Internal Examination I Answer Key DEPARTMENT OF CSE & IT. Semester: III Max.Marks: 100

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1 NH 67, Karur Trichy Highways, Puliyur C.F, Karur District Internal Examination I Answer Key DEPARTMENT OF CSE & IT Branch & Section: II CSE & IT Date & Time: & 3 Hours Semester: III Max.Marks: 100 Subject: CS6304 Analog and Digital Communication Staff Incharge: M.Kumar, AP/IT PART A (10 = 0Marks) 1. What is the need for modulation? (Nov/Dec 014) To reduce antenna height To multiplex more number of signals To reduce the noise and distortion To narrow banding the signal To reduce equipment complexity. State Carson s rule of FM bandwidth. (Nov/Dec 010) As per Carson s definition, the minimum bandwidth required for transmitting FM signal is that BW = [Δf + fm(max)] where Δf frequency deviation, fm(max) maximum modulating signal frequency. 3. Distinguish between FM and PM. (Nov/Dec 01) Frequency Modulation (FM) Phase Modulation (PM) The instantaneous frequency ωi is varied linearly with the message signal about an unmodulated carrier frequency ωc. The phase angle φ is varied linearly with the message signal about an un-modulated phase angle(ωct + θ0). The instantaneous frequency is The instantaneous frequency is i k c t f m(t) i ct k pm(t) kf= frequency sensitivity Kp= frequency sensitivity The general expression for FM wave is The general expression for PM wave is s( t) Acos[ c t k f t m( t) dt] s( t) Acos[ t k m( t)] c p 0

2 4. What are the sources of noise in communication systems? External Noises: Noises whose sources are external. External noise may be classified into Atmospheric noises Extraterrestrial noises Man-made noises or industrial noises Internal Noises: Noises which get generated within receiver or communication system. Internal noise may be put into following categories. Thermal noise or white noise or Johnson noise Shot Noise Transit time noise 5. Differentiate between coherent and non-coherent detection in demodulating AM wave. A Coherent detector has two inputs one for carrier reference signal and another one for modulated signal that is to be demodulated. For perfect demodulation, coherent detector requires the synchronized carrier for reference input. Otherwise, the performance of detection will be poor. A non-coherent detector has only one input ie., modulated signal. So synchronized carrier not required in non-coherent detector. Envelop detector is an example for non-coherent detection. 6. Define information capacity and bandwidth. (Nov/Dec 009)(Nov/Dec 010) Information capacity is a measure of how much information can be propagated through communication system. It is function of bandwidth and time. I α Bw.t Bandwidth is defined in general for digital communication system as fb = B logm. where fb is bit rate and M is number of levels used in digital communication systems. 7. What is M-ary encoding? (Nov/Dec 013) M-ary is term derived from binary. M simply represents a digit that corresponds to the number of levels possible for a given number of binary variables. For example, a digital signal with four possible conditions is an M-ary system with M = 4.

3 8. Draw the ASK and FSK signal for the binary signal s(t) = (April/May 010) (April/May 011) 9. Define the term Nyquist bandwidth. According to H.Nyquist, binary signal can be propagated through an ideal noiseless transmission medium at a rate equal to two times the bandwidth of the medium. The minimum bandwidth necessary to propagate a signal is called Nyquist bandwidth or Nyquist frequency. fb = B where fb is bit rate and B is Bandwith. 10. What is bandwidth efficiency? (May/June 014) Spectral efficiency, spectrum efficiency or bandwidth efficiency refers to the information rate that can be transmitted over a given bandwidth in a specific communication system. It is a measure of how efficiently a limited frequency spectrum is utilized by the physical layer protocol, and sometimes by the media access control (the channel access protocol). PART B (5 16 = 80Marks) 11. (a) Derive expression for AM wave. Define modulation index and express its value in terms of maximum and minimum voltage values of signal. Draw the spectrum and time domain signal of AM wave. (16 Marks) (Nov/Dec 009) (Nov/Dec 010)(April/May 011) Amplitude Modulation Definition: ( Marks) An AM signal is made up of a carrier (with constant frequency) in which its amplitude is changed (modulated) with respect to the signal (modulating signal).

