Questions related to Domestic Data Entry Operator under NSDC course.Full description
Big DataFull description
Full description
Full description
viva questions for Advanced communicationFull description
this is for 2 marks and 16 marks question papers for data communication.
its document about analog communication
advanced communication lab viva voceFull description
Viva Voce questions on communication systems
mobile comm. question bank
Light Fidelity Li Fi technology is a novel method of wireless communication network based on the use of visible LED lights between the violet 800 THz and red 400 THz to transmit data wirelessly. Li Fi has much broader spectrum compared to conventiona
Data Communication & Networking MCQ'SFull description
laser
WIRELESS DATA TRANSFER USING VISIBLE LIGHT COMMUNICATIONFull description
D
computer systems
Problems Q: A signal has fundamental frequency of 1000 Hz, What is the period? A: λ = /f = 1/1000 = 0.001 Sec. = 1 ms.
Q: Express the following in the simplest form you can? a) sin(2π ft - π ) + sin(2π ft + π ) b) sin 2π ft + sin(2π ft - π ). A: a.
sin (2π ft – π ) + sin (2π ft + π ) = 2 sin (2π ft +π ) or 2 sin (2π ft – π ) or –2 sin (2 π ft) b. sin (2π ft) + sin (2π ft – π ) = 0. Q: Sound may be modeled as a sinusoidal function. Compare the relative frequency and wavelength of musical note. Use 330 m/s as the speed of sound and the following frequencies for the musical scale. Note C D E F G A B C Frequency 264 297 330 352 396 440 495 528 A:
N = note; F = frequency (Hz); D = frequency difference; W = wavelength (m) Q: Find the period of the function f(t) = (10 cos t)2. A: We have cos 2x = cos x cos x = (1/2)(cos(2x) + cos(0)) = (1/2) (cos(2x) + 1). Then: f(t) = (10 cos t) 2 = 100 cos2t = 50 + 50 cos(2t). The period of cos(2t) is π and therefore the period of f(t) is π . Q: Given an amplifier with effective noise temperature of 10,000 K and 10-MHz bandwidth , what is the thermal noise level in dBW, may we except at output. A: N = 10 log k + 10 log T + 10 log B = –228.6 dBW + 10 log 104 + 10 log 107 = –228.6 + 40 + 70 = –118.6 dBW Q: What is the thermal noise level of a channel with a bandwidth of 10 kHz carrying 1000 watts of power operating at 50 °C. –23 –17 A: N = K T = 1.38 × 10 × (50 + 273) × 10,000 = 4.5 × 10 watts Q: What is the channel capacity for a teleprinter channel with a 300-Hz bandwidth and a signal-to-noise ratio of 3 dB, where the noise is white thermal noise?
A: Using Shannon's equation: C = B log2 (1 + SNR) We have W = 300 Hz, (SNR)dB = 3 Therefore, SNR = 100.3 C = 300 log2 (1 + 100.3) = 300 log2 (2.995) = 474 bps Q: A digital signaling system is required to operate at 9600 bps. a) If the signal element encodes a 4-bit word, what is the minimum required bandwidth of the channel? b) Repeat part (a) for the case of 8-bit. A: Using Nyquist's equation: C = 2B log2M, We have C = 9600 bps a.
log2M = 4, because a signal element encodes a 4-bit word Therefore, C = 9600 = 2B × 4, and B = 1200 Hz b. 9600 = 2B × 8, and B = 600 Hz Q: Given audio bandwidth of a telephone transmission facility, a nominal SNR of 56 dB (400,0000), and a certain distortion, what is the maximum channel capacity of a traditional telephone line. C = B log (1 + SNR ) A: Using Shannon’s formula: C = 3000 log2 (1+400000) = 56 Kbps 2
Q: Given a channel with an intended capacity of 20 Mbps, the bandwidth is 3 MHz. Assuming white thermal noise, what is SNR is required to achieve this capacity? A: C = B log2 (1 + SNR) 20 × 106 = 3 × 106 × log2(1 + SNR) log2(1 + SNR) = 6.67 1 + SNR = 102 SNR = 101 Q: If the received signal level for a particular digital system is -151 dBW and the receiver system effective noise is 1500 K, what is Eb/N0 for a link transmitting 2400 bps. A: (Eb/N0) dB = S dBW – 10 log R – 10 log T – 228.6 dBW (Eb/N0) = –151 dBW – 10 log 2400 – 10 log 1500 + 228.6 dBW = 12 dBW Q: If an amplifier has 30-dB voltage gain, what voltageratio does the gain represent? A: NdB = 30 = 20 log(V2/V1) V2/V1 = 1030/20 = 101.5 = 31.6 Q: An amplifier has an output of 20 W. What is the output in dBW? A: Power (dBW) = 10 log (Power/1W) = 10 log20 = 13 dBW
