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Telephone companies started providing their subscription withanalog services using analog network.
Later on Digital services were took their place. Nowadays carriers are thinkingabout changing their service lines digital too.
Analog services
These services are of two kinds
1-switch analog services
2-Analog leased services
1- Analog services are poor, slow and unreliable.
2- They often encountered problem during a call.
3- The net work it joins is called PSTN.
4- Its bandwidth is usually lies between 0 to 4000Hz.
5- It uses twisted pair cable to connect subscriber’s phone to the network viaexchange.
6- It is cheaper then Digital services.
7- In these services call is conveyed through switched lines or dedicatedlines.
Digital services
These services are largely offered nowadays because these are less sensitive ofnoise than analog.
These services are of three kinds.
1- Digital services provide better sound quality, fast and reliable services.
2- These are less sensitive than analog because signals are digital and noiseis analog so can be separated easily.
3- Provide data rate 56 kbps to 128 Mbps.
4- Use twisted pair cable, coaxial cable and fiber optical.
5- These are expensive.








Let us consider a signal with an initial intensity of I, that travelsthrough a medium and the final intensity is F. I have considered anelectromagnetic signal moving through an appropriate medium.

The attenuation is expressed in terms of decibels and given by the formula:

Attenuation = 10* log (I / F)

In the question the signal travelling through the transmission medium loses3/4 of its power.

If the initial power was I, the final power is I/4.

Substituting this in the formula for attenuation we get:

Attenuation = 10 * log (I / (I/4))

=> 10* log (4I/I)

=> 10* log 4

=> 10*0.6020

=> 6.02

Therefore the attenuation is 6.02 dB.

Dear Sir


Solution of Q2 is wrong. Please consult Lecture 28 in handouts.


Analog Services:

1- 1-There are many analog services but two of them are more important, Switched analog services and leased analogue services.

2- 2-In switched lines the callers used dial pad to dial the number which is mostly used in our homes while in leased lines (dedicated lines) there is no dial pad or dialing, the connectors connects permanently.

3- 3- Conditional lines are also be used to improve the quality of a line by signal distortion or delay distortion. In analog both noise and signal is analog.

4- 4- Analogue hierarchy, which is for maximizing the efficiency, telephone companies multiplex the signals from lower BW lines onto higher BW lines. And for this purpose, many switched or leased lines are combined into bigger channels and FDM is used for analog lines.

5- 5-And now a days Analog and digital both services are widely used in the world.

Digital Signals:

1- 1- Digital services are of three types, Switched/56, DDS and DS. These services are same as analog services but the difference is they are digital are they send the signals in digital format which helps in reduce noise.

2- 2- 56 switched services are like as analogue switched service, but it allows data rate higher than the analog switched service which is 56 kbps. In this service we use DSU for encoding which used by the service provider. And this type of services is costs much then analog because of the DSU.

3- 3-DDS service, it’s same as the analogue leased service and cheaper than 56 services because of no dial pads and its data rate is 64kbps.

4- 4- DS service, which is similar as analogue hierarchy, in analogue we used FDM and in this we used TDM for combining the lines into bigger once.

5- 5- Digital services are less sensitive then analog. In digital services, the noise and signal can easily be separated. So, digital services are best then analog services because of noise reduction and more reliable and fast and more data can be transferred as compared to the analogue services.

This means P2=(3/4)P1

–10log10(P2/ P1)= 10log10(0.75 P1/ P1)
=10(-0.125)= -1.25 dB


so the attenuation (loss of energy) in Desibels is -1.25

do u agree with me?

yup mani m agree with u... i calculated the same... but i dnt understand admin solution.. could sme1 guide me? that which 1 is correct?




b. Differentiate among all Address Resolution Techniques in a tabular format

The mechanism for performing address resolution depends on the protocol and hardware addressing schemes. However, there are three basic categories.


Table lookup

Closed-form computation

Dynamic message exchange


bindings or mappings are stored in a table in memory which is searched when necessary

a computer's hardware address can be computed from the protocol address using basic numeric operations

computers exchange messages across a network to resolve an address.


The table lookup approach requires an array where each element contains a pair (P, H) where P is a protocol address and H is the corresponding hardware address

Closed-form computation is often used in networks which have configurable addresses and values are chosen to optimise the translation from protocol address to hardware address

Dynamic message exchange usually consists of hosts exchanging information by agreeing to answer resolution requests for their addresses.

The chief advantage of the table lookup approach is generality - a protocol address can map to an arbitrary hardware address

Protocol address based on hardware address

Data link layer derives hardware address from protocol address


Use network to resolve IP addresses


Use a simple list containing IP address and hardware address for each host on net.

Computation involves basic Boolean and arithmetic operations

Message exchange with other computer(s) returns hardware address to source

Data link layer looks up protocol address to find hardware address

Data link layer derives hardware address from protocol address


Data link layer sends message requesting hardware address



TCP/IP can use any of the three address resolution methods and uses the Address Resolution Protocol (ARP) to ensure that all hosts use a common format for requests and responses. A request message contains an IP address and requests the corresponding hardware address; a reply message contains both the IP address and the corresponding hardware address

That’s means our P2=(1/4)P1

dB = 10log10(P2/P1)

10log10(P2/P1) = 10log10(0.25 P1/P1)

= 10(-0.15) = -1.5 dB


Is my answer is right?


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