Even for the pure tone modulation the FM spectrum consists of an in64257nite 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 S ID: 23189 Download Pdf

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Even for the pure tone modulation the FM spectrum consists of an in64257nite 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 S

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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. 1.2 1.1 Lock-in Demodulation as a Change to a Rotating Frame of Reference A lock-in ampliﬁer can ﬁnd the amplitude and phase of an input signal of the type ) = cos( (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), each one of length rotating at the same speed , in opposite directions: ) = Multiplying by 2 j stops the counter-clockwise rotating phasors [described by ] and makes the clockwise one precesses at )2 j j [2

Let us assume we can get rid of the fast rotating phasor by averaging/ﬁltering, indicated by the symbol : the averaged signal becomes )2 j j (1.2) Many thanks to Nikola Bucalovic for useful discussions and to Andrin Doll for the ﬁnal revision of this note.

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This is the demodulated signal. To make the connection with the existing literature, where the real (or or in-phase) and the imaginary (or or quadrature) components of the signal are shown, one notes that j cos( sin( ), from which jY where Re )2 j ) cos( cos Im )2 j ) sin( sin 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 j 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 eq. (1.1) and represent it as the sum of two phasors with equal length as before. Simple

trigonometry is suﬃcient to show that the projection on the axis is 2 cos( ) and has a null projection on the axis, see Figure 1.1a) and b). Figure 1.1: a) Decomposition of the input signal as two counter-rotating phasors of length . b) The sum of the two phasors is always along the axis. c) In the rotating frame, one phasor is at rest and the other rotating at 2 Imagine now to stand at the origin and to rotate counter-clockwise at the same speed : you will see one phasor at rest, at an angle with your forward direction ( is the angle at time = 0 of the input signal), while the other

is moving at 2 in the clockwise direction (the angle between you and the second phasor is time-dependent, , Figure 1.1c). After averaging the total signal, only the steady phasor remains: elementary trigonometry is suﬃ- cient to see that the steady phasor has projection on the axis equal to cos and on the axis equal to sin The factor comes from the decomposition of the real signal eq. (1.1) into the two vectors each of amplitude , of which one averages out in the rotating frame. Therefore, since one phasor only is detected, to account for the missing power, it is customary in lock-in

detection to display the components multiplied by 2 or 2 for peak or RMS amplitude respectively. This is the rotating frame approximation that NMR people like so much, applied to lock-ins!

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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) Re j with length and angle with the axis, the phasor rotated by an angle is Re Re j j , that is the product of the original phasor with the rotation operator

by an angle . The angle can also be a time-dependent angle, for instance ) = , in which case the phasor rotates at a constant speed. I can equally well think of j as a rotation of the observer: if the phasor in the previous example appears at an angle , this means that the observer has rotated by an angle , in the clockwise direction. So a rotation of the observer reference frame by an angle in the counter-clockwise direction multiplies by j , with the minus sign. The change to a rotating frame of reference I mentioned above, with an angle ) = , corre- sponds to multiplying the input signal

(each one of the two phasors) by j . But j is just cos sin ! 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 ) = j t. Think of ) as the sum of all components at diﬀerent frequencies The recipe for the Fourier transform has two steps: ﬁrst, rotate the reference frame at a speed j . Now, only the component is a steady phasor in this new rotating frame of reference. Second,

take an inﬁnitely long average, the time integration from to + so that all the moving phasors having frequencies diﬀerent from 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, ; 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, > : in the rotating frame, the signal slow moving phasor will appear moving counter-clockwise at a speed . As before, the fast one rotating at ) is averaged out. Using now the mathematical approach, we want to check the validity of the result: the multiplication by the reference signal j and the subsequent averaging of the 2 component gives )2 j j Ae [( a ccw rotating phasor at frequency : 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 ) = ) cos[ )] (1.3) with a constant carrier frequency and adequate ) and ). Never mind that I explain the Fourier transform in terms of the Fourier transform: all geeks like recursion.

