How are Throughput, Bandwidth, and Data Rate Related?
ฝัง
- เผยแพร่เมื่อ 25 ส.ค. 2020
- Uses IEEE802.11a as an example to explain the relationship between Throughput, Bandwidth and Data Rate.
Related videos: (see iaincollings.com)
• How is Data Sent? An Overview of Digital Communications • How is Data Sent? An O...
• How is Data Received? An Overview of Digital Communications • How is Data Received? ...
• What is a Cyclic Prefix in OFDM? • What is a Cyclic Prefi...
• How are Data Rate and Bandwidth Related? • How are Data Rate and ...
• How are Bit Error Rate (BER) and Symbol Error Rate (SER) Related? • How are Bit Error Rate...
Full Categorised list of videos and PDF Summary Sheets: iaincollings.com
.
![[MPD직캠] 베이비몬스터 치키타 직캠 4K 'LIKE THAT' (BABYMONSTER CHIQUITA FanCam) | @MCOUNTDOWN_2024.6.13](http://i.ytimg.com/vi/h99J4vWm7Ew/mqdefault.jpg)
I as a Telecommunication Engineer like your videos very much.
The way you put the information together is time saving for a lot of viewers
Thank you
Thanks for your comment. It's great to hear from people who like the videos. I'm glad you found it useful.
Short, concise yet retains all of the info needed!
Your channel deserves more attention!!
Best of luck
Thanks for your comment. Much appreciated!
Huge thanks from South Korea :)
Very clear and informative
Awesome Explanation
Viewers mind (who is watching) as a communication channel with lots of noise from other TH-cam videos (with unclear content in the same context) and your brief explanation as the IEEE802.11a protocol with highest throughput for beginners. hahaha excellent work.
I'm glad you were able to decode my message from in the noise :-)
Clear and concise description and basics with good explanation ! Loved it ! Thanks.
Glad it was helpful!
@@iain_explains Could you please make a video on CDMA throughput, Bandwidth and Data Rate ? That would be wonderful.
Thanks for the suggestion. I'll add CDMA to my "to do" list.
Very helpful and informative!!!
Glad it was helpful!
Great Explanstion
So lucidly and concisely elucidated.
Thanks Sir
I'm glad you liked it.
It is super useful to me. I am an engineer at communication system company in Taiwan and do service for HP in USA. We do a lot of Throughput measurement here. And now I know what is going on here.
I'm so glad it was helpful! Great to hear from people in the industry.
Great explanation. Thank you.
Glad it was helpful!
This is such good teaching, thanks Iain. I have studied this topic from both an engineering and a computer science perspective. Very interesting. Really appreciate the explanation of how to construct the equation!
Thanks for your nice comment. I'm glad you like the video, and my approach to explaining things.
Thank you… great explanation!
Glad it was helpful!
nice
Thank you for this interesting video. I really appreciated it!!!
I'm glad you found it useful.
Please Sir, have you written a book or technical paper in which we can fin these explanations? Or maybe you know a literature that explains these relathionships, please let me know. Thanks a lot!!!@@iain_explains
Thank you very much, Sir, for your VDO.
At the baseband (before multiply by the carrier frequency) , is there the negative frequencies? Because when we multiply by carrier, we will have both side band shifted, if it is there.
Yes, but that's taken into account when using a complex baseband representation. This video will hopefully help: "How are Complex Baseband Digital Signals Transmitted?" th-cam.com/video/0lkRJgnywkg/w-d-xo.html
hello Mr
what is the definition of air interface latency??
and THX
Thank you so much for the great and eye-opening explanation. Just a quick question, in which block is the source coding done?
This video should help: "How is Data Sent? An Overview of Digital Communications" th-cam.com/video/MAddbFfCsIo/w-d-xo.html
Huge thanks from North Korea
Glad you found the video helpful.
Thank you for your presentation I have a question , How can I calculate the ARQ retransmission rate and the TCP retransmission rate?
Great question. I'll put it on my list for a future video. It's not straightforward.
