Communications IB Paper 6 Handout 2: Analogue Modulation
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1 Communications IB Paper 6 Handout 2: Analogue Modulation Jossy Sayir Signal Processing and Communications Lab Department of Engineering University of Cambridge jossy.sayir@eng.cam.ac.uk Lent Term c Jossy Sayir (CUED) Communications: Handout 2 Lent Term 1 / 30
2 Outline 1 Introduction and Motivation 2 Analogue Modulation Amplitude Modulation Phase Modulation Frequency Modulation c Jossy Sayir (CUED) Communications: Handout 2 Lent Term 2 / 30
3 Introduction and Motivation So far... Now... We have studied some analog information sources We have studied communications channels (attenuation, noise and fading) How do we efficiently transmit these signals through the channel? Remember the following Fourier transform properties for a signal x(t) X(f ) and a channel impulse response h(t) H(f ): y(t) = h(t) x(t) Y (f ) = H(f )X(f ) s(t) = x(t) cos(2πf c t) S(f ) = 1 2 [X(f f c) + X(f + f c )] c Jossy Sayir (CUED) Communications: Handout 2 Lent Term 3 / 30
4 Introduction and Motivation Need for Modulation Communications channels are only able to transmit information (with low attenuation) over certain frequency bands In radio transmission, the size of the antennas is usually λ 2 = c 2f where λ is the wavelength of the signal and c is the speed of light. For example f = 300 Hz, λ 2 = 500 Km!!! We need to transmit at frequencies such that we have good propagation characteristics and small antenna size c Jossy Sayir (CUED) Communications: Handout 2 Lent Term 4 / 30
5 Introduction and Motivation A Simple Example: Combining the Convolution and Modulation Properties X(f) Information signal H(f) Transfer characteristics of channel B B f fc f Y (f) = H(f)X(f) Y (f) = H(f)S(f) no modulation modulation s(t) = x(t) cos(2πfct) f fc B fc fc + B f c Jossy Sayir (CUED) Communications: Handout 2 Lent Term 5 / 30
6 Introduction and Motivation What is modulation? Shaping one or multiple parameters of a carrier wave with the information signal x(t). f c carrier frequency s(t) = a cos(2πf c t + φ) Amplitude Modulation a = f [x(t)] Angle Modulation Phase Modulation φ = f [x(t)] Frequency Modulation φ = 2πf [x(t)]t c Jossy Sayir (CUED) Communications: Handout 2 Lent Term 6 / 30
7 Introduction and Motivation Modulated Signals Information signal Carrier wave AM wave PM wave FM wave c Jossy Sayir (CUED) Communications: Handout 2 Lent Term 7 / 30
8 Introduction and Motivation Types of Modulation Depending on the nature of the information signal x(t) we have Analogue modulation Digital modulation c Jossy Sayir (CUED) Communications: Handout 2 Lent Term 8 / 30
9 Amplitude Modulation AM Modulation f [x(t)] = a 0 + x(t) so that s AM (t) = [a 0 + x(t)] cos(2πf c t) We define the modulation index as m A = max t x(t) a 0 the percentage that the carrier s amplitude varies above and below its unmodulated level. c Jossy Sayir (CUED) Communications: Handout 2 Lent Term 9 / 30
10 Amplitude Modulation 3 AM signal with a 0 > max x(t), i.e., m A < s AM (t) m A < 1 desirable, think of extracting the information signal from the modulated signal by envelope detection. m A > 1 undesirable, phase reversals would appear, and recovering the information signal is more complex. t c Jossy Sayir (CUED) Communications: Handout 2 Lent Term 10 / 30
