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I have a wireless communication system that I am simulating in Matlab. I am performing some watermarking through slightly adjusting the phase of the transmitted signal. My simulation takes the original I (inphase) and Q (quadrature) values and adds in the watermark. I then have to simulate the resulting bit error rate after being transmitted. For now I just need to add varying amounts of thermal noise to the signal.

Since I have the signal represented as its I and Q channel it would be easiest to add AWGN(additive white Gaussian noise) to the I and Q directly. One thought was to add noise to both channels independently, but my intuition tells me that this isn't the same as adding it to the signal as a whole.

So how can I add noise to it when it is in this form?

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  • $\begingroup$ perhaps it could be of more help if you could give some details of the communication system you are simulating. $\endgroup$ – Rajesh Dachiraju Aug 18 '11 at 13:38
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    $\begingroup$ I would assume you just generate noise for both I and Q and then add them. I don't see why the noise would be correlated between the two. $\endgroup$ – endolith Aug 18 '11 at 13:52
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    $\begingroup$ @endolith, The noise difference would only appear in the mixer, besides that they should share their noise signals. $\endgroup$ – Kortuk Aug 18 '11 at 14:07
  • $\begingroup$ Are you saying that you would to add it to the quadrature multiplexed signal? $\endgroup$ – Phonon Aug 18 '11 at 14:11
  • $\begingroup$ @phonon, what do you mean by multiplexed? $\endgroup$ – Kortuk Aug 18 '11 at 14:31
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Yes, you can add AWGN of variance $\sigma^2$ separately to each of the two terms, because the sum of two Gaussians is also a Gaussian and their variances add up. This will have the same effect as adding an AWGN of variance $2\sigma^2$ to the original signal. Here's some more explanation if you're interested.

An analytic signal $x(t)=a(t)\sin\left(2\pi f t + \varphi(t)\right)$ can be written in its in-phase and quadrature components as

$$x(t)=I(t)\sin(2\pi ft) + Q(t)\cos(2\pi ft)$$

where $I(t)=a(t)\cos(\varphi(t))$ and $Q(t)=a(t)\sin(\varphi(t))$. If you wish to add AWGN to your original signal as $x(t)+u(t)$, where $u(t)\sim\mathcal{N}(\mu,\sigma^2)$, then you can add AWGN to each of the terms as

$$y_1(t)=\left[I(t)\sin(2\pi ft) + v(t)\right] + \left[Q(t)\cos(2\pi ft) + w(t)\right]$$

where $v(t), w(t)\sim\mathcal{N}(\mu/2,\sigma^2/2)$

Also note that because the in-phase and quadrature terms are additive, the AWGN can also be simply added to either of the two terms in the $IQ$ representation of $x(t)$ above. In otherwords,

$$y_2=I(t)\sin(2\pi ft) + \left[Q(t)\cos(2\pi ft) + u(t)\right]$$ $$y_3=\left[I(t)\sin(2\pi ft) +u(t)\right]+ Q(t)\cos(2\pi ft)$$

are statistically equivalent to $y_1$, although I prefer using $y_1$ because I don't have to keep track of which component has noise added to it.

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  • $\begingroup$ Since the signal has the noise it seems the noise would show up on both channels with the original magnitude but affected by the mixing process. I would think that the mixing process would affect the noise much more then addition of subtraction by splitting the signal. $\endgroup$ – Kortuk Aug 18 '11 at 14:34
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    $\begingroup$ Of course, if you had noise in the signal to begin with and then split it into its I-Q components, each will have noise associated with it. However, OP is talking about simulating it in MATLAB and he has the I and Q parts separately and wants to know how to add noise to these so as to simulate adding noise to the original signal. $\endgroup$ – Lorem Ipsum Aug 18 '11 at 14:42
  • $\begingroup$ good answer with lots of details, but fails to concisely answer the basic question -- OP: Ignore your intuition; adding WGN on the real axis with WGN on the imaginary axis results in complex WGN. Remember to scale by 3dB since variance of sum is double that of parts (stdv2 = 1.413 stdv1 ) $\endgroup$ – Mark Borgerding Aug 18 '11 at 14:46
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    $\begingroup$ @Yoda, you've got all the data, but you make the reader read through many equations before getting to the answer. I merely suggest putting your bolded part first, then provide the supporting details. $\endgroup$ – Mark Borgerding Aug 18 '11 at 14:55
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    $\begingroup$ @yoda, I was tired when I read this. Your answer is very astute. Thank you for taking the time! $\endgroup$ – Kortuk Aug 18 '11 at 19:26
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Kellenjb has not responded to queries from Rajesh D and endolith, and it is not easy to figure out what exactly he needs. But since I disagree with some of the details of the Answers given by yoda and Mohammad, I am posting a separate answer, where, with due apologies to Mark Borgerding, all the useful stuff appears at the very end after all the boring equations.

