The performance of AM and FM receivers. Editor: Xuanfeng Li Teacher: Prof. Xiliang Luo

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1 The performance of AM and FM receivers Editor: Xuanfeng Li Teacher: Prof. Xiliang Luo

2 The performance of AM receivers using Envelop Detection In a full AM signal, both sidebands and the carrier wave are transmitted, as shown by ss AAAA tt = AA cc 1 + kk AAAA mm(tt) cccccccππff cc tt Assume mm tt = 0, ββ AAAA 1 After band-pass filter, the narrow band noise nn ii tt is modeled as white Gaussian noise of zero mean and power spectral density NN 0 /2 nn ii tt = nn cc tt cccccccππff cc nn ss tt ssssss2ππff cc tt

3 The performance of AM receivers using Envelop Detection Then we have power of signal and noise respectively SS ii = ss 2 mm (tt) = AA2 cc 2 mm 2 (tt), NN ii = nn 2 ii (tt) = NN 0 WW 2 + AA cc B is bandwidth of band-pass filter. The input SNR is SS ii = AA cc NN ii 2NN 0 WW The input of demodulator is ss AAAA tt + nn ii tt = AA cc 1 + mm(tt) + nn cc tt cccccccππff cc tt nn ss tt sssssssππff cc tt = EE tt cos[2ππff cc tt + φφ(tt)] where EE tt = AA cc 1 + mm(tt) + nn cc tt 2 + nn 2 ss (tt) nn ss tt φφ tt = arctan[ AA cc 1 + mm(tt) + nn cc tt ] In order to simplify the analysis, we consider two special case AA cc 2 kk AAAA 2 mm 2 (tt)

4 The performance of AM receivers using Envelop Detection (a)large SNR It means AA cc 1 + mm(tt) nn cc 2 tt + nn ss 2 (tt) Thus EE tt = AA cc mm(tt) 2 + 2AA cc [1 + mm tt ]nn cc tt + nn cc 2 (tt) + nn ss 2 (tt) AA cc mm tt 2 + 2AA cc 1 + mm tt nn cc tt AA cc 1 + mm tt 1 + 2nn cc tt AA cc 1+mm tt AA cc 1 + mm tt 1 + nn cc tt AA cc 1+mm tt = AA cc 1 + mm tt + nn cc tt Since (1 + xx) 1 + xx 2, xx 1

5 The performance of AM receivers using Envelop Detection After eliminating DC component, we have SS oo = ss mm 2 (tt) = AA cc 2 mm 2 (tt), NN oo = nn cc 2 (tt) = nn cc 2 (tt)=nn 0 WW The output SNR is SS oo = AA2 2 cckk AAAA mm 2 (tt) NN oo 2NN 0 WW Then we have GG AAAA = SS oo/nn oo SS ii /NN ii = AA2 2 cckk AAAA mm 2 (tt) AA 2 cc +AA 2 2 = cc kk AAAA mm 2 (tt) mm 2 (tt) 1+kk AAAA 2 mm 2 (tt) < 1 Since ββ AAAA 1. And if mm tt = AA mm cccccccππff mm tt, mm 2 tt = 1 2 AA mm 2 μμ2 GG AAAA =, µ = kk 2+μμ AAAAAA 2 mm When 100% modulation, which means µ = 1, we get 1/3. Thus we know that envelop detector lowers the SNR.

6 The performance of AM receivers using Envelop Detection (a)small SNR It means Then Where EE tt = AA cc 1 + mm(tt) nn cc 2 tt + nn ss 2 (tt) AA cc mm(tt) 2 + 2AA cc [1 + mm tt ]nn cc tt + nn cc 2 (tt) + nn ss 2 (tt) 2AA cc [1 + mm tt ]nn cc tt + nn cc 2 (tt) + nn ss 2 (tt) = [nn 2 cc (tt) + nn 2 ss (tt)]{1 + 2nn cc tt AA cc [1+mm tt ] } nn 2 cc (tt)+nn 2 ss (tt) = RR(tt) 1 + 2AA cc 1+mm tt RR tt ccccccθθ(tt) RR(tt) + AA cc [1 + mm tt ] ccccccθθ(tt) RR tt = nn 2 cc tt + nn 2 ss tt, θθ tt = arctan nn ss tt nn cc tt Now the signal becomes part of noise and the system will no longer work.

