confusion over bandwidth, bitrate, data rate

Discussion in 'Computing and Networks' started by tscoco, Jun 14, 2012.

1. tscoco Thread Starter New Member

Jun 11, 2012
5
0
hello eveyone.
I want to know how much bandwitdh i need to allocate for indoor area.
so i thought i shoud look at data transfer value for video streaming (youtube, ...) and audio streaming in other to estimate the total minimum bandwidth i should allocate in an indoor environment where video or music can be played...

But i get confused with the notions of bandwidth , bitrate and data rate. I know that the bitrate permits to quantify the quality of an audio or a video. the more bitrate is used, the better is the quality.

For example, is knowing the bitrate for a MP3 could help me define the bandwidth required for an audio MP3 stream?

In other word, whats the relation and the difference between, bitrate, bandwidth and data rate.

thank you in advance for your help, it is driving me nutttt hahaha
coco

2. Papabravo Expert

Feb 24, 2006
11,137
2,175
The problem with your question is that is vague to the point of uselessness. There is no single answer unless you can provide more detail. Audio streams are continuous and usually given as a "sample rate" and a bit length. For example a Red Book CD, aka CDDA is commonly referred to as "16, 44.1" This means that each sample is 16 bits and there are 44,100 samples per second. You also have to know that there are two channels, left and right. If you do the multiplication
Code ( (Unknown Language)):
1.
2. 16 x 2 x 44,100 = 1.4112 MHz.
3.
That is the clock rate for moving the bits from one place to another. As an exercise tell me what the clock rate would be for "32, 192". that is 32 bits, 2 channels and 192,000 samples per second.

Video streams are a whole different matter. Digital downloads are yet another ball of wax because everything is broken up into packets and reassembled at the receiving end.

Try formulating a more precise set of questions e.g. what can I do with gigabit Ethernet, or firewire, or USB 3.0 and you might get some more intuitive feel for how things work.

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3. WBahn Moderator

Mar 31, 2012
20,229
5,755
As Papbravo says, your aren't providing enough information to expect a really good and applicable answer. With that in mind, here are some additional notions that might help.

Let's say that you want to move a certain amount of data, say 200MB of audio data, from point A to point B in 10 seconds.

The "data rate" is the amount of actual data that is transferred per unit time. In this case, you need an average data rate of 20MB/s or, equivalently, 160Mb/s. Hence, it is important not to get sloppy and understand the distinction between the unit 'B' (byte) and 'b' (bit). It matters!

It's also somewhat important to know if M is 1,000*1,000=1,000,000, 1,024*1,000=1,024,000, or 1,024*1,024=1,048,576. In most instances, the difference (~5% worst case) is small enough to get lost in the noise. But be aware of it -- and be aware that it is not always possible to know which one is being used, so use whichever is the worst-case for the present calculation if that's the situation. For simplicity, in the rest of this example we will use M=10^6 and k=10^3.

But your network is almost certainly NOT going to be transferring nothing but bits that are part of the data you are transferring. Depending on the protocol, you may be taking that 200MB and fragmenting them and spreading it out over, say 200,000 packets each of which is 1,500 bits long containing 1,000 bits of actual data and 500 bits of packet header information (sequence, routing, checksum, whatever). So to get your 200MB of data, the network is actually going to have to transfer 300MB of information with an average bit rate of 240Mb/s. On top of this, your are not going to be transmitting those bits nonstop, even if you are the only one on the network. The protocols are probably going to require some level of two way information in the form of acknowledgements and such, plus some packets are going to get corrupted and have to be resent and the transmitter is going to have to use some means of medium access control in order to allow for the possibility that other players are trying to access the medium to transmit their own data. Let's just throw a number at this and say that your system gets to transmit new data 50% of the time. That means that, when it is transmitting, it has to send information at the rate of 480Mb/s. Now, it probably isn't sending bits one at a time. It might be sending them, say, 4 bits at a time by using 16 different symbols. So these symbols only have to be transmitted at 8Msymbols/s. The number of symbols that are transmitted per unit time is the 'baudrate'; 1baud is 1 symbol per second.

