I have already answered these Computer Networking questions. I just wanted to co
ID: 668000 • Letter: I
Question
I have already answered these Computer Networking questions. I just wanted to confirm that they are correct. Please help
Explanation / Answer
10.
ADSL(Asymmetric digital subscriber line):
ADSL is broad band communication technology. ADSL technology allows to access internet (speed data transmission) over the existing copper line.
ADSL is the user(subscriber) favour technology. It gives high downstream data rate to the user and low upstream data rate.
It supports 1.5-9Mbps upstream
It supports 16-640Kbps downstream.
Thus, the ADSL can be characterized by : High speed to subscriber, low speed from the subscriber.
Therefore the correct option is (c).
11.
Although satellites provide high band width, there are some disadvantages with satellite communication. The two main service oriented disadvantages are
Propagation delay: There is a big time delay between two talks or two data transmissions.
Noise & Interference: Causes errors.
The main characteristic required for voice transmission applications is low propagation delay. For example, the live conversations required low delay.
Thus, due to propagation delays between sending and receiving equipment, the satellite communication is not suitable for voice transmission.
Therefore, the correct option is (a)
12.
VSAT (Very Small Aperture Terminal) :
VSAT is a small dish or antenna that sends and receives signals from satellite.
Thus, VSAT provides the connectivity to a satellite.
Therefore, the correct option is (a)
13.
XDSL & ADSL- Data transmission over telephone lines
LAP-D – ISDN protocol
HDLC- used for satellite communication (also used for OSI seven layer model supported networks)
Thus HDLC or an HDLC sub set is sued in satellite communication.
Therefore the correct option is (a)
14.
T1 line is a data transmission line. It has 24 channels. Each channel has a bandwidth of 64 Kbps.
Since T1 has 24 64Kbps channels, it is very costly. Another option for the users is opting for fraction T1. Thus, user can use only few channels of T1.
Thus, fraction T1 has multiples of 64Kbps channels.
Therefore, the correct option is (a)
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