= sin Qrfx) the computer is a nothing but a multiplier followed plication, the ASK output will be ASK output corresponding to binary 0 is ‘The digital signal from 1 signal. The ASK modulator is figure 9.20. Due to the multi transmitted, Theand aud Rate (Np) Fen gat bit (0 or 1) to represent one symbol the baud rate N,, will be same as bit rate R as shows, for ASK, we wse one gsion Le. ‘Thus, sransmission Bandwidth of the ASK Signal - Baud rate = Bit rate = idth of ASK signal is dependent on the bit rate R. Where bit ‘The bandwidth o! i . . LR Mis shown 9.20. For a bit rate of R bits/sec, the maximum bandwidth required for an signal afer = RHz. ‘he frequency spectrum of an ASK signal is shown in figure 9.22 which shows that the spectrum of the carrier frequency f, with upper and lower sidebands. consists t+. B= (1+) R ———+!_ poy 1 bandwidth ' KX 1 due to the use of 1 ‘ 1 ' ‘ 1 : the BPF, * Amplitude (4) : oC +. Brrx = 22 —————+| Fig. 9.22. Frequency spectrum of an ASK signal The transmission bandwidth BW of the ASK signal can be restricted by using a filter. The "sincted value of bandwidth is given as under : BW=(1+nyR fi es risa factor related to the filter characteristics and its value lies between 0 and 1. Here, “arrier frequency i.e., frequency of the sine wave being transmitted. %. '3. Bandwidth of ASK in terms of Baud Rate Por ; Teta s2# thown in figure 9.21, the baud rate = bit rate. ‘width of ASK in terms of bit rate is given by nisin satel si 77, = Bit rate, and f, = one bit interval. Tate Pt “oe ‘Ste and baud rate are equal for ASK, the expression for bandwidth is given by BW= gadue ay 2 2i = 3x 100 = 300 bits/sec, Gi) Baud rate = Number of signal units per second = 100 bauds/sec, EXAMPLE 9.5. An ASK transmitter transmits 5000 bits per second, Calculate the m bandwidth. Assume the transmission to be half duplex, Solution: Given: bit rate = 5000 bps, Mode : Half duplex ‘To find: Minimum bandwidth ‘The minimum bandwidth of ASK is given by BW (min) = N, For ASK, baud rate N, is equal to bit rate Thus, BW (min) = 5000 Hz EXAMPLE 9.6. The bandwidth of an ASK syste Solution: (3) For an ASK system, the baud rate is equal to the bandwidth, ‘Therefore, Baud rate N,, = 3000 bauds per sec, (2) For ASK, bit rate is equal to baud rate, Hence, Bit rate = 3000 bits/sec. EXAMPLE 9.7. Draw the fr ‘equency spectrum (1000 to 9000 Hz). Also, calculate the carrier: Bap between the bands in the two directions Solution: Given: System: ASK, Mode: Full duplex Bandwidth : 8000 Hz, 1, Bandwidth in each direction For a full duplex ASK, the bandwidth in each direction is given by, 8,000 No gap between bands, BW 2 = 10002 2. Carrier frequencies in each direction vd and forward transmission are: 1000 Hs to 5000 Hz and 6000 Hz to 9000 Hz respectively. ibe anes Seca placed at the center of each band, 4000 Therefore, f, (backward) = 1000+ == 8000'Hs 4000 _ aa §, forward) = 5000 +> =7000 Hs4, Merits and Demerits of ASK qhe advantage of vantage 38 that transmission: It is ust gA5. Pet Total Bandwidth —————+ ie ¥ Fig. 9.24. Frequency spectrum of duplex ASK system using ASK is its simplicity. It is easy to generate and detect. However, its it is very sensitive to noise, therefore, it finds limited a cords ia ad at very low bit rates, upto 100 bits/se. naa formance Comparison of AM and ASK Table 9.1. Performance ‘Comparison of AM and ASK No. Parameter of Comparison AM t ae Ce ao ] 1. | Variable characteristics of Amplitude Amplitude the carrier. 7 | Nature of modulating Wfodulating signal is | Modulating signal is digital signal analog 7 | Modulated signal shape. 