4 (4 Marks) Modulating signal (sine wave) and modulated carrier A carrier is described by v = Vc Sin ( c t + ) To amplitude modulate the carrier its amplitude is changed in accordance with the level of the audio signal, which is described by v = Vm Sin ( m t ) v = {Vc + Vm Sin ( m t )} * Sin ( c t ) = Vc Sin ( c t ) + Vm Sin ( m t ) * Sin ( c t ) Using Sin A * Sin B = ½ Cos (A - B) - ½ Cos (A + B) this becomes v = Vc Sin ( c t ) + ½ Vm Cos ( ( c - m) t ) - ½ Vm Cos (( c + m)t) (4 Marks) This is a signal made up of 3 signal components carrier at c (rad/s) Frequency is fc = c/ Hz upper side frequency c + m (rad/s) Frequency is ( c + m)/ = fm + fc lower side frequency c - m (rad/s) Frequency is ( c - m)/ = fm - fc Hz The bandwidth (the difference between the highest and the lowest frequency) is BW = ( c + m ) - ( c - m) = * m Rad/s ( = m/ Hz)

5 ( Marks) Amplitude (V) Lower side Carrier Upper side frequency frequency Angular Frequency c - m c Bandwidth = * m c + m Modulation Index = Em/Ec or (Vmax-Vmain)/(Vmax+Vmin) derived from AM wave (4 Marks) (b) (i) Write the evolution and description of SSB techniques in detail. (10 Marks) SSB Description: (5 Marks) Single sideband modulation is a form of amplitude modulation. As the name implies, single sideband, SSB uses only one sideband for a given audio path to provide the final signal. Single sideband modulation, SSB, provides a considerably more efficient form of communication when compared to ordinary amplitude modulation. It is far more efficient in terms of the radio spectrum used, and also the power used to transmit the signal. In view of its advantages single sideband modulation has been widely used for many years, providing effective communications, as well as forms being used for some analogue television signals, and some other applications. Single sideband modulation can be viewed as an amplitude modulation signal with elements removed or reduced. In order to see how single sideband is created, it is necessary to use an amplitude modulated signal as the starting point.

6 An amplitude modulated carrier showing sidebands either side of the carrier From this it can be seen that the signal has two sidebands, each the mirror of the other, and the carrier. To improve the efficient of the signal, both in terms of the power and spectrum usage, it is possible to remove the carrier, or at least reduce it, and remove one sideband - one is the mirror image of the other. A single sideband signal therefore consists of a single sideband, and often no carrier, although the various variants of single sideband are detailed below. Single sideband modulation showing upper and lower sideband signals It can be seen that either the upper sideband or lower sideband can be used. There is no advantage between using either the upper or lower sideband. The main criterion is to use the same sideband as used by other users for the given frequency band and application. The upper sideband is more commonly used for professional applications.

7 There are two methods used for SSB transmission. 1. Filter method. Phase shift method Filter method: (any one method description 5 marks)

8 Phase Shift Method: (ii) Obtain relationship between carrier and sideband power in an AM DSBFC wave and explain how power distribution takes place in AM DSB FC system. (6 Marks) (May/June 014) (Nov/Dec 010) In radio transmission, the AM signal is amplified by a power amplifier and fed to the antenna with a characteristic impedance that is ideally, but not necessarily, almost pure resistance. The AM signal is really a composite of several signal voltages, namely, the carrier and the two sidebands, and each of these signals produces power in the antenna. The total transmitted power is simply the sum of the carrier power and the power in the two sidebands PUSB and PLSB PT = PC + PUSB + PLSB You can see how the power in an AM signal is distributed and calculated by going back to the original AM equation: VAM = VCsin(πfct) + Vm Vm cos[π(fc-fm)t] - cos[π(fc+fm)t]

