Communication Theory: Amplitude Modulation and Single Side Band (SSB), Slides of Signals and Systems

Download Communication Theory: Amplitude Modulation and Single Side Band (SSB) and more Slides Signals and Systems in PDF only on Docsity! EE 361 Communication Theory Chapter 4 Amplitude Modulation Part A 1 m(t) = modulating message signal c(t) = carrier signal y(t) = modulated signal m(t) Modulator p(t) Modulating message signal Modulated signal baseband bandpass c(t) Carrier signal narrowband Amplitude Modulation: O(t) = 27 ft + Ao fi(t) = fe such that: yam (t) = glm(t)] cos(27 fet + Ao) Information contained in mr) is embedded in the time-varying amplitude. Frequency Modulation: a(t) = A, t O(t) = In fet + Ky | m(A)dd + Oo t yra(t) = A.cos(27 ft + 20K [ m(A)dX + Oo) 0 such that: Information contained in m(r) is embedded in the instantaneous frequency: J,(¢) is proportional to mr) Phase Modulation: such that: yppm(t) = Ac cos (27 fet + In Kpm(t) + 40) Information contained in m(f) is embedded in the generalized phase: 4(r) is proportional to m(7z) 2B - Posiisclf) anak. Bt M(f) \ “Bo. B f fe 0 fe Transmission Bandwidth lf the baseband message signal m(r) has the bandwidth B-Hz, then the DSB-SC signal @psg s-(4) requires a transmission bandwidth of 2B-Hz Generating the DSB-SC signal Use a product / non-linear / switching / ... modulator to generate Ppsp-sc() = m(Ae(2) Recovery of m(f) from @psp-scl4) If multiplying m(r) with c(t) = 4. cos w,¢ shifts M(f) to + f, what happens if we multiply @ysp-sc(4) with cos ws again ... 10  DSB-SC Generation: Non-linear modulator Non-Linear [output] = a [input] + [input]? Device Remember: Anon-linear device spreads the spectrum of its input. By controlling what is used as an input to the non-linear device(s) and by eliminating what is not needed we may generate the desired DSB-SC amplitude modulated signal. 11  DSB-SC Generation: Non-linear modulator Non-Linear Device ® p| Non-Linear Device a Y A - cos Wet _ Non-Linear Device a [output] = a [input] + b [input]? xt) 2(t) =a [m(t) + coswet] + b[m(t) + cos wet]? — a[—m() + cos wet] — b[—m(t) + cos wet]; =am(t) +4 coswet + bm?(t) + 2bm/(t) coswet + b cos” wet +am(t) — a coswet — bm?(t) + 2b m(£) cos wet — b cos? wet; = 2am(t) + 4bm(t) cosw,t. 12 DSB-SC Generation: Switching modulator TK. m(t) —+@}— | wi) 8 F[w(t)] =F[ m@o()] + | y THIF =~ =M(N)~F[0(0) _ jnuiet o(t)= 0 Cre =M(f)*¥Cn8(f — rfc) n=—0co = G.M(f - nf) m(t) M(S) TA mt +t 0 f w(t) A wif) te AC. fP 7 La t —]> LJ J of UU. -2f. . To keep these spectral components we need ... 15 DSB-SC Generation: Switching modulator wf) (% Bandpass mi{t) t — oO) Filter et “TA _ w(t) Yosp-sc(t) | vpsp-sc(t) M(f) vy A 0 B f wis) ACs IM, 2fe Ppsnsc(f) AC, A, fe 9 te 16  DSB-SC: Generation > Demodulation *« There are other AM generation techniques. Refer to the course reference text for further information and discussion. « We will use the terms mixing / frequency conversion / heterodyning interchangeably. These terms refer to the process: multiplication + bandpass filtering +« Now that we know how to generate DSB-SC amplitude modulated signals, let’s say