assume we can get rid of the fast rotating phasor by averaging/filtering, indicated by the symbol h i: the averaged signal becomes −j!S t j'S z(t) ≡ VS(t)2e = ASe (1.2) 1 Many thanks to Nikola Bucalovic for useful discussions and to Andrin Doll for the final revision of this note. 1 This is the demodulated signal. To make the connection with the existing literature, where the real (or X or in-phase) and the −j!S t imaginary (or Y or quadrature) components of the signal are shown, one notes that e ≡ cos(!St)− j sin(!St), from which z ≡ X + jY where −j!S t X = Re VS(t)2e = h2VS(t) cos(!St)i = AS cos 'S −j!S t Y = Im VS(t)2e = h−2VS(t) sin(!St)i = −AS sin 'S: This is equivalent to multipying the input signal by cos and − sin of the reference signal. Demodulation consists in two steps just enounced: • multiplication (or mixing) of the input signal by e−j!S t and • filtering of the mixed signal: the filters should remove at least the 2! component. (The type and implementation of the filters is not the subject of this note.) 1.1.2 Demodulation of the Signal: a Pictorial Representation I restate the demodulation process using a pictorial representation. Let us start with the input signal AS eq. (1.1) and represent it as the sum of two phasors with equal length 2 as before. Simple trigonometry AS is sufficient to show that the projection on the x{axis is 2 · 2 cos(!St) and has a null projection on the y{axis, see Figure 1.1a) and b). a) b) c) fS fS Y X AS Figure 1.1: a) Decomposition of the input signal as two counter-rotating phasors of length 2 . b) The sum of the two phasors is always along the x{axis. c) In the rotating frame, one phasor is at rest and the other rotating at 2!S. Imagine now to stand at the origin and to rotate counter-clockwise at the same speed !S: you will see one phasor at rest, at an angle 'S with your forward direction ('S is the angle at time t = 0 of 2 the input signal), while the other is moving at 2!S in the clockwise direction (the angle between you and the second phasor is time-dependent, −2!St + 'S, Figure 1.1c). After averaging the total signal, only the steady phasor remains: elementary trigonometry is suffi- AS cient to see that the steady phasor has projection on the x{axis equal to 2 cos 'S and on the y{axis AS equal to 2 sin 'S. 1 The factor 2 comes from the decomposition of the real signal eq. (1.1) into the two vectors each AS of amplitude 2 , of which one averages out in the rotating frame. Therefore, since one phasor only is detected, to account for thep missing power, it is customary in lock-in detection to display the components multiplied by 2 or 2 for peak or RMS amplitude respectively. 2This is the rotating frame approximation that NMR people like so much, applied to lock-ins! 2 1.1.3 The Explanation (and Some Easy Math) The rotating frame picture is equivalent to the mathematical approach: here is why. Complex exponents are a very practical way of dealing with rotations in a 2D plane: given a vector (phasor) Rejθ with length R and angle θ with the x{axis, the phasor rotated by an angle φ is Rej(θ+φ) = Rejθ · ejφ, that is the product of the original phasor with the "rotation operator by an angle φ". The angle φ can also be a time-dependent angle, for instance φ(t) = !St, in which case the phasor rotates at a constant speed. I can equally well think of ejφ as a rotation of the observer: if the phasor in the previous example appears at an angle θ + φ, this means that the observer has rotated by an angle −φ, in the clockwise direction. So a rotation of the observer reference frame by an angle φ in the counter-clockwise direction multiplies by e−jφ, with the minus sign. The change to a rotating frame of reference I mentioned above, with an angle φ(t) = !Rt, corre- sponds to multiplying the input signal (each one of the two phasors) by e−j!Rt. But e−j!Rt is just cos !Rt − j sin !Rt! This is why the cos and − sin multiplications to obtain the X (in-phase, the real part of the product) and the Y (the quadrature, the imaginary part) components. As a side note: The Fourier transformation is a demodulation with infinitely long averaging 1 Z +1 −j!Rt F (!R) = f(t)e dt: 2π −∞ Think of f(t) as the sum of all components at different frequencies !.3 The recipe for the Fourier −j!Rt transform has two steps: first, rotate the reference frame at a speed !R, f(t)e . Now, only the !R component is a steady phasor in this new rotating frame of reference. Second, take an infinitely long average, the time integration from −∞ to +1 so that all the moving phasors having frequencies different from !R average out. 1.1.4 Demodulation at a Different Frequency The pictorial representation is as correct as the mathematical approach and by no means an approx- imation of it. We will now only make use of the pictorial representation and predict what happens to the demodulated signal when the reference/demodulation frequency and the signal frequency differ, !R 6= !S; we then compare it to the math approach and show it is the same. Let us assume that the signal is advancing faster than the reference phase, !S > !R: in the rotating frame, the signal slow moving phasor will appear moving counter-clockwise at a speed !S − !R. As before, the fast one rotating at −(!S + !R) is averaged out. Using now the mathematical approach, we want to check the validity of the result: the multiplication by the reference signal e−j!Rt and the subsequent averaging of the 2! component gives A h i V (t)2e−j!Rt = ej(!S t+') + e−j(!S t+'S ) 2e−j!Rt = Aej[(!S −!R)t+'S ] S 2 a ccw rotating phasor at frequency !S − !R: this is exactly the result we obtained in the rotating frame. 1.2 Signal Modulation With the previous pictorial representation in mind, we can proceed to analyze signal modulation. Any generic input signal can be written as VS(t) = A(t) cos[!ct + '(t)] (1.3) with a constant carrier frequency !c and adequate A(t) and '(t). 3Never mind that I explain the Fourier transform in terms of the Fourier transform: all geeks like recursion. 3 1.5 a b 1.0 0.5 0.0 0.5 amplitude 1.0 1.5 0 20 40 60 80 100 20 40 60 80 100 time time Figure 1.2: a) Amplitude modulated and b) frequency modulated signal: note that in one case the amplitude changes with time, in the second the spacing between successive zero crossings. To understand amplitude and phase modulation, we change to the rotating frame of reference with !R = !c h i −j!ct j[!ct+'(t)] −j[!ct+'(t)] −j!ct j'(t) −j[2!ct+'(t)] VS(t)2e = A(t) e + e e = A(t)e + A(t)e where the 2 is for the peak amplitude.
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