9 where the first term is the carrier, the second term is the lower sideband, and the third term is the upper sideband. Now, remember that and are peak values of the carrier and modulating ine waves, respectively. For power calculations, rms values must be used for the voltages. We can convert from peak to rms by dividing the peak value by or multiplying by The rms carrier and sideband voltages are the VAM = Vc sin(πfct) + Vm Vm cos[π(fc-fm)t] - cos[π(fc+fm)t] The power in the carrier and sidebands can be calculated by using the power formula P = V /R where P is the output power, Vis the rms output voltage, and R is the resistive part of the load impedance, which is usually an antenna. We just need to use the coefficients on the sine and cosine terms above in the power formula: PT = ( Vc) R + ( Vm / R ) ( Vm / + R ) Vc = R Vm 8R Vm 8R Remembering that we can express the modulating signal Vm in terms of the carrier Vc by using the expression given earlier for the modulation index m = Vm/Vc, we can write Vm = mvc. If we express the sideband powers in terms of the carrier power, the total power becomes giving PT = Vc R ( mvc) 8R ( mvc) 8R Vc = R m Vc 8R m Vc 8R Since the term (Vc /R) is equal to the rms carrier power, it can be factored out, PT = Vc R 1 m 4 m 4 = Pc 1 m 1. (a) Define FM and PM modulation. Write down their equations. Describe suitable mechanism that can produce PM from FM modulator and vice versa. (16 Marks) (Nov/Dec 014) (Nov/Dec 009) (Nov/Dec 010)(April/May 010)(April/May 011) Angle Modulation Definition: ( Marks) Carrier angle is varied according to the slowly-varyingmessage signal Features:

10 It can provide a better discrimination (robustness) against noise andinterference than AM This improvement is achieved at the expense of increasedtransmission bandwidth In case of angle modulation, channel bandwidth may be exchangedfor improved noise performance Such trade-off is not possible with AM PM and FM waveforms: (4 Marks) PM and FM modulation: (4 Marks) Modulation index of FM: ( Marks) As per Carson s definition, the minimum bandwidth required for transmitting FM signal is that BW = [Δf + fm(max)] where Δf frequency deviation, fm(max) maximum modulating signal frequency.as per Bessel s function, the bandwidth is defined as the separation between sideband frequency which has minimum amplitude of 1% of maximum amplitude of carrier signal. FM and PM Equations derivations and spectrum equations: (4 Marks)

11 (b) (i) Draw the block diagram of superheterodyne radio receiver and explain the operation. What are the advantages of this receiver over TRF receiver? (10 Marks)(Nov/Dec 01)(Nov/Dec 013) Figure shows a block diagram with waveforms of a typical AM superheterodyne receiver developed to overcome the disadvantages of earlier type receivers. Let s assume you are tuning the receiver. When doing this you are actually changing the frequency to which the RF amplifier is tuned. The RF carrier comes in from the antenna and is applied to the RF amplifier. The output of the amplifier is an amplified carrier and is sent to the mixer. The mixer also receives an input from the local oscillator. These two signals are beat together to obtain the IF through the process of heterodyning. (Heterodyning will be further discussed later in this chapter and was covered in NEETS, Module 1, Modulation Principles.) At this time you should note the dotted lines connecting the local oscillator, RF amplifier, and the mixer. This is used on block diagrams and schematics to indicate GANGED TUNING. Ganged tuning is the process used to tune two or more circuits with a single control. In our example, when you change the frequency of the receiver all three stages change by the same amount. There is a fixed difference in frequency between the local oscillator and the RF amplifier at all times. This difference in frequency is the IF. This fixed difference and ganged tuning ensures a constant IF over the frequency range of the receiver. The IF carrier is applied to the IF amplifier. The amplified IF carrier is then sent to the detector. The output of the detector is the audio component of the input signal. This audio component is then passed through an audio frequency amplifier. The amplified audio component is sent to a speaker for reproduction. This allows you to hear the signal. You should note that a superheterodyne receiver may have more than one frequencyconverting stage and as many amplifiers as needed to obtain the desired power output.

12 HETERODYNING. As you know the intermediate frequency is developed by a process called heterodyning. This action takes place in the mixer stage (sometimes called a converter or first detector). Heterodyning is the combining of the incoming signal with the local oscillator signal. When heterodyning the incoming signal and the local oscillator signal in the mixer stage, four frequencies are produced. They are the two basic input frequencies and the sum and the difference of those two frequencies. The amplifier that follows (IF amplifier) will be tuned to the difference frequency. This difference frequency is known as the intermediate frequency (IF). A typical value of IF for an AM communications receiver is 455 kilohertz. The difference frequency is a lower frequency than either the RF input or oscillator frequencies. This lower frequency gives slightly better gain but does increase the chances of image frequency interference. Image frequencies will be discussed later in this chapter. DETECTION. Once the IF stages have amplified the intermediate frequency to a sufficient level, it is fed to the detector. When the mixer is referred to as the first detector, this stage would be called the second detector. The detector extracts the modulating audio signal. The detector stage consists of a rectifying device and filter, which respond only to the amplitude variations of the IF signal. This develops an output voltage varying at an audiofrequency rate. The output from the detector is further amplified in the audio amplifier and is used to drive a speaker or earphones. (ii) Explain the method of FM demodulation using PLL. (6 Marks)(Nov/Dec 013) Phase Locked Loop: ( Marks) The phase locked loop is a non-linear feedback loop. The control voltage to the VCO will endeavour to keep the VCO frequency locked to the incoming carrier, and thus will be an exact copy of the original message. PLL Demodulation of FM Block diagram: (4 Marks)