how we can undo what we have done ... demodulation 17  DSB-SC: Modulation and demodulation The complete modulation — transmission — demodulation chain for DSB- SC amplitude modulated signals: mit) Modulator Demodulator y(t) “© Channel > >| LPF cos 27 ft cos 27 ft Local Local Oscillator-1 Oscillator-2 e K'm(t) 20  DSB-SC: Coherent demodulation / detection We assumed TX RX Modulator Demodulator m(t) x “y Channel +(x) =| LPF em K'm(t) are synchronized: Im(t) = cos(2a fit + 1) with fi=fe=fe and oy = 2 la(t) = cos(27 fat + 62) What happens if they are not synchronized ... 21  DSB-SC: Coherent demodulation / detection To investigate effects of lack of synchronization let’s assume: TX Lin(t) = cos(27f,t) RX la(t) = cos(2a(f. + Af)t + d0) such that ypsp-sc(t) = Acm(t) cos 27 fet and yosp-sc(t)la(t) = Ac m(t) cos (2i fet) cos (27( fe +Afe+ oo). = * m(t) cos (27 Aft + oo) + ae m(t) cos(27(2f. + Aft + 6) 22  DSB-SC: Coherent demodulation / detection TTS ypsp-sc(t) la(t) = “ m(t) cos(27A ft + do) Fm eostrt2h + Af)t + ¢0) a Demodulator . Filtered out by the vpsp-sc(t) }—» LPF >| y(t) lowpass filter la(t) Local ‘Oscillator-2 Ac y(t) = 5 m(t) cos (QnA ft + oo) — Ae If we also assume Af= 0, demodulator output becomes: | y(t) = > m(t) cos @o 25  DSB-SC: Coherent demodulation / detection Demodulator A, Ypsp-sc(t) >()—>|_ LPF >) y(t)= 3 m(t) cos do la(t) lf Af=Oand Af < f. Local Oscillator-2 a » Ifweare lucky ¢,=0 =e u(t) = A'm/(t) + Ifwe are unlucky ¢,=7/2 mem y(t) =0 * Other values of d, mueeeeeeeel> varying degrees of attenuation Successful demodulation of DSB-SC amplitude modulated signals requires accurate synchronization of the two local oscillators: synchronous detection / coherent detection 26  Amplitude Modulation (AM) Modulated waveforms with suppressed carrier terms require fairly complex circuitry at the receiver to acquire phase synchronization, i.e., coherent detection, which makes the receivers expensive to manufacture. In applications where we have one or few transmitters and a much, much larger number of receivers (e.g. AM/FM radio broadcasting) it makes economic sense that the receivers are as simple as possible. To facilitate simple demodulation we consider the idea of transmitting a separate carrier term in the same frequency band as a DSB-SC amplitude modulated signal. Let's consider amplitude modulation signals of the form: @psp-sc + Carrier 27  AM: Demodulation Key elements of an AM signal yam(t) = [Ae + m(t)] coswet are the amplitude and envelope functions: amplitude = A, + m(t) envelope = |A, + m(t)| 30  AM: Demodulation amplitude = A. + m(t) envelope = |Ac + m(t)| ira  AM: Demodulation Asimple and inexpensive envelope detector can demodulate AM signals if: mai envelope « m(t) ==> envelope = E(t) = |A, +m/(t)| = [Ac + m(t)] = A(t) = amplitude uy [ A. +m(t)] >0 Conversely, if £(7) = [4, + m(A)] <0 for some +: + The envelope will no longer be a scaled and shifted version of the modulating signal m(7). * [4, + m(d] will experience phase reversals at zero crossings. + We have to use a synchronous/coherent detector as with DSB-SC amplitude modulated signals. 