Thursday, 29 December 2016

Computer Networks problems and solutions for UGC NET/GATE exam - Set 4

16.       Calculate the power of a signal with dBm = −30.
Solution
Sometimes the decibel is used to measure signal power in milliwatts. In this case, it is referred to as dBm and is calculated as dBm = 10 log10 Pm , where Pm is the power in milliwatts.
dBm = 10 log10 Pm => -30 = 10 log10 Pm
                            => log10 Pm = -3
                            => Pm = 10-3 mW
17.       The loss in a cable is usually defined in decibels per kilometer (dB/km). If the signal at the beginning of a cable with −0.3 dB/km has a power of 2 mW, what is the power of the signal at 5 km?
Solution
The loss in the cable in decibels is 5 × (−0.3) = −1.5 dB.
We can calculate the power as dB = 10log10P2/P1 = -1.5
log10P2/P1 = -0.15
P2/P1 = 10-0.15 = 0.71
P2 = 0.71P1 = 0.7 x 2 = 1.4 mW
18.       The power of a signal is 10 mW and the power of the noise is 1 μW; what are the values of SNR and SNRdB ?
Solution
The values of SNR and SNRdB can be calculated as follows:
                SNR = 10,000 μW / 1 μW = 10,000
                SNRdB = 10log1010,000 = 10log10104
                            = 10 x 4 = 40
Note:
In communication systems, noise is an error or undesired random disturbance of a useful information signal.
There are different types of noises.
Thermal - random noise of electrons in the wire creates an extra signal
Induced - from motors and appliances, devices act are transmitter antenna and medium as receiving antenna.
Crosstalk - same as above but between two wires.
Impulse - Spikes that result from power lines, lightning, etc.
Signal to Noise Ratio (SNR): To measure the quality of a system the SNR is often used. It is defined as the ratio of signal power to the noise power. It is usually given in dB and referred to as SNRdB.

The values of SNR and SNRdB for a noiseless channel are
SNR = signal power/0 = ∞
SNRdB = 10log10∞ = ∞
We can never achieve this ratio in real life; it is an ideal.
19.       Consider a noiseless channel with a bandwidth of 3000 Hz transmitting a signal with two signal levels. The maximum bit rate can be calculated as
Solution
C = 2 B log22n (C= capacity in bps, B = bandwidth in Hz)
= 2 x 3000 x log22 = 6000 bps
Note:
Nyquist Theorem: Nyquist gives the upper bound for the bit rate of a transmission system by calculating the bit rate directly from the number of bits in a symbol (or signal levels) and the bandwidth of the system (assuming 2 symbols/per cycle and first harmonic).
Nyquist theorem states that for a noiseless channel:
C = 2 B log22n
C = capacity in bps
B = bandwidth in Hz
2n = the number of signal levels
20.       Consider a noiseless channel with a bandwidth of 3000 Hz transmitting a signal with four signal levels (for each level, we send 2 bits). The maximum bit rate can be calculated as
Solution
C = 2 B log22n = 2 x 3000 x log24 = 12000 bps

Wednesday, 28 December 2016

Computer Networks problems and solutions for UGC NET/GATE exam - Set 3

11.       What is the bit rate for high-definition TV (HDTV)?
Solution
HDTV uses digital signals to broadcast high quality video signals.
The HDTV screen is normally a ratio of 16 : 9.
There are 1920 by 1080 pixels per screen, and the screen is renewed 30 times per second.
Twenty-four bits represents one color pixel.
1920 x 1080 x 30 x 24 = 1,492,992,000 bps = 1.5 Gbps
Note:
The TV stations reduce this rate to 20 to 40 Mbps through compression.
12.       What is the required bandwidth of a low-pass channel if we need to send 1 Mbps by using baseband transmission?
Solution
The answer depends on the accuracy desired.
a. The minimum bandwidth, is B = bit rate /2, or 500 kHz.
 
b. A better solution is to use the first and the third
    harmonics with  B = 3 × 500 kHz = 1.5 MHz.

c. Still a better solution is to use the first, third, and fifth
    harmonics with B = 5 × 500 kHz = 2.5 MHz.
Note:
A digital signal is a composite analog signal with an infinite bandwidth.

Baseband transmission of a digital signal that preserves the shape of the digital signal is possible only if we have a low-pass channel with an infinite or very wide bandwidth.

If the available channel is a bandpass channel, we cannot send the digital signal directly to the channel; we need to convert the digital signal to an analog signal before transmission.

In baseband transmission, the required bandwidth is proportional to the bit rate;
if we need to send bits faster, we need more bandwidth.

