Exercises

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

$\mathcal{P}_x = r_{xx}(0) = 4$. Verify: $\int_{-\infty}^{\infty} \frac{16}{4+(2\pi f)^2}\,df = \frac{16}{2\pi}\cdot\frac{\pi}{2} = 4$. $\checkmark$

$|r_{xx}(\tau)| = \left|\int P_x(f)\,e^{j2\pi f\tau}\,df\right| \leq \int |P_x(f)|\,df = \int P_x(f)\,df = r_{xx}(0).$$The inequality uses$P_x(f) \geq 0$to drop the absolute value and$|e^{j2\pi f\tau}| = 1$.

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

${P_x}_{y}(f) = 4\sin^2(\pi f) \cdot 2 = 8\sin^2(\pi f)$.

This is a high-pass spectrum: zero at $f = 0$, maximum at $f = 1/2$.

With $a = 1$ and $\omega = 2\pi f$, we can write $P_x(f) = \frac{1}{(1 + (2\pi f)^2)^2}$.

Using the known pair: $$r_{xx}(\tau) = \frac{1}{4}(1 + |\tau|)\,e^{-|\tau|}.$$

$r_{xx}(0) = 1/4$. Check: $\int \frac{df}{(1+(2\pi f)^2)^2} = \frac{1}{4}$. $\checkmark$

${P_x}_{y}(f) = (2\pi f)^2 \cdot 1 = (2\pi f)^2, \quad |f| \leq 5.$$

$\sigma^2_{y} = \int_{-5}^{5} (2\pi f)^2\,df = 4\pi^2 \cdot \frac{2 \cdot 5^3}{3} = \frac{4\pi^2 \cdot 250}{3} \approx 3290 \text{ W}.$$

Differentiation enormously amplifies high-frequency noise.

$P_x(f) = 5 + 2e^{-j2\pi f} + 2e^{j2\pi f} = 5 + 4\cos(2\pi f).$$

The minimum of $5 + 4\cos(2\pi f)$ occurs at $f = 1/2$: $P_x(1/2) = 5 - 4 = 1 > 0$. So $P_x(f) \geq 1 > 0$ for all $f$.

$P_x(f) = \frac{1}{1 - 1.8\cos(2\pi f) + 0.81}$. This peaks at $f = 0$ with value $1/(1 - 0.9)^2 = 100$ and has minimum $1/(1 + 0.9)^2 \approx 0.28$ at $f = 1/2$.

The periodogram from 128 samples will fluctuate wildly around the true PSD. Its variance is approximately $P_x(f)^2$ at each $f$.

Averaging 4 periodograms (each from 32 samples) reduces variance by roughly a factor of 4 but also reduces frequency resolution by a factor of 4 (the main lobe width goes from $1/128$ to $1/32$).

For any $\alpha \in \mathbb{C}$, ${P_x}_{z}(f) = {P_x}_{x}(f) + \alpha P_{xy}(f) + \alpha^* P_{yx}(f) + |\alpha|^2 {P_x}_{y}(f) \geq 0$.

This is a non-negative quadratic in $\alpha$. The discriminant condition gives $|P_{xy}(f)|^2 \leq {P_x}_{x}(f)\,{P_x}_{y}(f)$. Since $P_{yx}(f) = P_{xy}^*(f)$, we get $|P_{xy}(f)|^2 \leq {P_x}_{x}(f)\,{P_x}_{y}(f)$.

From mean-square calculus (Ch. 13): $r_{\dot{x}\dot{x}}(\tau) = -\frac{d^2}{d\tau^2}r_{xx}(\tau)$.

${P_x}_{\dot{x}}(f) = \mathcal{F}\{-\ddot{r_{xx}}(\tau)\} = (2\pi f)^2\,\mathcal{F}\{r_{xx}(\tau)\} = (2\pi f)^2\,P_x(f).$$

Differentiation in time corresponds to multiplication by $j2\pi f$ in frequency. The squared magnitude gives $(2\pi f)^2$. This is consistent with the LTI viewpoint: the derivative is an LTI operation with $\check{h}(f) = j2\pi f$.

$r_{xx}(\tau) = \int_{-1}^{1} e^{j2\pi f\tau}\,df = 2\operatorname{sinc}(2\tau)$.

$r_{xx}(1/2) = 2\operatorname{sinc}(1) = 2 \cdot \frac{\sin\pi}{\pi} = 0$.

Samples spaced by $1/(2 \cdot 1) = 1/2$ seconds are uncorrelated.

$\hat{P_x}(f) = \frac{1}{N}\sum_{n=0}^{N-1}\sum_{k=0}^{N-1} X_n X_k^*\,e^{-j2\pi f(n-k)}.$$

$\mathbb{E}[\hat{P_x}(f)] = \frac{1}{N}\sum_{n,k} r_{xx}[n-k]\,e^{-j2\pi f(n-k)}.KATEXPLACEHOLDER0END\mathbb{E}[\hat{P_x}(f)] = \sum_{m=-(N-1)}^{N-1} r_{xx}[m]\left(1 - \frac{|m|}{N}\right) e^{-j2\pi fm}.$$

This is the DTFT of the windowed autocorrelation with a Bartlett (triangular) window. As $N \to \infty$, the window approaches 1 for each fixed $m$, giving $\mathbb{E}[\hat{P_x}(f)] \to P_x(f)$.

