5 Analog carrier modulation with noise

Transcription

1 5 Analog carrier modulation with noise 5. Noisy receiver model Assume that the modulated signal x(t) is passed through an additive White Gaussian noise channel. A noisy receiver model is illustrated in Fig. 6. x(t) y(t) BPF z(t) Demodulator u(t) n W (t) f 0 B T Figure 6: Noisy receiver model n W (t) is the realization of a white random process N W (t) with power spectral density N 0 Watts/Hz, i.e. E [N W (t)n W (u)] = N 0 δ(t u) S N W (f) = N 0 The bandpass filter is designed to pass the modulated signal undistorted (and also amplify it in practice) but will cut noise power outside the frequency band of the modulated signal. The input of the bandpass filter is given by The output of the bandpass filter is given by y(t) = x(t) + n W (t) z(t) = x(t) + n(t) where n(t) is the realization of the bandpass random process N(t). Exercise: Draw the power spectral density of N W (t) and show that N W (t) has an infinite power. Exercise: Draw the power spectral density of N(t) and show that N(t) is a band-limited random process (band-limited noise) of power E [N (t)] = R N (0) = B T N 0. 56

2 Hence it is seen that the bandpass filter reduces the noise power. We assume that the message signal m(t) is the realization of a stationary random process of zero mean, whose power spectral density S M (f) is limited to a maximum frequency W. W is referred as the message bandwidth. Let M(t) be the corresponding random process. Example of the spectrum of M(t): (draw an example) For the modulation studied in this course (AM, DSB-SC, SSB, VSB, FM, PM), the modulated signal X(t) is a bandpass random process with bandwidth B T, whose spectrum is centered around f 0 with realization denoted x(t). f 0 is defined as follows f c, AM,DSB-SC, FM, PM f 0 = f c W, lower side-band SSB f c + W, Example of the spectrum of X(t): (draw an example) upper side-band SSB Assessing noise performance: Assume that the output of the demodulator can be written as u(t) = m(t) + w(t) where m(t) is a realization of the recovered message signal and w(t) is the noise component at the output of the demodulator. The effect of w(t) can be assessed by calculating the Output-Signalto-noise ratio. SNR 0 = Average power of the demodulated message signal Average power of the noise measured at the receiver output = P M P W The output SNR is meaningful only if there is an additive relationship between the signal and noise at the receiver output. So for FM or AM with envelope detection, we will consider low noise power at the receiver input to obtain an approximate additive relationship. 57

3 Since the output SNR depends on the modulation and the demodulation techniques used, to obtain a fair comparison of several demodulation techniques, we should use SNR 0 SNR I where SNR I is the Input-Signal-to-noise ratio defined as SNR I = Average power of the modulated message signal (at demodulator input) Average power of the noise measured at the demodulator input = P R P N = P X N 0 B T P R : Signal power at the receiver input (in absence of noise) = noiseless signal power at demodulator input P N : Average power of the filtered noise N(t) Since P N depends on the modulation through its transmission bandwidth, to obtain a fair comparison of noise performance for several modulation schemes, SNR I is not appropriate, and the baseband (channel) signal-to-noise ratio SNR b = SNR C should be used, where SNR b = SNR C = Signal-to-noise ratio without modulation 5. Baseband reference system A reference baseband system is obtained by passing a noisy message signal of power equal to the power of the modulated signal through a low-pass filter of bandwidth equal to that of the message signal as illustrated in Fig. 7. m (t) LPF v(t) n W (t) W Figure 7: Baseband reference system The output of the low-pass filter is given by v(t) = m (t) + n (t) where m (t) has the same power as the modulated wave. Then SNR b = SNR C = Average power of the modulated signal Average power of the noise in the message bandwidth = P M P N = P R N 0 W 58

