barcode macro excel free DATA AND SIGNALS in Software

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Example 342
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The bandwidth of a subscriber line is 4 kHz for voice or data The bandwidth of this line for data transmission can be up to 56,000 bps using a sophisticated modem to change the digital signal to analog
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Example 343
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If the telephone company improves the quality of the line and increases the bandwidth to 8 kHz, we can send 112,000 bps by using the same technology as mentioned in Example 342
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DATA AND SIGNALS
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Throughput
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The throughput is a measure of how fast we can actually send data through a network Although, at first glance, bandwidth in bits per second and throughput seem the same, they are different A link may have a bandwidth of B bps, but we can only send T bps through this link with T always less than B In other words, the bandwidth is a potential measurement of a link; the throughput is an actual measurement of how fast we can send data For example, we may have a link with a bandwidth of 1 Mbps, but the devices connected to the end of the link may handle only 200 kbps This means that we cannot send more than 200 kbps through this link Imagine a highway designed to transmit 1000 cars per minute from one point to another However, if there is congestion on the road, this figure may be reduced to 100 cars per minute The bandwidth is 1000 cars per minute; the throughput is 100 cars per minute
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Example 344
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A network with bandwidth of 10 Mbps can pass only an average of 12,000 frames per minute with each frame carrying an average of 10,000 bits What is the throughput of this network
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Solution
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We can calculate the throughput as Throughput = 12,000 x 10,000
=2 Mbps
The throughput is almost one-fifth of the bandwidth in this case
Latency (Delay)
The latency or delay defines how long it takes for an entire message to completely arrive at the destination from the time the first bit is sent out from the source We can say that latency is made of four components: propagation time, transmission time, queuing time and processing delay
Latency = propagation time + transmission time + queuing time + processing delay
Propagation Time
Propagation time measures the time required for a bit to travel from the source to the destination The propagation time is calculated by dividing the distance by the propagation speed
Propagation time = Dist:mce Propagation speed
The propagation speed of electromagnetic signals depends on the medium and on the frequency of the signaL For example, in a vacuum, light is propagated with a speed of 3 x 108 mfs It is lower in air; it is much lower in cable
SECTION 36
PERFORMANCE
Example 345
What is the propagation time if the distance between the two points is 12,000 km Assume the propagation speed to be 24 x 10 8 mls in cable
Solution
We can calculate the propagation time as 12000 x 1000 Propagation tIme = ' 8 24 x 10
=50 ms
The example shows that a bit can go over the Atlantic Ocean in only 50 ms if there is a direct cable between the source and the destination
Tra/lsmissio/l Time
In data communications we don't send just 1 bit, we send a message The first bit may take a time equal to the propagation time to reach its destination; the last bit also may take the same amount of time However, there is a time between the first bit leaving the sender and the last bit arriving at the receiver The first bit leaves earlier and arrives earlier; the last bit leaves later and arrives later The time required for transmission of a message depends on the size of the message and the bandwidth of the channel
Transmission time = Message size Bandwidth
Example 346
What are the propagation time and the transmission time for a 25-kbyte message (an e-mail) if the bandwidth of the network is 1 Gbps Assume that the distance between the sender and the receiver is 12,000 km and that light travels at 24 x 108 mls
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