How are Data Rate and Bandwidth Related? ("a super clear explanation!")
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- เผยแพร่เมื่อ 25 ก.พ. 2020
- Discusses the relationship between Data Rate and Bandwidth in digital communication systems, in terms of signal waveforms and frequency requirements.
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I'm a PhD student with one of my feet in communication. Your video series cleared up a lot of confusions I've been holding in my mind for a long time. Thank you dear sir. The community needs more great people like you.
Thanks for your nice comment. I'm so glad you've found the videos helpful. All the best with your PhD studies. Let me know if there are any other topics you might like me to cover.
@@iain_explains Thank you for your kind words. I'd really appreciate if you can cover degrees of freedom!
Yes it is very useful 👍
I m also a PhD scholar.
I do have my master‘s thesis defense tomorrow, your video really me in understanding the basics that may come as questions. Your video(s) helped me so much. Thank you very much an continue the great work, you really inspire !
Good luck with your thesis defence. I'm so glad you've found my videos helpful.
Thank you so much, i was so confused.
A very good explanation. Much appreciated 👌
Amazing explanation!
Talented educator
I have been trying to understand this for the longest. Thank you. I am grateful that I stumbled upon this video. Elaborate work.
Great. I'm glad you found it useful.
This video really helped me understanding these confusing concepts! Thank you so much :)
Glad it was helpful!
So do I ㅎㅎ
I really don't have words to thank you... you made it more than clear
Glad it helped!
A concise and broad explanation, that can be used as orientation to dive deeper in the corresponding topics (Interference, filtering, modulation, ...). Thank you so much!
Glad it was helpful!
I have listened to multiple videos to refresh my mind about communication. This channel was the best which answered many of my doubts. Thank you!
That's great to hear. I'm so glad you found the channel helpful.
Another great video that makes things super clear. Thank you Prof. Iain! :)
Glad you liked it!
This video is very helpful! Many thanks for such excellent explanation!
Glad it was helpful!
LOVE YOUR VIDEOS
I had 3 subjects about communications in college and we were always talking about this, never understood, just learned the formulas and got straight As. Now that I am writing my diploma, I wanted to know about this, bcs I one section I write about it. This video helped me sooo much!! Thank you very much Sir!
I'm so glad it helped. It's certainly something that people often confuse.
Hi Sir. Excellent video! Q: what would the data rate be for the last signal that is chopped off? Would this be the raised cosine filter? If so, would the data rate depend on alpha (for excess bandwidth)? Thanks!
clear, easy to understand, clever, helpful!! Thank you so much for thousand words!
Glad it was helpful!
Really good video! Clear and informative.
Glad it was helpful!
I continue to watch your videos because I always gain some knowledge from each. I really do appreciate them!
That's great to hear.
OMG thank you sir! extremely helpful!!!
Glad it helped!
you just helped me so much thanks a lot.
You are welcome!
Great video pal, Thanks for the comprehensive explanation!
Glad to help!
Excellent explanation! Subscribed.
Thanks and welcome
thank you, this was so useful. Big thumbs up !
Glad it was helpful!
Learned a lot, Thanks
Great. I'm glad you found it useful.
your explaining power is infinite....Hats off.
Thanks. I'm glad you've found the videos helpful.
This is a beautiful explanation. Amazing work. Thanks for this video. Also, nice touch with using a notebook and handwriting!
I'm so glad you liked the video, and my approach of aiming for a personal/human mode for my explanations.
Brilliant! Great explanation.
Glad it was helpful!
Mr Ian, you're an invaluable resource to the engineering community.
Thanks so much for your nice comment. I'm glad you like the videos.
Superb explanation..
Glad you liked it
im a telecom student from Ukraine. Thank you a lot for a super clear explanation!
My pleasure. Wishing you all the best.
Thanks a lot. Well explained :-) .
Glad it was helpful!
Hello Sir~ Your videos are always helpful and make lots of details clearly.
You really solve a lot of problems I had before!! Thank you for keeping doing this for all Communication Ppl :)
And I have a question about the notes in the video. (I'm not sure whether I'm right or wrong)
The example you used in the end of the video might be Amplitude Shift Keying and then you multiply the number of bits per symbol with the "data rate" you wrote in the middle of uppermost line in the beginning.
So I'm guessing whether the unit of "data rate" is supposed to be the symbol rate with unit "symbol per second".
Thank you !!!