4 5.3 SNR s for linear modulations with coherent detection Exercise: Draw a block diagram of a noisy receiver model with coherent detection. y(t) = x(t) + n W (t) (Received signal) z(t) = x(t) + n(t) (Output of bandpass filter) (6) x(t) = x I (t) cos(πf c t) x Q (t) sin(πf c t) n(t) = n I (t) cos(πf 0 t) n Q (t) sin(πf 0 t) Rice representation Note that lowercase letters denote realizations of the random processes denoted by capital letters. Exercise: Draw the power spectral densities of N I (t) and N Q (t) a) Input signal power: P R Let us assume that X I (t) and X Q (t) are jointly wide-sense stationary with constant means K I and K Q. Exercise: Verify that for AM, DSB-SC, SSB and VSB X I (t) and X Q (t) are jointly wide-sense stationary with constant mean. Hint: Recall that M(t) is assumed to a WSS random process with zero mean and WSS stationarity is preserved by linear time-invariant filtering. Note that X I (t) is zero mean except for AM modulation due to the DC component. 59

5 To find the received power P R, we set n W (t) = 0 in (6) and calculate the power of the signal left. If n W (t) = 0 then n(t) = 0 = z(t) = x(t) Hence P R = P Z = P X = lim T T E [ X (t) ] dt In order to simplify P R let us show that X(t) is a wide-sense cyclo stationary random process. E [X(t)] = E [X I (t)] cos(πf c t) E [X Q (t)] sin(πf c t) = K I cos(πf c t) K Q sin(πf c t) where K I and K Q are constants. Therefore the mean of X(t) is periodic with period /f c. E [X(t + τ)x(t)] = E [( X I (t + τ) cos(πf c (t + τ)) X Q (t + τ) sin(πf c (t + τ)) ) ( XI (t) cos(πf c t) X Q (t) sin(πf c t) )] = R XI (τ) cos(πf c (t + τ)) cos(πf c t) R XI X Q (τ) cos(πf c (t + τ)) sin(πf c t) R XQ X I (τ) sin(πf c (t + τ)) cos(πf c t) + R XQ (τ) sin(πf c (t + τ)) sin(πf c t) = R X I (τ) [cos(πf c (t + τ)) + cos(πf c τ)] R X I X Q (τ) [sin(πf c (t + τ)) sin(πf c τ)] R X Q X I (τ) [sin(πf c (t + τ)) + sin(πf c τ)] R X Q (τ) [cos(πf c (t + τ)) cos(πf c τ)] Hence it is seen that E [X(t + τ)x(t)] is periodic with period /f c. Hence the common period is /f c and X(t) is wide-sense cyclo stationary with power defined as P X = RX(0) a = fc E [ X (t) ] dt f c f c = R X I (0) cos(πf c 0) + R X I X Q (0) sin(πf c 0) R X Q X I (0) sin(πf c 0) + R X Q (0) cos(πf c 0) = R X I (0) + R X Q (0) (7) Note that (7) is only valid if X I (t) and X Q (t) are jointly wide-sense stationary with constant means. b) Input noise power: P N PN = N0BT 60

6 c) Input signal-to-noise ratio: SNR I P R SNR I = = W SNR C N 0 B T B T Recall that random processes with signal components are denoted by capital letters and their realizations are denoted by lowercase letters. AM: x I (t) = A c ( + µm n (t)) x Q (t) = 0 B T = W R XI (0) = E [ X I (t) ] = A c + A cµ E [ M n(t) ] + A c µe [M n (t)] = A c R XQ (0) = E [ X Q(t) ] = 0 SNR I = A c ( + kap M ) N 0 W = A c ( + µ P Mn ) 4N 0 W where P M is the power of M(t), P Mn is the power of M n (t) = P Mn = E [ M n(t) ] = = A c ( + k ap M ) 4N 0 W M(t) given by max m(t) P M (max M(t) ) = (max m(t) ) S M (f)df ( + µ P Mn ) Note that we assume that max m(t) is the same for all the realizations of M(t). If this condition is not satisfied, then the expression of SNR I should be expressed in terms of the amplitude sensitivity k a and P M instead of the modulation index µ and P Mn. DSB-SC: x I (t) = A c m(t) x Q (t) = 0 B T = W R XI (0) = E [ X I (t) ] = A ce [ M (t) ] = A cp M R XQ (0) = E [ X Q(t) ] = 0 where P M is the power of M(t) given by SNR I = A cp M 4N 0 W P M = E [ M (t) ] = S M (f)df SSB: x I (t) = A c m(t) x Q(t) = ± A c ˆm(t) B T = W 6

