Amplitude and Frequency/Phase Modulation - Zurich Instruments

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Pictorial representation of sidebands Electrical engineers know that the FM signal with ϕ(t) given by eq. (1.7) has a frequency spectrum consisting of an infinite number of sidebands at ω c ± nω m and the amplitude of the n-th sideband is given by the n-th Bessel function. Is there an intuitive picture of this? The demodulated signal d FM (t) = Ae jϕFM(t) has constant amplitude, so it describes a phasor that moves on a circle of radius A: let us take A = 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 d FM (t): an horizontal time independent phasor b 0 (h) = 1 and a vertical phasor that has the same time periodicity as ϕ FM (t) ∝ sin(ω m t). We now need to find its amplitude b 1 (h). Let τ ≡ ω m t: for small ϕ FM (τ), one has d FM (τ) = e jϕFM(τ) ≈ 1 + jϕ FM (τ) and the imaginary (i.e. vertical) component is approximated by b 1 (h) sin τ ≈ Im[d FM (t)] ≈ ϕ FM (τ) ≡ h sin τ. This small angle limit, usually taken for h < 0.2, is called narrow band FM by the engineers. a) b) c) d) Figure 1.4: The FM signal and its decomposition in terms of the Bessel functions. With large angle √ modulations (large h), this approximation fails, because the length of the approximated phasor b 2 0 (h) + b2 1 (h) sin2 τ ≈ 1 + h2 2 sin2 τ = 1 + h2 4 [1 − cos(2τ)] changes twice per cycle of the phasor, see Fig. 1.4b. One could compensate for this overshoot with the additional term b 2 (h) in the horizontal direction, Fig. 1.4c. Since the b 2 (h) correction has to take place twice per cycle, it must have a cos(2τ) dependence. We conclude that b 2 (h) cos(2τ) is the second set of sidebands at ±2ω m . However at τ = 0, π, b 2 (h) has a contribution towards positive x, and b 0 (h) = 1 − b 2 (h) must be smaller than 1 to make the unit phasor, see Fig. 1.4d. Therefore the conditions on b 0 (h) and b 2 (h) are { b 0 (h) + b 2 (h) = 1 when τ = 0 b 0 (h) + b 2 (h) cos(2τ) = cos ϕ FM (τ) −→ b 0 (h) − b 2 (h) = cos(h) when τ = π 6
a) b) Figure 1.5: The sideband phasors for a) b 1 (and the odd terms) and b) b 2 (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 J n (h), but we can note that also J 0 (h) has one maximum (its global) at h = 0. One could continue with the third order expansion, now in the vertical direction, which is odd in time, so its time dependence is b 3 (h) sin(3τ) and imagine that in general, the phasors must satisfy the following relations sin ϕ FM (τ) = ∑ n≥0 b 2n+1 (h) sin[(2n + 1)τ] cos ϕ FM (τ) = ∑ n≥0 b 2n (h) cos[2nτ] (1.8) Before showing that the b n (h) are related to the J n (h) Bessel function, one should note that the first vertical phasor is the sum of two counter-rotating complex phasors (Figure 1.5a), which at τ = 0 point one in the positive and the other in the negative x–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 τ = 0 are pointing in the same direction, +x. So: odd phasors have a π 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 ω c ± nω m with n ∈ N. The Relation with the Bessel Coefficients The b n (h) are in fact proportional to the Bessel coefficients. To show this, let us rewrite eq. (1.8): sin ϕ FM (τ) = ∑ b 2n+1 (h) sin[(2n + 1)τ] = ∑ b 2n+1 (h) [e ] (2n+1)τ − e −(2n+1)τ 2j n≥0 n≥0 cos ϕ FM (τ) = ∑ b 2n (h) cos[2nτ] = ∑ b 2n (h) [ e 2nτ + e −2nτ ] 2 n≥0 n≥0 Defining J n (h) ≡ (−1) n b n(h) 2 to take care of the negative coefficient in the sine, and J 0 (h) ≡ b 0 (h), one can write cos ϕ FM (τ) + j sin ϕ FM (τ) ≡ e jh sin(τ) = and multiplying both sides by e −jkτ and integrating gives 1 2π ∫ +π −π e −j(kτ−h sin τ) dτ = 1 2π +∞∑ n=−∞ 7 J n (h) ∫ +π −π +∞∑ n=−∞ J n (h)e jnτ (1.9) e j(n−k)τ dτ = J k (h) (1.10)
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Page 3 and 4: 1.1.3 The Explanation (and Some Eas
Page 5: pure a) PM b) d(t) pure AM Figure 1

Amplitude and Frequency/Phase Modulation - Zurich Instruments

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