When it's binary, the data rate equals the symbol rate. So in the video up until that point, I just used the term "data rate", and didn't mention the term "symbol rate". I didn't want to confuse things by introducing the extra "symbol rate" term.
@@iain_explains what you explained just reminds me of that in binary case and it makes sense. Thank you for spending your time replying this !!
This is an awesome video for seeing HOW data rate and bandwidth are related, but could you also do a video about WHY data rate and bandwidth are related? Something I'm trying to wrap my head around is WHY does increasing data rate need more physical space in the RF spectrum to accommodate that increased data rate? What is it about using more RF spectrum that allows more data? The way I think about it in my head is it's like parallel computing where by adding more wires (or frequencies) in a cable (bandwidth) allows you to simultaneously send more bits, and therefore achieve higher throughput. I'm sure this isn't actually why you need more RF spectrum for a higher data rate, but I think it helps explain what I'm trying to ask. Thanks for the video!
One way to see this is to think about sending the symbols at a higher rate. ie. with a smaller value of T. This means the sinc function in the frequency domain gets wider, which of course means you will be using more bandwidth. To go even deeper into it, recall that the Fourier Transform shows the frequency components that add together to construct the time-domain signal. If you want your time-domain signal to turn on and off more quickly, then that signal will naturally consist of higher frequency components (faster changes in the time domain equals higher frequency components in the signal, ie. wider bandwidth). For a video on this, see: th-cam.com/video/8V6Hi-kP9EE/w-d-xo.html
First of all thanks for the video . You mentioned that the sinc function will have a width of 2/T for the main lobe , however i think that the width must be (Pie)/T . In one of your videos you mentioned that the first zero crossing for the main lobe will be at Pie/(time period)
Ah, this is a topic that often confuses people. In this video, I am plotting the "real" frequency, f (in units of Hz). In the other video I was plotting the "angular" frequency, omega (in units of radians). And omega = 2 pi f . So that's where the "pi" comes from in that other video. Now, the factor of "2" depends on how the square function is defined. Sometimes it is defined between -T and T, and other times it is defined between -T/2 and T/2. One will have a factor of 2 in the Fourier transform, the other will not.
Thanks for clearing my doubt. I have became huge admirer of your videos, I cannot stop watching all your Playlist videos. They are awesome
Much needed. Thnx✌️
Wao... what an explanation! I am glad that I found your channel, but I'm sad that I found it so late.
I'm glad you found my channel too. The more subscribers the better. It helps to spread the word. I'm glad you've found the videos helpful.
@@iain_explains Hi Prof Iain, If time permits, can you please make some lectures on fixed point vs. floating point DSP. Regards
Thanks for the topic suggestion. I've added it to my "to do" list.
The one question I had is that for any given amount of spectrum bandwidth, how do you calculate the symbol rate?
For example, say you have 20 MHz of spectrum in 2.4 GHz. How do you figure out the symbol rate?
On a similar note, if you have the same 20 MHz of spectrum, but it's in the 60 GHz range, does the alter the symbol rate or is it still, ultimately, the same?
The symbol rate depends on the bandwidth and the pulse shape (it doesn't depend on the carrier frequency). See this video for more details about that: th-cam.com/video/Qe8NQx4ibE8/w-d-xo.html
@@iain_explains Thank you very much, I appreciate it!
Hey man, I don't even know how to thank you.
I wish you were my teacher even in elementary school
I'm so glad the videos are helpful to you.
The pointer can assume 12 different positions. There will be one position per
Second displayed. How high is the data rate??
Thanks for the video, was awesome. I have one question: why is the bandwidth of the time domain sinc function 1/T? How can we differentiate between cases 1 and 2 when determining bandwidth? Thank you
Thanks for your question. I've realised that I've never made a video about the Fourier Transform of the square waveform (from which we get the relationship you've asked about). I'll add it to my "to do" list. Basically though, the answer to your question can be seen from the duality property of the Fourier Transform.
@@iain_explains thank you very much, looking forward to the video!
Thank you for your excellent description of data rate and bandwidth.
How do you translate/explain of “8Hz data rate”?
Thank you
Sorry, “8Hz data rate” doesn't make sense.
hello ,
how can i increase data rate on xBee PRO S2C, i should get 250 kbps but i got only 30 kbps
hey professor really an amazing video thanks for it
my little question does the duration of the clock pulse which we are applying to the circuit has the same duration as that of data rate?
