Description : 12 voice channels are sampled at 8000 sampling rate and encoded into 8 bit PCM word. Determine the rate of the data stream. a. 354 kbps b. 750 kbps c. 768 kbps d. 640 kbps
Last Answer : c. 768 kbps
Description : How many voice band channels are there in a superjumbo group? A. 600 B. 900 C. 10,800 D. 3,600
Last Answer : C. 10,800
Description : ECE Board Exam March 1996 The number of voice band channels in master group per CCITT standard is __________. A. 300 B. 480 C. 600 D. 12
Last Answer : A. 300
Description : What carrier system multiplexes 96 voice band channels into a single 6.312 Mbps data signal? A. T1 carrier system B. T2 carrier system C. T1C carrier system D. T3 carrier system
Last Answer : B. T2 carrier system
Description : How many telegraph channels are there in the basic voice band channels? A. 24 B. 96 C. 48 D. 672
Last Answer : A. 24
Description : An FDM hierarchy which is formed by frequency – division multiplexing 10 super groups together for a combined capacity of 600 voice band message channels.
Last Answer : Mastergroup
Description : Which standard is utilized in intermodulation noise rates on PCM audio channels? A. CCIT Rec. G.151 B. CCIT Rec. G.172 C. CCIT Rec. G.190 D. CCIT Rec. G.190
Last Answer : B. CCIT Rec. G.172
Description : Used for transmission of PCM encoded time – division multiplexed digital signal.
Last Answer : T carriers
Description : The ratio of the monthly salaries of A and B is in the ratio 15 : 16 and that of B and C is in the ratio 17 : 18. Find the monthly income of C if the total of their monthly salary is Rs 1,87,450. A) Rs 66,240 B) Rs 72,100 C) Rs 62,200 D) Rs 65,800 E) Rs 60,300
Last Answer : Answer: A A/B = 15/16 and B/C = 17*18 So A : B : C = 15*17 : 16*17 : 16*18 = 255 : 272 : 288 So C’s salary = [288/(255+272+288)] * 1,87,450= Rs 66,240
Description : One of the reasons why FDM is being replaced by TDM is A. noise is amplified with voice when an FDM system is used B. it is difficult to place channels side by side C. there is more time than frequency D. Most available frequencies has been used
Last Answer : A. noise is amplified with voice when an FDM system is used
Description : Amount of Topic 10% for 4000 m2 if rate of clodinafop is 60 g/ha. a). 600 g b). 480 g c). 300 g d). 240 g
Last Answer : d). 240 g
Description : ____________ is the out-of-band signaling between toll central offices (Bell System Standard) A. 2,000 Hz B. 800 Hz C. 3,835 Hz D. 3700 Hz
Last Answer : D. 3700 Hz
Description : What is the phase delay of an 800 Hz voice signal if the phase shift is 15 degrees? A. 1.25 µsec B. 52 µsec C. 83.33 µsec D. 26 µsec
Last Answer : B. 52 µsec
Description : How many transponder channels are realized in the commercial C-band without frequency re-use? A. 24 B. 12 C. 36 D. 48
Last Answer : B. 12
Description : What is the quardband between transponder channels in the commercial C-band for satellite communications? A. 6 MHz B. 36 MHz C. 4 MHz D. 2 MHz
Last Answer : C. 4 MHz
Description : TV channels 2, 4 and 5 are known as ________. A. Mid band UHF B. Low band UHF C. High band VHF D. Low band VHF
Last Answer : D. Low band VHF
Description : Which of the following filters block FM radio band for TV channels (2 to 13)? A. High-pass filter B. Band reject filter C. Low-pass filter D. Band-pass filter
Last Answer : B. Band reject filter
Description : Minimum excretory urinary volume for waste products elimination during 24 hrs is (A) 200–300 ml (B) 200–400 ml (C) 500–600 ml (D) 800 ml
Last Answer : Answer : C
Description : Number of meiotic divisions required to produce 200/400 seeds of pea would be (a) 200/400 (b) 400/800 (c) 300/600 (d) 250/500
Last Answer : (d) 250/500.
