What are SNR and Eb/No?

[@jamesb5213]

Hi Professor, thank you so much for the video! A couple quick questions. I guess i dont quite understand how the word "power" is used both in the freq domain ( area of PSD) and just "power" in time domain (square of voltage). How is the area of the PSD (freq domain) different from a value in the power display (time domain), even though we call them both, "power?" Im a bit confused at 7:55. We are considering the energy of the noise wrt a time value of T. However, by "coloring in" the signal across all lobes for the symbol, it looks like we are considering the energy of the symbol wrt a time value of many T (since just the main lobe has a duration of 2T). Shouldn't the energy of the noise & symbol be compared with the same duration?

[@iain_explains]

The PSD is a power _density_ , so you need to add the area under the curve between two frequency values, in order to find the total power in that frequency band. This is analogous to the density of a gas in a balloon. To find the total mass of the gas in the balloon it would be necessary to multiply the density of the gas by the volume of the balloon. In the time domain, when you take the square of the signal you find the _instantaneous_ power at that time, t. The sum of the instantaneous power over all time equals the sum of the PSD over all frequencies. In answer to the second part of your question, you need to add all of the power in the transmitted waveform that is sent to represent a digital bit. This goes over multiple digital time periods (multiple T "slots"). For more details on this, see: "Pulse Shaping and Square Root Raised Cosine" kzbin.info/www/bejne/h5abf4Suac6Ve5o

[@eswnl1]

I wondered what the difference between SNR and E/N and I think it's because the SNR formula contains the signal bandwidth but for the E/N, it disappears because of the multiplication by T. So E/N is independent of bandwidth but not for SNR.

[@Julia-hu4xe]

To the little sketch on the left side where you draw the impulses (sinc shape). At that sampling time we can detect the current amplitude, that's clear. But we can also detect the phase at that time, is that right? Thanks.

[@iain_explains]

That picture is at baseband (so there is no phase). This video may help to see what I mean: "How are Complex Baseband Digital Signals Transmitted?" kzbin.info/www/bejne/Zp3Og32do96qock

[@Regi62]

I really liked this one. Thank you

[@iain_explains]

You're welcome!

[@pitmaler4439]

Hello, thank you for your interessting videos. I have 2 questions. 1. Isn't it more accourate to say: the area under the curve is the energy of the signal instead the power? 2. I read something about the Energy Signal. There you get integral (x(t)^2) from -infinite to +infinite. That refers to the time domain. One can say in the time- and the frequency domain: the area under the curve is the the enegy of the signal - is that right? (I only found the integration in the time domain. That confuses me a bit.) Thank You.

[@iain_explains]

I think you're getting confused between deterministic signals and random processes. A deterministic signal x(t) has a Fourier transform X(f), and the area under (x(t))^2 equals the area under (X(f))^2, and it equals the energy in the signal. In contrast, a random process takes random values at any given time t, and as such it doesn't have a Fourier transform. Instead, we define its "power spectral density", which is the density of power as a function of frequency. If you integrate the PSD over frequency, you get the total power. This video might help: "What is a Random Process?" kzbin.info/www/bejne/jWObXmxsds15aNU

[@jameerali2769]

Hi sir, I want to know the Eb/No to SNR when both FEC coding and repeating code is used.

[@manutauer]

Thanks, in the SNR equation, we have in the denominator the term N0*B. It is logical because it is a random signal. Like you said, we determine the p.d.f. and take the area under the curve. But we never have in the nominator something like S0*B. Is that because we have no random signal in the nominator? We cannot go the way over the p.d.f. with deterministic signals.

[@iain_explains]

It's an interesting observation, but your explanation is not correct. The signal is random. The reason we don't see a term like S0*B in the numerator is because the PSD of the signal is not flat over the bandwidth. So we write P on the numerator (we can't be more specific than that, in general). This is the integral of the PSD of the signal, over the bandwidth of interest.

[@manuelbeir576]

Thanks. The unity of the signal energy is (unity)^2, according to the formular e.g. V^2. But, how is that then related to the general unity of elec. energy, Joule? Is the energy another physical quantity than the signal-energy (both have apparently different unities)?

[@iain_explains]

Good question. Power = V^2/R . But it's common to say P = V^2, and implicitly assume everyone knows you're talking about a "unit resistance" (ie. R = 1 ohm). When we're talking about ratios of power in a communications receiver (eg. SNR), then the resistance is the same for the received "signal component" as it is for the received "noise component" (since it's the same receiver receiving both components), and so the R's cancel on the top and bottom of the ratio.

