I have a question on the equivalence of two models (complex vs. real) of a received signal in ISI channel. Yet its simplicity, I am a bit confused.

Model #1: Complex Assume we transmit a complex symbol $\overline{a}$ drawn from a high-order QAM constellation. On a vector form, we can write the receive signal as $$\overline{y} = \overline{G} \: \overline{a} + \overline{w},$$ where $\overline{G}$ is the channel matrix (it is real matrix of dimension $N \times N$) and $\overline{w}$ is the complex noise samples with zero mean and cross-correlation matrix $E\{\overline{w} \overline{w}^H\} =\sigma^2 \overline{G}$.

I would like to work with real-valued components instead of complex-valued components. That said, we can write the following real model of the received signal.

Model #2: Real $${y} = {G} \: {a} + {w},$$ where $y = \begin{bmatrix} \Re\{{\overline{y}}\}\\ \Im\{{\overline{y}}\} \end{bmatrix},$ $G = \begin{bmatrix} \Re\{{\overline{G}}\} & - \Im\{{\overline{G}}\}\\ \Im\{{\overline{G}}\} & \Re\{{\overline{G}}\} \end{bmatrix},$ $a = \begin{bmatrix} \Re\{{\overline{a}}\}\\ \Im\{{\overline{a}}\} \end{bmatrix},$ $w = \begin{bmatrix} \Re\{{\overline{w}}\}\\ \Im\{{\overline{w}}\} \end{bmatrix}$, $\Re$ and $\Im$ are the real and imaginary parts of complex numbers. In this particular question, $\Re\{\overline{G}\} = G$ and $\Im\{\overline{G}\} = 0$.

As can be seen, both models are equivalent and can be used at the receiver for detection. My question is what is the cross-correlation of the noise in the 2nd model in terms of $G$, i.e. what is $E\{w w^T\}$ in terms of $G$?

My trial is as follows: $$E\{w w^T\} = \begin{bmatrix} E\{\Re\{\overline{w}\}\Re\{\overline{w}^T\}\} & E\{\Re\{\overline{w}\}\Im\{\overline{w}^T\}\}\\ E\{\Im\{\overline{w}\}\Re\{\overline{w}^T\}\} & E\{\Im\{\overline{w}\}\Im\{\overline{w}^T\}\} \end{bmatrix}$$ In general, how can I relate this to $G$? My expectation is/was $E\{w w^T\} = \sigma^2 G$!!!

  • $\begingroup$ Why is the noise correlation matrix $\sigma_n^2G$? Normally, it is $\sigma_n^2I$. $\endgroup$ Apr 12, 2017 at 19:24
  • $\begingroup$ This is due to ISI channel. $\endgroup$
    – Noor
    Apr 12, 2017 at 19:28
  • $\begingroup$ I doubt that. If you have proper noise, then the noise covariance is always real-valued and hermitian. Your $G$ does not have this property. Can you describe (by editing your question), how you come up with this system model? With a complex $G$ as the covariance it doesn't make sense. It would be a very strange noise model, and needs clarification, if you really mean this. If you really mean it, I can take the effort to derive the real-valued for you. Note, that in this case you'd have $E[ww^T]\neq 0$ (which is already really uncommon). $\endgroup$ Apr 12, 2017 at 19:35
  • $\begingroup$ I think there is a misunderstanding here. $\overline{G}$ is not complex. It is real of dimension $N \times N$. Due to the conversion from complex-valued expression to real-valued expression, I used $G$ which is of dimension $2N \times 2N$, and $\Im\{\overline{G}\} = 0$. I have updated the question. $\endgroup$
    – Noor
    Apr 12, 2017 at 21:22

1 Answer 1


Even though I doubt your noise model, here's how you can do it:

Given the complex noise vector $\overline w$, with the covariance matrix $E[\overline w\overline w^H]=\sigma^2\overline{G}$.

Let's separate $\overline w$ into its real and imaginary part (I use a different notation, because I think this is faster to write and read): $\overline w=w_R+jw_I$, where $w_R, w_I$ denote the real and imaginary parts of $\overline w$. Insert this into the covariance definition:

$$\begin{align}E[\overline w\overline w^H]&=E[(w_R+jw_I)(w_R^T-jw_I^T)]\\&=E[w_Rw_R^T+w_Iw_I^T]+2jE[w_Rw_I^T]=\overline{G}\end{align}$$

Let's compare this to the expression we get, when explicitely separating the real and imaginary parts: $$\begin{align}E[ww^T]&=\begin{pmatrix}E[w_Rw_R^T] & E[w_Rw_I^T]\\E[w_Iw_R^T] & E[w_Iw_I^T]\end{pmatrix}=X\end{align}.$$

Hence, in order to define $X$ (which is the matrix you are interested in), we need to know $E[w_Rw_R^T], E[w_Rw_I^T]=E[w_Iw_R^T], E[w_Iw_I^T]$.