Description : A machine has an initial value of Rs. 5000, service life of 5 years and final salvage value of Rs. 1000. The annual depreciation cost by straight line method is Rs. (A) 300 (B) 600 (C) 800 (D) 1000
Last Answer : (C) 800
Description : Regeneration of molecular sieve requires it to be heated to a temperature of about __________ °C. (A) 80-120 (B) 200-300 (C) 600-800 (D) 1000-1100
Last Answer : (D) 1000-1100
Description : The round-trip propagation delay between two earth stations through a geosynchronous satellite is A. 500 to 600 ms B. 300 to 400 ms C. 600 to 700 ms D. 400 to 500 ms
Last Answer : A. 500 to 600 ms
Description : What is the minimum propagation delay of a geostationary satellite? A. 278 ms B. 239 ms C. 300 ms D. 600 ms
Last Answer : B. 239 ms
Description : What is the input impedance of a transmission line if its characteristic impedance is 300 Ω and the load is 600 Ω? Assume a quarter wavelength section only. A. 150 kΩ B. 150 mΩ C. 150 Ω D. 2 Ω
Last Answer : C. 150 Ω
Description : What is the baseband frequency of standard FDM basic supergroup? A. 312 to 552 kHz B. 300 to 600 kHz C. 60 to 2540 kHz D. 60 to 108 kHz
Last Answer : A. 312 to 552 kHz
Description : ____ system transmits frequency-division-multiplexed voice band signals over a coaxial cable for distances up to 4000 miles. A. T carrier B. X carrier C. L carrier D. F carrier
Last Answer : C. L carrier
Description : A Master group consists of (A) 12 voice channels. (B) 24 voice channels (C) 60 voice channels. (D) 300 voice channels
Last Answer : (D) 300 voice channels
Description : __________ carries digitally encoded user data. a) Traffic channels b) Control channels c) Signalling channels d) Forward channels
Last Answer : a) Traffic channels
Description : Which of the following types of channels moves data relatively slowly? a. wide band channel b. voice band challen c. narrow band channel
Last Answer : narrow band channel
Description : What are the datatransmission channels availablefor carrying datafrom one location to another? A. Narrowband B. Voice band C. Broadband D. All of theabove E. None of the above
Last Answer : All of theabove
Description : Which of the following types of channels moves data relatively slowly? A. widebandchannel B. voice band channel C. narrowband channel D. broadbandchannel E. None of the above
Last Answer : narrowband channel
Description : The basic voice band has how many octaves? A. 10 B. 4 C. 5 D. 3
Last Answer : B. 4
Description : Which causes a quantization noise in PCM system? A. Serial transmission errors B. The approximation of the quantized signal C. The synchronization between encoder and decoder D. Binary coding techniques
Last Answer : B. The approximation of the quantized signal
Last Answer : The approximation of the quantized signal
Description : What is the number of levels required in a PCM system with S/N ratio of 40 dB? A. 64 B. 128 C. 256 D. 512
Last Answer : B. 128
Description : ECE Board Exam March 1996 According to the Nyquist theorem, the sampling rate that can be used in a PCM system is ________ the highest audio frequency. A. once B. eight times C. twice D. thrice
Last Answer : C. twice
Description : Most satellites operate in which frequency band? A. 30 to 300 MHz B. 300 MHz to 3 GHz C. 3 GHz to 30 GHz D. Above 300 GHz
Last Answer : C. 3 GHz to 30 GHz
Description : The band of frequencies least susceptible to atmospheric noise and interference is: A. 30 – 300 kHz B. 300 – 3000 kHz C. 3 – 30 MHz D. 300 – 3000 MHz
Last Answer : D. 300 – 3000 MHz
Description : The extremely high frequency (EHF) band is in the radio spectrum range of _______. A. 30 to 300 kHz B. 30 to 300 GHz C. 3 to 30 MHz D. 3 to 30 GHz
Last Answer : B. 30 to 300 GHz
Description : Quantizing noise occurs in A. PCM B. PLM C. PDM D. PAM
Last Answer : A. PCM
Description : What is the minimum number of bits required in a PCM code for a range of 10,000? A. 12 B. 9 C. 14 D. 10
Last Answer : C. 14
Description : What is the quantization signal-to-noise ratio if an 8-bit PCM code is used? A. 40 dB B. 50 dB C. 60 dB D. 65 dB
Last Answer : B. 50 dB
Description : ________ is the method of encoding audio signals used in US standard 1544 kbit/s 24 channel PCM systems. A. Shannon’s law B. A-law C. Newton’s law D. Mu-law
Last Answer : D. Mu-law
Description : Steps to follow to produce PCM signal. A. Quantizing, sampling, and coding B. Sampling, quantizing, and coding C. Sampling, coding and quantizing D. Coding, quantizing and coding
Last Answer : B. Sampling, quantizing, and coding
Description : What is the reason why companding is employed in PCM systems? A. To overcome impulse noise in PCM receivers B. To allow amplitude limiting in the receivers C. To solve quantizing noise problem D. To protect small signals in PCM from quantizing distortion
Last Answer : D. To protect small signals in PCM from quantizing distortion
Description : The process of assigning PCM codes to absolute magnitudes A. Overloading B. All of these C. Quantizing D. Multiplexing
Last Answer : C. Quantizing
Description : ECE Board Exam March 1996 Which of the following pulse modulation systems is analog? A. Delta B. Differential PCM C. PWM D. PCM
Last Answer : C. PWM
Description : The most common method used for sampling voice signals in PCM systems.
Last Answer : flat top sampling
Description : An FDM hierarchy which is formed by frequency – division multiplexing five groups containing 12 channels each for a combined bandwidth of 240 kHz
Last Answer : Supergroup
Description : Daily requirement of phosphorous for an infant is (A) 240–400 mg (B) 1.2 gms (C) 800 mg (D) 800–1200 mg
Last Answer : Answer : A