Digital Communication

Transcription

1 T.Y. Diploma : Sem. V [ET/EN/EJ/EX/ED/EI] Digital Communication Time: 3 Hrs.] Prelim Question Paper [Marks : 100 Q.1(a) Attempt any THREE of the following : [12] Q.1(a) (i) Give the advantages and disadvantages of digital communication. [4] (A) Advantages of digital communication i) Noise immunity is more than analog communication. n. ii) Digital communication supports error detection and correction rection techniques. iii) It is easy to regenerate the digital signal than analog. iv) Digital signals are easy to store and manipulate. v) Digital communication is computable with advance data processing technique like digital technique process image processing. vi) Cost of digital communication system stem is low. Disadvantages i) The data rate of digital communication is very high. ii) Loss of information. r Q.1(a) (ii) List and explain various properties of Hamming Codes. [4] (A) Properties of Hamming Code i) Hamming codes are linear. ii) Number of bits in codeword is equal to n = 2 q 1 where e q = number of extra bit eg. if Q = 3 then n =? n = 7 iii) Number of message bit K should be always equal to 2 q 1 q iv) For the hamming code the minimum distance d min = 3. v) Hamming codes can generate systematic as well as non systematic code words. vi) Hamming codes are used to detect burst error. Q.1(a) (iii) What is companding? Draw the companding curves for PCM [4] system. (A) Companding By keeping step size constant we can perform non uniform quantization. This can be achieved by amplifying low level signals and alternating high level signal is called as compression at the receiver side, the signal is alternated at low level and amplified at high level is called as expansion. 1114/TY/Pre_Pap/Elec/2014/CP/Class/DC 1

2 : T.Y. Diploma DC The compression of signal at transmitter and expansion at receiver is combinely called as companding. Q.1(a) (iv) Draw the waveforms of ASK, FSK, PSK for binary input [4] (A) Given bit stream = with carr V V C V ASK V ASK V C 1 V FSK with carr V & C V C 1 V C 2 V PSK with carr V C 1 Compression at transmitter n f H f L f H f H f L Alternated Expansion at receiver Vidy nkar V in /TY/Pre_Pap/Elec/2014/CP/Class/DC

3 : T.Y. Diploma DC Q.1(b) Attempt any ONE of the following : [6] Q.1(b) (i) Discuss ASCII codes with suitable examples. [6] (A) ASCII Code : (American standard code for information interchange) Number of bits used to represent every symbol is 7. Number of different bit patterns = H H H : : 39H H A H B : : : : 5A Z H a H b : : 7AH z Example BABA C dy Bipolar +A NRZ A Q.1(b) (ii) Draw the block diagram for the generation of PAM. State the [6] function of each block. (A) Block diagram for the generation of PAM Pulse amplitude modulation is an analog to digital technique were the amplitude variations at input analog signal are converted into corresponding output digital signal. Hence here we make use of sampling, the block diagram is as shown below : 0y

4 : T.Y. Diploma DC Q.2 Attempt any FOUR of the following : [16] Q.2(a) List various properties required for line codes. [4] (A) Properties of line codes i) DC content ii) Signal power i) For a good line code DC content should be as low as possible. Split phase Manchester consist of very low DC component where as NRZ L and unipolar NRZ consists of large DC contents. If there is channel is less noisy then we use split phase Manchester. ii) For a good line code signal power should be as high as possible so that noise required to corrupt the large. NRZ L and unipolar NRZ has high signal power than split phase Manchester and RZ. Q.2(b) Discuss Multiplexing and its types. [4] (A) Multiplexing and its type Multiplexing is the technique which allows the simultaneous transmission of multiple signal data link. Multiplexing device is a physical line or a medium or a into logical segment called led as channels. Example cable TVy TV. D M E U M X U X FDM Multiplexing analog WDM Common communication Link digital TDM Synchronous TDM Asynchronous TDM /TY/Pre_Pap/Elec/2014/CP/Class/DC