7 R XI (0) = E [ XI (t) ] = A c 4 E [ M (t) ] = A c 4 P M ] R XQ (0) = E [ X Q(t) ] = A c 4 E [ ˆM (t) = A c 4 R ˆM(0) = A c 4 R M(0) = A c 4 P M SNR I = A c 4 P M N 0 W = A cp M 4N 0 W d) Output signal power By definition the center frequency of the bandpass filter can be written as f 0 = f c + α where B T α B T, depending on the modulation format. The output of the bandpass filter is given by z(t) = x I (t) cos(πf c t) x Q (t) sin(πf c t) + n I (t) cos ( π(f c + α)t ) n Q (t) sin ( π(f c + α)t ) = [ x I (t) + n I (t) cos(παt) n Q (t) sin(παt) ] cos(πf c t) [ x Q (t) + n I (t) sin(παt) + n Q (t) cos(παt) ] sin(πf c t) The output of the multiplier is given by v(t) = z(t) cos(πf c t) = [ xi (t) + n I (t) cos(παt) n Q (t) sin(παt) ] + xi (t) + n I (t) cos(παt) n Q (t) sin(παt) [ ] cos(4πf c t) [ xq (t) + n I (t) sin(παt) + n Q (t) cos(παt) ] sin(4πf c t) The output of the low-pass filter is given by u(t) = [ xi (t) + n I (t) cos(παt) n Q (t) sin(παt) ] (explain) Note that for AM, the actual demodulator includes a DC block after the low-pass filter, so the 6

8 demodulator output for AM is u (t) = A c k am(t) + n I(t) = A c µm n(t) + n I(t) since f 0 = f c = α = 0. The output signal power is obtained by assuming n W (t) = 0 thus n(t) = n I (t) = n Q (t) = 0. The output signal power is given by e) Output noise power u(t) = x I(t)αt) u (t) = A c k am(t) = A c µm n(t) AM P SO = R X I (0) 4 P SO DSB-SC,SSB, VSB = A ck a 4 P M = A cµ 4 P M n AM Let us consider separately AM, DSB-SC and SSB, VSB. For AM and DSB-SC, f 0 = f c so α = 0. To find the output noise power we assume m(t) = 0, therefore u(t) = n I(t) u (t) = n I(t) DSB-SC AM The output noise power is given by For SSB and VSB, α 0, u(t) = P N0 = R N I (0) 4 = N 0W [ ni (t) cos(παt) n Q (t) sin(παt) ] = [ n I(t) n Q(t) ] Since N I (t) and N Q (t) are zero mean WSS stationary, let us show that N I (t) and N Q (t) are zero mean wide-sense cyclo stationary and uncorrelated. [ ] E N I(t) = E [N I (t)] cos(παt) = 0 [ ] E N Q(t) = E [N Q (t)] sin(παt) = 0 63