13 Principle of operation: It is complicated by the fact that its performance is described by non-linear equations, the solution to which is generally a matter of approximation and compromise. Suppose there is an unmodulated carrier at the input. The arrangement is reminiscent of a product, or multiplier-type, demodulator. If the VCO was tuned precisely to the frequency of the incoming carrier, ω0 say, then the output would be a DC voltage, of magnitude depending on the phase difference between itself and the incoming carrier. For two angles within the 360 range the output would be precisely zero volts DC. Now suppose the VCO started to drift slowly off in frequency. Depending upon which way it drifted, the output voltage would be a slowly varying AC, which if slow enough looks like a varying amplitude DC. The sign of this DC voltage would depend upon the direction of drift. 13. (a) Explain about the indirect method or Armstrong method of generating wideband FM. (16 Marks) Explanation (4 Marks) (4 Marks) (4 Marks)

14 The crystal-controlled carrier oscillator signal is directed to two circuits in parallel. This signal (usually a sine wave) is established as the reference past carrier signal and is assigned a value 0.The balanced modulator is an amplitude modulator used to form an envelope of double side-bands and to suppress the carrier signal (DSSC). This requires two input signals, the carrier signal and the modulating message signal. The output of the modulator is connected to the adder circuit; here the 90 phase-delayed carriers signal will be added back to replace the suppressed carrier. The act of delaying the carrier phase by 90 does not change the carrier frequency or its wave-shape. This signal identified as the 90 carrier signal. Derivation of FM from PM (4 Marks) (b) Draw the block diagram of FSK receiver and explain the operation. Determine the (i) peak frequency deviation (ii) minimum bandwidth (iii) Baud rate for FSK signal with a mark frequency of 49KHz, space frequency of 51 KHz and input bit rate of Kbps. (16 Marks) (Nov/Dec 009) (8 Marks) It is complicated by the fact that its performance is described by non-linear equations, the solution to which is generally a matter of approximation and compromise. Suppose there is an unmodulated carrier at the input. The arrangement is reminiscent of a product, or multiplier-type, demodulator. If the VCO was tuned precisely to the frequency of the incoming carrier, ω0 say, then the output would be a DC voltage, of magnitude depending on the phase difference between itself and the incoming carrier. For two angles within the 360 range the output would be precisely zero volts DC. Now suppose the VCO started to drift slowly off in frequency. Depending upon which way it drifted, the output voltage would be a

15 slowly varying AC, which if slow enough looks like a varying amplitude DC. The sign of this DC voltage would depend upon the direction of drift. (i) Peak frequency deviation Δf = mark frequency space frequency / = (51K 49K)/ = 1KHz (3 Marks) (ii) Minimum Bandwidth = (Δf + fb) where fb bit rate (as per Carson s rule) BW = (1K+K) = 6KHz (3 marks) (iii) Baud rate = bit rate/n where N = No. of bits In FSK, N = 1, therefore Baud rate = Kbps/1 = Kbps. ( Marks) 14. (a) (i) Draw the block diagram of QPSK modulator and explain its operation. For QPSK modulator, construct the truth table, phasor diagram and constellation diagram. (10 Marks) (Nov/Dec 014) (Nov/Dec 009)(Nov/Dec 010)(Nov/Dec 01)(April/May 011) QPSK modulator: (4 Marks) QPSK modulator: (a) truth table; (b) phasor diagram; (c) constellation diagram (4 Marks)

