1B Paper 6: Communications Handout 2: Analogue Modulation

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1 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 is the process by which some characteristic of a carrier wave is varied in accordance with an information bearing signal A commonly used carrier is a sinusoidal wave, e.g., cos(πf c t). f c is called the carrier frequency. We are allotted a certain bandwidth centred around f c for our information signal E.g. BBC Cambridgeshire: f c = 96 MHz, information bandwidth KHz Q: Why is f c usually large? A: Antenna size λ c larger frequency, smaller antennas! / 3

2 Analogue vs. Digital Modulation Analogue Modulation: A continuous information signal x(t) (e.g., speech, audio) is used to directly modulate the carrier wave. We ll study two kinds of analogue modulation: 1. Amplitude Modulation (AM) : Information x(t) modulates the amplitude of the carrier wave. Frequency Modulation (FM): Information x(t) modulates the frequency of the carrier wave We ll learn about: Power & bandwidth of AM & FM signals Tx & Rx design In the last 4 lectures, we will study digital modulation: x(t) is first digitised into bits Digital modulation then used to transport bits across the channel 3 / 3 Amplitude Modulation (AM) Information signal x(t), carrier cos(πf c t) The transmitted AM signal is s AM (t) = [a + x(t)] cos(πf c t) a is a positive constant chosen so that max t x(t) < a The modulation index of the AM signal is defined as m A = max t x(t) a The percentage that the carrier s amplitude varies above and below its unmodulated level Why is the modulation index important? m a < 1 is desirable because we can extract the information signal x(t) from the modulated signal by envelope detection. 4 / 3

3 When modulation index > 1: Phase reversals occur x(t) cannot be detected by tracing the +ve envelope 5 / 3 AM Receiver - Envelope Detector s AM (t) C R L V out (t) On the positive half-cycle of the input signal, capacitor C charges rapidly up to the peak value of input s AM (t) When input signal falls below this peak, diode becomes reverse-biased: capacitor discharges slowly through load resistor R L In the next positive half-cycle, when input signal becomes greater than voltage across the capacitor, diode conducts again until next peak value Process repeats... Very inexpensive receiver, but envelope detection needs m A < 1. 6 / 3

4 CircuitDiagram AM waveinput Envelopedetectoroutput 7 / 3 Spectrum of AM Next, let s look at the spectrum of s AM (t) = [a + x(t)] cos(πf c t) S AM (f ) = F[s AM (t)] [ = F [a + x(t)] (ejπf c t + e jπf c t ] ) = a [δ(f f c) + δ(f + f c )] }{{} carrier + 1 [X (f f c) + X (f + f c )] }{{} information (F[.] denotes the Fourier transform operation) 8 / 3

5 Example S AM (f ) = a [δ(f f c) + δ(f + f c )] + 1 [X (f f c) + X (f + f c )] X(f) C W W f a / S AM (f) a / C/ C/ f c W f c f c + W f c W f c f c + W 9 / 3 Properties of AM s AM (t) = [a + x(t)] cos(πf c t) S AM (f ) = a [δ(f f c) + δ(f + f c )] + 1 [X (f f c) + X (f + f c )] 1. Bandwidth: From spectrum calculation, we see that if x(t) is a baseband signal with (one-sided) bandwidth W, the AM signal s AM (t) is passband with bandwidth B AM = W. Power: We now prove that the power of the AM signal is where P X is the power of x(t) P AM = a + P X 1 / 3

6 Power of AM signal P AM = lim T = lim T 1 T 1 T T T = a + P X + lim T [a + x(t)] cos (πf c t) dt [a + x(t)] 1 + cos(4πf ct) dt 1 T T [a + x(t)] cos(4πf c t) dt We now show that the last the last term is. cos(4πf c t) is a high-frequency sinusoid with period T c = 1 f c. g(t) = (a +x(t)) is a baseband signal which changes much more slowly than cos(4πf c t). Hence, with T = nt c, we have 1 T T g(t) cos(4πf c t) dt 1 ( Tc g() cos(4πf c t) dt + nt c Tc + T c g(t c ) cos(4πf c t) dt... + ntc ) g((n 1)T c ) cos(4πf c t) dt = (n 1)T c Hence P AM = a + P X. 11 / 3 Double Sideband Suppressed Carrier (DSB-SC) The power of AM signal is P AM = a }{{} carrier + P X The presence of a makes envelope detection possible, but requires extra power of a corresponding to the carrier In DSB-SC, we eliminate the a : We transmit only the sidebands, and suppress the carrier X(f) W W f S dsb sc (f) Lower Sideband Upper Sideband f c W f c f c + W f c W f c f c + W 1 / 3