7 The performance of FM receivers The incoming FM signal is defined as, ss FFFF tt = AA cc cos[2ππff cc tt + 2ππkk FFFF mm ττ ddττ] Still assume mm tt = 0 Then we have power of signal and noise respectively SS ii = AA cc 2 2, NN ii = nn ii 2 (tt) = NN 0 WW B is bandwidth of band-pass filter. The input SNR is SS ii NN ii = AA 2 cc 2NN 0 WW

8 The performance of FM receivers (a)large SNR Ignore the interaction between signal and noise under large SNR. Assume nn ii tt = 0, mm oo tt = kk dd kk FFFF mm tt kk dd is discriminator index, let it be 1. Thus we have SS oo = mm oo 2 (tt) = (kk FFFF ) 2 mm 2 (tt) Assume m tt = 0, the input of demodulator is AA cc cccccccππff cc tt + nn ii tt = [AA cc +nn cc tt ]cccccccππff cc nn ss tt sssssssππff cc tt = AA tt cos[2ππff cc + φφ tt ] Envelop AA tt = [AA cc + nn cc tt ] 2 +nn 2 ss tt Since SNR is large, the phase nn ss (tt) φφ tt = aaaaaaaaaaaa AA cc + nn cc (tt) aaaaaaaaaaaa nn ss(tt) AA cc nn ss(tt) AA cc

9 The performance of FM receivers Thus the output noise is nn dd tt = ddφφ tt dddd = 1 ddnn ss (tt) 2ππAA cc dddd The power spectrum density is So we have PP dd ff = ( kk dd ) 2 HH(ff) PP AA ss ff = ( 1 ) 2 ff 2 NN cc AA 0, ff < BB FFFF cc 2 NN 0 = WW WW WW PP dd ff dddd = ( 1 ) 2 ff 2 NN 0 dddd = 2NN 0WW 3 AAcc WW 3AA cc 2

10 The performance of FM receivers SNR at the output SS OO = 3AA2 cckk 2 FFFF mm 2 (tt) NN OO 2NN 0 WW 3 Then GG FFFF = 3kk FFFF 2 mm 2 (tt) WW 2 Example : Single-tone Modulation The modulated FM signal is defined by ss FFFF (tt) = AA cc cos[2ππff cc tt + mm ff ssssss2ππff mm tt] Where mm ff = kk FFFF ωω mm = ωω ωω mm = ff ff mm So we get mm tt = ff kk FFFF cccccccππff mm tt

11 The performance of FM receivers And mm 2 tt We have SS OO = 3AA2 cc( ff) 2 NN OO 4NN 0 WW 3 Where ββ = ff/ww is modulation index = ( ff)2 2 2kk FFFF = 3AA2 ccββ 2 4NN 0 WW GG FFFF = 3ββ2 2 For an AM system operating with a single-tone signal and 100% modulation, GG AAAA = 1 3 We just need to adjust ββ 0.5, there will be GG FFFF >GG AAMM

ANGLE MODULATION. U1. PHASE AND FREQUENCY MODULATION For angle modulation, the modulated carrier is represented by

ANGLE MODULATION. U1. PHASE AND FREQUENCY MODULATION For angle modulation, the modulated carrier is represented by [4.1] ANGLE MODULATION U1. PHASE AND FREQUENCY MODULATION For angle modulation, the modulated carrier is represented by xx cc (tt) = AA cccccc[ωω cc tt + φφ(tt)] (1.1) Where A ω c are constants the phase