Digital Communication MCQs1) What are the primary features of a transmitter?
Answer: d Explanation: A transmitter refers to an electronic device that produces a radio signal with the help of an antenna. The primary features which make a transmitter complex are higher transmitting power, higher clock speed, directional antennas. 2) The window gives a number of
Answer: c Explanation: The window gives the number of frames or bytes. 3) A speech signal, bandlimited to 8 kHz with peak to peak between +20 V to  20 V, and the signal are sampled at Nyquist rate, and the bits 0 and 1 are transmitted using bipolar pulses. Find the minimum bandwidth for distortionfree transmission in KHz?
Answer: b Explanation: While using polar pulses, the minimum bandwidth required is four times the theoretical bandwidth or Nyquist bandwidth. So the required bandwidth = 4 * Nyquist bandwidth = 4 × 2 FM = 4 × 2 × 8 = 64 KHz. 4) Which of the given device does the data compression?
Answer: d Explanation: Data compression refers to a reduction in the number of bits required to present the data. The compressing data can save memory, increase the file transfer rate, and minimize the cost for storage hardware and network bandwidth. Source encoder is a type of encoder which converts the digital signal or analog signal to a binary digit sequence. 5) According to Federal communications commission regulations, the maximum allowable frequency deviation is 40 kHz, for a TV signal, given that the percentage modulation of the audio portion is 50%. Find the frequency deviation of the audio signal in KHz?
Answer: b Explanation: Given; Maximum allowable frequency deviation ?f_{max} = 40 KHz Percentage modulation of the audio portion = 50% We know that the actual frequency deviation for the TV signal is calculated as 6) Which of the given filter has maximum flatness?
Answer: b Explanation: Butterworth filter refers to a type of filter whose frequency response is flat over the passband region. The low pass filter provides a constant output source from DC up to a cutoff frequency. If the signal is beyond the cutoff frequency, it simply rejects. 7) For an Amplitude modulation signal, the bandwidth is 20 kHz, and the highest frequency component present is 650 kHz. Find the carrier frequency used for this amplitude modulation signal?
Answer: a Explanation: Given; Bandwidth = 20 KHz FC + FM = 650 KHz We know that, BW = 2 FM = 20 KHz So, fm = 10 KHz Therefore, FC = 650 kHz  FM FC = 650  10 = 640 KHz 8) Space loss occurs due to a decrease in
Answer: c Explanation: Space loss occurs mainly due to a decrease in electric field strength. Due to a decrease in the electric field strength, there will be a reduction in the signal strength as a function of distance known as space loss. 9) Analog data with the highest harmonic at 40 kHz generated by a sensor has been digitized using 6 level PCM. Find the rate at which digital signal generated?
Answer: b Explanation: We know that, Nyquist rate = 2 × 40 kHz = 80 KHz 2^{n} ≥ 6 Thus, n = 3 So the bit rate = 80 × 3 = 240 kbps 10) A satellite receiver with a noise figure of 5.6 dB has a bandwidth of 24 kHz and comprises a preamplifier with a noise temperature 150 k and a gain of 40 dB. If the reference temperature is 300 k, find the equivalent noise temperature of the receiver?
Answer: d Explanation: Given; F_{2} = 5.6 dB = 3.63 Teq = 150 T = 300 We know that, = 1.5 + ( 2.63 × 10 ^{3} ) = 1.50 Therefore, the equivalent noise temprature Te = (F  1) T = (1.50  1) 300 = 150 K 11) In frequency modulation broadcast, the maximum deviation is 80 kHz, and the maximum modulating frequency is 20 kHz. In reference to Carson's rule, find the maximum required bandwidth?
Answer: b Explanation: Given; Maximum deviation = 80 KHz Maximum modulating frequency = 20 KHz We know that, According to Carson's rule, BW = 2 (fm + ?f) = 2 (20 + 80) = 200 KHz 12) If noise figure of a receiver is 1.8 at 20 oC, find its equivalent noise temprature?
Answer: c Explanation: Given; Noise figure of a recevier (NF) = 1.8 T_{O} = 273 + 20 = 293 K We know that, Equivalent noise temprature Teq = T_{O}( NF  1) = 293 (1.8  1 ) = 234.4 K 13) If a 120 V carrier peak changes from 170 V to 50 V by a modulating signal, find the modulation factor?
Answer: a Explanation: : Modulation factor also refers in terms of the values referred to the modulated carrier wave and is given as Where, V_{max} ? = Maximum values of the amplitude of the modulated carrier wave. V_{min} ? = Minimum values of the amplitude of the modulated carrier wave. Therefore, 14) RF carrier 15 kV at 1 MHz is amplitude modulated and modulation index is 0.8. Find the peak voltage of the signal?