Bandwidth requirements
13.       We have a low-pass channel with bandwidth 100 kHz. What is the maximum bit rate of this channel?
Solution
The maximum bit rate can be achieved if we use the first harmonic. The bit rate is 2 times the available bandwidth, or 200 kbps.
14.       Suppose a signal travels through a transmission medium and its power is reduced to one-half. This means that P2 is (1/2)P1. In this case, the attenuation (loss of power) can be calculated as
Solution
10log10P2/P1 = 10log10 0.5P1/P1 = 10log10 0.5
                                        = 10(-0.3) = -3 dB
A loss of 3 dB (–3 dB) is equivalent to losing one-half the power.
Note:
Attenuation means loss of energy -> weaker signal
When a signal travels through a medium it loses energy overcoming the resistance of the medium.
Amplifiers are used to compensate for this loss of energy by amplifying the signal.
To show the loss or gain of energy the unit “decibel” is used.
dB = 10log10P2/P1
P1 - input signal
P2 - output signal
15.       A signal travels through an amplifier, and its power is increased 10 times. This means that P2 = 10P1 . In this case, the amplification (gain of power) can be calculated as
Solution
10log10P2/P1 = 10log10 10P1/P1
                            = 10log10 10 = 10(1) = 10 dB

Tuesday, 27 December 2016

Computer Networks problems and solutions for UGC NET/GATE exam - Set 2

6.       A nonperiodic composite signal has a bandwidth of 200 kHz, with a middle frequency of 140 kHz and peak amplitude of 20 V. The two extreme frequencies have an amplitude of 0. Draw the frequency domain of the signal.
Solution
The lowest frequency must be at 40 kHz and the highest at 240 kHz. Figure below shows the frequency domain and the bandwidth.
7.       A digital signal has eight levels. How many bits are needed per level?
Solution
We calculate the number of bits from the formula,
Number of bits per level = log28 = 3
Each signal level is represented by 3 bits.
Note:
Two digital signals: one with two signal levels and the other with four signal levels.
8.       A digital signal has nine levels. How many bits are needed per level?
Solution
We calculate the number of bits by using the formula.
Number of bits per level = log29 = 3.17
Each signal level is represented by 3.17 bits. However, this answer is not realistic. The number of bits sent per level needs to be an integer as well as a power of 2. For this example, 4 bits can represent one level.

9.       Assume we need to download text documents at the rate of 100 pages per sec. What is the required bit rate of the channel?
Solution
A page is an average of 24 lines with 80 characters in each line. If we assume that one character requires 8 bits (ascii), the bit rate is
100 x 24 x 80 x 8 = 1,536,000 bps = 1.536 Mbps
10.       A digitized voice channel, is made by digitizing a 4-kHz bandwidth analog voice signal. We need to sample the signal at twice the highest frequency (two samples per hertz). We assume that each sample requires 8 bits. What is the required bit rate?
Solution
The bit rate can be calculated as
2 x 4000 x 8 = 64,000 bps = 64 kbps

Saturday, 24 December 2016

Computer Networks problems and solutions for UGC NET/GATE exam - Set 1

1.       The power we use at home has a frequency of 60 Hz. The period of this sine wave can be determined as follows:
Solution
T = 1/f = 1/60 = 0.0166 s
= 0.0166 x 103 ms = 16.6 ms
2.       The period of a signal is 100 ms. What is its frequency in kilohertz?
Solution
First we change 100 ms to seconds, and then we calculate the frequency from the period (1 Hz = 10−3 kHz).
100 ms = 100 x 10-3 s = 10-1 s
f = 1/T = 1/10-1 Hz = 10 Hz
= 10 x 10-3 kHz = 10-2 kHz
Note:
Frequency is the rate of change with respect to time.
Change in a short span of time means high frequency.
Change over a long span of time means low frequency.
3.       A sine wave is offset 1/6 cycle with respect to time 0. What is its phase in degrees and radians?
Solution
We know that 1 complete cycle is 360°. Therefore, 1/6 cycle is
1/6 x 360 = 60o
= 60 x 2π/360 rad = π/3 rad = 1.046 rad
Note:
Phase describes the position of the waveform relative to time 0.
Three sine waves with the same amplitude and frequency, but different phases
4.       If a periodic signal is decomposed into five sine waves with frequencies of 100, 300, 500, 700, and 900 Hz, what is its bandwidth? Draw the spectrum, assuming all components have a maximum amplitude of 10 V.
Solution
Let fh be the highest frequency, fl the lowest frequency, and B the bandwidth. Then
B = fh - fl = 900 – 100 = 800 Hz
The spectrum has only five spikes, at 100, 300, 500, 700, and 900 Hz (see Figure).
Note:
The bandwidth of a composite signal is the difference between the highest and the lowest frequencies contained in that signal.
5.       A periodic signal has a bandwidth of 20 Hz. The highest frequency is 60 Hz. What is the lowest frequency? Draw the spectrum if the signal contains all frequencies of the same amplitude.
Solution
Let fh be the highest frequency, fl the lowest frequency, and B the bandwidth. Then
B = fh - fl => 20 = 60 - fl => fl = 60 – 20 = 40 Hz

The spectrum contains all integer frequencies. We show this by a series of spikes (see Figure).

