Bandpass Modulation and Demodulation 1-Digital Communications-Lecture Slides, Slides of Digital Communication Systems

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Chapter 4: Bandpass Modulation and Demodulation ® docsity.com  Bandpass MOdulation and DEModulation —_— MODEM (Phase info | COHERENT NONCOHERENT | !N2 Phase ir required) required) | —— ce | = BINARY M-ary HYBRID BINARY || M-ary | |HYBRID | | if J if J ASK ASK APK(QAM) ASK ASK DPSK (OOK) (OOK) FSK FSK FSK PSK DPSK (QPSK, DPSK OQPSK) CPM CPM docsity.com  a Where for binary ASK (also known as ON OFF Keying (OOkK)) s,(t) = A,m(t)cos(@,t+¢), 0<t<T binary | Sy(t) = 0, O0<t<T binary 0 = Mathematical ASK Signal Representation a The complex envelope of an ASK signal is: g(t) = A.m(t) a The magnitude and phase of an ASK signal are: A(t) = A.m(‘), o(t)=0 a The in-phase and quadrature components are: x(t) = A,m(t) y(t) =9, the quadrature component is wasted. ® docsity.com  It can be seen that the “givin “win modulated is twice that —it , occupied by the source Upper sideband baseband stream impulse Lower sideband T,; Te*Ry, = Bandwidth of ASK a Bandwidth of ASK can be found from its power spectral density a The bandwidth of an ASK signal is twice that of the unipolar NRZ line code used to create it., i.e., 2 B= 2k, =—- 4 = This is the null-to-null bandwidth of ASK ® docsity.com  = Ifraised cosine rolloff pulse shaping is used, then the bandwidth is: 1 B=(1+r)RkR, >We rand r)R, = Spectral efficiency of ASK is half that of a baseband unipolar NRZ line code a This is because the quadrature component is wasted ® docsity.com  Noncoherent Receiver a Does not require a phase reference at the receiver a If we do not know the phase and frequency of the carrier, we can use a noncoherent receiver to recover ASK signal u Envelope Detector: y =i a r(t) —> Qf piece >| LPF b x AD >5(t) Envelope Detector he a The simplest implementation of an envelope detector comprises a diode rectifier and smoothing filter ® docsity.com  Frequency Shift Keying (FSK) In FSK, the instantaneous carrier frequency is switched between 2 or more levels according to the baseband digital data a data bits select a carrier at one of two frequencies a_ the data is encoded in the frequency Until recently, FSK has been the most widely used form of digital modulation;Why? a Simple both to generate and detect a Insensitive to amplitude fluctuations in the channel FSK conveys the data using distinct carrier frequencies to represent symbol states An important property of FSK is that the amplitude of the modulated wave is constant Waveform docsity.com  =» Analytical Expression S,) = — cos) a,f+¢), i1=O0,1,..M -1 0.(t) =[@,t+ o,f m(t)dT | d : Analog form Si = = fy + fam(t) = General expression is s,(t) = — cos( 2af,f+27iAft), i= O0,1,....M -1 Ss Where M=f,-f., fi = fo +iAf ® docsity.com  3. Coherent Detection of Binary FSK = Coherent detection of Binary FSK is similar to that for ASK but in this case there are 2 detectors tuned to the 2 carrier frequencies NA 2 Tu. ZA CcOSs(m at) = Recovery of fc in receiver is made simple if the frequency spacing between symbols is made equal to the symbol rate. ® docsity.com  Non-coherent Detection : One of the simplest ways of detecting binary FSK is to pass the signal through 2 BPF tuned to the 2 signaling freqs and detect which has the larger output averaged over a symbol period BPF Envelope Tuned @ f, Detector Sampler -——> Envelope 1 Detector - ! Time Sync docsity.com  Phase Shift Keying (PSK) In PSK, the phase of the carrier signal is switched between 2 (for BPSK) or more (for MPSK) in response to the baseband digital data With PSK the information is contained in the instantaneous phase of the modulated carrier Usually this phase is imposed and measured with respect to a fixed carrier of known phase — Coherent PSK For binary PSK, phase states of 0° and 180° are used Waveform: docsity.com  = We can also write a PSK signal as: s(t) = “Eco 0 20D) T M 2E 2r(i-1) . 2n(i-l) . = |_| cos ——— cos@,f-—sin———sin__ @f T M M = Furthermore, s,(t) may be represented as a linear combination of two orthogonal functions w,(t) and w(t) as follows s,(t) = VE 008, - VE sin =) w(t) Where y(t) = = cos| and y,(t) = = sin{o ® docsity.com  = Using the concept of the orthogonal basis function, we can represent PSK signals as a two dimensional vector S(t) (VE. cos Dy, JE, sin Dy, = For M-ary phase modulation M = 2‘, where k is the number of information bits per transmitted symbol = InanWM-ary system, one of M2 2 possible symbols, s,(t), ..., s,,(t), is transmitted during each T,-second signaling interval = The mapping or assignment of k information bits into M = 2* possible phases may be performed in many ways, e.g. for M=4 ® docsity.com  04 01 2 11 10 10 = Apreferred assignment is to use “Gray code” in which adjacent phases differ by only one binary digit such that only a single bit error occurs in a k-bit sequence. = Itis also possible to transmit data encoded as the phase change (phase difference) between consecutive symbols a This technique is known as Differential PSK (DPSK) = There is no non-coherent detection equivalent for PSK except for DPSK ® docsity.com  The signals are: 2E _ |2Es Ty Sy = ar cos(w,t) sS= op coset +a) =~ S S,= 2k, cos(@,f +7) =—- cos(@,1) Ss s 2E 37 2E, Ss, =,/— cos(a@,t + —) = |= sin(@.\1) YT, 2 17 Sy) =+ Js cos@,t, o—shift of 0° and 180° 8,,()=+ = sin@,t, shift of 90° and 270° Ss E,. = sin(@,t) s & docsity.com  = Wecan also have: Sopsk (t)= |= ope ED FE) =1,2,3,4 O<r<T, ® docsity.com  = One of 4 possible waveforms is transmitted during each signaling interval Ts a i.e., 2 bits are transmitted per modulation symbol — Ts=2T,) = In QPSK, both the in-phase and quadrature components are used = ThelandQ channels are aligned and phase transition occur once every T, = 27, seconds with a maximum transition of 180 degrees = From 2E . 22(i-1) Sooner (1) = “cos | 22f_.¢t + ——————_ ore (= 2 ft + AL s = As shown earlier we can use trigonometric identities to show that 2E. 2a(i-1) 2E, . | 2a(i-l) |. Sopor (tf) = .| —+ cos] ———— | cos(q@_.ft)— . }— sin] ———— |sin(@_¢ opsk (t) Vr a (@,t) r Wi (@,1) docsity.com