32  AM: Envelope detector Output of the envelope detector + has ripples ————— use a LPF + exhibits a d.c. component ===> use a d.c. blocking unit Envelope Detector Lowpass Filter DC-Block K'm(t) 35  AM: Modulation index Definition Modulation Index of an AM signal: dai wae where 1p = max |rn(£)| We can demodulate an AM signal using an envelope detector, if ==> E(t) = A(t) = [4, + m()] = 0, for all t or equivalently when ==> O< ysl 36  AM: Modulation index While the requirement 0 < «<1 will be a sufficient condition for demodulating AM signals using an envelope detector, AM broadcasting standards impose separate conditions on the modulation index based on positive and negative swings of the envelope. Let Amax = max A(t) = max [Ac +m(t)] Amin = min A(t) = min [Ac + m(t)] Definitions: 4 or errr —_ Arte Modulation Index: _ ra Positive Modulation Index: pa = Sus = Ae 5 F Ac errr Negative Modulation Index: p= A 37  AM: Single-tone modulation, an example Single-tone message signal m(t) = A,, cosw,,t such that the AM signal is: pam(t) = [ A. + A,, cos Wt | COS wet with wm < we Using the parameters: Mp = MAX. |m(t)| = Am. Amax = Ag + Am, Amin = A; _ Am evaluate: mp — Am Amax ~ Amin _ [Ae + Am] —[Ae - Am] Am UA.” AL ny oA, ~ A. — Amax —Ae — [Ae+ Am] — Ae _ Am a Ac “AL _ ie _ A, — [Ac — Am] _ Am pS A Az ~ Ae 40  AM: Single-tone modulation, an example Single-tone message signal m(t) = A,, cosw,,t such that the AM signal is: yam(t) = [ A. + Am COS Wm | COS Wet with wry, K we My Am OF a= Ae AT Both definitions of modulation index u give the same result. (i> 4 = be = Plt _ Pelee _ [Ac + Am] 7 [A. 7 An] _ Am 2A, 2A, Ac Amax —Ae _ [Ac +Am]—Ac _ Am Aco Ac _, Ae _ Amin _ Ae _ [Ac _ Am] _ Am Aco Ac _ All definitions of modulation indices including #, «., and 4 coincide since m(ft) is symmetric with respect to horizontal axis. 41  AM: Single-tone modulation, an example Single-tone message signal m(t) = A,,, cosw,,t such that the AM signal is: yam(t) = [Ae + Am COS Wnt | COS Wet with wn K we Case (i): 4,,/Ae <1 A(t) amplitude function E(t) envelope function E(t) = A(t) = [4, + A,,c08 w,,¢]20 such that F(r)~ m(7) =p ul perfectly suitable to use an envelope detector  AM: Single-tone modulation, an example Single-tone message signal m(t) = A,, cosw;,¢ such that the AM signal is: pam(t) = [Ac t+ Am COS wnt | cos wet with wy, < we Putting everything together we have Pam(f) =e [UF — fe) +0F +f.) +o [d(f- (fe + fm)) + 9(f — (fe — fm)) | +" [5(F + (fe fne)) + 6(F + (Fe + Frm) | 45  AM: Single-tone modulation, an example Single-tone message signal im(t) = A,,, cosw,,¢ such that the AM signal is: pam(t) = [| Ac +An COS Wnt | COS Wet With Wyn K We Alternatively, we can first expand ¢,,,(/) using trigonometric identities: pam(l) = A, cos wet + Ay, COS Wt COS Wel, -} COS Wet ie COS(We + Wm) ie COS(We — Wm) | —fe — fm fe —fe+ fm 0 fo — fm fe fe + fm 46  AM: Another example Consider the modulating signal m(r) and the corresponding AM signal: yam(t) = [Ae + m(t) | coswet A mi(t) vam(t)f An = Ac +08 A 0.8 ---} o4 aN Amin = Ac — 0.3 > rt —> ft -0.3 ---+ ------------- NY me First evaluate the parameters: Mp = Max jm(t)| = 