Let's do the comparison to $E[\overline w \overline w^H]$, which we have calculated before. We know that $G_I=0$ (i.e. the imaginary part is zero). Hence, $E[w_Rw_I^T]=0$, i.e. real and imaginary part are uncorrelated. What we dont know yet are $E[w_Rw_R^T]$ and $E[w_Iw_I^T]$. We only know that $E[w_Rw_R^T]+E[w_Iw_I^T]=\overline G$.

How to solve it? There are two possibilities:

1) (The more common assumption) We assume that real and imaginary parts of the noise have equal variance. This is often reasonble, e.g. when the noise is baseband noise and was obtained from downconversion a passband signal. Then, we have $E[w_Rw_R^T]=E[w_Iw_I^T]=\frac{1}{2}\overline{G}$.

2) Real and imaginary parts are not i.i.d. Then, the noise is not proper and the pseudocorrelation $T$ of the noise does not vanish. It is defined by $T=E[\overline w \overline w^T]$ and we calculate it to be (assuming $E[w_Iw_R^T]=0$) $$T=E[\overline w \overline w^T]=E[w_Rw_R^T]-E[w_Iw_I^T]$$ In order to accurately describe a non-proper noise, the value of $T$ must be known. Then, we have two unknowns ($E[w_Rw_R^T]$ and $E[w_Iw_I^T]$) and two equations $$\begin{align}\overline{G}&=E[w_Rw_R^T]+E[w_Iw_I^T]\\T&=E[w_Rw_R^T]-E[w_Iw_I^T]\end{align}$$ from which you can get the unknown via $E[w_Rw_R^T]=\frac{1}{2}(\overline{G}+T)$ and $E[w_Iw_I^T]=\frac{1}{2}(\overline G-T)$.

Given this derivation, I still doubt your noise model. To me it does not make any sense, that in the equation $y=\overline G \overline a+\overline{w}$, the noise should have variance $\sigma^2\overline{G}$ due to ISI. Can you ensure that $\overline G$ is Hermitian?

  • $\begingroup$ Thanks. A couple of points before accepting your answer. You said, $G_I = 0$, hence, $E\{w_R w_I^T\} = 0$. Should this expectation always equal to zero because real and imaginary parts of noise are independent? Or you are conditioning this on $G_I = 0$ such that the $\Re\{\overline{y}\}$ and $\Im\{\overline{y}\}$ are not tangled together? Another point, assuming proper noise RVs, then there is a constant 0.5 missing in the way I am defining $G$ in terms of $\overline{G}$? $\endgroup$
    – Noor
    Apr 13, 2017 at 13:08
  • $\begingroup$ First: You said $G_I=0$. Since $G_I=2E[w_Rw_I]$, we have $E[w_Rw_I]=0$. So, it follows from your assumption. In general, this does not always hold (i.e. it can happen that real and imaginary are correlated), but it would be very uncommon. Second: the dimension of $G$ is $2N\times 2N$. $\overline G$ has dimension $N\times N$, so both of your expressions dont fit. In fact, for your model you have $G=0.5 (I_2\otimes \overline G)$. ($\otimes$ denodes the Kronecker-Product and $I_2$ is the identity matrix of size 2). $\endgroup$ Apr 13, 2017 at 13:13
  • $\begingroup$ $G$ is defined as $G = \begin{bmatrix} \Re\{{\overline{G}}\} & - \Im\{{\overline{G}}\}\\ \Im\{{\overline{G}}\} & \Re\{{\overline{G}}\} \end{bmatrix}.$ Hence, $E\{w w^T\} = [0.5 \overline{G} \quad 0; 0 \quad 0.5\overline{G}] = 0.5 G$. BTW, $\overline{G}$ is Hermitian. $\endgroup$
    – Noor
    Apr 13, 2017 at 13:34
  • $\begingroup$ Yeah, sorry, I mixed up your $G$ with my $G$. I had defined $G$ to be the covariance of the real-valued $w$, you have defined it as the real-valued equivalent of $\overline G$. I have modified my answer, the matrix you are looking I named now $X$. $X$ consists of 4 blocks, and in my answer, I calculate the value of each. So, in my previous comment I actually meant $X=0.5(I_2\otimes \overline G)=0.5 G$. $\endgroup$ Apr 13, 2017 at 13:35

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