The bandwidth is the amount of the spectrum that must be utilized in order to get that 8Mbaud rate and that is a function of both the baud rate and the modulation scheme that is used. Rough rule of thumb is that you need 2Hz/baud.

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4. tscoco Thread Starter New Member

Jun 11, 2012
5
0
well, thank you for all those precisions. I was kind of expecting that my inexperience will be exposed
I might have not well explain my questions. Well, I will work on data transmission achieved by optical wireless communication, visible light communication. Through my experiment I will send an amount of data, low data rate but high data rate as well and i will analyse the quality transmission at the receiver (photodiode). But i wanted to bring life to those data rate.Let´s imagine I am able to transmit data at this rate: 300kbs(the lowest) and 20Mbps(at the highest). I would like to know what concrete application can be done at those speeds. That is why i wanted to know, for example what bandwidth is required to stream, for example 5min youtube video or to stream a MP3 music. I know it is pretty vague , because the video stream quality depends on the bitrate,the resolution, frame, the codec... same with the MP3 music(except for the resolution and the frame)
The general idea is, let´s assume a certain amount of bandwidth(3GB) is allocated for an indoor area, i would like to estimate the bandwitdh for streaming a video or an audio or for downloading an application through my mobile phoe, or for using skype(video call). All of this,to regulate and provide out of the the 3GB bandwidth i have, the right amount of bandwidth for that indoor environment (of course i assume that there is only one user).
Through all that i got confused between, the notion of "bitrates" , "bandwidth" and "data rates",
am i more clear?
Thank y´all again, i learn a lot...
tscoco

5. WBahn Moderator

Mar 31, 2012
20,229
5,755
I think you are still confusing some things.

3GB is not a bandwidth. It is a quantity of data. Similarly, whether the streaming You Tube video is 5min or 50min has little to no bearing on the bandwidth requirements. Bandwidth is related to how much data you can transfer PER SECOND. In a streaming application, you have to have enough bandwidth to keep up with the flow of data, whether it be for 10ms to 10 days.

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6. tscoco Thread Starter New Member

Jun 11, 2012
5
0
Hiii,
I see is see , it is like it is a huge puzzle in my head with a lot of information but not well put toghether...
Well WBahn, do you mean that the size of the video doesnt really matter for the ´bandwidth? i think it does.For example, lest consider with 2 HD youtube video with the same resolution , with the same codec but with different duration,one is 5minutes the other one is 20 minutes.
do you think they will have:
- the same size? (I dont think so...)
-The same bit rate? ( I dont think so...) The bit rate i am referring to is the bit rate necessary to encode a HD video , lets say 3Mbps...
But how can i evaluate the bandwidth i need to be able to keep up the flow of data for a streaming apllication for a given duration (let say 1 day for example)?

I also understood that for digital transfers , the data are fragmented into packets and reassembled at the receiving end. But is there a relation between those packets and the bit rate?

what is then the link between the bit rate necessary to encode my HD video and the bandwidth i will need to keep up a good transmission of the video? I understand it is not the same thing , the bandwidth can be adjusted some how depending on the apllication , but what is the bit rate then? It is not a constant value right? I also read about the dynamic data rate, can you explain it to me plizzz

sorry for asking so many question, i just really want to undertsand it. If you have some papers or documents to recommend me, i will really appreciate it,
thank you,
tscoco

Last edited: Jun 18, 2012
7. tscoco Thread Starter New Member

Jun 11, 2012
5
0
hiiiii Papabravo,
sorry i missed ur exercise, well,the clock rate "32, 192" would be 12,228 Mhz

I attached to this post a capture. Well the frequency he´s referring to, it the clock rate you are talking about?
And what about the "6 ch"? Does it mean that the transmission of the video is operated into 6 channels?

sorry if my question are dump

thank you,
tscoco

Last edited: Jun 18, 2012
8. WBahn Moderator

Mar 31, 2012
20,229
5,755
Perhaps a simple toy example will shed some light on things. As before, to keep things simple, we will use 1KB = 1000 bytes and 1MB = 1000KB.

