(A) Radio transmission
(B) Telegraphy
(C) Telephone
(D) None of the above
Ans:-B
Explanation:- Frequency shift keying is a method of transmitting digital signals.FSK was mostly used in telegraphy. The two binary states 0 and 1 are represented by a specific analog waveform in FSK. Nowadays, a modem converts the binary data into FSK for transmitting across telephone lines and does the reverse at the receiving end. There is one more method of transmitting digitals signals where the phase of a transmitted signal is varied to convey information. This is called Phase Shift Keying(PSK).
27. The baud rate is
(A) always equal to the bit transfer rate
(B) equal to twice the bandwidth of an ideal channel
(C) not equal to the signalling rate
(D) equal to half of the bandwidth of an ideal channel
Ans:-B
Explanation:-
The baud rate of a data communication system is the number of symbols per second transferred. It is equal to twice the bandwidth of an ideal channel.28.How much bandwidth is there in 1 micron of spectrum at a wavelength of 1 micron ?
(A) 300 MHz
(B) 3 GHz
(C) 300 THz
(D) 30 KHz
Ans:-D
29. Which of the following file transfer protocols use TCP and establishes two virtual circuits between the local and remote server ?
(A) FTP
(B) TFTP
(C) TELNET
(D) NFS
Ans:-A
Explanation:- There could be a ambiguity between the options A and B for this one. C and D are ruled out. So, what is the difference between FTP and TFTP. The purpose of both the protocols is to obtain files from a remote host. But TFTP stands for trivial file transfer protocol. It is not very reliable or secure. It uses the packet delivery service offered by UDP. On the other hand, FTP stands for File Transfer Protocol. It is a mechanism provided by TCP/IP. It uses the services offered by TCP. It is reliable and secure. It establishes two connections(virtual circuits)between the hosts, one for data transfer and another for control information. So the answer for this question is option A.
30. The threshold effect in demodulator is
(A) exhibited by all demodulator, when the input signal to noise ratio is low.
(B) the rapid fall on output signal to noise ratio when the input signal to noise ratio fall below a particular value.
(C) the property exhibited by all A.M. suppressed carrier coherent demodulator.
(D) the property exhibited by correlation receiver.
Ans:-B
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ReplyDelete28.How much bandwidth is there in 1 micron of spectrum at a wavelength of 1 micron ?
ReplyDelete(A) 300 MHz
(B) 3 GHz
(C) 300 THz
(D) 30 KHz
Solution:
delta_f = c * delta_lambda / (lambda*lambda) where delta_f is the bandwidth range, c is the light speed, delta_lamba is the wavelength range, and lambda is a certain wavelength. Therefore,
delta_f = 3*10e8 * 0.1*10e-6 / (10e-6 * 10e-6) = 3*10e13Hz = 30THz
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DeleteIt's 300 THz, as sprectum (given) is 1 micron, not 0.1 micron
DeleteEkta, the solution you mentioned is for 0.1 micron of spectrum, not for 1 micron. Check the link again:
ReplyDeletehttp://wiki.answers.com/Q/How_much_bandwidth_is_there_in_0.1_micron_of_spectrum_at_a_wavelength_of_1_micron
Very small units of length are measured in ‘microns’. One micron is one millionth of a meter, therefore, 1 micron is 0.001 mm.
Q-28 answer--C
ReplyDeletedelta_f = c * delta_lambda / (lambda*lambda) where delta_f is the bandwidth range, c is the light speed, delta_lamba is the wavelength range, and lambda is a certain wavelength. Therefore,
delta_f = 3*10e8 *10e-6 / (10e-6 * 10e-6) = 3*10e14Hz = 300THz