5 : T.Y. Diploma DC Q.2(c) What is need for delta modulation? Give its advantages & [4] disadvantages and applications. (A) Delta Modulation (DM) Need : In PCM N number of bits are transmitted per quantized sample which asks for large channel Bandwidth and signaling rate. This disadvantage can be overcome by using DM. DM transmits only one bit per sample instead of N therefore it extensively reduces signaling rate and channel Bandwidth. Advantages i) One bit codeword for output. ii) Simplicity of design for transmitter and receiver. iii) Low signaling rate iv) Low channel Bandwidth Disadvantage i) Slope overload present ii) Granular Noise Application i) Satellite transmission n System ii) Digital Communication Q.2(d) Write bandwidth requirement rement for DPSK, QAM, QPSK, BPSK. [4] (A) Bandwidth requirement ent of (i) DPSK Symbol duration = Ts = 2Tb 1 2 = Fs fb y= y= fs = fb 2 fb/2 +fb/2 fb Bandwidth = fb = fb 2 2 Bandwidth = fb (ii) QAM Bandwidth = fs (fs) = 2fs Bandwidth = 2Fb N = 2 Ts 2 = N.Tb 2fs fs 0 +fs +2fs 2fs

6 : T.Y. Diploma DC (iii) QPSK Bandwidth = fb/2 (fb/2) bandwidth = fb (iv) BPSK Bandwidth = (f0 + fb) (f0 fb) = f0 + fb f0 + fb Babdwidth = 2fb fb/2 fb/2 (f0 fb) f0 (f0 + fb) (f0 fb) f0 (f0 + fb) Q.2(e) Compare QPSK and QASK (4 points). [4] (A) Comparison between QPSK and QASK V Parameters QPSK QASK i) Type of Modulation Quadrature Phase Modulation Quadrature Phase and Amplitude Modulation ii) Location of signal On the circumference of Equally spaced and points circle placed symmetrical about origin. iii) Distance between d = 2 E for N = 2 d = 2 0.4E for N = 4 or b b the signal points N = 16 iv) Noise immunity Better than QASK More than QPSK v) Probability of error Less than QASK More than QPSK vi) Type of Synchronous Synchronous Demodulation vii) System complexity Less complex than QASK More complex than QPSK ank /TY/Pre_Pap/Elec/2014/CP/Class/DC

7 : T.Y. Diploma DC Q.3 Attempt any FOUR of the following : [16] Q.3(a) Explain the unipolar, polar and bipolar line codes. [4] (A) Unipolar Format This is also known as on-off signaling. In this, symbol 1 is represented by switching off the pulse. When the pulse occupies the full duration of symbol the unipolar format is said to be of the non-return to zero (NRZ) type. When it occupies onehalf of the symbol duration, it is said to be of the returntozero (RZ) type. When it occupies onehalf of the symbol duration, it is said to be of the returntozero (RZ) type. + 1 Binary The unipolar format offers simplicity of implementation. Polar Format In this, a positive pulse is transmitted for symbol 1 and negative pulse for symbol 0. It can be of the NRZ or RZ type. A polar wave form has no dc component, provided that the 0s and 1s in the input data occur in equal proportion ank Bipolar Format This is also known as pseudoternary signaling. In this positive and negative pulses are used alternately for the transmission of 1s, and no pulses for transmission of 0s. It can be of RZ or NRZ type. t t

8 : T.Y. Diploma DC * Important feature of bipolar format is that, in the absence of a dc component, even though the input binary data may contain long strings of 0s and 1s. * The pulse alternation property of bipolar format provides a capacity cit for in service performance monitoring such that, any isolated error which causes the deletion or creation of a pulse, will violet this property. * Bipolar format eliminates the ambiguity that may arise because of polarity inversion during the course of transmission. For these reasons, the bipolar format is used in the T1 carrier systems for digital telephony Q.3(b) Discuss Shannon s theorem em in brief. [4] (A) Shannon s Theorem Also known as source coding theorem. Statement ent Given a discrete memoryless source of entropy H, the average code word length for any source coding is bounded as L H. Explanation : For any source encoder, code efficiency is given as = H 100% L H entropy, H = M p log i 2 i1 L average code word length, L = i 1 p is length of the code in bits As such if L H, minimum value of L i.e. L min. = H. Under such circumstances, efficiency will be maximum. To increase efficiency variable length coding is done. eg. Huffman code. i M i1 p i i t /TY/Pre_Pap/Elec/2014/CP/Class/DC

