Exercises

$P_y(f) = \frac{N_0/2}{(1 + (2\pi f RC)^2)^2}$.

$\mathcal{P}_Y = \frac{N_0}{2} \int \frac{df}{(1 + (2\pi f RC)^2)^2} = \frac{N_0}{2} \cdot \frac{1}{2\pi RC} \cdot \frac{\pi}{2} = \frac{N_0}{8RC}$.

$B_N = \mathcal{P}_Y / N_0 = \frac{1}{8RC}$. Compare to the single-pole case $B_N = 1/(4RC)$: the second-order filter has half the noise bandwidth.

$P_x(f) = \frac{4 \cdot 2}{4 + (2\pi f)^2} \cdot 2 = \frac{16}{4 + (2\pi f)^2}$. Wait — let us be careful. $\mathcal{F}\{e^{-a|\tau|}\} = \frac{2a}{a^2 + (2\pi f)^2}$, so $P_x(f) = \frac{4 \cdot 4}{4 + (2\pi f)^2} = \frac{16}{4 + (2\pi f)^2}$.

$\check{h}(f) = \frac{1}{1 + j2\pi f}$, so $|\check{h}(f)|^2 = \frac{1}{1 + (2\pi f)^2}$.

$P_y(f) = \frac{16}{(4 + (2\pi f)^2)(1 + (2\pi f)^2)}$.

$r_{yy}(0) = \iint h(\alpha) h^*(\beta) r_{xx}(\alpha - \beta)\, d\alpha\, d\beta$. Using $|r_{xx}(\tau)| \leq r_{xx}(0)$: $$|r_{yy}(0)| \leq r_{xx}(0) \iint |h(\alpha)| |h(\beta)|\, d\alpha\, d\beta = r_{xx}(0) \left(\int |h(\tau)|\, d\tau\right)^2.$$

$g(t_0) = \int h(\tau) s(t_0 - \tau)\, d\tau = \int s^*(t_0 - \tau) s(t_0 - \tau)\, d\tau = \int |s(t_0 - \tau)|^2\, d\tau = \int |s(u)|^2\, du = E_s$.

$\text{SNR} = \frac{|\int \check{h}(f) \check{s}(f) e^{j2\pi f t_0}\, df|^2}{\int |\check{h}(f)|^2 P_N(f)\, df}$.

Write the numerator as $|\int [\check{h}(f) \sqrt{P_N(f)}] \cdot [\check{s}(f) e^{j2\pi f t_0} / \sqrt{P_N(f)}]\, df|^2$. By Cauchy-Schwarz: $\text{SNR} \leq \int |\check{s}(f)|^2 / P_N(f)\, df$. Multiplying by 2 for the standard definition: $\text{SNR}_{\max} = 2 \int |\check{s}(f)|^2 / P_N(f)\, df$.

Equality when $\check{h}(f) \sqrt{P_N(f)} \propto \check{s}^*(f) e^{-j2\pi f t_0} / \sqrt{P_N(f)}$, giving $\check{h}_{\text{opt}}(f) = c \cdot \check{s}^*(f) e^{-j2\pi f t_0} / P_N(f)$.

Outputting zero gives MSE $= \mathbb{E}[|X(t)|^2] = r_{xx}(0) = \sigma^2_{X}$.

Passing $Y$ unchanged gives $\hat{X} = X + N$, so $\text{MSE} = \mathbb{E}[|N(t)|^2] = \sigma^2_{N}$.

Since the Wiener filter minimizes MSE, $\text{MMSE} \leq \min(\sigma^2_{X}, \sigma^2_{N})$. In fact, $\text{MMSE} = \int \frac{P_x P_N}{P_x + P_N}\, df \leq \int \min(P_x, P_N)\, df \leq \min(\int P_x\, df, \int P_N\, df)$.

$P_x(f) = \frac{1}{1 - 1.8\cos(2\pi f) + 0.81}$.

$|\check{h}(f)|^2 = |1 - e^{-j2\pi f}|^2 = 2(1 - \cos(2\pi f)) = 4\sin^2(\pi f)$.

$P_y(f) = \frac{4\sin^2(\pi f)}{1.81 - 1.8\cos(2\pi f)}$. At $f = 0$: $P_y(0) = 0$. At $f = 1/2$: $P_y(1/2) = 4/(1.81 + 1.8) \approx 1.11$. The differentiator suppresses low frequencies and boosts high frequencies.

$|\check{h}(f)|^2 = e^{-2\pi (f/f_0)^2}$. Using $\int e^{-ax^2}\, dx = \sqrt{\pi/a}$ with $a = 2\pi/f_0^2$: $\int |\check{h}(f)|^2\, df = f_0 \sqrt{\pi/(2\pi)} \cdot \sqrt{2\pi} = f_0 \sqrt{\pi/2} \cdot \sqrt{2}$. Wait, let us compute directly: $\int e^{-2\pi f^2/f_0^2}\, df = \sqrt{\frac{f_0^2}{2\pi}} \sqrt{\pi} = f_0 / \sqrt{2}$.

Peak gain $|\check{h}(0)|^2 = 1$. So $B_N = \frac{1}{2} \cdot f_0/\sqrt{2} = f_0/(2\sqrt{2}) = f_0\sqrt{2}/4$. Actually, let me redo: $\int e^{-2\pi(f/f_0)^2} df$. Let $u = f/f_0$, $df = f_0\, du$. $= f_0 \int e^{-2\pi u^2}\, du = f_0 \sqrt{\pi/(2\pi)} = f_0/\sqrt{2}$. So $B_N = f_0/(2\sqrt{2})$.

$\text{SNR} = \frac{|\int \check{h}(f) \check{s}(f) e^{j2\pi f t_0}\, df|^2}{(N_0/2) \int |\check{h}(f)|^2\, df}$.