9 [ ] R N (t + τ, t) = E N I(t + τ)n I(t) I = R NI (τ) cos(πα(t + τ) ) cos ( παt ) periodic with period [ ] α R N (t + τ, t) = E N Q(t + τ)n Q(t) Q R N I N (t + τ, t) = E Q = R NQ (τ) sin ( πα(t + τ) ) sin ( παt ) periodic with period [ ] α N I(t + τ)n Q(t) = R NI N Q (τ) cos ( πα(t + τ) ) sin ( παt ) = 0 since N I (t) and N Q (t) are uncorrelated and zero mean Their average autocorrelation functions are given by R a 4α = α N I(τ) 4α R N (t + τ, t)dt = R N I (τ) cos(πατ) I R a (τ) = R N Q (τ) cos(πατ) = R N I (τ) cos(πατ) = R a N Q N I(τ) Since N I (t) and N Q (t) are uncorrelated and wide-sense stationary, u(t) in absence of message signal is also wide-sense cyclo stationary. Its autocorrelation function at (t, t) is given by R U (t, t) = E [ U (t) ] = 4 E [ N = 4 E [ N I (t) Hence the output noise power is ] I (t) + [ 4 E ] + [ ] 4 E N I (t) ] I (t) [ ] 4 E N I(t)N Q(t) N P NO = RU(0) a = 4 Ra (0) + N I 4 Ra (0) = N I Ra (0) = R N I (0) N I 4 = N 0B T 4 Exercise: Find the power spectral density of N I and N Q for SSB and verify that the output noise power for SSB is given by P NO = N 0W 4 64

10 f) SNR s AM: SNR I = SNR C SNR O = A ck ap M N 0 W = A c ( + k ap M ) 4N 0 W = A cµ P Mn N 0 W = A c ( + µ P Mn ) 4N 0 W SNR O = µ P Mn + µ P Mn SNR I = νsnr I = νsnr C where ν (ν ) is the modulation efficiency given by DSB-SC: SSB: ν = k ap M + k ap M = µ P Mn + µ P Mn SNR I = SNR C SNR O = A c 4 P M N 0 W SNR O = SNR I = SNR C = A cp M 4N 0 W = A cp M N 0 W SNR I = SNR C = A cp M 4N 0 W SNR O = A c P M 6 N 0 W 4 SNR O = SNR I = SNR C 5.4 SNR s for AM with envelope detection = A cp M 4N 0 W Draw a block diagram of a noisy receiver model with envelope detection. 65

11 The output of the bandpass filter is given by z(t) = x(t) + n(t) = A c ( + k a m(t)) cos(πf c t) + n I (t) cos(πf c t) n Q (t) sin(πf c t) The complex envelope of z(t) is given by The output of the envelope detector is z(t) = A c ( + k a m(t)) + n I (t) + jn Q (t) v(t) = z(t) = [ (A c ( + k a m(t)) + n I (t)) + n Q(t) ] / a) high SNR A c ( + k a m(t)) n I (t), n Q (t) The output of the envelope detector is given by [ ( ) ] / n Q (t) v(t) = (A c ( + k a m(t)) + n I (t)) + A c ( + k a m(t)) + n I (t) A c ( + k a m(t)) + n I (t) The output of the DC block is given by Hence the SNR s are given by SNR I = P R P N = SNR O = P S O P NO (same result as for coherent detection) u (t) = A c k a m(t) + n I (t) A c ( + k ap M ) N 0 W = A ck ap M N 0 W = νsnr I = νsnr C b) low SNR A c ( + k a m(t)) n I (t), n Q (t) = A c ( + µ P Mn ) 4N 0 W = A cµ P Mn N 0 W Show that in that case the output of the envelope detector is badly corrupted by noise. 66