Thursday, 15 December 2016

Computer Networks Multiple Choice Questions - Set 30

1.       Which of the following is the Layers of TCP/IP model?
(A) Physical, Network, Transport, Application
(B) Host to Network, Network, Presentation, Application
(C) Host to Network, Internet, Transport, Application
(D) Physical, Internet, Session, Application
Answer: C
2.       If either of the communicating device can ask for disconnection by sending DISCONNECT REQUEST TPDU to the other, and immediately disconnect without waiting for acknowledgement is called ...................
(A) Graceful disconnection               (B) Abrupt disconnection
(C) Greedy disconnection                  (D) Random disconnection
Answer: B
3.       Using POP3, the e-mails are stored at .................
(A) Users’ PC               (B) Router
(C) Gateway                 (D) Server
Answer: A
4.       What is the interface that allow web servers to talk to back end programs and scripts that can accept input and generate HTML pages in response?
(A) interior gateway interface            (B) common gateway interface
(C) exterior gateway interface           (D) None of these
Answer: B
5.       To view the pages on the Internet, they have to be installed on a ..................
(A) local server             (B) proxy server
(C) foreign server        (D) web server
Answer: D

6.       The software that enables the user to interact with the contents present on a web page?
(A) www                        (B) HTTP
(C) Browser      (D) URL
Answer: C
7.       The type of encoding in which manipulation of bit streams without regard to what the bits mean is .....................
(A) Destination encoding       (B) Entropy encoding
(C) Source encoding              (D) Differential encoding
Answer: B
8.       Multiplexing and Demultiplexing of Network connections is done by ...................
(A) Network Layer                    (B) Data Layer
(C) Data Link Layer                 (D) Transport Layer
Answer: D
9.       Two sides cannot attempt the same operation at the same time. This property is accomplished by ...................
(A) Session  Layer                   (B) Transport  Layer
(C) Physical Layer                   (D) Network Layer
Answer: A
10.    Number of layers in the OSI model?
(A) 5       (B) 4
(C) 7       (D) 8
Answer: C

Pages  29   30   31   32   33   34 

Wednesday, 14 December 2016

Computer Networks Multiple Choice Questions - Set 29

1.       The functions of internet Layer in TCP/IP are ....................
(A) Flow Control and Error Control
(B) Congestion Control and Flow Control
(C) Packet Routing and Flow Control
(D) Congestion Control and Packet Routing
Answer: D
2.       The protocols used in Host to network layer of TCP/IP model are ...................
(A) TEL NET and LAN
(B) ARPA NET and SAT NET
(C) PACKET RADIO and IP
(D) LAN and IP
Answer: B
3.       HTTP is the acronym of ...............
(A) Hyper Text Transfer Protocol
(B) Hyper Test Transfer Protocol
(C) Hyper Text Transport Protocol
(D) Hyper Text Transport Program
Answer: A
4.       In the earlier ARPA NET each node of network consisted
(A) TIP & a host            (B) BBN & a host
(C) IMP & a host          (D) SAP & a host
Answer: C
5.       Number of layers in TCP/IP model?
(A) 5       (B) 4
(C) 6       (D) 7
Answer: A

6.       Internet Layer in TCP/IP is ...................
(A) Connection Oriented
(B) Can be Connection Oriented and connection less
(C) Connection less
(D) Client Server type request
Answer: C
7.       The protocol defined by internet layer in TCP/IP is ......................
(A) TCP Protocol          (B) UDP Protocol
(C) SMTP                      (D) IP Protocol
Answer: D
8.       Two protocols defined in Transport Layer of TCP/IP?
(A) TCP and IP             (B) TCP and UDP
(C) UDP and IP            (D) None of these
Answer: B
9.       Which of the following is/are a connection oriented protocol(s)?
(A) TCP                                     (B) UDP
(C) TCP and UDP       (D) Neither TCP nor UDP
Answer: A
10.    UDP has the following properties.
(A) Connection oriented and reliable
(B) Connection Less and reliable
(C) Connection less and Unreliable
(D) Connection Oriented and Unreliable
Answer: C