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20 40 60 80 100 time 1.5 1.0 0.5 0.0 0.5 1.0 1.5 amplitude 20 40 60 80 100 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 )2 j )] )] j j [2 )] where the 2 is for the peak amplitude. Let us neglect the 2 term (the ﬁlters are selected such that ) and ) are not aﬀected) and consider the ﬁrst term only: ) = j (1.4) For the reader who has skipped the introduction about demodulation: eq. (1.3) describes a generic modulated signal and it can be written as ) = j [2 )] j We consider only the ﬁrst term, ) = j discarding for simplicity the factor as done in eq. (1.4): j accounts for what I refer to a rotation of the observer, as explained in Sect. 1.1.3. Eq. (1.4)

describes in general neither an amplitude nor a frequency modulated signal since both vary, see Fig. 1.3a. In an amplitude modulated signal, only the phasor length ) changes (the blue line in the ﬁgure) but not its phase: in other words, all the changes are along the phasor direction and not perpendicular to it. Phase modulation on the other hand would instead move the phasor on a circle (red line), with no changes to its radius. 1.2.1 Amplitude Modulation Amplitude and (narrow band) frequency modulations have equal Fourier power spectral densities: in fact their Fourier transforms

diﬀer by the sign of the sidebands but this disappears when squaring to obtain the power density. One can easily explain the sign by using eq. (1.4) and the rotating frame. A general AM signal has time independent phase oﬀset . Let us assume for simplicity = 0: AM ) = ) cos( rotating frame AM ) = AM ) (1.5) This means that in the rotating frame, the only time dependent term is the the phasor length.

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Figure 1.3: a) The phasor of a generically modulated signal can sweep the grey area. In pure amplitude modulation, only the length ) is time dependent, in pure phase

modulation, the phase only changes. The amplitude and phase modulations are not restricted to be a fraction of the radius and respectively. b) The decomposition of the sidebands of an AM signal: the sidebands must add up to have only a horizontal contribution: at = 0 they must have the same absolute value of the phase. To illustrate the spectral properties of this signal, we consider the simplest amplitude modulation, a pure tone modulation AM ) = 1 + cos( ) = 1 + j j with average amplitude 1 and amplitude modulation With reference to Fig. 1.3b, the modulation of the amplitude, cos( ),

originates from the two blue phasors of length h/ 2 counter-rotating with frequency : their sum, the thick blue arrow, is always along the axis. (The support carrier of length 1 is not shown in the Figure.) At = 0 they are both pointing outwards: this will be diﬀerent in the case of frequency modulation, see Sect. 1.2.2. These two phasors are the two sidebands of the AM spectrum: the phasor rotating ccw is the upper sideband at and the cw is the lower sideband at 1.2.2 Frequency Modulation A generic FM signal has the form FM ) = cos[ )] rotating frame FM ) = Ae j FM (1.6) with constant

Let us now consider the simplest case of frequency modulation FM ) = sin( ) (1.7) is also called the modulation index . The instantaneous frequency of the signal is the time derivative of the instantaneous phase FM ): ) = FM cos( hf is the peak frequency , the maximum frequency deviation. For reference, [1 + cos( )] cos( ) = cos( ) + cos( ) + cos( with and

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Pictorial representation of sidebands Electrical engineers know that the FM signal with given by eq. (1.7) has a frequency spectrum consisting of an inﬁnite number of sidebands at n and the amplitude of the -th

sideband is given by the -th Bessel function. Is there an intuitive picture of this? The demodulated signal FM ) = Ae j FM has constant amplitude, so it describes a phasor that moves on a circle of radius : let us take = 1. For a small phase swing, Fig. 1.4a, one can neglect the curvature of the circle and two terms only are suﬃcient to approximate FM ): an horizontal time independent phasor ) = 1 and a vertical phasor that has the same time periodicity as FM sin( ). We now need to ﬁnd its amplitude ). Let : for small FM ), one has FM ) = j FM 1 + j FM ) and the imaginary (i.e.