12 5.5 SNR s for angle modulation with noise x(t) y(t) BPF z(t) Angle Dectector u(t) LPF v(t) n W (t) f c B T W Figure 8: Noisy receiver model for angle modulation z(t) = x(t) + n(t) ϕ(t) = K = A c cos ( πf c t + ϕ(t) ) + n I (t) cos(πf c t) n Q (t) sin(πf c t) t Assuming no noise, we have m(τ)h(t τ)dτ v(t) = K m(t) (output of angle detector is proportional to m(t)) The lowpass filter is used to cut noise outside the frequency band of the recovered message signal. a) Channel signal-to-noise ratio: SNR C Input signal power: Assume first that m(t) is deterministic, then the input signal power P x = lim T = lim T T [ A c A c cos ( πf c t + ϕ(t) ) dt T = A c + A c lim T T dt + A c T cos ( 4πf c t + ϕ(t) ) dt cos ( 4πf c t + ϕ(t) ) ] dt Assume that f c and that ϕ(t) is slowly varying with respect to f c, then 4πf c t + ϕ(t) 4πf c t + ϕ 0 and P x A c + A c lim T T cos ( 4πf c t + ϕ 0 ) dt 67

13 A c + A c lim T A c T [ ( ) ( )] sin 4πfc T + ϕ 0 + sin 4πfc T ϕ 0 4πf c Assume now that m(t) is the realization (sample function) of a stationary zero mean Gaussian random process M(t) whose autocorrelation function is R M (τ). The angle modulated signal X(t) is now a random process which is neither wide-sense stationary nor wide-sense cyclo-stationary. However using generalization of Wiener-Khinchin theorem, its average autocorrelation function is defined as RX(τ) a = lim T T = A c lim T = A c + T R X (t + τ, t)dt E [ cos ( πf c (t + τ) + ϕ(t + τ) ) cos ( πf c t + ϕ(t) )] dt T { T lim E [ cos ( πf c (t + τ) + ϕ(t + τ) + ϕ(t) )] dt T T E [ cos ( πf c τ + ϕ(t + τ) ϕ(t) )] } dt = A c lim T T = A c lim T T E [ cos ( πf c τ + ϕ(t + τ) ϕ(t) )] dt R { e jπfcτ E [ e j[ϕ(t+τ) ϕ(t)]]} dt Since ϕ(t) is the output of a linear time invariant filter with input m(t), ϕ(t) is also a stationary Gaussian random process with autocorrelation function R ϕ (τ). It can be shown that for fixed t, Z(t) = ϕ(t + τ) ϕ(t) is a zero-mean Gaussian random variable with variance σ Z = R ϕ (0) R ϕ (τ) Therefore E [ e jz(t)] = E [ e jωz(t)] ω= = e ω σz = e [Rϕ(0) Rϕ(τ)] ω= And the average autocorrelation function of X(t) is given by RX(τ) a = A c lim T T = A c cos(πf cτ)a(τ) R { e jπfcτ e (Rϕ(0) Rϕ(τ))} dt 68

14 with The power of X(t) is given by a(τ) = E [ e j(ϕ(t+τ) ϕ(t))] = e [Rϕ(0) Rϕ(τ)] P X = R a X(0) = A c identical to the expression found when m(t) is deterministic. Input noise power: P N = N 0 B T Noise power in the message bandwidth: P N = N 0 W Input signal-to-noise ratio: Channel signal-to-noise ratio: SNR I = SNR C = A c N 0 B T A c N 0 W b) Output signal-to-noise ratio assuming a high Carrier-to-noise ratio: SNR O High carrier-to-noise ratio is defined as: A c r n (t) = n I (t) + n Q (t) The angle detector and the lowpass filter can be modeled as illustrated in Fig. 9. z(t) Phase Detector z (t) h o (t) v(t) Figure 9: Noisy receiver model for angle modulation where the filter h o (t) has transfer function given by H o (f) = {, H(f) f < W 0, else. The phase detector yields an output γϕ z (t) which is proportional to the phase of its input, where ϕ z (t) is the phase of z(t) defined as ϕ z (t) = arg [ z(t)] 69