16 A block diagram of a QPSK modulator is shown in Figure -17. Two bits (a dibit) are clocked into the bit splitter. After both bits have been serially inputted, they are simultaneously parallel outputted. The I bit modulates a carrier that is in phase with the reference oscillator (hence the name "I" for "in phase" channel), and the Q bit modulate, a carrier that is 90 out of phase. For a logic 1 = + 1 V and a logic 0= - 1 V, two phases are possible at the output of the I balanced modulator (+sin ωct and - sin ωct), and two phases are possible at the output of the Q balanced modulator (+cos ωct), and (-cos ωct). When the linear summer combines the two quadrature (90 out of phase) signals, there are four possible resultant phasors given by these expressions: + sin ωct + cos ωct, + sin ωct - cos ωct, -sin ωct + cos ωct, and -sin ωct - cos ωct. (Marks) (ii) Explain the bandwidth consideration of QPSK system. (6 Marks)(Nov/Dec 010) With QPSK, because the input data are divided into two channels, the bit rate in either the I or the Q channel is equal to one-half of the input data rate (fb/) (one-half of fb/ = fb/4). The worse-case input condition to the I or Q balanced modulator is an alternative 1/0 pattern, which occurs when the binary input data have a 1100 repetitive pattern. One cycle of the fastest binary transition (a 1/0 sequence in the I or

17 Q channel takes the same time as four input data bits. Consequently, the highest fundamental frequency at the input and fastest rate of change at the output of the balance.: modulators is equal to one-fourth of the binary input bit rate. The output of the balanced modulators can be expressed mathematically as (b) Explain in detail about 8 QAM transmitter and receiver. (16 Marks)(Nov/Dec 013)(April/May 010) 8-QAM is an M-ary encoding technique where M = 8. Unlike 8-PSK, the output signal from an 8-QAM modulator is not a constant-amplitude signal. The magnitudes of the I and Q PAM signals are always equal. Their polarities depend on the logic condition of the I and Q bits and, therefore, may be different. Figure shows the truth table for the I and Q channel -to-4-level converters; they are identical. (6 Marks)

18 8-QAM transmitter (5 Marks) 8 QAM Receiver: (5 Marks) The carrier recovery circuit reproduces the original reference oscillator signal. The incoming 8-QAM signal is mixed with the recovered carrier in the I product detector and with a quadrature carrier in the Q product detector. The outputs of the product detectors are 4-level PAM signals that are fed to the 4- to--level analog-to-digital converters (ADCs). The outputs from the I channel 4-to-- level converter are the I and C_bits, whereas the outputs from the Q channel 4-to-- level converter are the Q and _ C bits. The parallel-to-serial logic circuit converts the I/C and Q/ _ C bit pairs to serial I, Q, and C output data streams.

19 15. (a) What is known as Binary phase shift keying? Discuss in detail the BPSK transmitter and receiver and also obtain the minimum double sided Nyquist bandwidth. (16 Marks)(Nov/Dec 01)(April/May 010)(April/May 011) Phase Shift Keying Definition: (1 Mark) If the phase of the carrier (0) is varied proportional to the information signal, phase shift keying (PSK) is produced. M ary Encoding Definition: (1 Mark) M simply represents a digit that corresponds to the number of conditions, levels, or combinations possible for a given number of binary variables. Binary Phase shift Keying: The simplest form of PSK is binary phase-shift keying (BPSK), where N = 1 and M =. BPSK waveform: ( Marks) BPSK transmitter: (3 Marks)

20 Truth table, Phasor diagram and Constellation diagram: (4 Marks) BPSK Receiver: (3 Marks) Bandwidth: BW = fb (bit rate) ( Marks) (b) With neat schematic diagram, explain the balanced ring modulator of BPSK. (16 marks) The balanced modulator has two inputs: a carrier that is in phase with the reference oscillator and the binary digital data. For the balanced modulator to operate properly, the digital input voltage must be much greater than the peak carrier voltage. This ensures that the digital input controls the on/off state of diodes D1 to D4. If the binary input is a logic 1(positive voltage), diodes D 1 and D are forward biased and on, while diodes D3 and D4 are reverse biased and off (Figure -13b). With the polarities shown, the carrier voltage is developed across transformer T in phase with the carrier

21 voltage across T 1. Consequently, the output signal is in phase with the reference oscillator. If the binary input is a logic 0 (negative voltage), diodes Dl and D are reverse biased and off, while diodes D3 and D4 are forward biased and on (Figure 9-13c). As a result, the carrier voltage is developed across transformer T 180 out of phase with the carrier voltage across T 1. Faculty Incharge HoD/CSE&IT

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