Answer: d Explanation: Given; Modulation index (m)= 0.8 RF carrier = 15 kV We know that, Therefore the peak value of signal Vm = m × amplitude of carrier wave, Vc = 0.8 × 15 kV = 12 kV 15) A broadcast amplitude modulation radio transmitter radiates 140 kW when the modulation percentage is 75. Find the carrier power?
Answer: c Explanation: Given; Transmitted power P_{total} = 140 kW Modulation percentage = 75 = 0.75 We know that, 16) The power spectral density of a signal is
Answer: d Explanation: We know that the power of a signal is real and nonnegative quantity; therefore, the power spectral density of a signal is real, even, and nonnegative quantity. 17) A 1200 W carrier is amplitude modulated and has a sideband power of 400 W. Find the depth of the modulation?
Answer: b Explanation: Given; Pc = 1200 W Psb = 400 W We know that, power in sidebands is given as Or, m = √0.6 = 0.7746 = 0.775 18) A modulater is a device used to
Answer: d Explanation: Modulation refers to a lowfrequency signal process combined with a highfrequency signal to transmit it over a long distance. For example, the normal human voice signal constitutes low frequency to travel only to a short distance. Hence, to transfer it to a longdistance, usually, we need modulation. In other words, we can say that it is a device used to impress the information present in lowfrequency signals onto a radio frequency carrier. 19) If the two signals modulate the same carrier with different modulation depths of 0.4 and 0.8. Find the resulting modulation signal?
Answer: a Explanation: Given; Two different modulation depths m_{1} = 0.4 and m_{2} = 0.8 We know that, The total modulation index (depth) 20) A balanced modulator is used in the generation of which of the following signal?
Answer: b Explanation: A balanced modulator can be defined as a circuit in which two nonlinear devices are linked in a balanced mode to generate a DSBSC signal. 21) A given AM broadcast station transmits an average carrier power output of 50kW and uses a modulation index of 0.804 for sign wave modulation. Find the maximum amplitude of the output if a 60 ohm resistive load represents the antenna?
Answer: c Explanation: Given; Average carrier power output (Pc) = 50kW Modualtion index (m)= 0.804 Resistive load (R) = 60 ohm We know that, 22) Which of the given modulator is an indirect way of generating FM?
Answer: b Explanation: If we are talking about the direct method of FM generation, the VCO output frequency is controlled by the instantaneous value of the baseband signal. Reactance tube modulator facilitates it. In Armstrong modulation, FM is determined from PM, so it is an indirect method. 23) In a frequency modulation system, the maximum frequency deviation is 2000, and modulating frequency is 2kHz. Find the modulation index ??
Answer: a Explanation: Given; ?f = 2000 fm = 2 kHz = 2000 Hz We know that, 24) The maximum deviation allowed in a frequency modulation system is 120 kHz. If the modulating signal frequency is 20 kHz, find the bandwidth requirement as per Carson's rule?
Answer: c Explanation: Given; ?f = 120 kHz fm = 20 kHz We know that, The bandwidth requirement as per Carson's rule = 2 (1 + δ) fm 25) In FM modulation, when the modulation index increases, the transmitted power?
Answer: d Explanation: Frequency modulation refers to the process of modulating a sinusoidal wave frequency as per the amplitude of the baseband bearing signal. The signal can be digital or analog; in frequency modulation, the FM wave's amplitude is always constant, and transmitted power Pt is equaled toremains unchanged. 26) In a frequency modulated system, when the audio frequency is 600 Hz, and audio frequency voltage is 2.8 V, the frequency deviation ? is 5.6 kHz. If the audio frequency voltage is now increased to 7.4 V, find the new value of deviation?
Answer: b Explanation: Given; Modulation frequency , FM = 600 Hz Audio frequency voltage A_{m}= 2.8 V Frequency deviation ?f = 5.6 kHz We know that, ?f= K_{f}× A_{m} When amplitude of message signal is increased to 7.4 V That is, A'_{m} = 7.4 V So, new frequency deviation, ?f'= K_{f} × A'_{m} = 2 × 7.4 = 14. 8 kHz 27) In phase modulation, phase deviation is proportional to which of the following.
Answer: b Explanation: In phase modulation, Phase deviation varies linearly with the message signal frequency so that phase deviation is proportional to the message signal. 28) A system has a receiver noise resistance of 60 ohms, and it is connected to an antenna with an output resistance of 60 ohms. Find the noise figure of the system?
Answer: b Explanation: Given, Req = 60 ohm Ra = 60 ohm We know that, The noise figure is given as NF = 1 + Req/Ra = 1 + 1/1 = 2 29) Which type of noise reduction by limiters in FM receivers
Answer: c Explanation: Under the ordinary condition, limiters in FM receivers limit the impulse noise to the same extent as the random noise. 30) When a radio receiver is tuned to 600 kHz, its local oscillator provides the mixer with input at 1000 kHz. Find the frequency of other stations?
