Tutorials on Digital Communications Engineering

62 comments on “Tutorials on Digital Communications Engineering
  1. Jean-Benoit Larouche says:

    Hi! First of all, thanks a lot for all these tutorials, I am finishing my master degree in digital telecommunication and the tutorials helped me alot in different subjects! keep on the good work!

    At this moment, I am working on TCM and I read many times your tutorial on it. My goal is to prove that using TCM, with a 2/3 rate convolutional coder and a QAM-16 constellation mapping, I can acheive the same or better coding gain then a standard 8-PSK modulation scheme. My problem is to actually find a good 2/3 convolutional encoder with a good minimum squared distance. I tried the one in the tutorial, but with the suggested QAM-16 constellation mapping, the shortest path, is definitly too short to get a significant coding gain. Do you have any suggestion regarding a good convolutional encoder for QAM-16 TCM?

    I hope I was clear,

    Thanks a lot,

    JB

  2. idmond says:

    Hi, I really want to thank you for the effort you put in these amazing tutorials and I hope you keeping posting more and more of these tutorials. I just have one question about AM modulation:

    “In AM synchronous demodulation, Why we don’t divide m(t)coswt by cos(wt)
    instead of multiplying by cos(wt)?”

    When I was taking a class in Analog Communication… I recall that I was watching one of the video lectures on signals and systems from Berkeley University, the Professor was talking (in that video recording of the lecture) about amplitude modulation and why when we want to demodulate an AM modulated signal , in synchronous detection, we multiply the AM modulated signal = m(t)cos(wt) by cos(wt), where w is the carrier frequency, and why we don’t just divide by cos(wt) to get m(t) back, since this can be easily implemented by a simple divider circuit.

    well, he didn’t answer that question in the lecture, may be he just left it as an assignment for his students, I don’t know. but it just got me thinking about it, if it’s all about thinking mathematically, then it seems like it’s more intuitive and a whole lot easier if we just divide by coswt, why we go through all the trouble and multiply then we have to know the trig identity of (cos(a)cos(b)) and then put a BPF centered at w after the output…

    • Victor Levin says:

      While multiplication followed by a lowpass filter is equivalent to division, it is actually very different in implementation. The main problem with division is the issue of dividing by zero every period. This would cause amplitude spikes that would saturate the circuitry around the system. These can be avoided by syncing exactly the phases of the two cosines. Now instead of dividing by zero at t=2pi/w, there is cos(wt)/cos(wt) where both go to zero at the same time. Using L’hoptials rule, you can see that when t=2p/w, the output is 1 not infinity. But even if we managed to have perfect syncing, we still would have to build a divider circuit. These are more difficult to implement than a simple transistor mixer.

      Overall its just much simpler to build and implement the mixer to lowpass than a division circuit. But mathematically they are completely identical!

      • idmond says:

        Thanks for your reply Victor. I just don’t think that the L’hoptials rule will still work in circuit implementation. it is mathematically right, yes, but in practice, if the two cosines are in sync, then they eventually have to cross the value of zero at the same time, not just keep getting close to it, so you don’t get 1 here, but amplitude spikes.

    • Qasim Mahmood Chaudhari says:

      I was also confused about it for years. Then I realized that it is actually division if you consider the modulation/demodulation operations in complex lowpass equivalent domain.

  3. admin says:

    Divider circuits were not easy to build. That was the main impetus for doing it this way.

    • idmond says:

      why it isn’t that easy to build a divider circuit? I mean are there actually divider circuits that can divide by a function(not a number like arithmetic dividers) or we’re able to implement a divider circuit that can divide by a sinusoid function like cosine, but it’s much easier to implement a multiplier circuit than a divider one.

  4. MALAMMANAVARA RUDRESHA says:

    At this moment, I am working on fft/ifft for OFDM and I read many times your tutorial on it. My goal is to prove that using FFT/IFFT, with a 2/3 rate convolutional coder and a QAM-16 constellation mapping, I can acheive the same or better coding gain then a standard 8-PSK modulation scheme. My problem is to actually find a good 2/3 convolutional encoder with a good minimum squared distance. I tried the one in the tutorial, but with the suggested QAM-16 constellation mapping, the shortest path, is definitly too short to get a significant coding gain. Do you have any suggestion regarding a good convolutional encoder for QAM-16 TCM?

