Lecture 2. FOURIER TRANSFORMS AM and FM

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Lecture 2 FOURIER TRANSFORMS AM and FM We saw in the supplement on power spectra that the human range of hearing is concentrated in the range 4Hz to about 4Hz.

1 Lecture 2 FOURIER TRANSFORMS AM and FM We saw in the supplement on power spectra that the human range of hearing is concentrated in the range 4Hz to about 4Hz. That s where we d expect radio broadcasts to be, as well. But, check it out... the AM radio band is 535kHz to 75kHz, where khz means kilohertz or 3 Hz. FM is 88 to 8 MHz, where MHz is megahertz, or 6 Hz. Humans can t even hear those frequencies so what s the deal? The first part of the deal is that low-frequency radiation is easily absorbed by the atmosphere, so it s not a very good medium for transmitting signals over distances. People living on mountains learned this long ago they invent techniques like yodeling to transmit information. However, we re talking of using electromagnetic radiation, and that s the problem: electromagnetic signals have to be broadcast from antennae. Note, at right, a TV antennna. Why is it so tall? Because the theory of antenna design tells us that to broadcast a frequency ω, the antenna has to have height λ/2, where λ is the wavelength, the distance the wave travels in one cycle. That s c/ω where c is the speed of light, 86, miles/hr or miles/sec. A 4Hz signal would require a /4 sec to complete one cycle, so it has wavelength miles/sec /[4cycles/sec] = miles/cycle or 682 feet/cycle. An AM antenna at, say, 4 khz, would have a wavelength less by a factor of a thousand. In order to use radio frequencies for communication, then, we have to somehow embed low-frequency information (called the signal), into high-frequency radio waves (called the carrier wave) without substantially altering the carrier frequency or, at least, without changing it to a low-frequency wave. The technique of embedding a signal into a carrier is called modulation. A generic carrier can be written c(t) = A cos(2πf c t + φ) or Ae 2πi(f ct+φ) where A is real. We can modify A, f c, φ, leading to Amplitude Modulation, Frequency Modulation, and Phase Modulation. Each is workable, and each has its advantages.

2 We ll start with AM as it s a bit simpler. To transmit a pure tone at 4Hz, y s = A s cos(2πf s t) where f s = 4 is our frequency, you take that as the amplitude of a carrier wave, y c = cos(2πf c t), where we can assume f c is around khz. Trig identity: y c y s = A s cos(2πf s t) cos(2πf c t) = A s 2 [cos(2π{f c + f s }t) + cos(2π{f c f s }t)] Thus, the product is a wave with two frequencies, f c ±f s, which are going to be at, say,, ±4 Hz. This is well in the AM range, which is what we wanted: we modulated the carrier but it s still very high frequency. A bit of radio jargon: our modulation is called heterodyning. The signals at f c ± f s are called sidebands. And all of this is pretty useless unless we can recover the original signal from the modulated signal. Let m(t) be the signal we want to broadcast, and cos(2πf c t) be the carrier frequency. What goes out over the airwaves and comes in to the radio is m(t) cos(2πf c t). Now note the following cool trick: we multimply the signal by the carrier wave again! + [m(t) cos(2πf c t)] cos(2πf c t) = m(t) cos 2 cos(2π2fc t) (2πf c t) = m(t) 2 m(t) 2 + m(t) 2 cos(2π2f c t) The point here is that the original signal is now sitting at its original frequency, while another piece of it is modulated at twice the carrier frequency. This will produce sidebands at 2f c ± f s, but these sidebands will be very far away from any of the frequencies in m(t). This means you can recover m(t) simply by ignoring frequencies higher than any in m or, more precisely, by using a lowpass filter to eliminate them. Lab Problem Try this scheme with the selection from Billie Holliday s song Good Morning Heartache, sampled at 44Hz. What s the highest frequency in the song? Modulate it with a high frequency carrier wave cos(2πf c t), and demodulate using the scheme suggested above. How does it sound? There are a couple of problems with the above scheme. The first is it assumes that the carrier signal and the demodulating signal are in phase. What happens if they are out of phase? Try this with the demodulating signal cos(2πf c t + φ) where φ = π 3 ; π 2. The phase of the original signal is intrinsically unknowable, so the above can become a serious problem. There s also a potential problem with the carrier frequency: to reconstruct, you need to know it exactly. And, while the broadcaster may control the carrier wave cos(2πf c t) quite accurately, you may not be able to reconstruct that wave to the same accuracy. Assume that the broadcaster uses a carrier frequency f s, but you use a demodulating frequency of f s + f. Try reconstruction using values of f varying from one to five percent of f s. How does the demodulated signal sound? In real life, AM broadcasts may have a variation in carrier frequency could be up to 3Hz. Is this audible?