15 and z(t) is the complex envelope of z(t). For simplicity, let us assume that the proportionality constant is equal to. The output of the phase detector is given by ϕ z (t) = ϕ(t) + ɛ(t) The output of the filter h o (t) is given by = Km(t) h(t) + ɛ(t) v(t) = Km(t) h(t) h o (t) + ɛ(t) h o (t) Explain the chosen transfer function of h o (t) by considering first a noiseless case. Assume no noise, i.e. n(t) = 0, hence z(t) = ϕ z (t) = v(t) = What is the desired output v(t)? Consider now the noisy case again, then v(t) = Km(t) h(t) h o (t) + n o (t) (8) where n o (t) = ɛ(t) h o (t) Hence from (8) we have the equivalent baseband model illustrated in Fig. 0. m(t) K h(t) ϕ(t) ϕ z (t) h o (t) v(t) H(f) ɛ(t) Figure 0: Noisy receiver equivalent model with angle modulation. Expression of ɛ(t): z(t) = x(t) + ñ(t) = A c e jϕ(t) + r n (t)e jθn(t) z(t) = z(t) e jϕz(t) ϕ z (t) = ϕ(t) + ɛ(t) 70

16 Im ε(t) = phase error n(t) ~ = r (t) n z(t) ~ r (t) n θ n (t) ϕ(t) ε(t) ϕ(t) Ac ϕ (t) z Re Figure : Phasor diagram for angle modulation with noise Using the phasor diagram from Fig., show that tan ɛ(t) = r n (t) sin ( θ n (t) ϕ(t) ) A c + r n (t) cos ( θ n (t) ϕ(t) ) Assuming high carrier-to-noise ratio Therefore tan ɛ(t) ɛ(t) tan ɛ(t) r n(t) A c sin ( θ n (t) ϕ(t) ) 7

17 r n(t) A c sin ( θ n (t) ϕ(t) ) = r n(t) A c = n Q(t) A c sin ( θ n (t) ) cos ϕ(t) r n(t) A c cos ϕ(t) n I(t) A c E [ɛ(t)] = 0 sin ϕ(t) cos ( θ n (t) ) sin ϕ(t) since n I (t) and n Q (t) are zero mean. Assume that m(t) is the realization of a WSS random process M(t) independent of N(t), hence ϕ(t) is a WSS process independent of N(t). The autocorrelation of ɛ(t) is given by R ɛ (t + τ, t) = E [ɛ(t + τ)ɛ(t)] = { E [NQ (t + τ)n A Q (t)] E [ cos ( ϕ(t + τ) ) cos ( ϕ(t) )] c E [N Q (t + τ)n I (t)] E [ cos ( ϕ(t + τ) ) sin ( ϕ(t) )] where E [N I (t + τ)n Q (t)] E [ sin ( ϕ(t + τ) ) cos ( ϕ(t) )] + E [N I (t + τ)n I (t)] E [ sin ( ϕ(t + τ) ) sin ( ϕ(t) )]} = { RNI (τ) (E [ cos ( ϕ(t + τ) + ϕ(t) )] + E [ cos ( ϕ(t + τ) ϕ(t) )]) A c + R N Q (τ)(e [ cos ( ϕ(t + τ) ϕ(t) )] E [ cos ( ϕ(t + τ) + ϕ(t) )])} = R A NI (τ)e [ cos ( ϕ(t + τ) ϕ(t) )] c = A c R NI (τ)r { E [ e j(ϕ(t+τ) ϕ(t))]} since R NI N Q (τ) = 0 and R NI (τ) = R NQ (τ) = R N I (τ) a(τ) = R A ɛ (τ) c a(τ) = R { E [ e j(ϕ(t+τ) ϕ(t))]} Note that if M(t) is Gaussian, a(τ) = e [Rϕ(0) Rϕ(τ)]. From Fig. 0, the output signal v(t) is given by v(t) = Km(t) + n o (t) where n o (t) = ɛ(t) h o (t). Therefore the output signal power is given by P SO = K P M 7