    I hope I was clear,

    Thanks a lot,

  5. Dacosta says:

    Am studying analog anddigital communication,I really don’t understand the terms like QAM,FPSK,QPSK,FSK ETC

    • Charan L. says:

      Well then you should read some of these tutorials I have here, particularly the one on modulation which explains what these terms mean.

  6. priya rawal says:

    ma’am will u plz mail me a tutorial on ofdm

  7. NISHANT GHOSH says:

    “In AM synchronous demodulation, Why we don’t divide m(t)coswt by cos(wt)
    instead of multiplying by cos(wt)?”
    The divider circuit (or the system that divides m(t)coswt by cos(wt)) is non-invertible system. In the whole modulation and demodulation process we want a robust transmission of m(t).

  8. nishant ghosh says:

    Divider circuits are non-invertible systems. Here our system is just passing the message signal to the receiver terminal. If we are passing m(1) then we will get m(1) at the receiver but if we wish to transmit m(pi/2) then you cannot obtain m(pi/2) at the output…. you will have an indeterminate mathematical form which is not desired…

  9. Ben Thomas says:

    Hi,
    I have been an electronics engineer for nearly a decade. Although I was only recently steered towards this site by a friend, the experience of going through these tutorials was like meeting a friendly and extremely interesting stranger whose charm does not fade with time, thank you for this excellent content and wishing you the very best.

  10. Vishwas S Patel says:

    I have a basic doubt. Consider a sine wave, if you take two times the sampling rate according to nyquist theorem. If 2 samples taken in a single period is at 0degree and 180degree.. Then we end up having all zeroes.. Can you please explain it?

    • Charan L. says:

      Okay, what sine wave can you fit through these two points? Many, but the one with the smallest frequency is 1 Hz. And that/s all Nyquist is saying.

      Actually the 0 points mean that you have used a NYquist frequency that is exactly 2 timrs, but the definition says it needs to be larger than two tims the maximum frequency even if infinitesimally. Once you increase it a tiny bit, more than two times, you will not get those zeros and then can fit many harmonics through those points.

      Charan Langton

  11. shid says:

    Hi,Charan
    Can you explain more clearly about peak to average power ratio in matrix persepctive or other ?

  12. Just wanted to offer a sincere thanks for going to the trouble of putting together such a great site. You have a gift of making complex concepts easier to understand. This is a great use of the internet.
    Best regards. Jim

  13. Antriksh says:

    I and my students are very thankful to you Charan for your tutorials and sincere efforts to make the tutorials easy and quick explanatory.especially your tutorial on Viterbi has helped me a lot to understand and teach precisely .
    Hope to see some book on Communication soon.

  14. Karthik Ram KV says:

    Hi Charan,
    Your tutorial on OFDM was simply superb. A complex concept is explained very simple with comparisons to single truck shipment and multi truck shipment analogies and also water flow in faucet and shower analogies…. Hats off man…. Thanks a lot…

  15. Maruthi Prasad B N says:

    Good analogy towards the complex concepts…..!!
    found it worth sharing..
    thank u…

  16. Kris says:

    Hi,

    I don’t know whether this question can be asked here. But kinldy guide me, if you know the answer

    I have a 3 node system, with nodes in a line and the center node is taken as receiver equidistant to the other nodes.

    My objective is to get BER at receiver in Rayleigh fading channel, with one of the nodes as interferer to the other node’s transmission and BPSK modulation is used.

    With fading channel coefficients known at receiver as h1, h2, I tried to get BER using threshold detection as

    Assuming source transmits +1, BER is

    P(received signal<0)=P(|h1|+|h2|i+n<0) {where i:signal from interferer (+1, -1)};
    n: AWGN
    =P(n<-|h1|-|h2| given i=+1)*P(i=+1)+P(n<-|h1|+|h2| given i=-1)*P(i=-1)

    Finally I averaged over |h1| and |h2|.

    Then it is observed that BER with fading is lower than Ber without fading for high values of SNR.

    Kindly let me know whether this analysis is correct.

    Thanks in advance,

    Regards
    Kris

    • Charan L. says:

      Hi Kris,

      Not clear if you are going a simulation or a closed form solution.
      You will find the comparison of the closed form BER for a fading vs. a AWGN channel in my MIMO tutorial.
      And yes, you should get a higher BER for for same SNR under a fading channel.
      So you have some issue in your analysis.