3 Because the carrier frequency can be so variable, typical AM radio uses a variation on the above carrier/modulator method. Before explaining, it helps to visualize what the various waves look like; Below, left to right: carrier, modulator, modulated signal Check out the modulated signal, in blue. While that s what is sent and received, the signal we want to detect is the red carrier signal, called the envelope. It turns out that a relatively simple diode-capacitor-resistor circuit can do this, which is why AM ration became popular. The following diagram, from Taub & Schilling s Principles of Communications Systems, shows the circuit and what it is designed to do: Those with access to Labview ought to be able to implement this. One thing you notice immediately: the circuit has difficulty with zero crossings, which are all too common in our top picture. Fortunately, there s an easy fix:

4 Now, any trick that can read the red wave on top can compute the carrier frequency! And do we ever have such a trick... y=linspace(,, 4*24); w=sin(2*pi*8*y + cos(2*pi*3*y)); plot(w) z=abs(fft(w)); plot(z(:5))

5 cos(2π8t) cos(cos(2π3t)) cos(2π8t) 2 cos2 (2π3t) The cos 2 (2π3t) term generates cos(2π6t), and multiplied by cos(2π8t), we get sidebands at 2Hz and 4Hz. None of this is exactly right, but it does give a nice overview of the main spectral features. To make it exactly right, one can compute the spectra exactly! It gives rise to the Bessel functions, J n. Let s start. In phase modulation, PM, the modulating function m(t) is applied to the phase φ in c(t) = Ae 2πi(f ct+φ), changing it to c(t) = Ae 2πi(f ct+βm(t)) where β is the modulating factor. Frequency modulation is a bit stranger; we have c(t) = Ae 2πi(ν(t)+φ) If ν were linear, ν = ft+ψ then the frequency would just be the slope of ν. In FM, the instantaneous frequency is ν, and we set it so that ν = f c + βm(t). This changes FM to a form of PM, with β m as the phase. I want to compute the spectrum of a PM signal in a special case: m(t) = sin(2πf m t). Then c(t) = Ae 2πi(f ct+βm(t)) = Ae 2πi(f ct+β sin(2πf m t)) = Ae 2πif ct e 2πi(β sin(2πf mt)) Now e 2πi(β sin(2πfmt)) is a periodic signal of period /f m, so that it can be expanded in a Fourier series, so that c(t) = + e 2πi(β sin(2πf mt)) = + a n Ae 2πinf mt e 2πif ct = a n e 2πinf mt + a n Ae 2πi(f c+nf m )t so that the carrier signal now has sidebands at f c + nf m. However, if the a n are small, or better still, zero, with large n, then the sidebands will still be at f c, f c ± f m, f c ± 2f m,.... The A n are easily computed; they re where the J n are Bessel functions, and If the modulation factor J n (β) = e 2πi(β sin 2πt nt) dt = J n (2πβ) k= is small, we see ( ) k 2 k k! Γ(k + n + ) (2πβ)2k+n J n = ( ) n J n