9 : T.Y. Diploma DC Q.3(c) Describe WDM in detail. [4] (A) Wavelength Division Multiplexing (WDM) Power Fiber 1 Spectrum Fiber 2 Fiber 1 (i) Channels having different frequency ency ranges can be multiplexed on a long fiber. (ii) The only difference e with electrical FDM is that optical system is completely passive. (iii) Reason WDM is popular is that the energy on a signal factor is a few gigahertz width because it is impossible to convert. (iv) Hence wavelength division multiplexing is explained. Q.3(d) Compare between FHSS and DSSS (4 points). [4] (A) Comparison between DSSS and FHSS Parameters DSSS FHSS Vi i) Definition PN sequence of large Data bits are transmitted bandwidth is multiplied in different frequency with narrow band data slots which are changed by signal. PN sequence. 1 ii) Chip rate It is fixed, R c = R T c = max (R h, R s ) iii) Modulation technique Power BPSK c M-ary FSK V Tb iv) Processing gain PG = T = N PG = 2 t c Fiber 2 Spectrum Power Fiber 3 Spectrum a Shared fiber Fiber 3 Fiber 4 ar

10 : T.Y. Diploma DC v) Error probability P e = vi) Acquisition time Long 1 E erfc b 2 JT c P e = 1 e rb R c /2 2 Short vii) Effect of This system is distance Effect of distance is less. distance relative. Q.3(e) Explain multiplexing hierarchy (AT&T) for FDM system. (A) Multiplexing Hierarchy in FDM Consider example at telephony in which each voice channel is having range of r[4] 300H z to 3.4 KHz. Here we need to multiplex such n no of voice channel by modulating it with different subcarriers. Multiplexing hierarchy goes as follows. 12 rc Level 1 : Basic Group : [12 voice channels multiplied together] Level 2 : Super Group : [Upto 5 B.G mux together i.e.: upto 125 = 60 channels] Level 3 : Master Group : [Upto 10 S.Cr mux together i.e. upto 600 V.C.] Level 4 : Jumbo Group : [Upto 6 M.G. maximum together i.e. upto 3600 V.C.] S 1 1 S 1 2 S KHz F D M 24 = 48 KHz B Group 485 = 240 K yaf 1 F = 2400K 2 D M 1 F MG = 5 2 D K 1 M F JG D yasg channel M 5 amal Q.4(a) Attempt any THREE of the following : [12] Q.4(a) (i) List and explain different types of errors in data communication. [4] (A) Types of error : i) Single bit error or one bit error ii) Burst error Single bit error : if only 1 bit in transmitted bit sequence is changed because of noise then it is called as single bit error /TY/Pre_Pap/Elec/2014/CP/Class/DC

11 : T.Y. Diploma DC transmitted symbol C +A A Received bit seq code of C Burst error : If 2 or more than 2 bits are changed because of noise then it is called as Burst error. Q.4(a) (ii) Explain Baudot code with suitable example. e. [4] (A) Baudot code In baudot coding every symbol is represented by using 5 binary bits. 2 special codes are used : i) figure shift code ii) letter shift code By using 5 bit code we can generate only 32 different binary bit sequence but the number of characters acters in English text are more than 32 SO 2 special codes are used. Now codes attached ached like this : A B? : : Z Example A B? Letter ter shift special code idy1 1 Vid idy Figure shift code r Q.4(a) (iii) Define sampling theorem. List types of sampling techniques. [4] Draw the naturally sampled signal. (A) Sampling Theorem "A bandlimited signal can be reconstructed exactly if it is sampled at a rate atleast twice the maximum frequency component in it."

12 : T.Y. Diploma DC There are two types of sampling technique : (1) Natural sampling (2) Flat top sampling Natural Sampled output Q.4(a) (iv) Explain cyclic redundenry check with suitable diagram. Explain [4] CRC for data to be transmitted tted with division (A) Cyclic redundancy cheek technique is more powerful than parity check & check sum error detection. It is based on binary division. A sequence of redundant called CRC or CRC remainder is appended at the end of data unit such as byte. The resulting data unit after adding CRC remainder becomes exactly divisible by another predetermined ermined binary no. At the receiver, r, this data unit is divided by the same binary no. There is no error if this division does not yield any remainder. But a non-zero remainder indicates presence of error in received data unit Such an erroneous ous bit is then rejected. Procedure : Divide the data unit by predetermined divisor Obtain the remainder which is CRC CRC should have exactly bit less than divisor Append the CRC to the end n bit of the data unit and then divide it with predetermined divisor. Data Divisor CRC n bit code word Data CRC /TY/Pre_Pap/Elec/2014/CP/Class/DC