0.8, Amax = Ae + 0.8, Amin = Ae — 0.3  AM: Carrier and sideband power Definition: Let g(t) represent the time average of g(t) and let P, be the power delivered by the signal x(t) across the load ? x(t) P, = R Consider the AM signal @,),(7) modulating signal generated by the baseband message/modulating signal #(r) bandlimited to B,,-Hz. yam(t) = A, cosw.t + m(t) cos wet — —<—_-— carrier sidebands Objective: + Compute the power of the AM signal P,, and + Express it in terms of the carrier power P. and the sideband power P, 50  AM: Carrier and sideband power yam(t) = A, cosw,t + m() cos wet [ carrier sidebands —= A? cos? wet = A2/2 S A? cos? wet + 2A.m(t) cos? wet + m?(t) cos? wet 51  AM: Carrier and sideband power yam(t) = A, cosw.t + m(t) cos wet — pa carrier } sidebands P, = A? cos? wt +]2A,m(#) cos? wet m?(t) cos? wet because: f,. > B,, and m(t) = 0 2A, m(t) cos? wet = 2A, m(t) cos? wf = 0 52 Am This answer is true only when m(t) is single tone 55 Sideband Power: ---4-------------- 0 Carrier Power: 56 Carrier Power: P --»| (2 weer Ac/2 .------------------y0-- | ( 2 coop Torte A m/4 Seen een | -p oo | ~fe-fn © fet Fm 0 fe-fm 1 fettm 57  AM: Double sideband AM comparison Given a baseband modulating signal m(*) with baseband bandwidth 2,,-Hz consider the two amplitude modulation schemes we discussed so far: * DSB-SC: double sideband suppressed carrier AM * DSB-LC: double sideband with large carrier AM = AM es Se Bandwidth 2B.,, 2B,, Carrier no yes Demodulation coherent envelope or coherent Power efficiency 100% < 33% 60  Other AM techniques We will now study other variants of amplitude modulation techniques. Each of these AM systems has certain desirable characteristics that makes them suitable in specific applications. In particular, we will study the following AM techniques: * Quadrature Amplitude Modulation (QAM) * Single-Sideband Modulation (SSB) * Vestigial Sideband Modulation (VSB) But ... why are we still studying AM? 61  kHz | MHz | GHz | | tl fey Long Wave Radio FM 3G Satellite TV AM DAB Short Wave Radio GSM <= pereasing Range ; Decreasing Range =» reasing Bandwidth Increasing Bandwidth Source: International Telecommunication Union (ITU), Technology Research Programme: Research and Development at Ofcom 2004/05 62 Consider to baseband signals m,(7) and m.,(r) with baseband bandwidths of B,,-Hz each. Modulator voam(t) voam(t) = m4(t) cos wet + ma(t) sin wet. 65  These components are centered at 2/,and will be filtered out by the LPF. Modulator my (t) x cos wet CH —n/2 sinw,t x ma(t) a Channel NX ~ J-Demodulator K"'ma(t) Q-Demodulator 66 Modulator |-Demodulator my(t) K'm,(t) x cos ut ¥ OF Channel —7/2 | sin wet =(x) ma(t) K"'moa(t) . Q-Demodulator and similarly: | w(t) = ggam() sinwet =m, (t) cos wt sin wet + mo (t) sin? wet 1 1 1 — ae (t) sin 2w,t + gm” (t) = 3m (t) cos 2wet 67  QAM: Further Comments 1x64 File Tools @ @O br AM Constellation o |e ls BB 64-02 View Simulation Help ~ File Tools View (ai SP Constellation Simulation oS || & || Help » This is because transmitting both upper side band (USB) and lower side band (LSB) 71 Real valued signals will result in spectrum