In a typical communication system, the incoming signal is a bandpass signal of bandwidth $2B$ at center frequency $f_c \gg B$ Hz and can be expressed as $$ r(t) = I(t) \cos(2\pi f_c t) - Q(t) \sin(2\pi f_c t) $$ where $I(t)$ and $Q(t)$ are low-pass signals of bandwidth $B$ Hz and are referred to as the in-phase and quadrature components. Note the difference in signs and terminology from yoda'a writing: this way we can write $$r(t) = \text{Re}\left\{ [I(t) + jQ(t)] e^{j2\pi f_c t}\right\}$$ where $I(t) + jQ(t)$ is the complex baseband signal.

A local oscillator in the receiver generates signals $2\cos(2\pi f_c t + \theta)$ and $-2\sin(2\pi f_c t + \theta)$ but we assume perfect synchronization for simplicity so that the phase error $\theta = 0$. $I(t)$ and $Q(t)$ are recovered through two mixers (multipliers) and low-pass filters: $$ \begin{align*} r(t)[2\cos(2\pi f_c t)] &= I(t)[2\cos^2(2\pi f_c t)] - Q(t)[2\sin(2\pi f_c t)\cos(2\pi f_c t)]\\ &= I(t) + \left [ I(t) \cos(2\pi (2f_c)t) - Q(t) \sin(2\pi (2f_c) t) \right ]\\ r(t)[-2\sin(2\pi f_c t)] &= I(t)[-2\sin(2\pi f_c t)\cos(2\pi f_c t)] + Q(t)[2\sin^2(2\pi f_c t)]\\ &= Q(t) + \left [ - I(t) \sin(2\pi (2f_c)t) - Q(t) \cos(2\pi (2f_c) t) \right ] \end{align*} $$ where the double frequency terms (in square brackets) are eliminated by the low-pass filters which we assume to have sufficient bandwidth to pass $I(t)$ and $Q(t)$ without distortion.

Broadband noise is present in the front end of the receiver and the key questions that need to be answered are what happens in an actual receiver, and what must be done to simulate the reality.

  • In an actual system, the net result is that the outputs of the low-pass filters are $$ \begin{align*} x(t) &= I(t) + N_I(t)\\ y(t) &= Q(t) + N_Q(t) \end{align*} $$ where $N_I(t)$ and $N_Q(t)$ are independent zero-mean Gaussian random processes with common variance $$ \sigma^2 = \frac{N_0}{2}\int_{-\infty}^\infty \vert H(f) \vert^2 \mathrm df $$ where $H(f)$ is the common transfer function of the low pass filters. In particular, for each $t_0$, $N_I(t_0)$ and $N_Q(t_0)$ are independent zero-mean Gaussian random variables with variance $\sigma^2$. However, $N_I(t_0)$ and $N_I(t_1)$ need not be independent. The SNR can be taken to be the ratio of the signal power in $I(t)$ and $Q(t)$ to the noise variance.
  • In a quadrature down-sampling system or in a MATLAB simulation wishing to capture every nuance, "$r(t) + ~$ noise" is sampled $M$ times each cycle of the RF carrier at $f_c$ Hz, and so the $m$-th sample is $$ \begin{align*} r[m] &= r(m/Mf_c) + N[m]\\ &= I(m/Mf_c)\cos(2\pi (m/M)) - Q(m/Mf_c)\sin(2\pi(m/M)) + N[m] \end{align*} $$ where the $N[m]$'s zero-mean Gaussian random variables with common variance whose value depends on the SNR. These can be tracked through the mixers and the low-pass filters during the detailed simulation.
  • I do not recommend adding noise between the mixer outputs and the low-pass filter units. While there is noise introduced at that stage, this is typically overwhelmed by the noise from the front end that is coming through the mixers.
  • In some systems, A/D conversion is done at the output of the low pass filters. If more filtering is to be done (e.g. matched filtering), the sampling will typically be at a higher rate than $B^{-1}$ or the inverse of the filter bandwidth. If noise is introduced at this stage, then for each $m$, $N_I[m]$ and $N_Q[m]$ should be taken to be independent zero-mean Gaussian random variables, but whether $N_I[m]$ and $N_I[m+i]$ are independent or not requires a lot of thought and analysis, and details that are known to Kellenjb but not to us.
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  • $\begingroup$ Thanks, Dilip. Nice detailed, practically-focused answer. $\endgroup$ – Jason R Oct 5 '11 at 4:17
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Kellenjb,

The noise in both the I and Q are not in fact going to be gaussian. In fact they are going to originate from the same original noise vector. This is because there was only one noise vector to begin with at the receiver. So what is happening, is your signal comes into the receiver, where AWGN is added of course. Soon afterwards though, the receiver is going to project that (signal + noise) onto a sin basis, and onto a cosine basis, thereby giving you your I and Q components.

So now the noise in either branch is no longer gaussian, but are in fact, the product of a sin basis times orignal noise vector, and product of cosine basis times original noise vector.

The way I would recommend to simulate this, (are you doing all of this in baseband?), is to simply construct a sin and cosine basis, and simply multiply against (signal+noise), where 'signal' is your original signal of course, and then of course take it down to baseband after that. In fact once you filter for taking it down to baseband, your noise vectors are going to be non-white, and non-gaussian.

Hope this helps! :)

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