How are Complex Baseband Digital Signals Transmitted?
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- เผยแพร่เมื่อ 20 ก.พ. 2022
- Explains how complex baseband digital signals are transmitted, from both a a time domain and frequency domain perspective.
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Always wondered about this, finally makes sense to me. Thank you Professor Iain!
Glad it was helpful!
Nice video! It's always nice to come back here and review the basics after a long day of studying for my graduate level data communication classes
Glad you like the videos. It's always useful to refresh the basics, as you say.
Very good step by step explanation !
So clear for a French people. Every tricky technics becomes easy with you ☺️.
I would have liked to have you as a teacher. Thanks 🙏
Thanks so much. I'm glad you are finding the videos useful.
excellent explanation (as usual)
Glad you liked it
Thank you. I like to watch your videos. This time I think a key principal has been overlooked. The statment that sending 2 signal at the same time over the same baseband do not interfere each other may be missleading. The fact that one has been multplied with a sine and the other with a cosine, means that the signals are transmitted with a phase shift of pi/2.
I don't agree that it's misleading. It is exactly what is happening. Two signals are being sent at the same time, in the same bandwidth. The fact this is possible is because of the fact that sin and cos are orthogonal over a symbol period T, and that the digital data is constant over that time period. That's the whole point of the video. Yes, sin and cos are pi/2 phase shifted. But that fact alone does not explain why the quadrature mixer works.
This is the perfect mix of mathematical formulas and graphic intuition. Thanks
Glad you liked it.
Sir you are a world class researcher. You may explain modern access techniques like RSMA. Also you may explain fractional programming for communication systems to maximize data rate. As these techniques are state of the art as is your level.
Thanks for the suggestions. I've added them to my "to do" list.
Thank You. I have 2 short questions. In the second line, there is the complex signal in the frequency domain in the bb. In the time domain that signal would be with a complex part (z=x+jy), and in the first line, the real signal is without a complex part in the equation, is that correct?
To my mind that is the difference of a real and complex signal.
2. The equation that you show on the bottom (integral), is the equation of the orthogonal criterion of signals - is that correct?
Yes, that's correct (both things you mention).
thank you
You're welcome
Hi Iain, thanks for your video! I found some books making a complex signal by multiplexing the quadrature component with -sin wave, which is Q(t)*(-sin(2pf_ct)), but what you did is Q(t)*sin(2pf_ct). May I ask what is the difference between them?
The negative changes the phase by 180 degrees, that's all. It's still orthogonal to the cos waveform.
A short question, please. From the IF we convert down with a Digital Down Converter (DDC) to get to the baseband. Do we always digitize on that way? It is often said "the digital baseband" - can we also be in the analog baseband?
And within the bb, can we represent it in the time and frequency domain?
Thanks a lot.
The baseband equivalent signal model applies for both digital and analog signalling formats, however the output of the channel can only be represented at baseband in seperate I-and-Q form, for the digital format. When sending a baseband signal that is constant during each (digital) period T, the output of the channel separates nicely into I-and-Q, because sin() and cos() are orthogonal functions over a full wavelength. If the baseband signal is _not_ constant (which is the case for analog signals), then the baseband components are not orthogonal at the output of the matched filter receiver. See this video for more details: "What is a Baseband Equivalent Signal in Communications?" th-cam.com/video/etZARaMNN2s/w-d-xo.html
could you do a talk on the costas loop? I think its related but this performs carrier recovery.
Thanks for the suggestion. I'll add it to my "to do" list.
Wonderul explanation sir!
Why are we integrating the product of alpha cos squared over a time period T?
Hopefully these videos will help: "How are Correlation and Convolution Related in Digital Communications?" th-cam.com/video/We5q5FJcbcU/w-d-xo.html and "What is a Matched Filter?" th-cam.com/video/Ci-EjiMJo3I/w-d-xo.html
Hi Iain, first a lot of thanks, the videos really helping and good. I have a question, where did the Integral (in the receiving end) came from? when I learned this topic, it was similar but after the multiply by cos or sin there was LPF that killed the other component, so I little confused, thanks!
Yes, I mentioned the LPF at the 7:15 min mark of the video and drew the shape of the filter with the dotted line on the frequency domain graph. The LPF eliminates the high frequency "copies" but it still has an output that is a continuous time waveform. After that there still needs to be something at the output of the LPF to turn it into a digital sample of the input data. This is done with an integrator (to "add up the energy" over the digital time period T) and a sampler.
@@iain_explains Okay, thanks a lot
@@iain_explains Maybe a short continuation here. If the LPF eliminates the high frequency copies, why is the component with cos(4pi fc t) still within the integral? Thanks
Alright so i am still a bit confused about this subject. Would it be correct to say that we want to send a complex signal with a certain amplitude and phase (i.e. a point on the constellation diagram) but this is not possible with only one signal (because we can only create real signals). So to circumvent this problem we send two signals, one sine and one cosine. Each of which has a certain amplitude to create the desired point on the constellation diagram. We combine them and send them through the antenna. Then at the receiver we decombine them again into the seperate sine and cosine, read of their amplitudes and then use those two amplitudes to create the point on the constellation diagram?
Yes.
@@iain_explains Alright, thanks!