13 : T.Y. Diploma DC Data : Divisor : 1101 n = no. of divisor bits 1 = 4 1 = 3 divided = Divisor = CRC = 001 Code word = Data (+) CRC = Q.4(b) Attempt any ONE of the following : [6] Q.4(b) (i) Discuss Frequency Hopping (FH) spread spectrum. [6] (A) Frequency Hopping ping (FH) spread spectrum is a FM or FSK technique while DS spread spectrum, described in sec.2, is an AM (or PSK) technique. The signal to be frequency hopped is usually a BFSK signal although M-ary FSK, MSK or =id TFM can be employed. Considering that BFSK is used, we have that the original signal, before spread spectrum is applied, is s(t) = 2P cos ( t d(t) t ) s 0 where d(t) is the data to be transmitted. The FH modulation is then applied by varying the carrier frequency so that the resulting FH spread spectrum is v(t) = 2P cos ( t d(t) t ) ids s 0 Here the FH signal has a carrier frequency f i = i /2 which changes at the hopping rate f H H, i.e., the carrier frequency f i changes each T H seconds. The frequency chosen each T H is selected in a pseudo-random manner from a specified set of frequencies. Typically different frequencies are used to form this set. The primary advantage of FH is that it enables the transmitter to change its carrier frequency and thereby avoid an otherwise in-band interfering signal. For example, consider that the FH signal spends an equal time at each of 1000

14 : T.Y. Diploma DC frequencies f 1, f 2,, f Also assume that BFSK is employed. Then if the bit rate of the data is f b, the bandwidth used by the signal, at any carrier frequency f i, is B = 4f b. Now assume that there is an interfering signal having a bandwidth B = 4f b and a fixed center frequency f j. If frequency hopping were not employed and the interference were located at the transmitted ted signal's carrier frequency, i.e., j = i, then if the interfering signal power were sufficiently large the probability of error would be P e = 0.5 since under these circumstances we could do no better than to guess. (In a military system, the interferer determines the signal's carrier frequency and purposely transmits at that frequency to block or jam the communications.) By employing FH and using say 1000 frequencies, the probability of the same interferer causing an error is reduced to P e = 1/2000 = (note that thermal noise is ignored since it is considered to be a second-order effect when "jamming" is present). Q.4(b) (ii) Explain the properties of Hamming Code. [6] Define : (1) Hamming distance. (2) Minimum distance. (3) Weight of code words. (4) Error detection capacity. (5) Error r correction capacity. (A) Properties of Hamming Code i) Hamming codes are linear. ii) Number of q bits in codeword is equal to n = 2 1 where q = number of extra bit eg. if Q = 3 then n =? n n = 7 iii) Number of message bit K should be always equal to 2 q 1 q iv) For the hamming code the minimum distance d min = 3. v) Hamming codes can generate systematic as well as non systematic code words. vi) Hamming codes are used to detect burst error. i) Hamming Distance (d) : It is defined as the number of bits in which 2 valid codewords differ from each other. e.g. X 1 = X 2 = d = 3. ii) Minimum distance (d min ) : It is smallest hamming distance between any 2 valid codewords /TY/Pre_Pap/Elec/2014/CP/Class/DC

15 : T.Y. Diploma DC iii) Weight of codewords (W) : It is number of man zero elements of codeword provided. Codeword is not all zero codeword. d min = Min [W(x)] where X Error detection Capability : It is calculated by this, d min S + 1 Where d min = Minimum distance. S = Number of bits in error that can be detected. ed. e.g. d min = 3 3 S S S 2 upto 2 bits in the error can be detected. Error correction capability : it is calculated ated by this expression. d min 2t + 1 where t = number of bits in error that can be corrected. eg. d min = 3 3 2t t 2 2t t 1 this coding technique can correct upto 1 bit. Q.5 Attempt any FOUR of the following : [16] Q.5(a) Draw the waveforms for for : [4] (i) Split phase monchester. (ii) Differential monchester line code formats (A) For ve 0 ve +ve ve Split phase manchester t 0 Vi y differential manchester t

