Sampling Bandlimited Signals: Why are the Samples "Complex"?
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- Опубліковано 30 січ 2022
- Explains the process of sampling a band limited signal by down converting to baseband, and compares it to the sample rate required for a direct sampling approach.
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Very clear explanation, just a few thoughts: if you use a complex sinusoid to demodulate, there's no need for low pass filtering as the replicas at twice the frequency cancel themselves out. Then, the reason why we need sin and cos is because the signal at RF is not symmetrical around the carrier frequency. If it was, like in the case of an AM transmission, we could just demodulate with a cosine and the signal at baseband would be real.
My intention was to talk mainly about the general case of a general band limited signal (not symmetric in the passband, not carrier phase synchronised), and to use the AM radio example just as an illustration of one type of band limited signal. But I realise now that I forgot to make this point clear at the 2:40 min mark, before I started talking about sampling the bandlimited signal. Oh well. The things I've said in the video are all true. It's just that in the special case of AM, it's not all needed, you're right.
But just on one specific point: you're correct in saying that the "cos multiplied and LP filtered" signal would be real - but that's actually true for _any_ "cos multiplied and LP filtered" signal - even if the passband is not symmetrical. The same goes for _any_ "sin multiplied and LP filtered" signal. The low pass versions of each are always "real".
@@iain_explains fair enough. If the band pass signal is asymmetrical (ie it contains an "imaginary" component) and we only use a cosine to demodulate it, only the "real" part would end up at baseband, while the "imaginary" part would end up at twice the carrier frequency, pretty useless.
Yes, and that's why we need to demodulate with the sin wave too.
Thank you!
You're welcome!
Hello prof. could you make a video on RIS aided cell-free massive MIMO system.
Thanks for the suggestion. I've added it to my "to do" list.
Always very thank you !!!! Today I reading a sampling theory book from beyond bandwidth. If possible please explain what it is.
Sorry, I'm not sure what you're referring to exactly.
Sir please make a video on FFT
Are there particular aspects you'd like to know about? I ask this because the FFT function is exactly the same as the DFT function (the difference is just that the FFT is implemented in a more efficient way). I've got videos on the channel about DFT already, which might help you, if you haven't seen them already. For example: "Discrete / Fast Fourier Transform DFT / FFT of a Sinusoid Signal" ua-cam.com/video/lwQTNcWtN7w/v-deo.html and "How does the Discrete Fourier Transform DFT relate to Real Frequencies?" ua-cam.com/video/pIFz84oj9cA/v-deo.html and others that you will find at iaincollings.com
Excellent video prof Iain. One thought. Issue of high sampling of bandpass signals may not be related to using sin and cos for demodulation I feel. Bandpass signals can be sampled at much lower rate (say for e.g. at twice the bandwidth under certain conditions) .
I think you're referring to sub-sampling or under-sampling techniques. Yes, if the passband is conjugate symmetric then you can sample at a low rate, chosen such that there are aliased copies centred on f=0, and also chosen to ensure that the other aliased copies do not overlap. This is a way to avoid multiplying by the cos and sin. If the passband is not conjugate symmetric, then the sampling frequency needs to be twice as high as the conjugate symmetric case (which is equivalent to needing twice as many samples, which is equivalent to needing the same number of complex valued samples - nothing comes for free).
Thanks a lot anyway for the wonderful videos you make. Makes learning from first principles much enjoyable
That's great to hear. I'm glad you like the videos.
sir can i conclude that when we are transmitting the signal its a real signal i.e is either a sine or cosine component but at the receiver due to delay and fading and constructive and destructive interference its become a complex-valued signal which means both sine and cosine components are present in the signal at the receiver therefore we have to multiply it with sine and cosine wave both
Hopefully this video answers your question: "How are Complex Baseband Digital Signals Transmitted?" ua-cam.com/video/0lkRJgnywkg/v-deo.html
Sir, In OFDM systems we complex sample exactly at the same rate as bandwidth? but here we sample at twice the bandwidth! confused!
Ah, this is a common confusion. In this video I am talking about sampling the band limited signal in a way that would enable me to exactly reproduce the band limited signal if I wanted to. ie. you need to sample at twice the highest frequency component of the complex baseband signal. In OFDM (or any other digital communications waveform), the task is to "sample" the signal in order to recover the digital data that the waveform represents. This is a different sampling task. It does not allow you to fully reconstruct the OFDM waveform - only to extract the "data". So therefore you do not need to sample as fast - only at the rate of the data symbols.
Hello,
Why do I need the sin multiplication? If I did the modulation with cos wave so I returned back to my original signal(they look the same on the frequency domain)
Well, firstly, the video is intended to be more general than just for RF communication signals. It's for sampling any band limited signals. I just use RF modulation as an example. And secondly, what you're saying only works if you have _exactly_ accounted for (and cancelled) the phase offset between the transmitter oscillator, the receiver oscillator, and whatever other phase rotations come about from the channel. These days phase adjustments are mostly done in the digital domain after sampling - so it is necessary to sample both the cos component and the sin component.
@@iain_explains Thank you. VERY helpful and insightful videos!!
Thanks. I'm glad you like them.
@@iain_explains
I think that's regarding coherent and non-coherent demodulation?
When we use SDR to get weather satellite data, we used a costas loop after sampling because there was no carrier reference. Not sure but I think its because just the IQ baseband was not good enough.
@@iain_explains but sir what if we are transmitting only real signals. like case of BPSK where our signal is real than we not need to multiply with sine at the receiver end despite of the phase rotation occuring due to the delay of channel . what can be reason for this
Sir after 6:12 nothing is clear. Please explain
Hopefully this video will help: "Orthogonal Basis Functions in the Fourier Transform" ua-cam.com/video/n2kesLcPY7o/v-deo.html
So, this wouldn't apply to FM RF?
FM uses a slightly different demodulation method (since it's not a single carrier system). For example, for FSK, it most commonly uses a bank of matched filters (one matched to each of the frequencies).
I think this is how software defined radios work.
2:33 Unclear. Why are you adding your carrier to your modulator? Unclear what "sample rate" means with AM.
I didn't add the carrier to the modulator. I simply indicated that the highest frequency component in the modulated signal is f_c+f_m
There are gaps in the explanation. Please do not skip the intermediary steps.