End sem 7; week 2 winter session 2025
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# Homework 5 - Aidan Sharpe
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## Problem 1
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In a binary communication system, let the receiver test statistic of the received signal be $r_0$. The received signal consists of a polar digital signal plus noise. The polar signal has values $s_{01}=A$ and $s_{02}=-A$. Assume that the noise has a Laplacian distribution:
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$$f(n_0) = \frac{1}{\sqrt{2} \sigma_0} e^{-\sqrt{2}|n_0|/\sigma_0}$$
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where $\sigma_0$ is the RMS value of the noise, $f(n_0)$ is the probability density function (PDF), and $n_0$ is the signal. In the case of a PDF of $s_{01}$ and $s_{02}$, replace $n_0$ by $r_0-A$ and $r_0+A$. The shape of the PDF for $s_{01}$ and $s_{02}$ is the same. Find the probability of error $P_e$ as a function of $A/\sigma_0$ for the case of equally likely signaling and $V_T$ having the optimum value.
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Given the two PDFs corresponding to $s_1$ and $s_2$, the probability of a bit error is the same as the area of the intersection of the two PDFs, as seen in figure \ref{error_pdfs}.
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The curve traced out by this area is
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$$p(r_0) = \frac{1}{2}\left[f(r_0 | s_1) + f(r_0 | s_2) - \|f(r_0 | s_1) - f(r_0 | s_2) \|\right]$$
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to find the area under the curve (the probability of a bit error $P_e$), we integrate $p(r_0)$ for all values of $r_0$. Since $f(r_0 | s_1)$ and $f(r_0 | s_2)$ are PDFs, the area under each must be unity. Therefore, distributing the $\frac{1}{2}$ term, we are left with:
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$$P_e = 1 - \frac{1}{2} \int\limits_{-\infty}^\infty \left|f(r_0 | s_1) - f(r_0 | s_2)\right| dr_0$$
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Finally, we expand and simplify the integral to:
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$$P_e = 1 - \frac{\sqrt{2}}{4\sigma_0} \int\limits_{-\infty}^\infty \left|e^{-\sqrt{2} |r_0-A| /\sigma_0} - e^{-\sqrt{2} |r_0 + A|/\sigma_0}\right| dr_0$$
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After evaluating the intergral, we find that $P_e = e^{-\sqrt{2}A/\sigma_0}$.
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## Problem 2
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A digital signal with white Gaussian noise is received by a receiver with a matched filter. The signal is a unipolar non-return to zero signal with $s_{01}=1$[V] and $s_{02}=0$[V]. The bit rate is 1 Mbps. The power spectral density of the noise is $N_0/2=10^{-8}$[W/Hz]. What is the probability of error $P_e$? Assume the white Gaussian noise is thermal noise.
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For a unipolar signal received by a receiver with a matched filter, the probability of error is given by:
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$$P_e = Q\left(\sqrt{\frac{A^2 T}{4 N_0}}\right)$$
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where $A = 1 - 0 = 1$ is the amplitude and $T= 1[\mu$s]. Therefore, $P_e = 2.03 \times 10^{-4}$.
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