Math Lane Proof — KaTeX Build-Time Rendering

Source A: Mathematical Foundations for IoT — The Derivation

(content/capstone/mathematical-foundations.qmd, lines 165-231)

Free-space radio begins with transmit power $P_t$ spread over area $4\pi d^2$:

$S = \frac{P_tG_t}{4\pi d^2}$

An antenna collects with effective aperture:

$A_e = \frac{G_r\lambda^2}{4\pi}$

So Friis is:

$P_r = P_tG_tG_r\left(\frac{\lambda}{4\pi d}\right)^2$

and free-space path loss is:

$\mathrm{FSPL} = \left(\frac{4\pi d}{\lambda}\right)^2$

Because $\lambda = c/f$:

$\mathrm{FSPL}_{dB} = 20\log_{10}(d) + 20\log_{10}(f) + 20\log_{10}\left(\frac{4\pi}{c}\right)$

Decibels come from power ratios:

$L_{dB} = 10\log_{10}\left(\frac{P_2}{P_1}\right)$

so multiplied gains and losses become addition:

$P_{rx,dBm} = P_{tx,dBm} + G_{tx,dB} + G_{rx,dB} - L_{path,dB}$

Sampling repeats a spectrum every $f_s$ hertz:

$X_s(f) = \sum_{k=-\infty}^{\infty}X(f-kf_s)$

Non-overlap requires:

$f_{max} \leq f_s - f_{max}$

therefore:

$f_s \geq 2f_{max}$

Quantization step size is:

$q = \frac{V_{FS}}{2^N}$

Uniform rounding noise has mean-square value:

$\overline{e^2} = \frac{q^2}{12}$

For a full-scale sine wave:

$\mathrm{SNR}_{dB} = 10\log_{10}\left(1.5 \times 2^{2N}\right)$

$\mathrm{SNR}_{dB} \approx 6.02N + 1.76$

For an RC low-pass filter, the capacitor law $i = C\,dV/dt$ and Ohm’s law give exponential settling with:

$\tau = RC$

and cutoff:

$f_c = \frac{1}{2\pi RC}$

Battery runtime is charge divided by average current:

$I_{avg} = \frac{\sum I_i t_i}{\sum t_i}$

$t_{hours} = \frac{\mathrm{capacity}_{mAh}}{I_{avg,mA}}$

Worked numbers (same source, lines 237-242 — dense mixed inline math + prose, exactly the “up to 36 instances in one file” density the pilot flagged):

• Shannon ceiling: $C = B\log_2(1+\mathrm{SNR}) = 125000\log_2(11) = 432429$ bit/s $\approx 432$ kbit/s
• Link budget: $P_{rx} = 14 - 120 = -106$ dBm, and margin $= -106 - (-137) = 31.0$ dB, or $10^{31/10} = 1.26\times10^3$ times the minimum receiver power
• Sampling: the chapter’s $360$ Hz vibration content needs $f_s \geq 2(360) = 720$ Hz; a $2$ kHz choice gives $2000/720 = 2.78\times$ the minimum and a $1000$ Hz Nyquist frequency
• Sixteen-bit data stream: ideal quantization SNR $= 6.02(16) + 1.76 = 98.1$ dB, while the data rate at $2$ kHz is $2000 \times 16 = 32000$ bit/s
• RC anti-alias filter: for $f_c = 400$ Hz, $RC = 1/(2\pi 400) = 398$ us; with $R = 10$ kohm, $C = 398$ us$/10000 = 39.8$ nF
• Battery record: $I_{avg} = 160(2/60) + 0.8(58/60) = 5.33 + 0.773 = 6.11$ mA, so $2000/6.11 = 328$ h $= 13.7$ days

Source A continued: Calculus, Exponentials, Logs

(same file, lines 429-620)

Notation: $\frac{dx}{dt}$ means “the rate at which x changes with respect to time t”

The PID formula uses derivatives:

$P(t) = K_p \cdot e(t) + K_i \cdot \int e(t)dt + K_d \cdot \frac{de(t)}{dt}$

• D term ($\frac{de}{dt}$): How fast is the error changing? (prevents overshooting)