[@ΚωνσταντίνοςΧορόζογλου]

Thank you very much for your video Professor! I think that I am confused abiut the meaning of the No. What values take the No? If the transmitter transmitts with a constant power for each bit, then what influence the values of Eb/No? Why in the BER curves versus Eb/No we guve several values to Eb/No. Of what it is influenced? Thank you very much! I hope you understand my question 😊

[@zahraazaidan1152]

thanks sir ...I'm studying based on your videos

[@stellatauer761]

Thanks, so when I increase the bandwidth, the ratio between signal and noise power doesn't change? Is that correct? (Because the area under the curve increases for the signal and the noise) I somewhere read that the SNR does not increase with the bandwidth, because the noise increases faster with B than the signal. Perhaps that is wrong.

[@iain_explains]

It depends on whether you increase your data/symbol rate at the same time as increasing the bandwidth. If you keep the data/symbol rate the same, then there is no point in increasing the bandwidth, because all you will be doing is including more noise.

[@dansbrazil]

Great video, very well explained. The symbol time, Ts is 1/W. Can I see the symbol time by just looking the sinc function in the time domain?

[@iain_explains]

The sync functions for each data symbol overlap in time, offset by T, as it shows at the 4min30sec mark of the video. The resultant signal is the addition of all the sinc shaped signals, and you can notice that at each integer multiple of T there is only one sinc function that is not equal to zero (since the sinc function has zero-crossings at multiples of T). For more details on pulse shaping, check out: kzbin.info/www/bejne/h5abf4Suac6Ve5o

[@knavaneethan79]

Sir thank you. You contribute much to get a nice job . Thank you 😊😊😊😊😊

[@iain_explains]

I'm glad it was helpful.

[@amitbora6336]

Hi professor, I have a question regarding a few papers where I saw the BER being plotted against the transmit power in dBm. All these years I had seen only BER vs SNR plots but can you please explain how to plot BER vs the transmit power? In this case, do we need to fix the noise power? And is it even fair to compare BER w.r.t. transmit power, which we usually fix it while obtaining BER plots? Thank you!

[@iain_explains]

That's a great question. Many researchers effectively ignore the relationship between the BER-v-SNR curves and the real practical transmit power levels, by assuming that the channel is "normalised to unity gain" (either exactly, or "on average" in the case of fading channels). By making this assumption ("normalisation"), it effectively means that the transmit power and the receive power are the same in their model. In the case you're talking about, where they plot BER-v-P_trans, then yes you're right, they would need to be assuming a particular noise power level at the receiver (unless they are assuming an interference-limited multi-user situation, where the noise is negligible compared to the interference - but then they would have to state the interference level).

[@amitbora6336]

@@iain_explains Thank you professor for the reply. I tried this on MATLAB and it worked. I did the following - First, I varied P_t _dbm= [-20:5:20]dbm so that P_t_linear = 10^(P_t_dbm - 30)/10. I then fixed SNR_db = constant so that SNR_lin = 10^(SNR_db/10). Now, I defined the variance of the noise as sigma_2 = E_s*P_t/SNR, where E_s is the symbol energy. Following this, I obtain the output as y = sqrt(P_t*E_s)*H*x + n, where n is the AWGN and I generate n as n = sqrt(sigma_2/2)*(randn(1,N) + i*randn(1,N)). When I plot the BER vs the varying P_t in dbm, I get the desired results. Thank you so much once again!

[@iain_explains]

Great. I'm glad it helped.

[@adamkimbrough6483]

Video on C/N0 in the context of GPS next?

[@iain_explains]

Thanks for the suggestion. I'll put it on my "to do" list.