with conjugate symmetry An asymmetric spectrum will result in a complex signal. Let us split M(f) into its USB and LSB components M+(f) and M-(f) If, m+(t) →M+(f) and m-(t) →M+(f), then 72 Here, Hh(f) is the transfer function of the Hilbert Filter defined by, 75 Hilbert Transform Hy(f) = —jsen(f) = ee if f > 0; +j, iff <0. cos 27 fot Ln fF cos(2r fot — >) = sin 27 fot J» (nit) Every frequency component in 7(r) experiences 7/2-radians phase shift 76 SSB Considering only the Upper Side Band M(f) M_(f+ fe) M4(f — fe) bf. fe—Bm fe +Bm ssp, (f) = HM_(f + fe) 77 However, it may not be easy to design a wideband phase shifter that shifts the Phase of every frequency component exactly by 90o Compare with QAM 80 Generation of SSB Signals: Selective Filtering Method Why should 77(7) have a frequency hole? If there is no frequency hole: M(f) DSB-SC } / . f Without a frequency hole we ° ft —— need an ideal BPF ... ! SSB, g t 0 f.-Bm fe f+ Ba? 81  Generation of SSB Signals: Selective Filtering Method Why should 777(7) have a frequency hole? If there is a frequency hole: M(f) Mf) JAIN of Ba f 0 Bm f DSB-SC / } / Dp oe 0 Tey Transition Band ... [Hop()| on GOOD ! ° Cint oo po 82  DSB Transmit both sidebands, 5, = 2B, High | [ one complete SB ] Compromise —_ + [ a trace (vestige) of the other SB ] SSB Transmit one sideband, B; = B Low m 85  [ one complete SB] + [a trace (vestige) of the other SB ] M(f) bpsr-sc(f) This part of the USB is missing. Sysp(f) This part of the LSBis retained.  Generation of Vestigial Side Band Signal Same selective filtering approach as used in generating SSB signals Two step process i. Generate a DSB-SC signal; ii. Filter out and shape the sidebands using an appropriately designed BPF. m(t) VSB Modulator Ypsp-sc(t) |}—» (x) —+| H,,(f) COS Wel pe pvsn(t) 87  s(t) = pvsn(t)[2coswet | S(f) = ®vsp(f — fe) + Ovsp(f + fe) bysp(f) | LN 2 0 Sento fe fet Bm , dysn(f — fe) . LA, Sysp(f + fe) aS 90 Gysp(f) = K [M(f —f.) +M(f + fe)] Hu(f) VSB Modulator | m(t) Spsp-sc(f) = K [M(f - fe) +M(f + fe) Y¥(f) = SU) Hips (Pf) 91  Choosing 7,(7) S(f) = ®vsn(f — fc) + ®vsn(f + fe) -2f. a -B, 10: Bm 2fe Analyze Y(f): Y(f) = S(f) Hips (Pf) = [Ovsa(f — fe) + Oven (f+ fe) ] Hips (Ff) =K [ [M(f ~2f-) + M(f)] Hel f = fe) + [M(f) + M(f + 2fe)] Hof + fe) | Hops (f) = K | [Ho(f — fo) + Hol f + fo)] MUN) + [M(f + 2fc) Hol f + fo) + M(f — 2fe) Ho(f — fe)] | Hins(f) 92  Choosing 7,(/) S(f) = ®vsn(f — fc) + Pvsn(f + fe) —2fe - Br 0 Analyze Yj): Y(f) = Sf) Hips (f) = [@vsn(f — fe) + Pvse(f + fe) | Hips(f) =K | [M(t —2fe) + MAS - fe) + [M(f) + MU +2f0)) Hof + fe) | Hof) =K | [Hu(f - fo) + Hol f + fo] MUP) + [M(f + 2fe) Hel f + fe) + IM(f — 2fe) Hu( f — fe) | — Hips (Ff) 95  Choosing H,(/) S(f) = ®vsa(f — fe) + Ovse(f + fe) —2fc a / Analyze Y(f): Filtered out by the LPF. Y(f) = S(f) Hips (F) + [M(f +2.) Hullo) ME =2I-SHo(F — fe) |] Hips (0) = K" [Ho(f ~ fe) + Hol + fe)] M(A) —_—“ A (f — fe) + Hu(f + fe) = constant —Bn <= f= Bm 96  Choosing 4,(/) H(f — fe) + Ay(f + fe) = constant —-By,<f< Bm The above condition implies => Hf) symmetric for | f—f.1 <= 4, I,(f) symmetry 0 NL’ “f U t fe—te fe fot fe 97