      Charan Langton

  17. Jorge says:

    Hi Charan,

    Awesome tutorials. I’ve been working on designing an OFDM scheme for a a channel. We were told to assume that a channel bandwidth of 1MHZ is available.

    The channel has the following multipath profile:
    t(us) 0 1 2 3 4 6
    P(db) 0 -10 -20 -0 -20 -20

    I am asked to determine the required frequency spacing for the subcarriers such that each subchannel experiences a flat channel. Determine the resulting OFDM symbol period and Cyclic prefix to ensure no ISI among OFDM symbol.

    So far this is what I’ve done:

    my delay spread is : 1.49us
    Bc0.9: 13.3 KHz, therefore my subcarrier frequency spacing is 13 KHz.
    Number of subcarriers : 76.92
    My OFDM symbol period is: 76.92us

    when it comes to finding out the CP, i’m not sure if I can assume it to be 5us.

    Am I on the right track here or really off it?

    HELP PLS.

    • Charan L. says:

      Your CP needs to be just a little greater than your delay spread.

      Charan

      • Jorge says:

        Thanks a lot Charan, your reply brought me back to life.

        I actually thought that CP needed to be greater than the second moment of the power delay. It makes more sense now.

        Can I conclude my required frequency spacing for the sub-carriers to be 13 KHz ( as I got a Bc0.9= 13.3 KHz)and a a symbol period of 76.92us, so that each sub-channel will experience a flat channel?

        Thanks for your help and advice.

        • Charan says:

          hi Jorge, the frequency spacing should be more than the coherence bandwidth. Can you calculate that and see if it is near the 13 kHz. The other thing is that you need a power of two number of carriers so if your total bandwidth divided by the 13 kHz is not the power of 2 number, then you will have to zero pad those extra carriers and can not count them as carrying data.

          Charan

          • Jorge says:

            Hi Charan,

            I’ve been working on calculating that frequency spacing. I came across an info which states that the sub-carrier spacing is the inverse of the symbol time less the guard period; and that the guard time must be 2-4 times the delay spread. Is this accurate?

            Btw, I thought that for flat fading the sub-carrier spacing needed to be below than the coherence bandwidth.

            Anyway, I’m a bit confused, but i’ll keep trying to calculate that sub-carrier spacing.

            Thanks

          • Charan L. says:

            I am going to assume that this is the max delay spread. Now the coherence BW is 1/DS = 666666 Hz. (I had a 2pi factor before, ignore) This is less than your bandwidth of 1 Mhz, so the channel will experience fading, but not a lot.

            So any spacing for the sub-carriers less than this amount will do.
            The number of carriers will depend on the standard you want to use. If you use 802.11, and you choose 48 carriers, this will give you a spacing of 20.8Khz.
            For 48 carriers, we get Tn = 48mus.
            Cyclic prefix we can set as 1/8 of above or 6 mus.
            Symbol time would be 48 + 6 = 54 mus.
            Throughput is 1 msps minus the CP overhead.

            Charan

  18. Charan says:

    Coherence bandwidth relates more to total symbol rate. It is directly related to delay spread.
    2pi/d. For the delay spread, you do not need four times the delay spread, that would be adding a lot extra overhead. No more than x2 would be my choice.
    Yes, the spacing needs to be less, not more than coherence bandwidth not more. I said it wrong.

    Your coherence BW comes to about 4 MHz, but you are allocated 1 MHz. So you are limited in spacing by how much data you want to transmit. The maximum is 1 Msps which would require app. 1 Hz spacing and 1 kHz of data would need app 1 kHz spacing. (minus the CP overhead in each case). Hope this helps.

    Are you using a standard OFDM or can you create anything you want?

    Charan Langton

    • Jorge says:

      Thanks again Charan,

      I’m using standard OFDM. All I have to work with is the multipath profile, which is :

      t(us) 0 1 2 3 4 6
      P(t)(dB) 0 -10 -20 0 -20 -20

      And of course, a channel bandwidth of 1 MHz available. I began by determining the required spacing for the subcarriers such that each subchannel experiences a flat channel.

      For this I calculated the first moment of the power delay profile (taking into account the multipath profile), which came to 1.51us. Then I calculated the second moment of the power delay profile, which resulted in: 4.53us.