11 Amplitude Modulation AM Spectrum We denote the spectrum of s AM (t) = [a 0 + x(t)] cos(2πf c t) by S AM (f ) = F[s AM (t)] (F[.] denotes the Fourier transform) and it is given by S AM (f ) = F[s AM (t)] = a 0 F[cos(2πf c t)] + F[x(t)] F[cos(2πf c t)] = a 0 2 [δ(f f c) + δ(f + f c )] + 1 }{{} 2 [X(f f c) + X(f + f c )] }{{} carrier information c Jossy Sayir (CUED) Communications: Handout 2 Lent Term 11 / 30
12 Amplitude Modulation X(f) B B f SAM(f) Lower sideband Upper sideband fc B fc fc + B fc B fc fc + B c Jossy Sayir (CUED) Communications: Handout 2 Lent Term 12 / 30
13 Amplitude Modulation Properties of AM 1 From the spectrum calculation, we see that the resulting AM modulated signal s AM (t) occupies a bandwidth B AM = 2B since both sidebands are transmitted. 2 The transmitted power is P AM = a P x 2 = P c + 2P sb where P c = a2 0 2 is the carrier power and P sb = Px 4 required to transmit one sideband. is the power c Jossy Sayir (CUED) Communications: Handout 2 Lent Term 13 / 30
14 Amplitude Modulation Improving AM: Double Sideband Suppressed Carrier (DSB-SC) The carrier transmission wastes power since only a fraction of the total power goes to transmit the information message DSB-SC transmits both sidebands but not the carrier: it uses the modulation property of the Fourier transform directly. Bandwidth of DSB-SC the same as AM Power of DSB-SC B DSB-SC = 2B P DSB-SC = P x 2 = 2P sb which improves on AM since the carrier is not transmitted. c Jossy Sayir (CUED) Communications: Handout 2 Lent Term 14 / 30
15 Amplitude Modulation Improving AM: Single Sideband Suppressed Carrier (SSB-SC) By symmetry we could obtain one sideband from the other, so transmission of both sidebands is not strictly necessary SSB-SC transmits only one sideband and does not transmit the carrier Bandwidth of SSB-SC half of AM or DSB-SC! Power of SSB-SC B SSB-SC = B P SSB-SC = P sb c Jossy Sayir (CUED) Communications: Handout 2 Lent Term 15 / 30
16 Amplitude Modulation X(f) Information signal B B SAM(f) f AM fc fc f SDSB-SC(f) DSB-SC fc fc f SSSB-SC(f) SSB-SC fc fc f c Jossy Sayir (CUED) Communications: Handout 2 Lent Term 16 / 30
17 Phase Modulation PM Modulation We modulate the instantaneous phase of the carrier signal yielding the PM modulated signal θ i (t) = 2πf c t + φ x(t) s PM (t) = a 0 cos(θ i (t)) = a 0 cos (2πf c t + φ x(t)) where φ is the phase deviation or modulation index of PM. Analogue PM is rarely used in practice PM has most of the properties of FM We will study FM in some detail c Jossy Sayir (CUED) Communications: Handout 2 Lent Term 17 / 30
18 Frequency Modulation FM Modulation We modulate the instantaneous frequency of the carrier signal f i (t) = f c + k f x(t) which translates into an instantaneous phase equal to t θ i (t) = 2πf c t + 2πk f x(τ)dτ yielding the FM modulated signal t s FM (t) = a 0 cos(θ i (t)) = a 0 cos (2πf c t + 2πk f 0 0 ) x(τ)dτ c Jossy Sayir (CUED) Communications: Handout 2 Lent Term 18 / 30
19 Frequency Modulation s FM (t) t c Jossy Sayir (CUED) Communications: Handout 2 Lent Term 19 / 30
20 Frequency Modulation FM Properties 1 Constant transmitted power P = a Nonlinearity: FM(x 1 (t) + x 2 (t)) FM(x 1 (t)) + FM(x 2 (t)) 3 FM more robust to noise than AM, since the message is hidden in the frequency and not in the amplitude 4 Bandwidth penalty with respect to AM c Jossy Sayir (CUED) Communications: Handout 2 Lent Term 20 / 30
