ECE 359 Spring 2003 Handout # 16 April 15, SNR for ANGLE MODULATION SYSTEMS. v(t) = A c cos(2πf c t + φ(t)) for FM. for PM.

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1 ECE 359 Spring 23 Handout # 16 April 15, 23 Recall that for angle modulation: where The modulation index: ag replacements SNR for ANGLE MODULATION SYSTEMS v(t) = A c cos(2πf c t + φ(t)) t 2πk f m(t )dt φ(t) = k p m(t) k f m max W β = k p m max for FM for FM If β 1, then we have narrowband angle modulation with bandwidth 2W Hz If β > 5 we have a wideband angle modulation with bandwith given by Carson s rule: B c = 2W (1 + β) Hz Now consider sending a wideband angle modulated signal on an AWGN channel as shown below AWGN Channel Receiver v(t) y i (t) BPF f c ± B c /2 y 1 (t) Angle demodulator y 2 (t) LPF ±W y o (t) α n(t) As before, we begin the SNR analysis by writing equations for the outputs at various stages of the receiver y i (t) = αv(t) + n(t), and y 1 (t) = αv(t) + n BP (t) where n BP (t) is a sample path of a bandpass WGN process with spectral height N /2 and bandwidth of B c Hz centered around f c We can write N BP (t) = N c (t) cos(2πf c t) N s (t) sin(2πf c t) where N c (t) and N s (t) are independent lowpass WGN processes with spectral height N and bandwidth B c /2 We can rewrite n BP (t) as n BP (t) = e n (t) cos(2πf c t + θ n (t)) where e n (t) = n 2 c(t) + n 2 s(t), ( ) and θ n (t) = tan 1 ns (t) n c (t) c V Veeravalli, 23 1

2 Thus y 1 (t) = cos(2πf c t + φ(t)) + e n (t) cos(2πf c t + θ n (t)) = r(t) cos(2πf c t + φ(t) + n φ (t)) For small noise, ie N B c = A2 cα 2 2N B c > 1, we showed in class using phasors that n φ (t) e n(t) sin(θ n (t) φ(t)) Now we make the assumption that φ(t) varies slowly wrt θ n (t) This assumption holds for wideband signals since n c (t) and n s (t) have bandwidth B c /2 that is much larger than the bandwidth of φ(t) (which equals W ) Thus n φ (t) e n(t) sin θ n (t) cos φ e n(t) cos θ n (t) sin φ = n s(t) os φ n c(t) αa sin φ This means that N φ (t) is approximately WSS with zero mean and ACF given by R Nφ = R N s cos 2 φ + R N c sin 2 φ = R N c ie, N φ (t) is a lowpass WGN process with bandwidth B c /2 Hz and spectral height The output of the angle demodulator y 2 (t) is given by: φ(t) + n φ (t) = k p m(t) + n φ (t) y 2 (t) = ] [φ (t) + n φ (t) 1 2π where n φ (t) is derivative of n φ(t) Output SNR : For PM, N A 2 c α2 = k f m(t) + 1 2π n φ (t) for FM (1) y 2 (t) = k p m(t) + n φ (t) where N φ (t) is (approximately) a WSS, zero mean, process with PSD S Nφ (ω) = N A 2 c α2 if f B c /2 Recall that two key assumptions were used in arriving at this approximation for N φ (t) These are: ➀ The angle modulated signal is wideband, ie, β > 5 ➁ The signal-to-noise ratio at the output of the front-end BPF is large, ie, N B c = 2N W (β + 1) = Γ 2(β + 1) > 1 Now y o (t) = [y 2 (t)] LPF = k p m(t) + n o (t) c V Veeravalli, 23 2

3 where n o (t) is n φ (t) passed through a W Hz LPF filter Since B c > 2W, and hence S No (f) = H LPF (f) 2 S Nφ (f) = N P n,o = 2 S No (f)df = 2N W Since the signal component of y o (t) is k p m(t), it is clear that P s,o = k 2 p Furthermore, = A2 cα 2 2 Γ = A2 cα 2 2N W Thus Γ o = k2 p 2N W = k2 p Γ = β 2 p where m n (t) = m(t)/m max is the normalized message signal (m max ) 2 Γ = β2 p n Γ (2) Note that Γ o grows with β 2 p, and for large enough β p we can have Γ o larger (or even much larger) than Γ However, the tradeoff of bandwidth for SNR cannot be obtained indefinitely Eventually the small noise assumption (see ➁ above) is violated, and (2) no longer holds In fact, for very large β p, Γ o actually decreases with β p and eventually becomes worse than Γ Output SNR for FM: For FM, where n 2 (t) = 1 2π n φ (t) y 2 (t) = k f m(t) + n 2 (t) and where the PSD of N o (t) is given by S N2 (f) = 1 N f 2 if f B (2π) 2 (2πf)2 A S Nφ (f) = 2 cα 2 c /2 y o (t) = [y 2 (t)] LPF = k f m(t) + n o (t) S No (f) = H LPF (f) 2 S N2 (f) = N f 2 A 2 c α2 since W < B c /2 Hence P n,o = 2 S No (f)df = 2N f 2 df = 2N W 3 3 Also, it is clear that P s,o = k 2 f and Γ = /(2N W ) Thus Γ o = A2 cα 2 2N W 3kf 2 W 2 = 3k2 f (m max) 2 W 2 (m max ) 2 Γ = 3β2 f (m max ) 2 Γ = 3β2 f n Γ (3) c V Veeravalli, 23 3

