I'm working on an exercise but I'm stuck at the question where the noise becomes colored, can someone help me ?

Let $\{\psi_1(t) = \mathbf{1}\{t \in [0,1)], \psi_2 = \mathbf{1}\{t \in [1,2]\}\}$ be an orthonormal basis. We are sending $$ w_1(t) = \psi_1(t) + \psi_2(t), \quad w_2(t) = \psi_1(t) - \psi_2(t)\\ w_3(t) = -\psi_1(t) + \psi_2(t), \quad w_4(t) = -\psi_1(t) - \psi_2(t) $$ across an AWGN channel of power spectral density $N_0 / 2 = 1$. The received waveform is then $R(t) = w_i(t) + N(t)$. At the receiver we observe $Y_1 = \langle R, \psi_1 \rangle$ and $Y_2 = \langle R, \psi_2 \rangle$.

a) Conditioned on i-th message is being sent, what is the distribution of $(Y_1, Y_2)$.

I have answered the following

$$ Y_1 = \langle R, \psi_1 \rangle = \langle w_i, \psi_1 \rangle + \langle N, \psi_1 \rangle \sim \mathcal{N}(\langle w_i, \psi_1 \rangle, 1) \\ Y_2 = \langle R, \psi_2 \rangle = \langle w_i, \psi_2 \rangle + \langle N, \psi_2 \rangle \sim \mathcal{N}(\langle w_i, \psi_2 \rangle, 1) $$

b) What is the MAP decision region and the corresponding error probability of this transmission scheme?

For this, I have :

By looking at the waveform, we see that we use $4$-QAM and the MAP rule is just minimum distance decoding since we have an AWGN channel. The error probability is

$$ \mathbb{P}_e = 2Q(1) - Q^2(1) $$

Now, we will assume that the noise process $N(t)$ is Gaussian but not white, and, for all $t$,

$$ \mathbb{E}[N(t)N(t-\tau)] = \begin{cases} 1-|\tau|, & |\tau| < 1 \\ 0, & \text{otherwise}. \end{cases} $$

The receiver does not know of this new noise model, hence it still computes $(Y_1, Y_2)$ and forms its decision based on the decision region that you have developed in b).

c) What is the error probability of the receiver under this new noise model? It is sufficient to give express the error probability as an integral with an explicit integration region.

And here I'm stuck, I really don't know what to do... any help ?


1 Answer 1


To solve this problem, it is necessary to understand what exactly is meant by $\langle R, \psi_1 \rangle$ and $\langle R, \psi_2 \rangle$. These are the outputs of correlators or matched filters and we have that \begin{array}{rllll} Y_1 &= \displaystyle{\int_0^1} R(t)\cdot 1 \,\mathrm dt &= \displaystyle{\int_0^1} (-1)^{b_0} + N(t)\,\mathrm dt &= (-1)^{b_0} + \displaystyle{\int_0^1} N(t)\,\mathrm dt &= (-1)^{b_0} + Z_1\\ Y_2 &= \displaystyle{\int_1^2} R(t)\cdot 1 \,\mathrm dt &=\displaystyle{\int_1^2} (-1)^{b_1} + N(t)\,\mathrm dt &= (-1)^{b_1} + \displaystyle{\int_1^2 N(t)}\,\mathrm dt &= (-1)^{b_1} + Z_2 \end{array} where $Z_1$ and $Z_2$ are zero-mean Gaussian random variables. What are their variances? \begin{align}\operatorname{var}(Z_i) &= E[Z_i^2] \\ &= E\left[\int_{i-1}^i N(t)\,\mathrm dt\int_{i-1}^i N(\tau)\,\mathrm d\tau\right]\\ &= \int_{i-1}^i\int_{i-1}^i E[N(t)N(\tau)]\,\mathrm dt\,\mathrm d\tau\\ &= \int_0^1\int_0^1 1- |t-\tau|\, \mathrm dt\,\mathrm d\tau. \end{align} Now consider the inner integral $\displaystyle{\int_0^1 1- |t-\tau|\, \mathrm dt}$ for some fixed $\tau \in [0,1]$. At $t=0$, the integrand has value $1-\tau$. As $t$ increases, the integrand increases linearly till it reaches value $1$ at $t=\tau$. As $t$ increases beyond $\tau$ towards $1$, $t-\tau$ is a positive number and so the integrand decreases linearly to value $\tau$ at $t=1$. From this, it is easy to work out (do the math for yourself, for crying out loud!) that $$\displaystyle{\int_0^1 1- |t-\tau|\, \mathrm dt = \frac 12+ \tau -\tau^2}$$ (quick check: value is $\frac 12$ at $\tau=0$ and at $\tau = 1$). It follows that \begin{align} \operatorname{var}(Z_i) &= \int_0^1\int_0^1 1- |t-\tau|\, \mathrm dt\,\mathrm d\tau\\ &= \int_0^1 \frac 12 + \tau -\tau^2 \,\mathrm d\tau\\ &= \left . \frac{\tau}{2} + \frac{\tau^2}{2} - \frac{\tau^3}{3}\right\vert_0^1\\ &= \frac 23. \end{align} The symbol error probability for the receiver can be worked out from this result. Since the noise variance is smaller, we get a smaller error probability. Note that the receiver is oblivious to the fact that $Z_1$ and $Z_2$ are in fact correlated random variables. Taking the correlation into account and making a joint decision on the two bits transmitted instead of individual decisions on the bits would reduce the error rate even more, but that is a whole 'nuther question.

  • $\begingroup$ Awesome work thank you :)) $\endgroup$
    – Bozu
    Commented Jun 19, 2022 at 19:22

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