One question here for this example, the symbol which is represented by four bits is multiplied by cos and sine ? or there’s 2 bits (four levels ) for cos signal and the other 2 bits for sine
In general, QAM constellation points are mapped to the data bits according to either grey coding (one bit change between neighbouring constellation points), or maximising the hamming distance along a trellis (as in trellis coding). See "What is Trellis Coding?" th-cam.com/video/rnjy4_gXLAg/w-d-xo.html
Thank you. Such a nice video! Can I ask why complex baseband is used rather than real baseband? What are the advantages?
Complex baseband is not real. It's just a convenient way to mathematically represent what's going on in "passband" communications (eg. wireless communications over a given allocation of radio frequency bandwidth).
Could you present a carrier recovery of high order QAM? that would be great! Thank you.
Thanks for the suggestion. I've added it to my "to do" list.
Thank you for your video and your careful lecture! May I ask another question? This question has puzzled me for a long time. I completely agree with what you said, that a complex signal is actually composed of two real signals. In the frequency domain, the baseband signals and the passband signals shifted to the carrier point are conjugate symmetric. But why is the spectrum not symmetric when they are added together to form the so-called complex signal? How can we visually explain all of this in the frequency domain?
I'm not sure exactly what you mean, but this video might help: "What is Negative Frequency?" th-cam.com/video/gz6AKW-R69s/w-d-xo.html
That's a good question which I had in my mind too. The point is, when you sum up these two components, you need to multiply imaginary part (modulated by sine) by "j". Also note that these are not just symmetric, they are conjugate symmetric. When you sum them as I + Q*j, you will see that they are not conjugate symmetric anymore. I leave a simple example from MATLAB below:
>> a2 = [-1 -1 1 1 -1];
>> a1 = [-1 1 1 1 -1];
>> fftshift(fft(a2))
ans =
1.2361 -3.2361 -1.0000 -3.2361 1.2361
>> fftshift(fft(a1))
ans =
-0.3820 + 1.1756i -2.6180 + 1.9021i 1.0000 + 0.0000i -2.6180 - 1.9021i -0.3820 - 1.1756i
>> fftshift(fft(a1+1i*a2))
ans =
-0.3820 + 2.4116i -2.6180 - 1.3340i 1.0000 - 1.0000i -2.6180 - 5.1382i -0.3820 + 0.0605i
14:07 I'm sorry I don't understand why you approximate the integral of cos to 0. It's going to be 0 only if T is a multiple of 1/fc, am I wrong?
When T is a multiple of 1/f_c the integral is _exactly_ zero. I briefly mention this at 15:52 but I should have made it more explicit. However, even if it wasn't an exact multiple of 1/f_c, then the integral would be approximately zero, since 1/f_c is always much smaller than T, so the portion of the wavelength in the time between 0 and T that doesn't "cancel out" (ie have balanced positive and negative components), will only be for a very short time period, and therefore contribute only a very small amount to the integral.
I LOVE YOU!
Thanks, I'm glad you're finding the videos helpful.
very helpful sir
i have a request video on topic OQPSK offset quadrature phase shift keying
thanks
Thanks for the suggestion. It's on my "to do" list.
Incredible
Thanks. Glad you liked it.
@@iain_explains you're welcome. I always keep watching your channel. Yesterday, I tired to examine the orthogonality between signals using MATLAB. You inspired me to do that.
Nice one! That's great to hear.
@@iain_explains Actually, I'm switching from computer engineering to the communication. I tried to upload a short video on my channel demonstrating an example about the orthogonality of the subcarriers in ofdm. I would really appreciate if you give me a feed back, probably, there might be some missing point in it. 😊
th-cam.com/video/yAoAvw-RVfQ/w-d-xo.html
the beef is at the end... it all has to do with the phase and intergrating over the symbol T time
Yes, that's right.
This is a question that has bothered me for quite some time but i've never looked it up: Can imaginary signals transmitted in another way than this, or is it simply physically not possible to emit an imaginary signal?
Well, the important thing to remember is that all signals are real (even if we call some of them "imaginary"). This video will hopefully help: "Is the Imaginary Part of QAM Real?" th-cam.com/video/6asDtzaVjbQ/w-d-xo.html
Why there is (2 * pi) in cos (2 * pi * fc * t) function?
Radial frequency (radians per second) equals 2pi times the frequency in Hz (cycles per second).
I came here from the "OFDM and the DFT" video. I understand the principle of OFDM, but I don't think the IFFT operation gives the full picture. It's not clear to me why OFDM block diagrams usually show the IFFT as the key operation and nothing else. Separating real and imaginary is a completely non-linear operation, and this is not reflected in the block diagrams. In other words, when you create a complex baseband signal using the IFFT, it's not possible to shift it up to a carrier and make it real through a linear operation. Is this correct?
Those block diagrams you're talking about are only showing the baseband perspective. The block diagram I showed in this video shows what's needed to actually send the signal in the passband.
@@iain_explains Thank you very much for the quick response! in real world algorithms, is the OFDM IFFT and the up-conversion fused into a single operation, or does it stay separate as drawn here?
What do you mean by "fused into a single operation"? The IFFT is a digital operation performed in a digital ASIC. The up-conversion is an analog operation performed using mixers, filters, amplifiers, etc.
@@iain_explains Ok, that makes sense. I was just wondering if, at a base station receiver for example, the carriers are down-converted digitally using an FFT as well.
isn't that true? "even if the baseband signal is real or complex the band pass signal is always real".
Not wanting to sound too pedantic, but real signals are real, and complex signals are not real.