Let's say that a particular video format requires 1KB per frame and is streamed at 10 frames per second. Now let's say that you want to watch two video clips, one after the other, by streaming them. The first if 1 minute long and the second is 1 hour long. The total amount of data, X, in the first clip will be:

$
X = \left( 1 \ \frac{min}{clip} \right) \left( 60 \ \frac{s}{min} \right) \left( 10 \ \frac{frames}{sec} \right) \left( 1 \ \frac{KB}{frame} \right) \ X = 600KB/clip
$

Similarly, there will be

$
Y = 36MB/clip
[\tex]

in the second clip since it is 60 times as large.

However, because you are streaming them, you only have 60 seconds to transfer the data for the first clip while you have 3600 seconds to transfer the data for the second clip. So the rate at which data must be transfered for each clip is:

R_X = \left( 600 \ \frac{KB}{clip} \right) \left( \frac{1 clip}{60s} \right) = 10 \frac{KB}{s}
\
R_Y = \left( 36 \ \frac{MB}{clip} \right) \left( \frac{1 clip}{1hr} \right) \left( \frac{1000KB}{MB} \right) \left( \frac{1hr}{60min) \right) \left( \frac{1min}{60s) \right) = 10 \frac{KB}{s}
" alt="
R_X = \left( 600 \ \frac{KB}{clip} \right) \left( \frac{1 clip}{60s} \right) = 10 \frac{KB}{s}
\
R_Y = \left( 36 \ \frac{MB}{clip} \right) \left( \frac{1 clip}{1hr} \right) \left( \frac{1000KB}{MB} \right) \left( \frac{1hr}{60min) \right) \left( \frac{1min}{60s) \right) = 10 \frac{KB}{s}
" />

While the size of the clip affect the total amount of data that has to be transferred, it does not affect the rate because, in any give amount of time, you only need to transfer the data needed to play that the data associated with that same amount of time's worth of the recording.

The data rate we are talking about here is also known as the throughput, or how much application data can be put through the link every second. So you want to figure out what the total throughput needs are for all of the various applications that will be using the link and then multiply by some safety factor (such as 2, but the more the better) in order to keep some breathing space and to allow the system to catch up if it gets behind without impacting the playback. Then you want to find a wireless link that can support that total throughput. Don't just look at the bps number on the side of the box. That's the raw bitrate and includes all of the link and network overhead. You want to find out what is the reasonable throughput for application data using that link. It will depend on the protocols used, but many protocols have roughly similar overheads and if you've given yourself enough breathing space the differences won't matter." alt="
Y = 36MB/clip
[\tex]

in the second clip since it is 60 times as large.

However, because you are streaming them, you only have 60 seconds to transfer the data for the first clip while you have 3600 seconds to transfer the data for the second clip. So the rate at which data must be transfered for each clip is:

$
R_X = \left( 600 \ \frac{KB}{clip} \right) \left( \frac{1 clip}{60s} \right) = 10 \frac{KB}{s}
\
R_Y = \left( 36 \ \frac{MB}{clip} \right) \left( \frac{1 clip}{1hr} \right) \left( \frac{1000KB}{MB} \right) \left( \frac{1hr}{60min) \right) \left( \frac{1min}{60s) \right) = 10 \frac{KB}{s}
$

While the size of the clip affect the total amount of data that has to be transferred, it does not affect the rate because, in any give amount of time, you only need to transfer the data needed to play that the data associated with that same amount of time's worth of the recording.

The data rate we are talking about here is also known as the throughput, or how much application data can be put through the link every second. So you want to figure out what the total throughput needs are for all of the various applications that will be using the link and then multiply by some safety factor (such as 2, but the more the better) in order to keep some breathing space and to allow the system to catch up if it gets behind without impacting the playback. Then you want to find a wireless link that can support that total throughput. Don't just look at the bps number on the side of the box. That's the raw bitrate and includes all of the link and network overhead. You want to find out what is the reasonable throughput for application data using that link. It will depend on the protocols used, but many protocols have roughly similar overheads and if you've given yourself enough breathing space the differences won't matter." />

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9. tscoco Thread Starter New Member

Jun 11, 2012
5
0
$

Hallo,

Thank U VErrrrrrry much, i understand much more now....$