1 0 1 |e hate : 4 | Variation in the carrier Continuous variation in Carrier ON or OFF depending | amplitude, accordance with the on whether a Jor is to be — amplitude of modulating transmitted. ignal esos signal. di tH seal sidebands Two Two Benin; r ae , GeoR z ee Foor Poor Snel i oa Radio broadcasting | Date Transmission at ee eel ral Hon ~Sction Method | Binvelo Bn1 second ae Bitrate = 5, Baud rate = 5 (b) Representation of digital signal using FSK. Fig. 9.25, 2. FSK Generation Let us consider the FSK Benerator shown in figure 9.25, Tt consists of two oscillators which Produce sine waves at frequencies f, and fy Tespectively. The oscillators outpus are applied to the inputs of two multipliers (product modulators), and the data bit d(t) may be described as under ; ‘The other input to the two multipliers are the “nals b, and po, The relation between p,, Po Data bit to be transmitted Value of d(t) | binary 0 =I x binary 1 #1 0 When, a binary 0 is to be transmitted, py = 1 and Pi =. therefore, the output of the fmt modulator only is present and the frequency of the transmitiet signal is f,. Si , when a Present and the frequency of the transmitted signal is f,. The binary FSK signal is mathematically binary 1 is to be transmitted, p) = 0 and p, = 1, therefore, the output of the other multiplier only is represented as under : Vepsx (®) = Po. sin (nit) + p, sin 2nf, = @5) Se ——sath of FSK Signal Sane of FSK signal is dependent on the pulse width, o bit ate R= Vf, and the 952. width te eee the requenses f, and f,, as shown in figure 9.26. The maximum bandwidth of. gen ven DY 8 = (ev®)(e22 Baar= (6+ 9)-(673 06) = (f,-f+R) ere thobandiwidth can be restricted by using a bandpass filter (BPF). The restricted bandwidth isgven bY: R B= Ih fl +495 ~ OD ‘ed to the filter characteristics and its value lies between 0 and 1 ere, ris the factor relat is kept at least 2R/3, Substituting this value in equation (9.7), we ‘esparation between f, and fy Bt (9.8) 2 oR 2R+R=— 4 3 sy 3 This shows that FSK requires larger bandwidth than ASK and PSK (to be discussed next). 153, Bandwidth for FSK in terms of Baud Rate Tor FSK also, bit rate is equal to baud rate. We can imagine ‘xbination of two ASK spectrums centered at frequencies fy and f, Amplitude See Ne imate Woe Nhe See a sana max = the FSK spectrum to be as shown in figure 9.27. 1 Nye -f) +} Bandwiath Fig. Illustration of spectrum of FSK ——is not preferred for the high speed modems because with ao T™ aay sae ‘This dntreases tha/haiih alii a z x ~ andwidth required ress signal. ines have a very low bandwidth, it i i Jephone lines have a very width, it is not Possible to satisfy: bandwidth: @ lade of FSK at higher speed. Therefore, FSK is preferred only Ge low speed modems. ag PHASE SHIFT KEYING (PSK) Y f fact, phase shift keying (PSK) is the most efficient of the three modulation methods. Lae sed for high bit rates. In PSK, phase of the sinusoidal carrier is changed according = em pit to be transmitted. Figure 9.29 shows the simplest form of PSI ® K called Binary PSK SK), The carrier phase is changed between 0 and 180 by the bipolar digital signal. A bipolar Gi unal is used to represent the digital data form the DTE. 4 0 Bipolar " 1 1 i 1 1 1 180° vi ' ae 4 sin (2n ft) ' —sin(2: toh 180° : aH | t shift 1 i 7 ' t ' | Tbaud 1 baud T baud 1 baud baud | a | 1 1 second Tir Qaaeaee Tr Toe Bit rate = 5, Baud rate = 5 Bit rate = Baud rate Fig. 9.29. Binary phase shift keying (BPSK) The BPSK signal can be represented mathematically as under : ang Vapsk (t) Vapsx (t) Combini = ming the two conditioy ns, in (2m f,t), when binary 0 is to be represented ~sin (2nf,t) ‘in (2x f,t + m), when binary 1 is to be represented. we can write