This channel covers essential elements of the topics Signals, Systems, and Digital Communications, which are fundamental to so much of Engineering and Modelling.
I am a Professor and an IEEE Fellow, and I'm passionate about teaching students from a practical understanding perspective.
For a full categorised list of Videos and Summary Sheets, goto: www.iaincollings.com
Is this a history lesson no mathematical explanation of how we arrive from the time domain to the frequency domain no explanation of the food you serious, which is basically the representation of any periodical function
This video is focused on giving an intuitive explanation of the Fourier Transform. If you want the maths, then there are a great many textbooks that provide those details.
I think pdf is a term for continuous random variables which shows likelihood for each point. So probability mass function is a more accurate term in discrete random variables
Thank you for explaining! I have a question in 10:30. My english skill is not good so i don't understand the 'h' and 'root p' in equation. Could you tell me more detail about that..?
The 'h' is the channel gain (which will be a complex number with magnitude between 0 and 1), and the "root p" is the square root of the transmit power.
I think that the three first rules can be replaced by the clockwise rule ! 🙂. When the current is going away from me then the magnetic field is clockwise. When the current is coming towards me Then magnetic field is counterclockwise. And vice versa ! 🙂
Yeah, unfortunately you only have a 1/3 or 33.3% probability to NOT get the car. No one disputes the 1/3 or 33/3% probability TO get the car; it's the probability to NOT get the car that is in debate. TO get the car 1/3 Door A + 1/3 Door B + 1/3 Door C = 1 car to NOT get the car 1/3 Door B/C + 1/3 Door A/C + 1/3 Door A/B = 1 No car 1/3 car + 2/3 NO car = null equation Having 2 doors doesn't automatically mean 2/3 or 66% probability. They both combined to contain a total of 1/3 or 33.3% probability for you to NOT get the car. Your probability to pick one of those two doors is only 1/6 or 16.6%. And from a simple logic point of view, at no point in time do you have ANY probability to get Monty's no-car door. And where and when is that remaining 1/3 or 33.3% to NOT get the car if 2/3 or 66.6% of the time you don't get the car?! So 1/3 or 33.3% to get the car, and 1/6 or 16.6% + 1/6 or 16.6% to not get the car means the Monty Hall Problem is a 50/50 guess to either stay of switch. 🤙
Sorry, but you're clearly not understanding the game. Perhaps try watching the video again. If you get someone else to play the game show host, and try the game for yourself, you'll soon realise that if you change doors then you'll get the car 2 times out of every 3. Try it. You'll see.
@@iain_explains Iain - you seem like a decent fella and I certainly thank you for your time and reply. The ⅓-⅔ CULT claims that since you are picking from the ⅔ or 66.6% no car side, your chance to NOT get the car is twice the probability of you getting the car. Two is bigger than one so you are twice as likely to get the car if you would always switch. But at NO TIME do you have a chance to EVER pick 33.3% of that no car probability, Monty's door. Monty has that 33.3% no car probability. It is there before you even make the pick, yes I know that. But you have ZERO probability to get it. It can be in either of the no car doors and it still doesn't matter because you don't even have a 1% probability to pick it. Even though you are picking from the 66.6% side, you are only picking from 33.3%. So how does 33.3% car and 33.3% no car not a 50/50 guess? Then there are also these 100 door analogy, where if you pick one door and the host opens 98 doors to reveal no car, leaving only one no car door. And the claim is that wouldn't your car be more likely to be in the 99 doors that you didn't pick than the one you DID pick? Dude, you had ZERO probability to EVER pick those 98 no car doors, which means you have the 1/98 probability door you picked, and the 1/98 probability door that you didn't pick. You should always switch because you are now 99% more likely to have not picked the car?! You NEVER had access to the 98% the host opened. You have and always had access to only the 1% you picked, and the 1% you didn't pick. When this paradox first came about, these types of media platforms didn't exist to discuss it in. There were only biased media outlets promoting Vos Savant, leaving equal numbered of thousands, if not more, qualified mathematicians and university professors disagreeing. Now that we can clearly discuss it, my only purpose is to get it right. Thanks again, dude. No hard feelings. 🤙
Thank you so much, Iain! You’re making clear convolutions to me. What would be the procedure to resolve this with the convolution theorem? I mean, I tried it but would we be able to obtain one result to different intervals of t? I'm very confused about it. Thank you.