$\rho = \text{SNR} / (2E_s/N_0) = \frac{|\int \check{h} \check{s} e^{j2\pi f t_0}\, df|^2}{\int |\check{h}|^2\, df \cdot \int |\check{s}|^2\, df}$. By Cauchy-Schwarz, this is $\leq 1$ with equality iff $\check{h}(f) \propto \check{s}^*(f) e^{-j2\pi f t_0}$.

$r_{xd, y}(\tau) = \mathbb{E}[X(\Delta) Y^*(\Delta - \tau)] = r_{xy}(\tau)$, but the desired signal is $X$ at time $\Delta$, so $P_{x(\Delta), y}(f) = P_x(f) e^{j2\pi f \Delta}$ (the time shift introduces a phase).

$\check{h}_{\text{opt}}(f) = P_{x(\Delta), y}(f) / P_y(f) = \frac{P_x(f) e^{j2\pi f \Delta}}{P_x(f) + P_N(f)}$.

The predictor is the Wiener filter multiplied by a phase shift $e^{j2\pi f \Delta}$ that accounts for the prediction horizon. As $\Delta \to 0$, it reduces to the standard Wiener filter.

By orthogonality, $\text{MSE} = r_{xx}(0) - r_{\hat{x}\hat{x}}(0)$, since the error is orthogonal to the estimate.

$r_{\hat{x}\hat{x}}(0) = \int P_{\hat{x}}(f)\, df = \int |\check{h}_{\text{opt}}(f)|^2 P_y(f)\, df$.

$\text{MMSE} = r_{xx}(0) - \int |\check{h}_{\text{opt}}(f)|^2 P_y(f)\, df$. The MMSE equals the signal power that the filter cannot capture.

$E_s = 2\int_0^T (1 - t/T)^2\, dt = 2T \int_0^1 (1-u)^2\, du = 2T/3$.

$\text{SNR}_{\max} = 2E_s/N_0 = 4T/(3N_0)$.

$\check{h}_{\text{opt}}(f) = P_x(f) / (P_x(f) + P_N(f))$. Since $P_x(f) \geq 0$ and $P_N(f) \geq 0$, the numerator and denominator are both non-negative, and the denominator $\geq$ the numerator. Hence $0 \leq \check{h}_{\text{opt}}(f) \leq 1$.

$P_{dy}(f) = \check{w}(f) P_{xy}(f) = \check{w}(f) P_x(f)$.

$\check{h}_{\text{opt}}(f) = P_{dy}(f) / P_y(f) = \check{w}(f) P_x(f) / (P_x(f) + P_N(f))$. This is the standard Wiener filter multiplied by the smoothing kernel's frequency response.

Integral over $|f| \leq (1-\alpha)/(2T)$: $2T \cdot (1-\alpha)/(2T) = 1-\alpha$.

In each rolloff band (width $\alpha/(2T)$ on each side), the integrand averages to $T/2$. Total contribution: $2 \cdot (T/2) \cdot \alpha/(2T) \cdot 2 = \alpha$ (accounting for both sides). Actually, the integral of the half-cosine taper over each transition band gives $T \cdot \alpha/(2T) \cdot 1/2 \cdot 2 = \alpha/2$ per side, total $\alpha$. Wait — more carefully: $\int = 1-\alpha + \alpha = 1$. Peak gain is $T$, so $B_N = \frac{1}{2T} \cdot 1 = 1/(2T)$.

$B_N = 1/(2T)$ regardless of $\alpha$. The raised-cosine filter has the remarkable property that its noise bandwidth equals the Nyquist bandwidth, independent of the rolloff factor.

$P_{xy}(f) = C^*(f) P_x(f)$.

$\check{h}_{\text{eq}}(f) = \frac{C^*(f) P_x(f)}{|C(f)|^2 P_x(f) + P_N(f)}.$$

At frequencies where $|C(f)|^2 P_x(f) \gg P_N(f)$, $\check{h}_{\text{eq}}(f) \approx 1/C(f)$ (zero-forcing equalization). At frequencies where noise dominates, the equalizer suppresses the output (noise regularization). This MMSE equalizer avoids the noise amplification problem of the zero-forcing equalizer $\check{h}(f) = 1/C(f)$.

$[P_y(f)] = [\check{h}(f)]^2 \cdot [P_x(f)] = 1 \cdot \text{W/Hz} = \text{W/Hz}$. The output PSD has the same units as the input PSD, as expected.

$P_N(f) = {N_0}_{1}/2 + {N_0}_{2}/2 = ({N_0}_{1} + {N_0}_{2})/2$.

$\mathcal{P}_N = ({N_0}_{1} + {N_0}_{2}) B_N$.

Stage 1 adds noise power $P_1 = k_B T_0 B_N (F_1 - 1)$ at its input. Stage 2 adds $P_2 = k_B T_0 B_N (F_2 - 1)$ at its input, which referred to the overall input is $P_2 / G_1$.

Total input-referred added noise: $P_1 + P_2/G_1 = k_B T_0 B_N [(F_1 - 1) + (F_2 - 1)/G_1]$.

$F_{\text{total}} = 1 + (F_1 - 1) + (F_2 - 1)/G_1 = F_1 + (F_2 - 1)/G_1$.

$\mathbf{r}_{yx}(f) = \mathbf{H} P_x(f)$.

$\mathbf{R}_{yy}(f) = P_x(f) \mathbf{H} \mathbf{H}^H + \sigma^2 \mathbf{I}$.

By the matrix inversion lemma: $\mathbf{g}(f) = \frac{P_x(f)}{P_x(f) \|\mathbf{H}\|^2 + \sigma^2} \mathbf{H}$. This is the LMMSE beamformer, which combines the $M$ channels optimally.