vertical) component is approximated by ) sin Im[ FM )] FM sin τ. This small angle limit, usually taken for h< 2, is called narrow band FM by the engineers. Figure 1.4: The FM signal and its decomposition in terms of the Bessel functions. With large angle modulations (large ), this approximation fails, because the length of the approx- imated phasor ) + ) sin 1 + sin = 1 + [1 cos(2 )] changes twice per cycle of the phasor, see Fig. 1.4b. One could compensate for this overshoot with the additional term ) in the horizontal direction, Fig. 1.4c. Since the ) correction has to take place twice

per cycle, it must have a cos(2 ) dependence. We conclude that ) cos(2 ) is the second set of sidebands at However at = 0 , ) has a contribution towards positive , and ) = 1 ) must be smaller than 1 to make the unit phasor, see Fig. 1.4d. Therefore the conditions on ) and ) are ) + ) cos(2 ) = cos FM ) + ) = 1 when = 0 ) = cos( ) when

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Figure 1.5: The sideband phasors for a) (and the odd terms) and b) (and the even terms). The blue arrow is the direction of the modulation. At this point, we still do not know we are dealing with the Bessel functions ), but we can note that also

) has one maximum (its global) at = 0. One could continue with the third order expansion, now in the vertical direction, which is odd in time, so its time dependence is ) sin(3 ) and imagine that in general, the phasors must satisfy the following relations sin FM ) = +1 ) sin[(2 + 1) ] cos FM ) = ) cos[2 n ] (1.8) Before showing that the ) are related to the ) Bessel function, one should note that the ﬁrst vertical phasor is the sum of two counter-rotating complex phasors (Figure 1.5a), which at = 0 point one in the positive and the other in the negative axis direction. This also holds

for all odd phasors. On the other hand, for the even phasors, which are horizontal, they are the sum of two phasors (Figure 1.5b) which at = 0 are pointing in the same direction, + . So: odd phasors have a phase diﬀerence between the upper and lower sideband, even phasors are in phase. This distinguishes the AM sidebands from the FM (ﬁrst set of) sidebands. Eq. (1.8) describes all sidebands at n with The Relation with the Bessel Coeﬃcients The ) are in fact proportional to the Bessel coeﬃcients. To show this, let us rewrite eq. (1.8): sin FM ) = +1 ) sin[(2 + 1) ] =

+1 (2 +1) (2 +1) cos FM ) = ) cos[2 n ] = n n Deﬁning 1) to take care of the negative coeﬃcient in the sine, and ), one can write cos FM ) + sin FM jh sin( jn (1.9) and multiplying both sides by jk and integrating gives k sin ) (1.10)

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which is the integral generating function of the Bessel coeﬃcients 0.4 0.2 0.0 0.2 0.4 0.6 0.8 1.0 amplitude Figure 1.6: The ﬁrst ﬁve Bessel functions. Note the zero crossing of ) at = 2 4048. Sidebands, the Hardcore Math While I am at it, I will also show the usual mathematical treat- ment: this requires some

higher math. To get the frequency spectrum, one starts from the JacobiAnger expansion jx sin jn ) = ( 1) We are going to consider a slightly more complicated phase modulation, ) = sin( ) + the FM signal is ) = cos[ sin( ) + Writing jh sin( jn [( n n (1.11) and taking the real part gives ) = ) cos[( n n From http://en.wikipedia.org/wiki/Bessel function . This should also solve the Bessel equation 00 ) + hJ ) + ( ) = 0? From the Laurent series x/ 2)( /t with the substitution i for sin or ie i for cos. From https://ccrma.stanford.edu/ jos/mdft/FM Spectra.html See also

http://mathworld.wolfram.com/BesselFunctionoftheFirstKind.html

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