Answer: a Explanation: Given; Turned frequency, FC = 600 kHz Local oscillator frequency, f = 1000 We know that,2 Local oscillator frequency, f = FC + f_{if} f_{if }= f  _{FC} = 1000  600 = 400 kHz Therefore, the image is given as f_{c }+ 2 f_{if} = 600 + 2 × 400 = 600 + 800 = 1400 kHz 31) What is the automatic frequency control voltage of the FM transmitter VCO?
Answer: b Explanation: VCO stands for the voltagecontrolled oscillator. An oscillator refers to a circuit that producers an AC signal, usually of high frequency. A voltagecontrolled oscillator is an oscillator whose frequency is controlled by voltage. It ranges from 1 MHz with onevolt input to 1 MHz per volt to 10v/10MHz. 32) Without any filtering, a broadcast station at 1.800 kHz is heard together with another station at 2800 kHz on a superheterodyne receiver. Find the value of employed IF?
Answer: c Explanation: Given Image frequency = 2800 kHz Signal frequency, fs = 1800 kHz We know that the IF (Intermediate frequency), 33) If the turnon energy loss in a transformer is 55 MJ and turnoff energy loss is 78 mj. If the mean power loss is limited to 220 W, find the maximum switching rate that can be achieved?
Answer: b Explanation: Given; P = 220 W We know that, Energy loss = Mean power loss, P × time duration 34) Analog data having the highest harmonic at 40 kHz generated by a sensor has been digitized using 6 level PCM; find the generated digital signal rate?
Answer: d Explanation: Given L = 6 FM = 40 kHz Since the number of quantization levels, L = 2^{n} Therefore, L= 6 , 2^{n }= 6 or n = 3 So 3 bits are required to represent each sample, So the rate of a digital signal generated is given as Rb = n × 2 × FM = 3 × 2 × FM = 3 × 2 × 40 = 240 kbps 35) In a delta modulation scheme, the step height is 80 mV, and the step width is 2 ms. Find the maximum slope that the staircase can track?
Answer: c Explanation: Given; Step height = 80 Mv Step width = 2 ms We know that, The maximum slope that the staircase can track in a delta modulation scheme is given as 36) Frequency frogging is used in carrier system to
Answer: b Explanation: Frequency frogging means the interchanging of frequencies. It performs specific tasks such as avoiding oscillations and feedback to reduce crosstalk and correct for a highfrequency response slope in the transmitter line. 37) A TDM link has 25 signal channels, and each channel is sampled at 7 kHz. If each sample is represented by 8 bits and contains an additional bit for synchronization, find the total bit rate for the TDM link?
Answer: c Explanation: Given; Number of channel N = 25 Sampling frequency, fs = 7 kHz Number of bits, n = 8 Additional bit C = 1 (Smaplw contains one additional bit) Bit rate, R_{b} = N × n × fs + Cfs = 25 × 8 × 7 + 1 × 7 = 1407 kbps 38) Output data ratio of a 4 bit PCM  TDM system sampling 16 voice channels, comparing these using µ  law at the rate of 6 kHz and with an Iframe alignment word, is
Answer: b Explanation: Given; n =4 N = 16 a = 1 Fs = 6 × 10^{3} We know that, Each frame alignment consists of a synchronizing bit in addition to data, so the equation is given as As we know, Ts = 1/fs, therefore = n (N + a) × fs = 4 (16 + 1) × 6 × 10^{3} = 4.08 × 10^{3}bit/sec 39) 23 voice channels are sampled uniformly at a rate of 10 kHz and then time division multiplexed. The sampling process uses flattop samples with a two µs duration. The multiplexing operation includes the provision of synchronization by adding an extra pulse of 2 µs duration. Find the spacing between successive pulses of the multiplexed signal?
Answer: b Explanation: Given; Number of voice signals = 23 Synchronization pulses = 2 Total number of pulses in one rotation = 23 + 2 = 25 We knonw that, The time duration for one time frame, So time duration utilized by 25 pulses = 100 µs Therefore, the time utilised by each pulse Since actual duration of each pulse is 2 µs So, the spacing between successive pulse = 4  2 = 2 µs 40) The unmodulated carrier power in an AM transmitter is 8 kW. A sinusoidal modulating signal modulates this carrier. The maximum percentage of modulation is 60 %. If it is reduced to 50 %, find the maximum unmounted carrier power in (kW) that can be used without overloading the transmitter?
Answer: a Explanation: Unmodualted carrier power Pc = 8 kW Maximum modulation index m_{max} = 0.6 We know that, total power is given as 41) For an Amplitude modulation signal, the bandwidth is 10 kHz, and the highest frequency component present is 450 kHz. Find the carrier frequency used for this amplitude modulation signal?
Answer: a Explanation: Given; Bandwidth = 10 KHz fc + fm = 450 KHz We know that, BW = 2 fm = 10 KHz So, fm = 5 KHz Therefore, fc = 50 KHz  fm fc = 450  10 = 440 KHz
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