      Having these results I then estimated the delay spread, which resulted in: 1.5us. Then, my coherence bandwidth: BC0.9: 13.3 KHz (and BC0.5: 133.3 KHz). I only took into account BC0.9, hence I assumed a frequency spacing of 13 KHz.

      When you say that my coherence bandwidth comes to 4Mhz, is there something I’m miscalculating? or not seeing when applying the formula?

      I know asking this may be silly, but I rather ask than regret later. Sorry for dragging this issue.

      Thanks a lot Charan.

      • Charles says:

        By using Wc:2pi/d you’ll get 4.18 Mhz, around 4Mhz, but When using Bc0.9 formula according to you multipath the result is different. I’m fairly new to OFDMA, so that’s all I can do to help you. Wait to c what Charan has to say about it.

        Sorry Pal.

    • vishal says:

      i studied bandwidth so much that i dont even know what it is anymore!
      for instance 1.laser has higher bandwidth than led’s
      2.optical fibre bandwidth is 10^20 hz something ,which has many channels than copper wire
      3.copper wire and optical fibre are just single conduit ,how can they have may channels .
      these are silly questions but i studied so much on this ,,, i am not able to get insight on this
      please clear my doubts

      • Charan L. says:

        It helps to think of bandwidth as lanes with finite widths on a freeway. Let’s say that the freeway is 500 feet wide. We have six lanes, one on the left edge is narrow and only 20 feet wide. Then we have a wide lane which is 150 feet wide, then we have three more at 100 feet wide. Then we have one at the right edge which is also 20 ft wide. The total space taken up by these lanes is 20+150+3(100)+30 = 500. Now give these lanes name.
        Lane 1: 0 – 20 ft, A Band
        Lane 2 – 20 – 170 ft, B Band
        lane 3 – 170 – 270, C Band
        Lane 4 – 270 – 370 ft, D Band
        Lane 5 – 370 – 470 ft, E Band
        Lane 6 – 470 – 500 Ft, F Band

        The freeway is the electromagnetic spectrum, for 0 Hz to infinity Hz. We divide it into lanes of certain frequency band and those are called bands. Each media is capable of operating in only a few of these bands. Energy can travel in wires only in the lower bands, wireless can operate in pretty much all. The regulatory bodies control at which frequency you can transmit and assign you a small part of a band, and that is the bandwidth that you are assigned. You as an operator can then further subdivide the allowed band if you want, or use all in one channel.
        For example, a satellite operator is allowed to use a band of 12 GHz to 13 GHz for downlink. He can take this 1000 MHz and then subdivide into ten 100 MHz channels. Then he would design his system for 100 MHz (often called transponders) and then place these like lanes on a freeway next to each other separated by 100 MHz center frequency. Each of these sub-bands then has a center frequency and a bandwidth, which is of course 100 MHz, the width.

        A media has a bandwidth “capability” but then there is the bandwidth that you are allowed to use. So one is the characteristic of the transmission media and the other is where you are allowed to operate.

        The allowed frequencies are allocation by the ITU and best thought of as lanes on a road where you may drive and the media capability is best thought of as the with of the car that travels in that lane. The car with width is the media capability and it is always less than the allowed band in which it operates.

        In this picture the colored bands are the allowed ranges.

        http://static.ddmcdn.com/gif/cell-phone-radiation-spectru.gif

        Charan Langton

  19. josephat says:

    i will be glad to see the analysis of blind channel equalization algorithms, especially adaptive one basing on stochastic gradient methods. honestly i have to thank you a lot for your support in academic success.

  20. Naveen Manikandan Periasamy says:

    I’m looking for the mathematical expression for Pulse Width Modulation. I believe it is something close to the following,

    Y(t)=X((t/W)+Theta) (Y is modulated signal, X is square pulse, W is width and Theta is phase shift)

    Please correct me if I’m wrong.

    Naveen

  21. Kaustubh joshi says:

    Hi Charan,

    I would like to thank you for this information and congratulate you on all the effort you must have taken to provide this information.
    I sincerly thank you again.
    Regards,
    KJ

  22. sowmiya krishnagopal says:

    hai mam i am doing my BE final year.I am gng to do project on “energy efficiency power adapative spatial diversity” the ppaper wasx published in IEEE .kindly give me an idea how to make changes in this paper mam.

