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I want to know the difference between Bandwidth and propagation speed. How are they associated with each other i.e. the impact on bandwidth if propagation speed varies or the impact on speed if bandwidth varies. I have gone through this difference between propagation speed and bandwidth in digital communication still I am unable to understand it very clearly. Could it be possible to explain with some other example like may be with water pipes etc.

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    $\begingroup$ The very first sentence of the answer you reference is "They are unrelated." Varying one has no effect on the other. You will need to say what you find unclear or what you don't understand otherwise another answer may be just as unclear to you. $\endgroup$ – Derek Elkins May 18 '17 at 23:33
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These two terms are actually completely unrelated. Bandwidth is a way of expressing how much "stuff" you can send on a network link at a time. Think of it like a highway. If you have only one lane on the highway, then you have a bandwidth of 1 car per some unit of time (let's say seconds). So every second, only 1 car can pass through that particular segment of roadway. But if you widen the road to have 4 lanes, then you can get 4 cars per second. It's literally the width of the frequency band you can send data over.

Now propagation speed is how quickly something can propagate, or move a certain distance. Consider the cars again. If they are driving 60 mph, then then it will take one hour to travel 60 miles. Regardless of how many lanes are on the highway, each car will still take one hour to travel that whole distance. Cars in the front and the back of each lane will take one hour. By the way, the reverse is true as well: increasing the speed of each car has no affect on how many you can fit side by side on any stretch of road.

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Think of a bucket chain.

  • The bandwidth is the number of buckets you pass along every minute (or second, or ...).
  • The propagation speed is how long it takes to pass a bucket along the chain,
    • which is ${(number \; of \; buckets \; in \; the \; chain) \over bandwidth}$.
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