7 DSB-SC ctd. The transmitted DSB-SC wave is s dsb-sc (t) = x(t) cos(πf c t) X(f) W W f S dsb sc (f) Lower Sideband Upper Sideband f c W f c f c + W f c W f c f c + W How to recover x(t) at the receiver? Phase reversals cannot use envelope detection 13 / 3 DSB-SC receiver DSB-SC Receiver: Product Modulator + Low-pass filter s dsb sc (t) v(t) Low-pass filter ˆx(t) cos(πf c t) 1. Multiplying received signal by cos(πf c t) gives v(t) = x(t) cos (πf c t) = x(t) }{{} low freq. + x(t) cos(4πf ct) }{{} high freq.. Low-pass filter eliminates the high-frequency component Ideal low-pass filter has H(f ) = constant for W f W, and zero otherwise 14 / 3

8 Properties of DSB-SC s dsb-sc (t) = x(t) cos(πf c t) S dsb-sc (f ) = 1 (X (f + f c) + X (f f c )) Bandwidth of DSB-SC is B dsb-sc = W, same as AM Power of DSB SC is P dsb-sc = P X (follows from AM power calculation) DSB-SC requires less power than AM as the carrier is not transmitted But DSB-SC receiver is more complex than AM! We assumed that receiver can generate locally generate a frequency f c sinusoid that is synchronised perfectly in phase and frequency with transmitter s carrier Effect of phase mismatch at Rx is explored in Examples paper 15 / 3 Single Sideband Suppressed Carrier (SSB-SC) DSB-SC transmits less power than AM. Can we also save bandwidth? x(t) is real X ( f ) = X (f ) Need to specify X (f ) only for f > In other words, transmission of both sidebands is not strictly necessary: we could obtain one sideband from the other! X(f) W S ssb sc (f) W f Upper Sideband f c W f c f c f c + W Bandwidth is B ssb-sc = W, half of that of AM or DSB-SC! Power is is P ssb-sc = P X 4, half of DSB-SC 16 / 3

9 Summary: Amplitude Modulation X(f) Information signal W W f S AM (f) AM f c W f c f c + W f c W f c f c + W S dsb sc (f) DSB-SC f c W f c f c + W f c W f c f c + W S ssb sc (f) SSB-SC f c W f c f c f c + W 17 / 3 You can now do Questions 1 5 on Examples Paper / 3

10 Frequency Modulation (FM) In FM, the information signal x(t) modulates the instantaneous frequency of the carrier wave. The instantaneous frequency f (t) is varied linearly with x(t): f (t) = f c + k f x(t) This translates to an instantaneous phase θ(t) given by θ(t) = π t t f (u)dt = πf c t + πk f x(u)du The modulated FM signal t s FM (t) = A c cos(θ(t)) = A c cos (πf c t + πk f A c is the carrier amplitude k f is called the frequency-sensitivity factor ) x(u)du 19 / 3 Example What information signal does this FM wave correspond to? s FM (t) t (a) a constant, (b) a ramp, (c) a sinusoid, (d) no clue / 3

11 FM Demodulation At the receiver, how do we recover x(t) from the FM wave? (ignoring effects of noise) t s FM (t) = A c cos (πf c t + πk f ) x(u)du The derivative is ds FM (t) t = πa c [f c + k f x(t)] sin (πf c t + πk f dt ) x(u)du The derivative is a passband signal with amplitude modulation by [f c + k f x(t)] If f c large enough, we can recover x(t) by envelope detection of ds FM(t) dt! Hence FM demodulator is a differentiator + envelope detector d F Differentiator: dt jπf (frequency response). See Haykin-Moher book for details on how to build a differentiator 1 / 3 Properties of FM t s FM (t) = A c cos (πf c t + πk f ) x(u)du Power of FM signal = A c, regardless of x(t) Non-linearity: FM(x 1 (t) + x (t)) FM(x 1 (t)) + FM(x (t)) FM is more robust to additive noise than AM. Intuitively, this is because the message is hidden in the frequency of the signal rather than the amplitude. But this robustness comes at the cost of increased transmission bandwidth What is the bandwidth of the FM signal s FM (t)? The spectral analysis is a bit complicated, but we will do it for a simple case... where x(t) is a sinusoid (a pure tone) / 3