16 : T.Y. Diploma DC Q.5(b) What do you understand by channel capacity? How can it be increased? (A) Channel Capacity Data Rate R = n.f s Where, n = Number of bits/sample. f s = Number of sample/sec. Channel capacity is defined as maximum possible data rate is with which information can be convey on channel with minimum error. ror. Q.5(c) List the different error detecting methods. Describe checksum [4] method with suitable example. (A) Different error detection schemes Repetition codes Parity bits Checksums Cyclic redundancy checks (CRCs) Cryptographic hash functions Checksums Most of error detection techniques uses a process known as checksum to generate an error-detection character. The character results from summing all the bytes of a messagem sagem together, discarding and carry over from the addition. Again, the process is repeated at the receiver and the two checksums are compared. A match between receiver checksum and transmitted ted checksum indicates good data. A mismatch indicates an error has occurred. curred. This method, like CRC, is capable of detecting single or multiple errors in the message. The major advantage of checksum is that it is simple to implement in either hardware or software. The drawback to checksum is that, unless you use a fairly large checksum (16- or 32-bit instead of 8-bit), there are several data-bit patterns that could produce the same checksum result, thereby decreasing its effectiveness. It is possible that if enough errors occur in a message that a checksum could be produced that would be the same as a good message. This is why both checksum and CRC error-detection methods do not catch 100% of the errors that could occur, they both come pretty close. Example : What is the checksum value for the extended ASCII message Help!? Solution : The checksum value is found by adding up the bytes representing the Help! characters: [4] /TY/Pre_Pap/Elec/2014/CP/Class/DC

17 : T.Y. Diploma DC H e l p ! Checksum checksum error-detection process that uses the sum of the data stream in bytes. The hardware solution relies once more on exclusive OR gates, which perform binary-bit addition. Each 8-bits of data are exclusive e ORed with the accumulated total of all previous 8-bit groups. The final accumulated total is the checksum character. Q.5(d) Explain FDMA system with schematic diagram. Compare FDMA and [4] TDMA. (A) FDMA (Frequency division multiple access) This technique is based an FDM tech In FDMA available bandwidth is shared by all the station Each station is allocated with a particular frequency band to send its data. Example Station 1 Guard Bands Station 2 Station 3 Station 4 f 0 f 1 f 2 f 3 f 4 f 5 freq f 1 f 2 f 3ala Available B.W. In above diagram frequency band f 0 to f 1 is reserved for station 1, f 2, to f 3 is reserved for s 2 etc. Guard bands are also provided in between the adjacent frequency slot to avoid crosstalk. Features Overall channel band width is shared by multiple users therefore no of users can transmit their information simultaneously. Guard band are provides because : i) To avoid cross talk ii) Impossible to achieve ideal filtering to separate different users Power efficiency different users. Synchronization is not required

18 : T.Y. Diploma DC Advantages : All stations can operate continuously and simultaneously. Power required for transmission depends on the no of channels being transmitted. SNR is improved because of Fm Synchronization is not required. Disadvantages : Each channel on earth station can used only a part of total satellite B.W. Inspite having guard bands adjacent channel interference nce is present. Because of the use of Fm, Bw required & therefore less no of channels can be accommodated in available Bandwidth. Parameter FDMA TDMA Technique Sharing of overall B.W. of Sharing of overall time os satellite transponder satellite transponder Syndrome Not required Code word Not required red Not required Power Less Full power n is poss efficiency Guard time & Guard band required Guard time required Guard band anrequired dy Q.5(e) Describe working of DSSS system in detail. [4] (A) DSSS system (Direction Sequence (DS) Spread Spectrum) A Direct Sequence (DS) spread spectrum signal is one in which the amplitude of an already modulated signal is amplitude modulated by a very high rate NRZ binary stream of digits. Thus, if the original signal is s(t), where s (t) = 2P d(t) cos 0 t (1) s (a binary PSK signal), the DS spread spectrum signal is v (t) = g(t) s(t) = requiredla ala an nk TDMAk 2P g(t) d(t) cos s 0 t (2) where g(t) is a pseudo-random noise (PN) binary sequence having the values 1. The characteristics of g(t) are extremely interesting and are discussed in some detail in Ranging using DS spread spectrum. Here we merely assume that g(t) is a binary sequence as is the data d(t). The sequence g(t) is generated in a deterministic manner and is repetitive. However, the sequence length before repetition is usually extremely long and to all intents and purposes, and without serious error, we can assume that the sequence is /TY/Pre_Pap/Elec/2014/CP/Class/DC

