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The moment of inertia about a specified axis measures how a mass distribution resists changes of angular velocity about that axis. For point masses $m_i$ at perpendicular distances $r_{\perp i}$, or a continuous distribution with mass density $\rho$,Its SI unit is $\mathrm{kg}{\mathrm{m}}^2$. The distance is to the entire axis, not to one point on it: moving a point mass along the axis leaves its contribution unchanged. The moment of inertia depends on both the distribution and the chosen axis; it is not determined by total mass alone. About an axis fixed in an inertial frame, a rigid body with larger moment of inertia has smaller angular acceleration under the same net axial torque, and greater rotational kinetic energy at the same angular speed.^[ref]

For a simple model, put four equal point masses of total mass $M$ on light rigid spokes at radius $a$. Their moment of inertia is $M{a}^2$. Doubling $a$ quadruples the moment of inertia without changing $M$. The illustrated placements $a=R/3$ and $a=R$ give $M{R}^2/9$ and $M{R}^2$. Actual spokes and finite-sized weights contribute their own moments of inertia; the point mass model neglects them. Repositioning weights during rotation also breaks the fixed-distribution assumption, so it requires a separate work and angular momentum calculation.

The moment of inertia of a uniform disc follows by dividing it into concentric annuli: $dm=(2M/R^2)rdr$ and $I={\int }_0^R{r}^2(2M/R^2)rdr=M{R}^2/2$. For an ideal thin ring every mass element is at $R$, so $I=M{R}^2$. A ring of finite radial thickness instead uses the moment of inertia of a uniform annulus, $I=M(R^2+b^2)/2$, with inner radius $b$. A moment of inertia is thus a compact summary of radial placement rather than a complete description of shape. For several joined pieces, their moments of inertia add only when each is calculated about the same axis.^[ref]

A physical pendulum makes the moment of inertia measurable through timing. Martín Monteiro, Cecilia Stari, Cecilia Cabeza and Arturo C. Martí used a metal bar with suspension holes $1.0\mathrm{cm}$ apart and an attached smartphone. They report bar length $1.199\mathrm{m}$, bar mass $0.2518\mathrm{kg}$ and phone mass $0.1311\mathrm{kg}$. Changing the suspension hole changes the pivot axis. Their readouts were phone gyroscope angular velocity records, analyzed by autocorrelation or a fit to a sinusoid, and an independent photogate; the logged records for the sinusoid fits lasted $10\mathrm{s}$. They observed agreement between the three timing methods and the geometry-based model, including a minimum of oscillation period as the suspension point varied. These are reported apparatus details and observations, not measurements made for this article.^[ref]

To infer a moment of inertia from such a measurement, determine the total mass $M_{\mathrm{tot}}$ including every attached part, locate the assembly's centre of mass $G$, and measure its distance $h>0$ below the horizontal pivot $O$. Record successive crossings in the same direction, or the elapsed time $t_N$ for $N$ complete oscillations, giving oscillation period $P=t_N/N$. The physical pendulum model with a fixed pivot, a uniform gravitational field, negligible friction and the small-angle approximation givesThe crossing times, masses and geometric distances are the measured inputs; $I_{O,\mathrm{inferred}}$ is an inference conditional on this model, while an integral over the assumed shape supplies an independent model prediction. The parallel axis theorem converts the inferred pivot moment of inertia to one about a parallel axis through $G$. The controls and interpretation of that comparison are described there. An axial measurement determines one scalar moment of inertia, not the complete inertia tensor needed for general rotation.

The Nyquist–Shannon sampling theorem says that a bandlimited function can be recovered exactly from equally spaced samples, provided the sampling rate is high enough. It is the bridge between a continuous waveform and a sequence of numbers: the missing values between samples are determined by the restriction on its frequencies.

Here is a precise finite-energy version. Let $f\in L^2(\mathbb{R})$, meaning ${\int }_{\mathbb{R}}|f(t){|}^{2}dt<\infty$, and suppose its Fourier transform, defined through the Plancherel theorem, is zero outside $[-2\pi B,2\pi B]$. Thus $B>0$ is the largest allowed ordinary frequency, in cycles per second, rather than an angular frequency. Set $T=1/f_s$, where $f_s$ is the number of samples per second. If $f_s\ge 2B$, the continuous representative of $f$ is recovered byThe quotient is the normalized sinc function, with value one when its argument is zero. Each shifted sinc function is one at its own sampling time and zero at every other sampling time. Its coefficient is therefore exactly the required sample value. For this L2 space formulation, the reconstruction converges in the L2 space norm and uniformly in $t$. The critical rate $f_s=2B$ is allowed for this class; choosing $f_s>2B$ leaves a gap between the occupied frequencies and the sampling boundary.

For a numerical example, a bandlimited function with no frequencies above $3$ cycles per second can be reconstructed from its samples every $0.1$ seconds: $f_s=10>6$. Exact reconstruction uses the whole sequence indexed by $\mathbb{Z}$. A finite recording and a finite sum of sinc functions introduce truncation error; joining adjacent samples by straight lines is generally a different reconstruction.

The obstruction is aliasing. At sampling rate $f_s=10$, the two distinct waveformshave identical samples at $t=n/10$, becauseThe samples alone cannot tell whether the waveform oscillates twice or eight times per second. If we know in advance that all frequencies are below $5$, the $8$-cycle waveform is excluded and the ambiguity disappears. These pure cosines are an illustrative aliasing example, rather than members of the finite-energy L2 space class in the stated theorem.