[@sebastiancardenas438]

Bro, thanks. A question, because when i increase EbNo decrease the BER(bit error rate)

[@iain_explains]

You might like to watch the following video, which explains how BER and SNR are related kzbin.info/www/bejne/rKWtZ6B3rpirg8U

[@xinliu5261]

Hi Professor, I got a confusing question, let me describe it for you, and hope I can get the solution. The thing is when I tried to use BI-AWGN channel as my simulation channel, I am a bit confused about the conversion of SNR, Es/N0, and Eb/N0. Let us say, suppose without coding, normally with higher modulation(QPSK, 16QAM) two-dimensional transmission the conversion relation is SNR=Es/N0=REb/N0. So I can get the corresponding BER versus SNR and Eb/N0 curve. HOWEVER, in the BI-AWGN case where BPSK is used, I can not figure out the relation between them. I searched some references, is it correct to analyze it in this way? That is for the BI-AWGN channel, the Shannon capacity is 0.5log2(1+SNR)=0.5log2(1+2REb/N0), so the conversion between them is SNR=2Es/N0=2REb/N0? Hope you can give me some thoughts and suggestions. Look forward to hearing from you. Many thanks, Xin.

[@iain_explains]

Perhaps you might like to watch the video again. It exactly answers the question you are asking.

[@xinliu5261]

@@iain_explains Thank you for your reply. I watched the video but the link between the SNR and Eb/N0 is not given. Hence I derived it further, which gives me SNR=Es/N0. However， I saw a relation like this SNR=2Es/N0 for BIAWGN in some books and references. And for higher two-dimensional modulation, SNR=Es/N0. That confused me a bit. One of the books I saw is Channel coding by William E. Ryan.

[@iain_explains]

The relationship between power and energy is simple: Energy per symbol (E_s) equals Power (P) multiplied by the time per symbol (T). Sometimes a factor of 2 appears in some definitions, purely depending on how they define N_0. Some people define the "height" of the power spectral density of the noise to be N_0, rather than the definition I used which was N_0/2. This is simply a notation issue, since the real parameter that is important is the variance of the Gaussian distribution of the noise, ie. sigma^2. The "N_0 people" will have a definition of the relationship between N_0 and sigma^2 that has a factor of 1/2 in it, compared to the "N_0/2 people". For more details, see: "What is Noise Power in Communication Systems?" kzbin.info/www/bejne/laLRZYWwgteLeMU

[@JoeDombroski]

Is the term ENR a term used for Eb/No ?

[@iain_explains]

No - not that I've heard.

[@aritradeb1935]

Sir please give the book name from which we can read all these concepts

[@aarzoosingh8316]

hello sir hope you are doing fine my question is that if we decrease the duration of bit than bit energy is reduced than why is it so that if we are transmitting at higher rates with reduced bit time we require higher energy??

[@iain_explains]

Sorry, but you're question is not very clear. Perhaps you're confusing Power with Energy? You might like to watch: "Signal Power and Energy" kzbin.info/www/bejne/bXqciXiedtqjasU

[@meghanav763]

Thank you so much sir for this valuable video

[@iain_explains]

You're welcome. I'm glad it helped.

[@HarryWoo]

Can we filter thermal noise? You used filter to white noise spectrum and outside of filter bandwidth, noise is reduced in your drawing.

[@iain_explains]

The thermal noise is added to the received signal in the receiver amplifier (for electronic amplifiers that we use today - it's just what happens, there's nothing you can do about it, unless you use expensive low noise amplifiers - and even those are not completely noise free). So yes, we can filter thermal noise - it's happens in the receiver matched filter. For more details see this video (an in particular the comment in the description below the video, which talks about the bandwidth issue): What is White Gaussian Noise (WGN)? kzbin.info/www/bejne/h5e4goCwfcuVrtU

[@iain_explains]

Also, you might like to watch this video for more details: "What is Noise Power in Communication Systems?" kzbin.info/www/bejne/laLRZYWwgteLeMU

[@dimitrisv.1729]

I understand the difference between the two terms. But in high SNR regions, Eb/N0 value is also high, right?

[@iain_explains]

Yes.

[@userx001]

Good demonstration, thank you! My question is : In some cases, SNR may equal to Eb/No?

[@iain_explains]

Yes, SNR = Eb/No when using binary modulation with a sinc pulse shape. With sinc pulse shaping, the signal power is flat across the bandwidth (it is a square function in the frequency domain), and therefore Es = P/W. In the binary case, Eb=Es, and therefore P = Eb W. So if you substitute this into the SNR equation, you are left with Eb/No.

[@userx001]

@@iain_explains Thank for your reply, I've got it. thanks!