19 : T.Y. Diploma DC truly random, i.e., there is no correlation at all between the value of a particular bit and the value of any other bits. Furthermore, the bit rate f c of g(t) is usually much greater than the bit rate f b of d(t). As a matter of fact the rate of g(t) is usually so much greater than f b, we say that g(t) "chops the bits of data into chips", and we call the rate of g(t) the chip rate f c, retaining the words, bit rate, to represent f b. To see that multiplying the BPSK sequence s(t) by g(t) spreads the spectrum we refer to figure 1 which shows a data sequence d(t), a pseudo-random (often called a pseudo-noise or PN) sequence g(t) and the product sequence g(t) d(t). Note that (as is standard practice) the edges of g(t) and d(t) are aligned, that is, each transition in d(t) coincides with a transition on g(t). The product sequence is seen to be similar to g(t) indeed if g(t) were truly random, the product sequence would be another random sequence g'(t) having the same chip rate f c as g(t). Since the bandwidth of the BPSK signal s(t) is nominally 2f b the bandwidth of the BPSK spread spectrum signal v(t) is 2f c and the spectrum has been spread by the ratio f c c/f b. Since the power transmitted by s(t) and v(t) is the same, i.e., P s, the power spectral density G s (t) is reduced by the factor f b /f c. Fig. 1 : (a) The waveform of the data but stream d(t). (b) The chipping waveform g(t). (c) The waveform of the product g(t) d(t). Vid lankar To recover the DS spread spectrum signal, the receiver shown in figure 2 first multiplies the incoming signal with the waveform g(t) and then by the carrier V2 2 cos 0 t. The resulting waveform is then integrated for the bit duration and the output of the integrator is sampled, yielding the data d(kt b ). We note that at the receiver it is necessary to- regenerate both the sinusoidal carrier of frequency 0 and also to regenerate the PN waveform g(t).

20 : T.Y. Diploma DC Fig. 2 : A BPSK communication system incorporating a spread spectrum technique. Q.6 Attempt any FOUR of the following : [16] Q.6(a) Explain BPSK, draw its waveforms. [4] (A) BPSK : Binary phase shift keying In the BPSK phase of a carrier is modified ied according cording to value of the symbol. b(t) b(t) phase shift transmitted ted signal o 2Pcos.(2ft) c o 2P cos. (2 ft c +V 0 o VV 180 o T b In general transmitted BPSK signal is. S(t) = b(t). 2P. cos(2 f t) c bdyal a Binary bit Seq. Bipolar NRZ encoder b(t) ac 2P cos (2fc t) Working of transmitter of BPSK s(t) /TY/Pre_Pap/Elec/2014/CP/Class/DC

21 : T.Y. Diploma DC Bipolar NRZ encoder : given information signal will consists of binary bits 1 & 0 this binary bit sequence is converted into bipolar NRZ signal by this block output of bipolar NRZ encoder is b(t). Where, b(t) = +1 when bit = 1 b(t) = 1 when bit = 0 Modulator : bipolar b(t) is used as modulating signal and 2P cos (2 ft) is c used as carrier. The modulator multiplies 2 input signal to generate BPSK signal S(t) = b(t). 2P.cos (2 f t) BPSK waveforms : Binary bit +1 b(t) 0 1 2P (2f t) c 0 BPSK signal s(t) o 180 o 1 o 0 o 180 o 0 o c Q.6(b) Draw the block diagram of M-ary FSK transmitter and receiver. [4] (A) Mary Frequency Shift Keying (MFSK) In MFSK, we use group of N bits, which gives rise to M different symbols. M = 2 N Each symbol duration is T S = N T b an Each symbol uses different frequency, hence, MFSK uses frequencies f 1, f 2, f M. Vidy180o 0 o The probability of error is minimized of the frequencies f 1, f 2, f M are selected so that M signals are mutually orthogonal. One commonly used arrangement simply provides that the carrier frequency be successive even harmonics of the symbol frequency f S = 1/T S. t t t