Thank you for the video . It was very insightful . However ,I did not understand that after dividing the total bandwidth (20Mhz) by 64 ( # of sub carriers) we get 312.5 KHz . Subsequently you wrote symbol rate is 1/312.5 Khz . My question here is that how symbol rate is related to the bandwidth ? For example consider a binary symbol for 10 millisecond duration . The Fourier transform of that will be a sinc function with main lobe width of 2/10 ms . Inverse of that is 5ms . So if we inverse the bandwidth we get lesser symbol duration. But in your case you took the inverse of subchannel bandwidth and symbol duration is specified as inverse of that (3.2 microseconds) . Can you comment on whether my understanding is correct or Am i missing something ?
Yes, you're right, a square pulse width of T seconds corresponds to a sinc function in the frequency domain of width 2/T Hz (th-cam.com/video/ZBSvMbO0mPQ/w-d-xo.html). However we don't ever send "ideal" square pulse shapes. Instead we send raised cosine pulse shapes (or approximations to them). The "ideal" raised cosine pulse shape is the sinc pulse shape, which only requires 1/T Hz bandwidth. See "Pulse Shaping and Square Root Raised Cosine" th-cam.com/video/Qe8NQx4ibE8/w-d-xo.html
Hello thanks for this video, I have a question if my wireless Access Point is operating on 80MH bandwidths and if I change it to 40MHz will my average throughout go down ? From your video it seams it would go down as per formula throughout is directly proportional to data rate and data rate will go down with Chang in bandwidths. Please clarify
There are lots of factors/parameters that might change in your modem as a result of changing the bandwidth, but in general, yes, if you use less bandwidth, then it will reduce the throughput. If the bandwidth is the only thing that changes, then it is a roughly linear relationship. Halve the bandwidth, and the throughput will halve. Check out this video for more insights: "How are Data Rate and Bandwidth Related?" th-cam.com/video/ZBSvMbO0mPQ/w-d-xo.html
Very Informative video it is!
but I'm having some confusion. Why are you taking 64 subcarriers at one place and 48 subcarriers in the other? Doesn't legacy LTE with 180 kHz bandwidth only can carry 12 subcarriers of 15 kHz size?
As I say in the video, the numbers I used relate to IEEE802.11a WiFi over a 20 MHz channel. Other systems that also use OFDM, such as LTE, use different bandwidths and different numbers of carriers. One of the reasons for the differences is that cellular/mobile standards like LTE need to handle user mobility with potentially fast fading channels, and so they need to allocate more resources (eg. sub channels) to do channel estimation.
Very informative video. Love your videos sir. I have a question, how do I get the details formula of this throughput formula? like how to get the value of MAC contention rate, ARQ retransmission rate to get the throughput value? thank you sir
Great questions, ... but no simple answers, sorry. There are lots of variables that come into it. Many research papers have been written on these topics.
Hi Ian, thanks for this video. Why symbol rate is not used in the data rate formula? Wouldn’t the data rate be symbols/s x bits/symbols x # of subcarriers? 6:05
The symbol rate _is_ used in the data rate formula. It is 1/(4 microseconds)
@@iain_explainsgot it! Thank you. In another video you mentioned bandwidth is 2/T and data rate is 1/T. In this case wouldn’t the symbol rate be 1/312.5/2 us ?
Thanks for the video. I have some queries :
1) FEC is alternatively named as channel coding?
2) Is not HARQ a part of MAC layer? why have you put it as a différent layer between MAC and IP?
3) You mentioned about two factors while calculating the throughput. a) MAC contention and b) due to ARQ that you described to be linked to the channel. Not very clear about their distinction. Can you please explain again? The factor linked to ARQ is there because it means retransmission of the same packet that effectively reduces the throughput ? If yes, then this can be due to MAC congestion, no?Or what I am saying is true for MAC HARQ which is different from ARQ mentioned here ? Does ARQ mentioned here indicate the Radio link layer ?
1) Yes
2) Yes, HARQ is part of the MAC layer, but ARQ is above the MAC layer, in what is called Radio Link Control (for mobile comms). I didn't mention HARQ in this video.
3) Yes, the factor linked to ARQ is there because it means retransmission of the same packet that effectively reduces the throughput. This can be because of a number of reasons, fundamentally it's because the FEC was not able to overcome the errors that happened during the transmission (for whatever reason).
@@iain_explains thanks. So, ARQ can be due to MAC congestion as well, e.g., when two different transmissions collide due to the same radio resources,no?