[@roseblue3938]

thank you for detailed video. just ask, if i want to plot BER (as x axis) and SNR (as y axis) should i still use line plot or semilogy plot? thanks in advance

[@Darknight38850]

Hello, I'am trying to compare SNR of a real signal (sqrt(2*P)*cos(2*pi*f*t)) with real white noise (sigma²) VS the SNR of a complex signal (sqrt(P)*exp(2*i*pi*f*t)) with complex white noise (sigma²/2 on I and sigma²/2 on Q). When I compute the global SNR (mean(abs(signal²))/mean(abs(noise²)), I have same values for real and IQ signal. But when I extract the signal using a FFT, I have same noise power density (power computed in a canal where the signal is not) for both real and IQ signal, but the power of real signal (computed in a FFT canal where the signal is) is divided by two compared to IQ signal (because half of the power is in negative frequencies). Then SNR estimation of real signal is divided by two compared to IQ signal. I don't understand why noise level in a given fft canal is not divided by two too. Where am I wrong and what can I do to estimate the SNR in a fft canal properly ? Thank you

[@iain_explains]

Sorry, I'm not sure exactly what you mean by "extract the signal using a FFT". However, if you've got noise power in both the real and imaginary components of a complex signal, but you've only got signal power in the real component, then when you put them together and do an FFT, you'll be adding two lots of noise power, and only have one lot of signal power. To put it another way, if you know that your signal is purely real, then you shouldn't be using a complex signal model.

[@Darknight38850]

@@iain_explains Hello Extract the signal using a fft means using it as a filter bank. For instance 1024 points fft does 1024 frequency filters. We obtain the time signal by shifting the input signal by something about 512 or even 1024 points. I'm not adding complex noise to real signal. I am comparing on the first hand a real signal with real noise and on the other hand a complex signal with complex noise. I simply manage power such that signal power is the same and noise power is the same in both cases. So SNR is also the same in both cases. Then I extract the signal using the fft, and lose half of the real signal power but not the noise. So half of the RSB...

[@Darknight38850]

Let's try to be clearer about the issue. In fact what does not correspond to your presentation is the SNR reading using the fft of a real signal with real noise. In the double sided case, when we often write N0/2 on the PSD, we suspect that in a fft bin we have only half on the noise power such as we have half of the signal power. In practice for a given real signal with given SNR = PS/PN, at sampling frequency FS and using nfft points for the fft, we have ~PS/2 in the positive bin corresponding to signal frequency, and PN/nfft of noise power in any bin (SNR in the bin = PS/2*nfft/PN). Which means SNR is reduced by a factor 2 compared to the complex situation where we still have PN/nfft of noise power in the bin and PS of signal power (SNR in the bin = PS*NFFT/PN). Whereas before the FFT, we have SNR = PS/PN in both cases...

[@malifsyahputranasution7650]

sir.. what is the relationship between SNR (signal-to-noise ratio) and variance? matlab have function awgn (for additive, white, gaussian noise) with input SNR and can return variance as the result. Instead of asking us to give variance as the input for function awgn, matlab then asking us to give the SNR as the input, and based on the SNR, the appropriate variance can be obtained. for example, SNR 30 dB gives us variance equal 0.001, SNR 20 dB gives variance equal 0.01, for now I just understand, if SNR high, variance will be low, and vice versa.. I mean the correlation in mathematical representation like your explanation in this video, thank you sir, this question just in case you can explain because your video make me clear about the 'white' in awgn function matlab, but just one more question in my head

[@iain_explains]

Hopefully this video helps: "What is Noise Power in Communication Systems?" kzbin.info/www/bejne/laLRZYWwgteLeMU

[@malifsyahputranasution7650]

@@iain_explains thank you sir

[@siamakehsani744]

How is that AM when you shift frequency? It is FM

[@iain_explains]

You might like to watch this video: "Amplitude Modulation AM Radio Signal Transmission Explained" kzbin.info/www/bejne/Y4G6mGBmoJeiodU

[@tuongnguyen9391]

oh would plotting the bit error rate against the snr make any sense for us ?

[@iain_explains]

Both metrics are valid. The best one to use depends on what you are trying to show or compare. You can map from one to the other.

[@falcon-xm3uo]

Hello - Please make a video on SNR/EsNo/EbNo - Thanks

[@iain_explains]

Not sure what you are asking for. This video is on exactly that topic!

[@falcon-xm3uo]

@@iain_explains you are right! sorry my bad

[@eeliasselias549]

Thank you inmensly

[@iain_explains]

Glad it helped.

[@panduwilantara3070]

GOOD

[@GallacherDwight-w4w]

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