What are Doppler Shift, Doppler Spread, and Doppler Spectrum?

Sdílet
Vložit
  • čas přidán 21. 08. 2024
  • Explains the relationship between Doppler Shift, Doppler Spread and Doppler Spectrum for mobile communications channels.
    Related videos: (see: iaincollings.com)
    • How Does a Doppler Shift Affect Digital Communications? • How Does a Doppler Shi...
    • Why is Doppler a Problem for OFDM? • Why is Doppler a Probl...
    • What are Flat Fading and Frequency Selective Fading? • What are Flat Fading a...
    • What are Fast Fading and Slow Fading? • What are Fast Fading a...
    • What are Decibels and why are 3dB and dBm important? • What are Decibels and ...
    • What is Rayleigh Fading? • What is Rayleigh Fading?
    • What is Rician Fading? • What is Rician Fading?
    Full categorised list of videos with pdf Summary Sheets: iaincollings.com
    .

Komentáře • 55

  • @fardadansari1885
    @fardadansari1885 Před 3 lety +4

    Professor , just saying thanks is so limited appreciation against your precious effort for make us digest the very complex concept in appearance and so simple in meaning. Really appreciated
    Fardad Ansari , Communication system master student from Iran

    • @iain_explains
      @iain_explains  Před 3 lety +1

      Thanks for your nice comment. I'm glad the explanations have helped.

  • @ayushsoni8983
    @ayushsoni8983 Před 3 měsíci +1

    Nicel explaination

  • @mmk904
    @mmk904 Před rokem

    Thank you very much, this video helps me understand these terms more clearly. Thank you.

  • @ibrahimshikdaher7551
    @ibrahimshikdaher7551 Před 3 lety

    So clear and descriptive videos from you

  • @surbhirani5528
    @surbhirani5528 Před 2 lety

    Neatly Done , thank you sir .
    from INDIA ☺

  • @caddyrecords
    @caddyrecords Před rokem

    Is the Doppler effect presented here (and also almost in all wireless text books) representing a low speed (v

    • @iain_explains
      @iain_explains  Před rokem

      Sorry, I'm not a physicist, so I'm not exactly sure what you mean by "relativistic analysis". The equation is a straightforward calculation of the change in frequency of a signal that is travelling at the speed of light, c, when there is a change in distance between the transmitter and receiver at the speed v cosθ (ie. this is the speed in the direction of the wave's propagation between the transmitter and receiver, when one or other of them is travelling at the speed v).

  • @eshwarnetha802
    @eshwarnetha802 Před 3 lety

    Very Nice Explaination!

  • @bunty55555
    @bunty55555 Před 3 lety

    Thanks for your effort , very useful

  • @ramyaranjanbarik681
    @ramyaranjanbarik681 Před 2 lety

    Very lucid explanation .For 5G NR systems ,one of reason of changing subcarrier spacing is ,Wider subcarrier spacing would provide better resistance to such increased Doppler shifts at higher frequency bands. May you help to relate it ,how.

    • @iain_explains
      @iain_explains  Před 2 lety

      Increased Doppler shift means there will be increased Doppler spread (for a given scatterer scenario). This will lead to the sinc-shaped subcarriers in OFDM spreading out, and interfering with their neighbours in non-orthogonal ways. Wider subcarriers means there will be less inter-subcarrier interference (as a percentage of the subcarrier bandwidth). This video might help: "How are OFDM Sub Carrier Spacing and Time Samples Related?" czcams.com/video/knjeXo3VZvc/video.html

  • @ngolisaoran8166
    @ngolisaoran8166 Před 3 lety +1

    hello. Can you make a video about how to measure IRR (image rejection ratio) in reality? Because when we manufacture a chip receiver in reality, and it is a "black block", we only know the input and ouput signal, and we must use a professional machine to measure IRR. And I know that they measure by put RF signal to chip, they measure, and put Image signal to chip, and measure, like superposition method, then calculate IRR. Can you discuss about this method, detaily? thank you and hope you're always the best. Merci beaucoup!

    • @iain_explains
      @iain_explains  Před 3 lety +1

      Thanks for the suggestion. I've added it to the "to do" list.

  • @anonyymi3822
    @anonyymi3822 Před 3 měsíci

    What are some practical implications of these phenomena?