Partly, yes, but only if the MAC wasn't able to resolve the congestion itself. The MAC protocol aims to take care of the congestion on the channel, (eg. using successively increasing backoffs in the case of CSMA/CA in WiFi).
Thanks
Hi Iain. Big fan, your videos are very very good. Can you please make a video on bits per symbol??
Glad you like the videos. What do you want to know about that topic in particular? Does this video answer your question? "What is a Constellation Diagram?" th-cam.com/video/kfJeL4LQ43s/w-d-xo.html
@@iain_explains thanks for replying. The video on constellations did clear my doubts.
Superb video Iain. Greetings from Spain!
Just one comment, I prefer to calculate data rate with Nyquist formula, you know:
Vmax = 2*B*log2(V) so with your example it would be = [2*312,5*log2(6) ]*48*3/4 = 56 Mbps. :D
The so-called Nyquist bit rate is an ideal situation where there is assumed to be zero noise in the channel, and where the signal can take on a finite number of levels. You need to think carefully about how you use it when considering real channels like the ones I am talking about in this video. One point to make is that the formula you've written does not take into account the time needed to send the cyclic prefix. Another point, is you seem to be using 6 levels, when QAM has 8 levels (in each of the real and imaginary directions/dimensions).
Sir, would you please explain why symbol rate equals bandwidth?
For explanations, you might like to check out these videos on my channel: "How are Bit Error Rate (BER) and Symbol Error Rate (SER) Related?" th-cam.com/video/du-sExIUV-Y/w-d-xo.html and "How are Data Rate and Bandwidth Related?" th-cam.com/video/ZBSvMbO0mPQ/w-d-xo.html
One Question: How can we attain the "packet error rate" by having the "symbol error rate", considering FEC? Suppose, the packet has 10 symbols, and SER is 10^-5. Then, the packet error rate will be [(1- 10^-5)^10] without having FEC. I cannot, unfortunately, calculate that rate when we have FEC.
Great question. The relationship you're asking about depends on the type of FEC that is being used. There are a great many research papers written on this topic for all of the many FEC schemes proposed over the years. The relationships are almost never straightforward, and often can't be written in "closed form". In many cases people have developed upper bounds, or approximations, or simulation based relationships.
Very informative video , thanks .
You mean GoodPut = throughput + L2+L3+L4 overhead.?
Edit : GoodPut = throughput - (L2+L3+L4 overhead)
No, I didn't mention "goodput" in the video. Throughput is the actual amount of data that is successfully sent/received over the communication link. In the example in my video, it is the end-to-end rate of sending data. Goodput is a less-well defined concept at the application layer, which is higher up in the protocol stack. There will be many reasons why an application is not able to make use of all of the Throughput that is available to it on a link or over a network, which means that the Goodput is always less than the Throughput (but not because of layer 1-4 issues - they are already taken into account in the Throughput).
in general the BW is 2/symbol_duration, but why is BW = 1/T in ofdm?
Well, "in general" is a subjective phrase. Hopefully this video will help: "How are OFDM Sub Carrier Spacing and Time Samples Related?" th-cam.com/video/knjeXo3VZvc/w-d-xo.html
Excuse me.I have a question for why the time duration of cyclic prefix is 0.8ms
The 0.8us CP is a design choice in the standard. It is 25% of the 3.2us symbol time. It was decided that this would be enough to account for the maximum delay spread that could be expected in most scenarios/environments/channels where WiFi would be deployed. This video may give more insights into the propagation path effects: "What are Flat Fading and Frequency Selective Fading?" th-cam.com/video/KiKPFT4rtHg/w-d-xo.html
Can I request you create a video about HARQ?
Thanks for the suggestion. It's already on my "to do" list. I think I'll bump it up the priority order.
The math at time 4:08 for bitrate calculation for 802.11 wifi seems to be not giving 54mbps.
Anyone tried calculating and share the formula if worked.
I just put the numbers straight into my calculator, and it gives 54 Mbps. Perhaps you should check again.
@@iain_explains
6 * 4 * 0.75 * 4000000 = 864000000.
864000000/16 = 54000000. Which is 54mbps.
I don't know why i did divide with 16.
I don't get why you used 48 sub-carriers to calculate the data rate? If it's not because of the headers, then why?
And how would you find the channel capacity?
Some of the sub-carriers are used for sending pilot symbols for channel estimation at the receiver.