6 If the modulation factor β is small, we see J n (2πβ) (2πβ)n Γ(n + ) and, as this decreases geometrically with n, the sidebands where the signal is strong are simply the first few sidebands f c, f c ± f m, f c ± 2f m,..., f c ± kf m for small k. Now why exactly did we do this computation for the very very special modulated signal m(t) = β sin(2πf m t)? You might want to take a general m, expand it in a Fourier series, and see how that affects the computations above. If you decide to go ahead, keep in mind that the modulating signal is band-limited, so that it will have only a finite number of Fourier coefficients. However that goes, there s another important application for the theory just developed. It arose in early engineering times, when clocks were susceptible to slight errors. A sinusoidally oscillating error in timing gave rise to clock jitter, and it provided a slightly inaccurate measurement of signals. If you re sampling a trig function, this gives rise to a small phase shift in the function. Modern digital clocks have phase-lock circuits to minimize jitter, but modern experiments do not. Every experimental measurement has a slight error built in; this will give rise to some error in frequency computations. The true value is then the carrier, with amplitude J ; the first sideband has amplitude J. Then the signal-to-noise ratio, in db, is J (2πβ) 2 log J (2πβ) or, approximately 2 log (πβ). In our research, we try to identify the peak of the R-wave in the heartbeat. We apply wavelet filters to the data, which can shift the peak by an observed amount of roughly 4 milliseconds, or about -48db. Professor Bovrik in EE at UT Austin provides an example of AM/FM in image processing. Another application we have in mind is to real-life signals, as measured in labs. AM, FM, PM are all carefully and artificially constructed to accomplish certain purposes, and to do this, they change only one parameter at a time. But real-life signals can mix all three. Let s start with Matlab, generating a signal with AM and PM: Here, the signal: >> y=linspace(,, 4*24); >> w=y.^2.*sin(2*pi*8*cos(pi*y).*y);

7 No surprise: the amplitude increases quadratically, while the frequency -- well, it isn t very clear what the frequency does. The FFT will tell us, though: the spectrum here is significantly more complicated, and we re likely to have a very hard time sorting I had a carrier frequency of 8Hz, and now I seem to have some significant sidebands. I d expect sidebands at, 4, 7 and 2Hz, and I get those -- though not with decreasing amplitude! (You can try this with a similar signal -- here we ve chosen beta = ; try choosing beta =. or., and see what the spectrum looks like). We seem to have trouble sorting out the AM and the FM effects. This is where our new trick comes in. If our signal were complex valued, it d look like s(t) = A(t)e iφ(t), and A(t) would be the amplitude modulation, while the phase would be in φ. The miracle is that there s a function which can take a real-valued s and embed it into an analytic signal; it s called the Hilbert transform, and is available as hilbert in Matlab. Then hilbert(s) = A(t)e iφ(t) = A(t), which gives the modulating function. Here goes: >> h=hilbert(w); >> plot(abs(h)) >> hold >> plot(-abs(h)) >> plot(w, r ) The red curve is an approximation to the envelope of the signal s. Note by the way that the Hilbert transform is a smoothing: operator: the bumps in the signal have been reduced.

8 Below, the true envelope, y.^2, plotted in red, against the Hilbert, in green: Not bad! But there s a very obvious problem, the way green and red diverge near the end of the data. We ve mentioned this issue at the very beginning, when we talked about running averages. The divergence is called an edge effect: filters have a difficult time near the beginnings and ends of data, because -- well, because it s like falling over a cliff; the data ends and suddenly there s nothing to average. Lab Problem: There are two standard ways to reduce edge effects (at least two!); each involves putting some extra data near the edges. One method is called zero-padding: instead of starting with the data, you put some zeroes at the beginning or end. The second method is called reflection: you reflect the data about the end, as though you had an even or odd function. Try these. For the zero padding, you can try y=linspace(,, 4*24); z=y.^2.*sin(2*pi*8*y + cos(2*pi*3*y)); w = [zeros(, ) z]; Now try the Hilbert transform and plotting. For the reflection, the sine function is automatically an odd function, so you could easily let y=linspace(-,, 8*24); z=y.^2.*sin(2*pi*8*y + cos(2*pi*3*y)); Which method gives a Hilbert transform that diverges least from the known envelope y^2?

1B Paper 6: Communications Handout 2: Analogue Modulation

1B Paper 6: Communications Handout : Analogue Modulation Ramji Venkataramanan Signal Processing and Communications Lab Department of Engineering ramji.v@eng.cam.ac.uk Lent Term 16 1 / 3 Modulation Modulation