ฉันจะเพิ่ม AWGN ในสัญญาณ I และ Q ได้อย่างไร?

ฉันมีระบบสื่อสารไร้สายที่ฉันจำลองใน Matlab ฉันกำลังทำการลายน้ำผ่านการปรับเฟสของสัญญาณที่ส่งเล็กน้อย การจำลองของฉันใช้ค่า I (inphase) และ Q (quadrature) ดั้งเดิมและเพิ่มในลายน้ำ ฉันต้องจำลองอัตราความผิดพลาดของบิตที่เกิดขึ้นหลังจากถูกส่ง สำหรับตอนนี้ฉันแค่ต้องเพิ่มสัญญาณรบกวนความร้อนจำนวนมากในสัญญาณ

เนื่องจากฉันมีสัญญาณที่แสดงเป็นช่อง I และ Q มันจะง่ายที่สุดในการเพิ่มAWGN (เสียงรบกวนแบบเกาส์สีขาวเสริม) ลงใน I และ Q โดยตรง หนึ่งความคิดคือการเพิ่มเสียงให้กับทั้งสองช่องอย่างอิสระ แต่ปรีชาญาณของฉันบอกฉันว่านี่ไม่เหมือนกับการเพิ่มไปยังสัญญาณโดยรวม

ดังนั้นฉันจะเพิ่มเสียงรบกวนเมื่ออยู่ในฟอร์มนี้ได้อย่างไร

ใช่คุณสามารถเพิ่ม AWGN ความแปรปรวนแยกแต่ละสองแง่เพราะผลรวมของสอง Gaussians ยังเป็นแบบเกาส์และผลต่างของพวกเขาเพิ่มขึ้น สิ่งนี้จะมีผลเช่นเดียวกับการเพิ่ม AWGN ของความแปรปรวน2 σ 2ให้กับสัญญาณดั้งเดิม ต่อไปนี้เป็นคำอธิบายเพิ่มเติมหากคุณสนใจ$\sigma^2$$2\sigma^2$

สัญญาณการวิเคราะห์สามารถเขียนได้ในองค์ประกอบในเฟสและการสร้างพื้นที่สี่เหลี่ยมจัตุรัสเป็น$x(t)=a(t)\sin\left(2\pi f t + \varphi(t)\right)$

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

ที่และQ ( T ) = ( T ) บาป( φ ( T ) ) หากคุณต้องการเพิ่ม AWGN ไปยังสัญญาณดั้งเดิมของคุณเป็นx ( t ) + u ( t )โดยที่คุณ( t ) ∼ N ( μ , σ $I(t)=a(t)\cos(\varphi(t))$$Q(t)=a(t)\sin(\varphi(t))$$x(t)+u(t)$$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.

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.

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! :)