21 Frequency Modulation What information signal does this FM modulated signal correspond to? s FM (t) t (a) a constant, (b) a ramp, (c) a rectangular pulse, (d) no clue c Jossy Sayir (CUED) Communications: Handout 2 Lent Term 21 / 30
22 Frequency Modulation Bandwidth of FM Signals Consider FM modulation of a tone, x(t) = a x cos(2πf x t), then f i (t) = f c + k f a x cos(2πf x t) θ i (t) = 2πf c t + k f a x f x sin(2πf x t) We define f = k f a x frequency deviation and m F = f f x modulation index, which represents the maximum phase deviation, i.e., the maximum departure of the angle θ i (t) from 2πf c t of the carrier. Then the FM signal becomes s FM (t) = a 0 cos (2πf c t + m F sin(2πf x t)) c Jossy Sayir (CUED) Communications: Handout 2 Lent Term 22 / 30
23 Frequency Modulation Bandwidth of FM Signals Using cos(a + B) = cos A cos B sin A sin B we can write s FM (t) = a 0 cos(2πf c t) cos (m F sin(2πf x t)) a 0 sin(2πf c t) sin (m F sin(2πf x t)) Now using Fourier series we write (C n = J n (m F ) are Bessel functions of the first kind) cos (m F sin(2πf x t)) = C 0 + C 2n cos(4nπf x t) sin (m F sin(2πf x t)) = n=1 C 2n 1 sin(2(2n 1)πf x t) n=1 c Jossy Sayir (CUED) Communications: Handout 2 Lent Term 23 / 30
24 Frequency Modulation Bandwidth of FM Signals Using the relationships sin A sin B = 1 2 [cos(a B) cos(a + B)] and cos A cos B = 1 2 [cos(a B) + cos(a + B)] we obtain s FM (t) = a { 0 2 cos(2πf c t) 2 C 1 [cos(2π(f c f x )t) cos(2π(f c + f x )t)] + C 2 [cos(2π(f c 2f x )t) + cos(2π(f c + 2f x )t)] C 3 [cos(2π(f c 3f x )t) cos(2π(f c + 3f x )t)] } +... c Jossy Sayir (CUED) Communications: Handout 2 Lent Term 24 / 30
25 Frequency Modulation Bandwidth of FM Signals Putting everything together and using J n (m F ) = ( 1) n J n (m F ) s FM (t) = a 0 n= J n (m F ) cos(2π(f c + nf x )t) which creates sidebands at harmonics of f x, i.e., it expands the bandwidth beyond f c + f x (that of AM) S FM (f ) = a 0 2 n= J n (m F ) [δ(f f c nf x ) + δ(f + f c + nf x )] c Jossy Sayir (CUED) Communications: Handout 2 Lent Term 25 / 30
26 Frequency Modulation 1 Fixed n 0.8 n=0 0.6 n=1 n=2 0.4 n=3 n=4 n=5 J n (m F ) m F c Jossy Sayir (CUED) Communications: Handout 2 Lent Term 26 / 30
27 Frequency Modulation 0.5 Fixed m F = J n (m F =5) n c Jossy Sayir (CUED) Communications: Handout 2 Lent Term 27 / 30
28 Frequency Modulation Example: FM spectrum of a pure tone with m F = 5. c Jossy Sayir (CUED) Communications: Handout 2 Lent Term 28 / 30
29 Frequency Modulation Bandwidth of FM Signals: Carson s rule Carson proposed the following rule to estimate the effective bandwidth of an FM-modulated tone (the absolute bandwidth is infinite, as shown by our calculations in previous slides) ( B FM = 2 f + 2f x = 2 f ) m F For general signals x(t) of bandwidth B, the generalised Carson s rule gives B FM = 2( f + B) c Jossy Sayir (CUED) Communications: Handout 2 Lent Term 29 / 30
30 Frequency Modulation Example BBC Radio Cambridgeshire: f c = 96 MHz and f = 75 khz. Assuming the voice/music signals have B = 15 KHz, we have m F = = 5 and B FM = 2( f + B) = 2( ) = 180kHz, while B AM = 30kHz Note that FM has better quality (larger SNR robustness against noise). c Jossy Sayir (CUED) Communications: Handout 2 Lent Term 30 / 30
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