No. It is possible to send multiple data bits per clock pulse. This video will hopefully help: "What is a Constellation Diagram?" th-cam.com/video/kfJeL4LQ43s/w-d-xo.html
For the time domain 3rd option of signal ( limited in time) you mentioned the BW is >1/T..in frequency domain...will it be >2/T..as per transformation described for pure pulse signal ..option 1 ?
Yes, technically you're right. For any time-limited signal, the frequency range will be infinite. In practice though, we always have to send time-limited signals, so we accept that there will be frequency side lobes, and we try to design the signal shape so that the side lobes are small/negligible. And then we describe/define the "bandwidth" to be the frequency range where there are non-negligible components. Sometimes this is the 3dB frequencies (see my video th-cam.com/video/ZMgd2obTZyc/w-d-xo.html). Also you might like to watch my video on pulse shaping: th-cam.com/video/Qe8NQx4ibE8/w-d-xo.html
Great video sir
Thanks. Glad you liked it.
Hi..thanks for the video.. but can you say how BER, SNR are related to get the maximum symbol rate for a transfer function.. in think that will really help ..thank u
Have you watched my video on BER and SNR for PSK and QAM? th-cam.com/video/vtJ6mAy3xMc/w-d-xo.html and
Thank you so much sir for these types of videos.. really appreciate it...i have one doubt at 7:16, if bandwidth is > 1/T then is it safe to assume that bandwidth= 2/T? If so will the bandwidth remain same for any modulation sheme(like bpsk, 16qam,64qam and so on)?
No. The shape of the pulse, and the rate that pulses are sent, determine the bandwidth. There is no reason the bandwidth needs to be exactly integer multiples of 1/T. For example, the square pulse at the top of the page has an infinite bandwidth.
That when system speed slow enough but at high speed that techniques is not efficient i think and if we utilize speed as advantage the split bandwidth base on two time variable ( time domain and time sequencing then instead of voltage data we transmit by pulse like telegraph but extremely fast then it does not effect too much on electrical system because normally that system is protected against eddy current already so only router at substation read
Hi Iain, this does not apply to QAM modulation methods used for mobile communication, right? Because in these cases each bit is mapped to a frequency amplitude and phase signal (thus I do not need to send an electric pulse). Am I correct?
I'm not sure you're understanding what modulation is, exactly. The only difference between what I've shown in this video, and QAM, is that in the case of QAM the "electric pulse" (as you call it) is modulated by a carrier waveform (and that there is another orthogonal waveform sent at the same time - one on a cos wave, and the other on a sin wave ... plus there are multiple amplitudes, not just the on-off situation I talked about here). Everything else is the same. These videos might help: "What is a Constellation Diagram?" th-cam.com/video/kfJeL4LQ43s/w-d-xo.html and "How are Complex Baseband Digital Signals Transmitted?" th-cam.com/video/0lkRJgnywkg/w-d-xo.html
In the top example, I assume “passband” means following modulation. At baseband, would it only be half of the sinc, ie 10 MHz with fc = 0?
It depends on how you define "bandwidth". Some people define "baseband bandwidth" and "passband bandwidth" differently. More insights here: "What is a Baseband Equivalent Signal in Communications?" th-cam.com/video/etZARaMNN2s/w-d-xo.html
Thank you so much Prof. That is really helpful. I have a question here, I am performing real test using OFDM system in wireless communications system, with number of sub-carriers N and modulation order M, I am using sampling rate t = 1GSample/s, what will be here the data rate? Lets ignore the guard interval and overhead packets and so on, I mean we consider that all sub-carriers are carrying information bits, log_2(M) each sub-carrier. can we consider here the bandwidth B = t, or it should be B = t/2 ? I mean if we could decided the correct B, it will be straightforward to calculate the data rate in that case. Thanks again prof.
Hopefully this video will help: "How are OFDM Sub Carrier Spacing and Time Samples Related?" th-cam.com/video/knjeXo3VZvc/w-d-xo.html
Hi Sir
What will be signal bandwith if Signal is
case1. Train of pulses with 1Mbps (Mega bits per second)
case2. PRN code with 1Mcps (Mega chips per second)
I wanted to know why the Fourier transform of a band limited signal of bandwidth B is symmetric around 0 meaning why does the spectrum extends to the negative frequency range ?
