### Pset 2, Abubakar Abid (It may take a second for the LaTeX-javascript library to load...)

1a. Deriving the Poisson Distribution from the Binomial. We start from the binomial formula $$p_n(x) = \frac{n!}{(n-x)! x!}p^x(1-p)^{n-x}$$ Taking the log of each side, we have $$\ln{p_n(x)} = \ln{n!}-\ln{(n-x)!} -\ln{x!} + x\ln{p}+(n-x)\ln{(1-p)}$$ For large $$n$$ and small $$x$$, we can approximate this as: $$\ln{p_n(x)} = n\ln{n}-n-(n-x)\ln{(n-x)}+(n-x) -x\ln{x}+x + x\ln{p}+(n-x)(-p)$$ Because $$n \gg x$$, $$\ln{p_n(x)} = x\ln{n} - (x\ln{x}+x) + x\ln{p}-(np)$$ We exponentiate each side, resulting in: $$p_n(x) = \frac{n^x}{x!}\frac{p^x}{e^{np}}$$ Letting $$N = np$$, $$p_n(x) = \frac{N^x e^{-N}}{x!}$$ 1b. Factorial Moments of the Poisson Distribution. The factorial moments are given by $$\langle{x(x-1)(x-2)...(x-m+1)}\rangle = \sum_{x=0}^\infty{\frac{e^{-N} N^x}{x!}x(x-1)(x-2)...(x-m+1)}$$ Simplifying $$= e^{-N} \sum_{x=0}^\infty{\frac{N^x}{(x-m)!}}$$ Let $$k = x-m$$, $$= e^{-N} \sum_{x=0}^\infty{\frac{N^{k+m}}{k!}}$$ $$= N^m e^{-N} \sum_{x=0}^\infty{\frac{N^{k}}{k!}}$$ $$= N^m e^{-N} e^{N} = N^m$$ 1c. Fractional Error in Poisson Distribution. From 1b, we can set $$m=1$$ to have that $$\langle x \rangle = N$$ We can set $$m=2$$ to have that $$\langle x(x-1) \rangle = \langle x^2-x \rangle = \langle x^2 \rangle - \langle x \rangle = N^2$$ meaning that $$\langle x^2 \rangle = N^2 + N$$ Since $$\sigma^2 = \langle x^2 \rangle - \langle x \rangle^2 = N^2 + N - N^2 = N$$ So that $$\sigma = \sqrt{N}$$ and 3.19 follows.

2. From Problem 1c, it follows that for large $$N$$, we would require $$100^2=10,000$$ samples to have a less than 1% error rate and would require $$10^{12}$$ samples to have an error rate of less than one part in a million. For visible light at 600 nm, this corresponds to wattages of $$10^4 h c/\lambda\ \approx 3.3*10^{-15}W$$ and $$10^4 h c/\lambda\ \approx 3.3*10^{-7}W$$

3a. Input Voltage. The average value of voltage resulting from noise will be $$\sqrt{\langle V_n^2 \rangle} = 4kTR\Delta f \approx 1.82 \mu V$$ For the SNR to be more than 20dB, then $$\frac{V_{in}}{V_n} > 10$$ so $$V_{in} > 18.2 \mu V$$ 3b. Capacitor Size. By the equipartition theorem, we know that the thermal noise in any degree of freedom is $$\frac{1}{2}kT$$, so the noisy voltage on a capacitor satisfies: $$\frac{1}{2}kT = \frac{1}{2}C\langle V_n^2 \rangle$$ Solving for $$C$$, we find that $$C = 1.2nF$$ 3c. Current Shot Noise. The shot noise of a current source is given by $$\sqrt{\langle I_n^2 \rangle} = \sqrt{2 q I \Delta f}$$ $$0.01 = \sqrt{\langle I_n^2 \rangle}/I = \sqrt{\frac{2 q \Delta f}{I}}$$ Solving for $$I$$, we get that $$I = 64 pA$$ 4a. Differential Equation. $$\begin{bmatrix} \dot{p_0} \\ \dot{p_1} \\ \end{bmatrix} = \begin{bmatrix} 1-\alpha & \beta \\ \alpha & 1-\beta \\ \end{bmatrix} \begin{bmatrix} p_0(t) \\ p_1(t) \\ \end{bmatrix}$$ 4b. Solving the Differential Equation. To solve this differential equation, we must find the eigenvalues and the eigenvectors: $$\det \begin{bmatrix} 1-\alpha - \lambda & \beta \\ \alpha & 1-\beta - \lambda \\ \end{bmatrix} = 0$$ This gives us the quadratic equatio: $$\lambda^2 + (\beta + \alpha - 2)\lambda + (1 - \alpha - \beta) = 0$$ Which has the solutions: $$\lambda = 1, 1 - \beta - \alpha$$ And the eigenvectors are: $$\begin{bmatrix} \beta \\ \alpha \\ \end{bmatrix}, \begin{bmatrix} 1 \\ -1 \\ \end{bmatrix}$$ This gives $$p_0(t) = \beta e^t + e^{(1-\alpha-\beta)t}$$ $$p_1(t) = \alpha e^t - e^{(1-\alpha-\beta)t}$$ 4c. Autocorrelation Function. To derive the autocorrelation function, we evaluate $$\langle x(t)x(t+\tau) \rangle$$ $$\langle [\alpha e^t - e^{(1-\alpha-\beta)t}][\alpha e^{t+\tau} - e^{(1-\alpha-\beta)(t+\tau)}] \rangle$$

4d. Power Spectrum. Once we have the autocorrelation function, deriving the power spectrum is easy, we simply take the Fourier Transform and it should turn out to be a Lorentzian

4e. Bandwidth. Once we have the Lorentzian, we would evaluate the frequencies at which it takes half it's maximal value.

4f. 1/f Noise. To derive this, we carry out the integral $$S(f) = \int_0^\infty \frac{2\tau}{1+(2\pi f \tau)^2}p(\tau) d\tau$$ for $$p(\tau) \propto \frac{1}{\tau}$$ Gives us the desired power spectrum