    • @iain_explains
      @iain_explains  Před 3 měsíci

      Here's two examples: "How Does a Doppler Shift Affect Digital Communications?" czcams.com/video/9HqlfqBDwpY/video.html and "Why is Doppler a Problem for OFDM?" czcams.com/video/mB0GF9uKC48/video.html

  • @maazawan7629
    @maazawan7629 Před 11 měsíci

    Thank you so much professor!
    Any comments on how to overcome the doppler spectrum issue specially in a multipath environment when there is contribution from other components of doppler shifts other than the most direct 0 degree or the 180 degree path...

    • @iain_explains
      @iain_explains  Před 11 měsíci

      When you have Doppler spread, you need to make sure that other signals on neighbouring carriers are sufficiently far apart. If they're not, then you'll need to implement inter-carrier interference rejection, or just put up with a lower SINR. These videos give more insights: "How Does a Doppler Shift Affect Digital Communications?" czcams.com/video/9HqlfqBDwpY/video.html and "Why is Subcarrier Spacing Bigger in 5G Mobile Communications?" czcams.com/video/uyDvjpPn8ms/video.html

  • @sarahrok9555
    @sarahrok9555 Před 2 lety

    that you for this

  • @marcinrybakowski8926
    @marcinrybakowski8926 Před 3 lety

    Hello, thank you for very good videos. The video explaining channel state information could be interesting.

    • @iain_explains
      @iain_explains  Před 3 lety

      Thanks for the suggestion. I've added it to the "to do" list.

  • @jonesy3516
    @jonesy3516 Před 9 měsíci

    Dear Ian, thank you very much for your very clear explanations. If I understand correctly, a channel with a single tap cannot have a Doppler spread (Jake's spectrum). However, I think I have seen simulation models of channels with a signle tap in which the tap was also assigned a jake spectrum, or channel models such as the TU6, in which a jake spectrum could be assigned to each individual tap. Wikipedia says "Let the scatterers be uniformly distributed around a circle at angles alpha_n with k rays emerging from each scatterer.", which would also mean that a single tap could have a Jakes spectrum. en.wikipedia.org/wiki/Rayleigh_fading#Jakes's_model
    Can a single tap have a Jakes spectrum or do I always need several taps? Thank you very much for your help!

    • @iain_explains
      @iain_explains  Před 9 měsíci

      It depends on the model. Some channel models have a finite number of "scatterer delays", where each one is assumed to be made up of a collection of individual scatterers (each with a different amplitude and gain, and potentially a different doppler shift - since the scatterers that are grouped together in time/delay don't need to be physically located near to each other). In this case, each of the "scatterer delays" (or "channel taps") would have a Doppler spread.

  • @Kashmiri.111
    @Kashmiri.111 Před rokem

    I think It's not always necessary that the receiver is moving for the Doppler shift to take place in some case the source could also be moving
    I have some questions:
    What is θ?
    If it's the angle b/w the incident wave and the direction of motion (of either the source/receiver).
    Then what would happen when the receiver is stationary and the source is moving away from the receiver? when θ would equal to 180° and cos180 is -1 and Doppler shift would have the lowest possible value, we know when source would move away from the receiver the apparent frequency will be less than actual frequency.
    Therefore in this case at θ = 180°, apparent frequency < actual frequency
    Now let's consider another case where the source is stationary and the receiver is moving towards the source, then the θ = 180° and the Doppler shift has the lowest possible value, but we know when receiver is moving towards the source the apparent frequency > actual frequency.
    Now the question is, how is it possible that when θ = 180° (in first case) the apparent frequency was less than actual and in second case when θ = 180° the apparent frequency is greater than actual frequency?
    Is there any relation between the Doppler shift and the apparent frequency?
    Is it correct to say: Doppler shift is inversely proportional to apparent frequency?
    Also Bd(i.e Doppler spread) should equal to 2 |fd| , because doppler shift could have negative value if θ is 180 ,in that case Bd will have negative value but doppler spread can't be negative. Therefore the modulus is used.
    I hope i am clear and i don't mean to challenge you in any way, I'm just curious with these questions in my mind. Maybe the concept is not clear to me(replayed this video for atleast 20+ times).

    • @iain_explains
      @iain_explains  Před rokem

      Any movement of any element will cause Doppler shift: the receiver, the transmitter, or even just the reflectors in the channel. I don't use the term "apparent frequency". It sounds like a vague term to me. I think it's better to stick to well defined terms, otherwise things can get confusing.