Sorry, I'm not sure what you're asking. Have you watched my other videos: "Convolution in 5 Easy Steps" ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-aMaw4EumwyE.html and "What is Convolution? And Two Examples where it arises" ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-X2cJ8vAc0MU.html
The equations in this video are for the baseband representation. There are many more details that would need to be added, if I were to include the full transmission chain, including pulse shaping filters, D-to-A converters, amplifiers, analog filters, ... etc. One of those things would also be a matched filter at the receiver - matched to the pulse shape. For more details on the baseband model, see: "How are Complex Baseband Digital Signals Transmitted?" ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-0lkRJgnywkg.html and "What is a Baseband Equivalent Signal in Communications?" ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-etZARaMNN2s.html
Hi professor, IEEE channel model D(typical office) has max delay of 390ns, and 802.11ac short guard interval(should be the length of cyclic prefix?) is 0.4us which is 400ns and is longer than max delay. Does it mean model D won't create any inter-symbol-interference? But why are we still seeing WiFi throughput drop when testing when testing with channel emulator and applying model D?
Yes, there won't be any ISI between the OFDM symbols in that case, however that doesn't mean that the channel is flat across the spectrum. There will still be an effect from the multipath, causing some sub channels to have low gain. I'm planning to make a video about this in the next week or so. Keep a lookout.
Thank you so much for the clear explanation. I would appreciate it if you can have a video about mmwave channel model vs. Terahertz channel model. How are they different?
Thanks for the topic suggestion. I've got it on my "to do" list. I assume you're already seen the videos I already have on the topic? See iaincollings.com
They need to be integer, in order to maintain the orthogonality between the subcarriers. Hopefully this video explains it: "OFDM Waveforms" ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-F6B4Kyj2rLw.html
2:04 If only the first element is nonzero, the output of IDFT has to be the DC value, not the sinusoid of fundamental frequency. The subcarriers range from C_0 to C_{N-1}. C_0 will correspond to the DC value at the output and C_1 will give the fundamental sinusoid of period = N. Please respond if I am wrong.
Try running these lines of code in Matlab: X=zeros(1,40); % Frequency domain vector X(2)=1; % Put data into the first subcarrier (only) ... note that X(1) corresponds to f=0 x=ifft(X); % Use IFFT to generate time domain signal stem(real(x)) % plot the time domain signal
Thank you for the great explanation, I really find it insightful. About the definition of the f(x,y) on the top of the page, should that not be f(x,y) = P(X = x, Y = y) because else you'll have to take the integral and calculate the cumulative distribution over x<=X<=x+dx and y<= Y <= y+dy?
P(X = x, Y = y) equals zero for all x and all y. This is because it is a density function. The probability of any particular x or y (ie. exact, to infinite precision accuracy) is zero (since there are infinite possible values). Hopefully this video will help: "What is a Probability Density Function (pdf)?" ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-jUFbY5u-DMs.html
I have studied my Phd proficiency exam with your videos and they help me so much that i can get the best result:) thanks for your clear and perfect explanations in which i could get the basics of the subject as well main ways of thinking the subject:) Thanks a lot for your guidance.
L is the number of ISI terms. L times the symbol period equals the delay spread. For more details, see: "What is Intersymbol Interference ISI?" ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-I087FUvW2ys.html
Yes, that's right. Interestingly, I've never noticed that before. It's obviously potentially confusing, once it's pointed out. Thanks for alerting me to it. I've added a note to the description below the video.
Here are some videos you might like to watch, that discuss examples of randomness in communication systems: "What is Gaussian Noise?" ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-VIvYxnhkvvc.html , "What is Rayleigh Fading?" ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE--FOnYBZ7ZfQ.html , "How is Data Sent? An Overview of Digital Communications" ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-MAddbFfCsIo.html