12 FM modulation of a tone Consider FM modulation of a tone x(t) = a x cos(πf x t). We have f (t) = f c + k f a x cos(πf x t) θ(t) = πf c t + k f a x f x sin(πf x t) f = k f a x is called the frequency deviation f is the max. deviation of the carrier frequency f (t) from f c β = k f a x f x = f f x is called the modulation index β is the max. deviation of the carrier phase θ(t) from πf c t Then the FM signal becomes s FM (t) = A c cos (πf c t + β sin(πf x t)) 3 / 3 The spectrum of the FM signal We want to understand the frequency spectrum of s FM (t) = A c cos (πf c t + β sin(πf x t)) We can write where s FM (t) = Re [A ] c e jπf c t + jβ sin(πf x t) = Re [ s(t)e ] jπf ct s(t) = A c e jβ sin(πf x t) s(t) is periodic with period 1/f x Can express using Fourier series (Fundamental frequency f x ) s(t) = c n e jπf x nt n= 4 / 3

13 FM Spectrum ctd. You will show in Examples Paper 8 that the Fourier series coefficients of s(t) are where J n (β) = 1 π c n = A c J n (β) π π e j(β sin u nu) du J n (.) is called the nth order Bessel function of the first kind Thus s(t) = A c e jβ sin(πf x t) = c n e jπf x nt = A c n= n J n (β)e jπf x nt. 5 / 3 Therefore s FM (t) = Re [ ] [ s(t)e jπf ct = Re = Re n= [ n= = A c n= c n e jπf x nt e jπf ct ] A c J n (β) e jπ(f c +nf x )t ] J n (β) cos(π(f c + nf x )t) Taking Fourier Transforms, the spectrum of the FM signal is S FM (f ) = A c n= J n (β) [ δ(f f c nf x ) + δ(f + f c + nf x ) ] 6 / 3

14 Plots of Jn (β) vs β n n n n n 7 / 3 Example What is the spectrum of the FM signal when x(t) is a pure tone and the modulation index β = 5? Jn (β) vs n for β = Jn(β) n 8 / 3

15 Example ctd. The spectrum is S FM (f ) = A c n= J n (5) [ δ(f f c nf x ) + δ(f + f c + nf x ) ] X(f) -f x f x S FM (f) -f c f c 9 / 3 Bandwidth of FM signals To summarise, when x(t) has only a single frequency f x, the spectrum of the FM wave is rather complicated: There is a carrier component at f c, and components located symmetrically on either side of f c at f c ± f x, f c ± f x,... The absolute bandwidth is infinite, but... the side components at f c ± nf x become negligible for large enough n Carson s rule for the effective bandwidth of FM signals: 1. The bandwidth of an FM signal generated by modulating a single tone is ( ) B FM f + f x = f β. For an FM signal generated by modulating a general signal x(t) with bandwidth W, the bandwidth B FM f + W (Recall: for any FM wave, f is the frequency deviation around f c ) 3 / 3

16 Example BBC Radio Cambridgeshire: f c = 96 MHz and f = 75 khz. Assuming that the voice/music signals have W = 15 khz, we have and the bandwidth β = f W = = 5 B FM = ( f + W ) = ( ) = 18 khz, while B AM = W = 3 khz FM signals have larger bandwidth than AM, but have better robustness against noise. 31 / 3 Summary: Analogue Modulation Amplitude Modulation with information signal of bandwidth W AM modulated signal: Bandwidth W, high power, simple Rx using envelope detection DSB-SC: Bandwidth W, lower power, more complex Rx SSB-SC: Bandwidth W, even lower power, Rx similar to DSB-SC Frequency Modulation with information signal of bandwidth W : FM signal has constant carrier amplitude constant power Bandwidth of FM signal depends on both β and W Can be significantly greater than W Better robustness to noise than AM the information is hidden in the phase 3 / 3

Communications IB Paper 6 Handout 2: Analogue Modulation

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