    • @Kashmiri.111
      @Kashmiri.111 Před rokem

      @@iain_explains sure and what about theeta? Is that the angle between direction of arrival of wave and direction of motion of mobile receiver? I had a thought of it might be the angle between the direction in which the receiver receives the signal and the Relative velocity vector of source wrt receiver, am i right?

    • @iain_explains
      @iain_explains  Před rokem +1

      Yes, it is the angle between the direction in which the receiver receives the signal and the relative velocity vector of source wrt receiver.

    • @Kashmiri.111
      @Kashmiri.111 Před rokem +1

      @@iain_explains thanks professor

  • @herry8377
    @herry8377 Před 10 měsíci

    If the transmitter doesn't move, but there are some high mobility vehicles between BS and UE.
    Would it cause the doppler shift in this situation?

  • @hydersyed7998
    @hydersyed7998 Před 3 lety

    Good explanation with a real time example. Is there a video on your channel for "Delay spread and Coherent Bandwidth"?

    • @iain_explains
      @iain_explains  Před 3 lety

      Delay spread and Coherent Bandwidth are on my "to do" list. I haven't got around to them yet, partly because they are somewhat arbitrarily defined, so there's not much to say about them. I'll give it some more thought.

    • @iain_explains
      @iain_explains  Před 3 lety

      You might like to watch my video on ISI: "What is Intersymbol Interference ISI?" czcams.com/video/I087FUvW2ys/video.html

    • @hydersyed7998
      @hydersyed7998 Před 3 lety +1

      @@iain_explains Sure I will watch. Even ISI is a bit difficult for me.

  • @maniys
    @maniys Před 2 lety +2

    I wish u more subs Professor

  • @se2702
    @se2702 Před 2 lety

    I was recently looking at a transmitted microwave signal on a FFT display. The signal was generally a clean trace form but with occasional slanted doppler shifts caused by aircraft scatter. When it began to precipitate (in the form of heavy sleet) the signal pattern became constantly wider on the display due to multipath. Was this doppler spread that I was seeing at that time? Thanks! You are the best.

    • @iain_explains
      @iain_explains  Před 2 lety +1

      Yes, the sleet will most likely have caused scintillation. Multiple paths coming in from different angles will each have their own Doppler shifts, which gives an overall Doppler Spread.

  • @davey4617
    @davey4617 Před 2 lety

    If the car was travelling toward the transmitter, it would create the higher frequency components travelling toward and then lower then it goes past it. But when the car is directly under the transmitter is there any special result. Do the higher frequency signals add up to cancel out or something when the car reaches essentially at 0 displacement from the transmitter. As displacement is 0, I guess velocity is 0. Would this affect the signal received by the car? Hmm confused myself I think.

    • @iain_explains
      @iain_explains  Před 2 lety +1

      If the car is directly under the base station, then its velocity towards/away from the base station antenna is zero, along the direct line-of-sight path (it's not moving up or down, only horizontally on the earth). It's the relative velocity between the transmitter and receiver that matters - in the direction of the radio wave propagation. If there are reflected propagation paths (eg. off nearby buildings), then there will be doppler shifts along those paths, because their directions will include horizontal components.

    • @davey4617
      @davey4617 Před 2 lety

      @@iain_explains I sort of understand your point but I still don't fully understand. In my head I think of the car under the base station at that point in time, the same as if a person were just stood there.

  • @bkquanvu3228
    @bkquanvu3228 Před rokem

    a beautiful car at 6:20 ::))

  • @samthou7334
    @samthou7334 Před 2 lety

    Sorry can you explain what was Doppler spread again?

    • @iain_explains
      @iain_explains  Před 2 lety +1

      I'm not sure what you're asking, sorry. My explanation is in the video. Maybe try rewinding and watching again - sometimes things click together on a second viewing. Or perhaps you could be more specific in your question?

    • @samthou7334
      @samthou7334 Před 2 lety

      My question is" what is Doppler spread"?
      Is it just the range of frequencies of Doppler shift?

    • @iain_explains
      @iain_explains  Před 2 lety +1

      Sorry, maybe I'm missing something here? The answer is "yes". I explained exactly the answer to your question at the 4:35 min point of the video.

    • @samthou7334
      @samthou7334 Před 2 lety

      Ok...i just wanted to make sure😅 because i have been seeing so many different definitions in the internet.. anyway thank you for the clarification 👍