4 Again Γ o grows with β 2 f, and for large enough β f we can have Γ o larger (or even much larger) than Γ, as long as the low noise assumption of ➁ holds, ie, as long as Γ 2(β f + 1) > 1 Threshold Effect in Angle Modulation If condition ➁ does not hold, the angle modulation sytem is said to be in threshold For fixed Γ, β > Γ 2 1 the system is in threshold For fixed β, Γ < 2(β + 1) the system is in threshold In threshold, equations (2) and (3) do not hold, and in fact, Γ o is smaller than the right hand side of these equations When the system is deep into threshold, Γ o is worse than Γ So, before applying (2) or (3), we must check to make sure that the system is not in threshold ag replacements Pre-emphasis and De-emphasis (PD) in FM Recall that for FM, where the PSD of N o (t) is given by y o (t) = k f m(t) + n o (t) S No (f) = N f 2 Thus, high frequencies in the signal (those close to W ) see more noise than low frequencies at the output of the demodulator We can exploit the non-white nature of the noise to reduce the noise power at the output via filtering However, a filter applied to y o (t) will distort the message signal We can get around this problem by prefiltering m(t) as shown below: Channel Receiver m(t) h p (t) FM mod y i (t) BPF y 1 (t) FM y 2 (t) LPF y 3 (t) f c ± B c /2 demod h d (t) ±W y o (t) α n(t) The filters with impulse responses h p (t) and h d (t) are the pre-emphasis and de-emphasis filters, respectively They are usually chosen to be first-order Butterworth low-pass and high pass filters with ( ) H d (f) 2 1 = ( ) 2 H p (f) 2 f 2 = f f f with h p (t) h d (t) = δ(t) The cut-off frequency f is chosen to be less than W Note that FM with PD is a general angle modulation scheme which lies somewhere in between FM and PM c V Veeravalli, 23 4

5 The output of the LPF (before de-emphasis) is given by: y 3 (t) = k f m(t) h p (t) + n 3 (t) where n 3 (t) has the same PSD as n o (t) without PD that we studied before, ie, S N3 (f) = N f 2 Key Point: m(t) sees both filters and is hence unaffected by PD, whereas n(t) sees only h d (t) Hence, with PD y o (t) = k f m(t) h p (t) h d (t) + n 3 (t) h d (t) = k f m(t) δ(t) + n o (t) = k f m(t) + n o (t) where n o (t) is n 3 (t) filtered by h d (t) Thus 1 N S No (f) = H d (f) 2 ( ) f 2 2 S N3 (f) = 1+ f f Hence the noise power at the output is P n,o = 2 S No (f)df = 2N f 3 Recall that without PD (assuming system is not in threshold) and P s,o is the same with and without PD Thus Γ o,fm-pd = Γ o,fm (P n,o ) FM = 2N W 3 3 (P n,o ) FM (P n,o ) FM-PD = Γ o,fm It is easy to show that Γ o,fm-pd > Γ o,fm if f < W [ ( )] W W tan 1 f f ( W f ) 3 [ ( )] 3 W W f tan 1 f Example For FM broadcasting, B = 15 khz, β = 5 and f = 2π 21 rad/s If the average-topeak power ratio is 5 and Γ = 3 db, we can show that Γ o,fm-pd =4574 db (after cheking that system is not in threshold) and that Γ o,fm-pd =591 db Thus we get almost a 3 db improvement in SNR over DSB-SC and even more over conventional AM This is why FM radio sounds so much better than AM radio! c V Veeravalli, 23 5

Solution to Chapter 4 Problems

Solution to Chapter 4 Problems Problem 4.1 1) Since F[sinc(400t)]= 1 modulation index 400 ( f 400 β f = k f max[ m(t) ] W Hence, the modulated signal is ), the bandwidth of the message signal is W = 00