You might like to try watching my video "What is negative frequency?" th-cam.com/video/gz6AKW-R69s/w-d-xo.html
Hello, thanks for sharing. Can I ask you a good book on digital communications? for advanced learner, thank you
This is the best, in my opinion: J.G. Proakis, “Digital Communications”
hello, thank you for your videos, they are wonderful, although i have a question regarding translating from data rate to bandwidth, All literature on the internet they tend to divide data rate by 2 to get to frequency bandwidth , for example 56gps pam2 , it is translated into 28ghz , 224gbs pam4 into 56 gas and so forth, please if you can explain that, it is very confusing to me
Ah yes, this factor of 2 often confuses people. In my video I talked about the passband bandwidth (assuming the time domain waveform was being modulated up to a carrier, for example for wireless communications). The bandwidths you are referring to in your question, are baseband bandwidths (assuming the time domain waveform is exactly what I drew on the left hand side, without any carrier modulation). Passband bandwidths are 2 x Baseband bandwidths. These videos might help: "Amplitude Modulation AM Radio Signal Transmission Explained" th-cam.com/video/-PWg-0k2oks/w-d-xo.html and "What is a Baseband Equivalent Signal in Communications?" th-cam.com/video/etZARaMNN2s/w-d-xo.html
@@iain_explains Thank you so much for the explanation, now it clear.
@@iain_explains according to what you have said, baseband bandwidth would be 1/T for thé first signal. But he is asking why it is 1/2T, no?
Is the bottom example related to the raised cosine filter?
Yes, that's right. More details here: "Pulse Shaping and Square Root Raised Cosine" th-cam.com/video/Qe8NQx4ibE8/w-d-xo.html
My question is, In standard Ethernet with transmission rate of 20Mbps, the length of the cable is two meters and the size of frame is 1024 bits, the propagation speed is of a signal is 2/10^8m/s. What is the total percentage of time medium occupied but not used by station?
Sounds like this is one of your assignment/test questions. Sorry if this sounds a bit direct/rude, but I'm not here to do your homework for you. If you've got an "understanding" question, then I'll be more than happy to answer it.
Thanks a lot for your wonderful video. I am wondering how the power of modulated digital signal related to the RF carrier power🤔
They are the same. The "modulated digital signal" is (by definition) the transmitted signal that has been generated by multiplying the RF carrier sinusoidal wave with the data-carrying baseband wave. That's what "modulation" means. The baseband signal has "modulated" the carrier waveform.
@@iain_explains I am wondering three components as digital baseband signal, rf carrier and modulated signal, the power spectrum density of modulated signal can be regard as the shift of basedband to RF. However, I am confused by the power relation of these three, as we always use the word'transmit power'. Does it refer to some kind of combination of baseband signal power and rf power? Many many thanks.
"Transmit Power" refers to the power of the transmitted (modulated) signal. The baseband signal has a certain power (depending on the electronics that implements/generates it) and it gets multiplied by a sinusoidal carrier wave (which has a certain power depending on the electronic circuitry that generates it). The resultant waveform is then amplified by a transmit amplifier, with a certain gain (possibly adaptive, depending on the implementation). It might go through a transmit filter, which potentially alters the power. And then it gets sent by an antenna that has a certain antenna gain ... and so on. All of these things are just scaling factors that get measured/calibrated/adjusted in practice. The only two things that really matter are that the radiated power is below the legal limits, and that by the time the signal reaches the receiver, that the received power is high enough above the receiver noise power.
@@iain_explains Got it, Thank you so much for these general and detailed explanation, Prof. Iain, wish you a nice weekend!
Great. I'm glad I've been able to help.
Is it possible having a data rate that exceeds bandwidth? When is bandwidth the upper bound for data rate?
If there is zero noise, then it is theoretically possible to have an infinite data rate over any finite bandwidth. You might like to check out my video on capacity: What are Channel Capacity and Code Rate? th-cam.com/video/P0WY96WBUyA/w-d-xo.html
Professor, The problem you discussed at @5:46.
Every practical signal is time limited and signal don't exist on t < 0. in that case why do we need to worry if signal starts infinite time ago. My understanding is signal being infinite is just theoretical concept. please correct me
In answer to your question: "why do we need to worry", I think you've already answered it yourself. If the signal you want to implement needs to start infinite time ago, but you are dealing with practical signals that are time limited and don't (can't) exist on t
Hello sir I want to know if u could explain synchronous and asynchronous for me
Thanks for the suggestion. I'll add it to the to-do list.
hii lain,
when we are sending bits in the time domain by sinc function than why have you considered the bit duration as 2Tb is it because we need to extend the duration of bits we are sending when we send in form of sinc pulses
Sorry, I don't know what you mean. I don't say the bit duration is 2Tb.