  23. Zeeshan says:

    hello…
    i am zeeshan and pursuing master in broadcast technology. sir i got confused how a transport stream get divided in transponder bandwidth in DVB-S.

    • Charan L. says:

      Its usually a bit stream and it is demultiplxed, which is a sort of a decimating process. I am not sure of your question, so thats that best I can do.

      Charan Langton

  24. Abhishek says:

    Hi, your website has really helped a lot in understanding of digital communication. Tutorial of raised cosine pulse shaping and isi Are explained in much simple and clear form. Thanks a lot for your efforts.

  25. geeta gupta says:

    why intereference is minimum in case of orthogonal signals?

  26. geeta gupta says:

    why intereference is minimum in orthogonal signals?

  27. Oluwole Oyetoke says:

    Hello, pleasse do you have a good write up also on demodulation just like you have on demodulation.

  28. Dave Kelleher says:

    Charan, simlpy the best explanation of modulation techniques I have come across in my 20 years in Broadcast engineering. Others tend to leave out the realisation of the modulators, however your step by step and diagramtic approach is simply magic.

    dave

  29. Shobhit says:

    Can u please help me with constellation diagram 3pie/8 shifted 8-psk. how to calculate average value and minimum value of the envelope

  30. shuceb says:

    can yuo plz help me tutorial about carrier frequeny offset

  31. Re.y says:

    I’ve read the ofdm pdf . that was great amazing but for continuing i need more …??!
    i download many pdfs but yours was really helpful
    (sorry if i made mistake about writing)

  32. KB. Lee says:

    Thank you for your effort

  33. RAGHUVEER.D.S says:

    EXTREEMELY INFORMATIVE AND INTERESTING TUTORIALS IN DIGITAL MODULATIONS PRINCIPLELS UNDERSTANDING AND THE FUNDAMENTAL CONCEPTS ARE MADE VERRY CLEAR-I AM AN ELETRICAL ENGINEER TRYING TO UNDERSTAND THE FUNDAMENTALS IN COMMUNICATION THEORY AND PRACTICALS IN THIS ERA OF COMMUNICTIONS.-I AM EXTREEMELY THANKFULL TO YOU FOR PROVIDING THIS INSIGHT INTO DITAL MODULATIONS SO CLEARLY AND IN SIMPLE CONCEPTUAL MANNER WITH APPROPRIATE DIAGRAMS AND WAVE FORMS-I AM HIGHLY OBLIGED TO YOUR TUTORIALS PUBLISHED ON LINE.-THANK YOU IMMENSELY.

  34. amir says:

    Love your website , you should become teacher/professor
    Regards
    Amir

  35. mahdi says:

    Thanks a lot!
    I found it fantastic.

  36. Anirudha Kulkarni says:

    Hi,
    First of all I must say that, all the chapters written on this website is a great resource.
    Many thnaks for expaling digital communication in such a simple way.

    I was going through Inter Symbol Interference and Root Raised Cosine Filter chapter.

    I have one doubt in the example of satellite transponder on page 11.
    Probably I am missing something here to understand it.

    The doubt is,

    When we use QPSK ( or any PSK ) , we use the sine and cosine waveforms to send the signal which is already in bit stream form.

    The bit stream form is has square shape pulses.

    But QPSK modulated signal will have portion sine and cosine waveforms.

    So the question is why do you say that the “unshaped” QPSK is square pulsed.
    Pardon me if I am not able to understand the context of “unshaped”.

    In any case, if a signal is digitally modullated then the bits would be sent in the form of sine and cosine only.

    So why do you say that pulse shaping is required on the square pulse before sending.

    Regards,
    Anirudha

  37. Muhammad Altaf says:

    Firstly i want to thank you for putting together such good tutorials. I am doing my masters in Communications and Signal Processing. This forum has helped me a lot since the commencement of my degree. I have a request if you can take out some time to prepare a tutorial on FBMC (Filter Bank Multi Carrier) modulation scheme. Currently, I am about to start a project on this topic but can not find a specific and to the point tutorial on this topic. Your effort would be really appreciated.
    Thank You

  38. omar says:

    its understanding to good

  39. Subodh says:

    Hi Charan,

    The information on Fourier analysis was really awesome. Big Fan !!!!

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