For a single realization with known $\Phi$, the periodogram is $\hat{P_x}(f) = \frac{A^2}{4N}|D_N(f - f_0)|^2 + \frac{A^2}{4N}|D_N(f + f_0)|^2 + \text{cross terms}$, where $D_N(f) = \sum_{n=0}^{N-1} e^{-j2\pi fn} = e^{-j\pi(N-1)f}\frac{\sin(N\pi f)}{\sin(\pi f)}$ is the Dirichlet kernel.

The Dirichlet kernel has a main lobe of width $\sim 2/N = 1/32$ centered at $f_0$. The peak value is proportional to $N$, so $\hat{P_x}(f_0) \approx A^2 N / 4$, which grows with $N$ — reflecting the delta function in the true PSD.

The sidelobes of $|D_N(f)|^2$ produce spectral leakage: apparent power at frequencies away from $f_0$. Windowing (Hamming, Hann) reduces the sidelobe level at the cost of widening the main lobe.

$P_x(f) = \frac{1}{|1 - 1.5 e^{-j2\pi f} + 0.7 e^{-j4\pi f}|^2}$.

The poles of the transfer function are at $z = 0.75 \pm j\,0.3708$, which have angle $\pm \arctan(0.3708/0.75) \approx \pm 0.459$ rad $= \pm 0.073$ cycles. The PSD peaks near $f \approx 0.073$. The process has a resonant frequency determined by the complex pole angles.

For $(X_0, \ldots, X_{N-1})$ with covariance $\mathbf{R}_N$ (Toeplitz): $h(X_0, \ldots, X_{N-1}) = \frac{N}{2}\log(2\pi e) + \frac{1}{2}\log\det\mathbf{R}_N$.

Szegő's theorem states that for a Toeplitz matrix $\mathbf{R}_N$ with symbol $P_x(f)$:

$$\frac{1}{N}\log\det\mathbf{R}_N \to \int_{-1/2}^{1/2} \logP_x(f)\,df$$

as $N \to \infty$ (assuming $\logP_x$ is integrable).

$\bar{h} = \lim_{N\to\infty}\frac{1}{N}h(X_0,\ldots,X_{N-1}) = \frac{1}{2}\log(2\pi e) + \frac{1}{2}\int_{-1/2}^{1/2}\logP_x(f)\,df.$$

$r_{xy}(\tau) = \mathcal{F}^{-1}\left\{\frac{3}{1 + j2\pi f}\right\} = 3\,e^{-\tau}\,u(\tau),$$where$u(\tau)$ is the unit step function.

The cross-correlation is one-sided (causal), which occurs when $Y(t)$ depends on past values of $X(t)$ but not future values — consistent with a causal LTI relationship.

${P_x}_{w}(f) = \sum_m \sigma^2\,\delta[m]\,e^{-j2\pi fm} = \sigma^2$.

$\int_{-1/2}^{1/2} \sigma^2\,df = \sigma^2 = {r_{xx}}_{w}[0]$. $\checkmark$

The triangle $\Lambda(f) = \max(0, 1 - |f|)$ is the convolution of $\operatorname{rect}(f)$ with itself. By the convolution theorem:

$$r_{xx}(\tau) = \operatorname{sinc}^2(\tau).$$

Since $\operatorname{sinc}^2(\tau) \geq 0$ for all $\tau$, the autocorrelation is indeed non-negative. This is consistent with the PSD being the square of a Fourier transform (the rect function acts as a "spectral factor").

$5 + 4\cos(2\pi f) = 1 + 4 + 4\cos(2\pi f) = |1 + 2e^{-j2\pi f}|^2$. (Expand: $|1 + 2e^{-j2\pi f}|^2 = 1 + 4 + 4\cos(2\pi f) = 5 + 4\cos(2\pi f)$.)

$P_x(f) = \frac{|1 + 2e^{-j2\pi f}|^2}{|1 - 0.5e^{-j2\pi f}|^2} = \left|\frac{1 + 2e^{-j2\pi f}}{1 - 0.5e^{-j2\pi f}}\right|^2.$$So$\check{h}(f) = \frac{1 + 2e^{-j2\pi f}}{1 - 0.5e^{-j2\pi f}}$and$\sigma^2 = 1$.

The pole is at $z = 0.5$ (inside the unit circle), so the filter is causal and stable. The zeros are at $z = -2$ (outside) and $z = -1/2$ would give the minimum-phase factor.

$\int_{-\infty}^{\infty} \frac{df}{1 + (2\pi f RC)^2} = \frac{1}{2\pi RC}\cdot\pi = \frac{1}{2RC}.$$

$B_n = \frac{1}{|\check{h}(0)|^2}\int |\check{h}(f)|^2\,df = \frac{1}{1}\cdot\frac{1}{2RC} = \frac{1}{2RC}$.

Wait — this is the two-sided integral. The noise bandwidth convention uses $B_n = \frac{\int_0^{\infty}|\check{h}(f)|^2\,df}{|\check{h}(0)|^2} = \frac{1}{4RC}$.

$\sigma^2_{y} = N_0 \cdot B_n = \frac{N_0}{4RC}$.

$P_x(f) = \frac{3}{2}[\delta(f-2) + \delta(f+2)] + 2.$$

The discrete part $\frac{3}{2}[\delta(f-2)+\delta(f+2)]$ comes from the periodic component at $\pm 2$ Hz. The continuous part $P_x(f) = 2$ is a white noise floor. The process is a sinusoid embedded in white noise.