@@iain_explains @4:13 the sinc pulse drawn is having the main loab frequencies from (-T to T) which will make it duration of 2Tb thats what i was refering
OK, I see. Why do you write "Tb"? That's the symbol for Terra-Bit (10^12 bits). Anyway, no, the "duration" is in fact infinite! The waveform goes from negative infinite time to positive infinite time. But anyway, due to the zero crossings, you can send one symbol every T seconds. That's the time between the zero-crossings. When there is a zero-crossing, you can centre another pulse at that time, and not interfere with this pulse.
I recently discovered your TH-cam series on digital communications. Thanks so very, very much for the informative work. However, regarding this video, i.e., "How are Data Rate and Bandwidth Related," I can't seem to gain an intuitive (i.e., physical and non-mathematical-formula-based) understanding of two aspects of the video. As to the first such aspect, can you offer an intuitive explanation of why the time-based square wave (signal, baseband) requires a spread/range of frequencies to adequately depict the frequency-based bandwidth/passband? In other words, if a sine wave can be described as a single frequency (as I saw it was the case in another of your videos), why (physically) does it take a range of frequencies to "build" the square wave. As to the second aspect of your video for which I can't find an intuitive understanding, what is (again, sorry to keep this drum, physically) it about the square wave that requires a greater and greater range of frequencies to achieve a greater and greater bit rate for the transmission of the square save signal? Again, deepest thanks for your videos. Charlie
Welcome to the channel. I'm glad you like it. I've been meaning to make a video on this for ages. It's on my "to do" list, but I think I should lift it up the priority order. Basically, all (or almost all) signals can be constructed by adding together sinusoids of various amplitudes, phases and frequencies - that's what the Fourier Transform tells us (that's the brilliance of Fourier's construction, back in the 1821). Slowly changing signals can be constructed from sinusoids with low frequencies. The more "peaky" the signal, the higher the frequency components will be. So, when you think about it, if you want to "construct" a perfectly square corner in a signal, you would need sinusoidal components that oscillate infinitely fast. I definitely need to make a video about this, but until then, this video may provide extra insight: "What is the Gibbs Phenomenon?" th-cam.com/video/Ol0uTeXoKaU/w-d-xo.html
@@iain_explains Thank you again. You’re reply is most helpful and the Gibbs Phenomenon video is also very valuable. I look forward to you’re recording (ad , of course, prioritizing) a video that more directly addresses the questions in my original post and does so with an emphasis on offering an intuitive answer to those questions. Charlie
Hi @Charlie Uniman , I just made a video to answer your first question in this comment thread. Hope it helps: "Why are Sinc and Square a Fourier Transform Pair?" th-cam.com/video/ZcTWLwXGql0/w-d-xo.html
@@iain_explains Thank you very much. I can’t wait to view the video tomorrow. Charlie
then what is the difference b/w the spectral efficiency and data rate? that is [bit/sec/Hz] means?
Thanks for the question. I'm planning to make a video on spectral efficiency, so keep an eye out for it. In summary, the spectral efficiency is the data rate (in bit/sec) divided by the bandwidth (in Hz) over which it is transmitted (where the "data rate" is defined as the rate of sending "information bits"). For more details on the different "rates" see this video: "How are Throughput, Bandwidth, and Data Rate Related?" th-cam.com/video/IY6fDYwC2fU/w-d-xo.html
@@iain_explains ok thanks i ll checking
For example - For some random data input if 1 bit period =4n sec, then Bit rate = 250Mbps and Bandwidth is 500Mhz. Is my understanding is correct, please let me know if I am wrong.
As a general rule of thumb, yes (for binary modulation). The bit period is the inverse of the bit rate (for binary modulation). But the bandwidth is not so clear cut (as this video explains). The bandwidth relationship depends on the choice of the pulse shape (in the time domain). I show three examples in the video. Perhaps this more detailed video might help: "Pulse Shaping and Square Root Raised Cosine" th-cam.com/video/Qe8NQx4ibE8/w-d-xo.html
The nyquist data rate = 2B log2M..for bpsk, we have B = data rate/2...???