Notation: $\int f(x)dx$ means “sum up all the tiny pieces of f(x)”

Total energy consumed by an IoT device:

$E_{total} = \int_0^T P(t) \, dt$

$E_{cycle} = (50mW \times 10ms) + (0.01mW \times 990ms) = 0.5mJ + 0.0099mJ \approx 0.51mJ$

The exponential function: $y = e^x$ where $e \approx 2.718$

• After $1\tau$: Value drops to 37% of initial ($e^{-1} \approx 0.37$)

$\tau = R \times C = 100\Omega \times 10\mu F = 1ms$

Definition: If $b^x = y$, then $\log_b(y) = x$

• $\log_{10}$ (common log) - used in dB calculations
• $\ln$ or $\log_e$ (natural log) - used in signal processing
• $\log_2$ (binary log) - used in information theory (bits)

$dB = 10 \log_{10}\left(\frac{P_{out}}{P_{in}}\right)$

$C = B \log_2(1 + SNR)$

$dBm = 10 \log_{10}(P_{mW})$

Source A continued: Vectors and Matrices

(same file, lines 687-773)

Notation: $\vec{v} = \begin{bmatrix} x \\ y \\ z \end{bmatrix}$ or simply $[x, y, z]$

Magnitude (Length): How “big” is the vector?

$|\vec{v}| = \sqrt{x^2 + y^2 + z^2}$

Example: Accelerometer reads $[0.5, 0.3, 0.8]g$

$|\vec{a}| = \sqrt{0.5^2 + 0.3^2 + 0.8^2} = \sqrt{0.98} \approx 0.99g$

Notation:

$M = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$

Rotating sensor data to align with a reference frame:

State vector might be $[position, velocity, acceleration]$:

$\vec{x} = \begin{bmatrix} p \\ v \\ a \end{bmatrix}$

State transition predicts next state:

$\vec{x}_{k+1} = F \cdot \vec{x}_k$

Describes uncertainty in measurements:

$P = \begin{bmatrix} \sigma_x^2 & \sigma_{xy} \\ \sigma_{xy} & \sigma_y^2 \end{bmatrix}$

Example: Transforming a 3D point (a wide equation — deliberately kept in this proof page to test the overflow-protection rule in TestMathLayout.astro; see its <style> comment):

$\begin{bmatrix} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \end{bmatrix} \begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix} = \begin{bmatrix} 2 \\ 4 \\ 6 \end{bmatrix}$

Source B: Fresnel Zones and Deployment

(content/networking-core/net-prop-fresnel.qmd, lines 74-119 — a real “net-prop-*” file, the task brief’s own example)

At a point with distances (d_1) and (d_2) from the two antennas, the first-zone radius is

The zone is widest near the path midpoint. Lower frequencies have longer wavelengths and therefore larger Fresnel radii for the same geometry.

ITU-R P.526 uses clearance equal to 60% of the first-zone radius as a boundary for the diffraction zone. Treat (0.6r_1) as a useful screening target, not as a guarantee of availability or delivery.

For wavelength (lambda), the radius at one location along the path is

where (d_1) and (d_2) use the same distance unit and the result follows that unit.

With distance in kilometres and frequency in gigahertz, a convenient form is

At the midpoint of a total path (D), (d_1=d_2=D/2), so

for (D) in kilometres, (f) in gigahertz, and radius in metres.

Worked Path-Profile Decision

(same file, lines 231-241)

Consider a hypothetical 5 km path at 915 MHz with an obstacle 2 km from the transmitter and 3 km from the receiver.

so the 60% screen is about 11.9 m at that location.

Suppose the direct path line is 18 m above the common datum there and the surveyed obstacle top is 9 m. The geometric clearance is 9 m, or

The obstacle enters the 60% screen.

Coverage check: \begin{align}

(content/coap/coap-api-design.qmd, lines 845-861 verbatim — estate-wide grep of content/*/*.qmd for \begin{[a-zA-Z*]+} turns up exactly three environments in use: aligned (35), align (14), bmatrix (13; all exercised above via Source A). This block is a real align instance, picked from coap since it overlaps this lane’s hand-picked module.)

CON (Confirmable) message energy:

Annual comparison (525,600 messages):