22 : T.Y. Diploma DC Thus we have, f Kf 1 S f K 2 2 fs f K 2M2 f 3 S Now we can write equation of MFSK signal as V t P t 2P cos t P t 2P cos t... P t 2P cos t MFSK 1 S 1 2 S 2 m S M where P 1 (t), P 2 (t), P m (t) are NRZ unipolar signals out of which only one is +1 for every T S (= NT b ) sec. Transmitter b (t) V MFSK (t) In the figure, when N bits of b(t) enter into serial to parallel converter, they appear simultaneously ly at the output of serial to parallel converter. These bits are given to DAC which generates an analog voltage V m where m = 1, 2, M. That is, the output of DAC may have M different levels. This voltage V m is applied to frequency modulator which generates a sine wave of constaint amplitude but the frequency is dependent on input voltage V m. Receiver er Serial to parallel converter BPF 1 BPF 2... BPF M 1 2 : N 2.dy Nbit DAC Envelope detector Envelope detector... Envelope detector V m m = 1, 2,.. M Select largest output Frequency Modulator Nbit output (ADC) V MFSK (t) anka Vi yde... b 1 b 2 b N /TY/Pre_Pap/Elec/2014/CP/Class/DC

23 : T.Y. Diploma DC PSD of MFSK signal V MFSK (t) = P t 2P cos t P t cos t... P t 2P cos t 1 S m S m P 1 1 M M 8 NP T S b 2 2 Sa f f NT Sa f f NT 1 b 1 b 8... S G MFSK (t) = ffff... ff ff Sa f fmnt Sa b f fmnt b P S 8 G MFSK (f) NP T Q.6(c) Draw the block diagram of ADM transmitter. Describe its working [4] with waveforms. (A) Adaptive delta modulation (ADM) In the adaptive delta modulator if there is the rate of change of input signal is very high we will increase the level. And if the rate of change of input signal is low then we will decrease the level. x (kt s s) x (kt s ) + e (nts) S 8 b f 1 f b / N f 1 Step size calculation Quantizer f 1 f f f M laf + delay t s f M + f b / N f b B W M 2 N ank + accumulator f

24 : T.Y. Diploma DC amplitude 0 2 ADM Receiver Receivers sign + + Step size delay t s calculation Advantages : i) It reduces slop overload error by increasing step size. ii) It reduces granular noise by decreasing ladela step size. Disadvantages i) ADM transmitter receiver are complex and ii) Cost of ADM is high. 2 2 Q.6(d) State applications of s.s. modulation (any four). [4] (A) Applications of Spread Spectrum Techniques (i) For combining the intentional interference. (ii) For reducing the unintentional interference. (iii) For suppressing the interference due to the multipath interference. (iv) In the low probability of intercept application as explained in section. (v) Due to large bandwidth of a spread spectrum signal can be recognized only by the authorised user. All other receiver consider this as noise. Q.6(e) Explain quantization in detail. [4] (A) Quantization Process It is a process of approximation or rounding off. It converts sampled signal into approximate quantized signal which consists of finite no. of pre-decided voltage levels. > < LPF time ankar ar /TY/Pre_Pap/Elec/2014/CP/Class/DC

25 : T.Y. Diploma DC Each sample value at input of quantizer is approximated to nearest predicted voltage level. These standard levels are called quantization levels. Amplitude V L s s s s s s s s Input signal x(t) is assumed to have peak-peak swing of V L to V H volts. This entire voltage range has been divided into Q equal intervals each of size 'S'. S is called step size V V H L S = Q Quantization noise i.e. Q = 8 for nw At the center of these ranges the quantization levels q 0, q 1, q 2,, q 7 are placed. x q (t) is quantized version of x(t) which is available at the output of q quantizer. When x(t) is in the range of 0, the corresponding to any value of x(t) the quantizer output will be equal to 'q 0 '. Similarly for all the values of x(t) in 1, the quantizer output is q 1. Thus in each range from 0 to 7 signal x(t) is rounded off to nearest quantization level and quantized signal is produced. In PCM, coded no. is transmitted for each level sampled in modulating signal. If we do not use quantizer block in PCM then this will need a large no. of bits per word which will increase bit rate and finally bandwidth. Quantization has effect of reducing this infinite no. of levels to relatively small nos. which can be transmitted without difficulty. Quantized signal x q (t) Input signal x(t) q 7 q 6 q 5 q 4 q 3 q 2 q 1 q 0 nkq 0 Time