It depends on what you mean by the "nyquist data rate". I think you mean the data rate that the Nyquist data transmission theorem tells us is the highest rate possible, assuming you are using ideal sinc function pulse shaping. In which case, B is the bandwidth of the low pass base-band equivalent channel. I think I'll add this topic to my video "to do" list, to help clarify it.
Hi, I'm confused as to what these sidelobes represent. Could you explain this or direct me towards a good resource that explains?
Square shaped time-domain digital signals have a power spectral density that is a squared sinc function (in the frequency domain). Hopefully these videos will help explain more: "Visualising the Fourier Transform" th-cam.com/video/U7ii8agAhIs/w-d-xo.html and "What is Power Spectral Density (PSD)?" th-cam.com/video/DoSLMEEo1Y0/w-d-xo.html
And make sure you check out my webpage that has a categorised listing of all the videos on my channel: iaincollings.com
@4:54 There would be interference among the multiple signals right?
Actually no. Notice that at the sampling times T, 2T, 3T, ... there is only ever a single sinc pulse that is non-zero (at those exact sampling times). It might help to watch this video that explains the pulse shapes more: "Pulse Shaping and Square Root Raised Cosine" th-cam.com/video/Qe8NQx4ibE8/w-d-xo.html
@@iain_explains ok. If sampling period T then there won't any interference. Thank you.
why the minimum band width required is Rb/2 whereas it shoud be Rb insted since when we using sinc pulse in the input we are getting a rectangular pulse in the frequency and thats the minimum required band width . i hope you are understanding my doubt i have always had this confusion about why Rb/2 .thanks!
Sorry, I'm not sure what you're referring to. I don't use the symbol "Rb" anywhere in my video.
@@iain_explains Rb is nothing but 1/Tb [where Tb is duration of bit]
Did you watch my video? In the video, I explain that if a sinc pulse shape is used, then the bandwidth is 1/T. I don't know where you are getting Rb/2 from. I don't say it anywhere in my video.
Has anyone told you that you look like Bruce willis...
Great content.. (y)
No, that's a new one. I've been told I look like other people, but not him. Maybe I could be a movie star after all. 🤩
Sir, my name is Dheeraj. I am pursuing Ph.D first year. Sir, my research area is closely related to Information Theory. Sir, I have a doubt regarding rate. Some times Rate is mentioned to be some x bits per channel use. Sir, Here what exactly does per channel use mean? If we say rate is 100bits/sec, is it same as 100bits/channel use. Sir, I have been trying to understand this for a while now but I am not able to figure it out. Sir, why exactly are we shifting to bits/channel use domain while we have rate expressed in bits/second? Sir, please explain me this if you have free time.
It depends on the channel model. For example, for the binary symmetric channel (BSC), it tells you the probability that a digital "one" is received in error, but it doesn't tell you anything about the rate at which you send the "ones" and "zeros" (the assumption is that it has already been taken into account in the calculation of the probability of error). So for the BSC, the capacity is expressed in terms of the number of bits that can be reliably transmitted per input bit, or in other words, per "channel use". The way you use the channel has already been defined in that case - it is by sending either a "one" or a "zero". For more general channel models, such as a band limited AWGN channel, where the input distribution is not restricted to just "ones" and "zeros", the units of capacity are bits per second, since the equation includes the bandwidth in Hz (ie. per second).
@@iain_explains Thank you very much sir. I understood the difference between the two
I guess the top sinc function is magnitude only? Had it not been we could see the duality more easily.
Yes, you're right. It's the convention though, to plot the magnitude in the frequency domain, since there is generally a complex (phase) component in that domain. I probably should have made that more clear in the video.
1) If the bandwidth is infinite and absolutely no noise, how fast can we send/receive data?
2) If the media is of infinite bandwidth but with some noise, how fast can we send/receive data? Assuming that your device is fast enough.
1) infinitely fast. 2) infinitely fast. This video gives more details: "What are Channel Capacity and Code Rate?" th-cam.com/video/P0WY96WBUyA/w-d-xo.html
We should just abolish schools, and get best videos in every field, including this, and just give them to the students!
Possibly. The future is coming faster than many people realise.
While the video is great, there seems to be a problem. At 0.44 secs you say the formula is 2/T and at 2.35 secs you say it is 2timesT?????
You might want to listen to it again. at 2:35 I say that it is "two times one on T" ... that means 2 x (1/T) which equals 2/T
Banks a lot(thats minionese)...
Glad you liked it.