18 where P M is the power of the message signal m(t). The output noise power is P NO = E [ n o(t) ] = S No (f)df = H o (f) S ɛ (f)df = W W H o (f) S ɛ (f)df since H o (f) = 0 for f > W. Therefore we only need to find the expression of S ɛ (f) for f W. Let A(f) = F {a(τ)} Exercise: Assume that A(f) 0 for f > B T and draw a typical graph of A(f). Let us assume that W B T and A(f) 0 for f > B T, then S ɛ (f) = F {R ɛ (τ)} = S A NI (f) A(f) c = N 0 A c B T B T A(f λ)dλ since R ɛ (τ) = R N I (τ) a(τ) A c = N 0 A c N 0 A c N 0 A c B T +f B T +f B T A(u)du (u = f λ) A(u)du assuming f W and W B T B T A(u)du = N 0 a(0) = N 0 A c A c assuming that f W and A(f) 0 for f > B T f W Therefore since H o (f) = for f W, the output noise power is given by H(f) P NO = Output signal-to-noise ratio: W W H o (f) S ɛ (f)df = N 0 A c W W df H(f) SNR O = N 0 A c K P M df H(f) W W 73

19 = K P M W W SNR C using SNR C = A c df N 0 W H(f) = W β P M W H(W ) (max m(t) ) W W SNR C df H(f) using β = K H(W ) max m(t) = λβ P Mn SNR C = λ(β + )β P Mn SNR I using B T = W (β + ) and SNR C = B T W SNR I where λ is defined as P M (max m(t) ) { H(W ) λ = W W W } df H(f) is the average-to-peak power ratio of the message signal or equivalently and P Mn = the power content of the normalized message signal. Example: m(t) = A m cos(πf m t) A m 0 = P M = A m The normalized message signal is m n (t) = m(t) max m(t) = cos(πf mt) = P Mn = Frequency modulation: Thus H(f) = jπf neglecting δ(f) λ = = { ( ) W (πf) df πw W W { W f df} = 3 W 3 W } Phase modulation: PM H(f) = = λ = See part f of section 4.4 for the justification of the form of H(f). 74

20 c) Output signal-to-noise ratio SNR O assuming a low Carrier-to-noise ratio: Low Carrier-to-noise ratio: A c r n (t) Let us write ϕ z (t) = θ n (t) + ɛ (t) Interchanging the role of x(t) and ñ(t), it can be shown that ϕ z (t) θ n (t) + A c r n (t) sin( ϕ(t) θ n (t) ) Exercise: Draw a phasor digram of z(t) at two time instants t and t such that the phase of the noise θ n (t ) θ n (t ) + π. Then the phase ϕ z (t) will also change by π since it is dominated by the phase θ n (t). If during an interval [t, t ], ϕ z (t) changes by π, the output signal may have a rapid degradation during [t, t ] causing what is called FM clicks for FM modulation. For FM, H o (f) = jπf corresponding to a differentiator. v(t) = Km(t) + dɛ(t) dt Exercise: draw dɛ(t) for an FM modulated signal when due to the noise ϕ dt z (t) changes by π during certain intervals corresponding to low-carrier-to-noise situations. We define a threshold value of SNR C such that the high-carrier-to-noise ratio SNR O formula applies. The threshold is defined as the minimum SNR C or equivalently carrier-to-noise ratio SNR I yielding an output signal-to-noise ratio that is not significantly deteriorated from the value predicted by the usual output signal-to-noise ratio formula. Threshold SNR C value for FM: SNR CT = 0(β + ) 75

21 SNR IT = 0 In [, p. 338], SNR IT = 0 to correspond to practical observations, but the other threshold value is more often used in text books. FM with tone modulation: For FM with tone modulation, i.e. m(t) = cos(πf m t) it can be shown that SNR O = 3 β SNR C + β { } (9) π SNR C exp (β + ) SNR C If high carrier-to-noise ratio is assumed in (9), then we obtain SNR O 3 β SNR C which agrees with the high carrier-to-noise ratio formula found previously (λ = 3 and P Mn = ). 76