Exactly at a sampling boundary, the function class matters. At rate $f_s$, the sine $\sin (2\pi (f_s/2)t)$ has zero values at every $n/f_s$. It is a nonzero waveform indistinguishable from zero on that grid, but has infinite energy and is excluded by the finite-energy hypothesis above. This explains why a rule intended to cover arbitrary pure sinusoids should demand a strict inequality. For spatial data, replace seconds by distance: the Nyquist spatial frequency is $1/(2p)$ for sample pitch $p$.

The proof idea is short: the samples are Fourier coefficients of the Fourier transform restricted to a sampling-sized frequency interval. Because a bandlimited function has no contribution outside that interval, its Fourier series determines the whole Fourier transform; Fourier inversion then gives the sinc function expansion. See sampling expansion by periodic Fourier projection for this connection.

Claude E. Shannon's Communication in the Presence of Noise (1949), Section II, Theorem 1 and equation (7), gives the sampling statement and reconstruction using shifted sinc functions in a communication setting. Shannon explicitly notes earlier mathematical forms and credits Nyquist with emphasizing the sampling interval in telegraphy.^[ref]

Choose both waves as sinusoids with a common zero phase and amplitude $A$. The trigonometric sum identity givesThe slowly varying wave envelope has constant phase when $\delta x-ϵt$ is constant. Its speed is $ϵ/\delta$, and in the close-frequency limit the group velocity isIt describes the motion of a narrow-band envelope in a medium where the constituent waves retain a coherent packet interpretation; it is not an unconditional signal-speed theorem in strongly absorbing or anomalously dispersive media. The carrier phase velocity is $v_p=\omega /k$. In a nondispersive medium, $\omega =vk$ with constant $v$, so $v_g=v=v_p$.

A nonstationary process has a statistical law that changes with the time origin. In the weak sense, this includes a time-dependent mean or variance, or a covariance that depends on the two times separately rather than only on their lag. A strictly stationary process requires invariance of every finite-dimensional distribution under a common time shift.

In the preceding plots, series $1$ suggests a stochastic trend or changing level, and series $4$ suggests a deterministic seasonal mean. These are the intended nonstationary examples. A plot alone cannot prove either conclusion: in particular, a stationary random-phase sinusoid can have a seasonal-looking trace and an oscillatory ACF. The interpretation of series $4$ as nonstationary concerns a seasonal mean tied to calendar time.

Three standard responses are to remove a fitted deterministic trend or seasonal mean; to use regular or seasonal differencing for an appropriate trend or seasonal component; and to stabilize a changing variance by a transformation or explicit seasonal scale model. For example, logarithms or a Box–Cox transformation can address level-dependent variance. The operation should match the source of nonstationarity; differencing a varying variance does not generally make it stationary.

The Sturm oscillation theorem gives three nodes of a wavefunction for the fourth bound state. reflection symmetry and nondegeneracy of one-dimensional bound states make it odd: one node of a wavefunction is at the centre, and the other two lie at opposite positions. In each flat allowed region the wavefunction is a sinusoid, with $k=\sqrt{2m(E-V)}/\hslash$; its wavelength is shortest in the central, deepest region. Outside the well it decays exponentially, with opposite signs on the two sides. Both $\Psi$ and ${\Psi }^{\prime }$ remain continuous at every finite step, as required by wavefunction matching at a finite potential step; only the curvature jumps. In particular, the wavefunction need not vanish at a step.

This original wavefunction matching at a finite potential step illustration uses freely chosen well dimensions and energies. It shows the required parity, node count, wavelength ordering and decaying tails; the source drawing supplies no numerical scale from which to fix exact node positions.

Sampling at $t_n=n/f_s$ makes a frequency $f$ indistinguishable from $f+k{f}_s$, since the extra phase is $2\pi kn$. For a real sinusoid, reflection to $|k{f}_s-f|$ also gives an alias, up to phase. For example, a $50\mathrm{Hz}$ cosine sampled at $80\mathrm{Hz}$ has exactly the samples of a $30\mathrm{Hz}$ cosine. A measured low-frequency feature can therefore originate above the instrument's Nyquist frequency.

To reconstruct an arbitrary baseband signal limited to bandwidth $B$, sample strictly above $2B$ and use an analogue anti-aliasing filter before the sampler. A practical filter needs a transition band, so merely choosing a nominal rate $2B$ is insufficient. frequency components already folded together cannot in general be separated by later digital filtering. aliasing from insufficient sampling is distinct from spectral leakage caused by a finite observation window.

For electric potential difference and electric current varying as sinusoids, choose the phasor conventionThe electrical impedance isIts modulus gives the voltage-to-current amplitude ratio, and its complex argument gives the voltage-minus-current phase difference. For ideal components,These follow from Ohm's law $V=RI$, the capacitor law $I=CdV/dt$, and the inductor law $V=LdI/dt$. In phasor notation a time derivative becomes multiplication by $i\omega$, turning the linear ordinary differential equations of the electrical circuit into algebraic equations. Electrical impedances add in a series circuit; their reciprocals, the electrical admittances, add in a parallel circuit. This computes both amplitudes and phase differences in the steady response to sinusoids.

A phasor is a complex amplitude representing a sinusoid at one specified angular frequency: $a(t)=\mathrm{Re}(\hat{A}{e}^{\mathrm{i}\omega t})$. Its modulus gives the peak amplitude and its complex argument gives the wave phase in this convention. Both may instead use root mean square magnitudes if that normalization is stated.

The University of Colorado Boulder's Lab 3: Filters, §§7.1–7.4, specifies measuring the resistor and capacitor, then driving the assembled RC low-pass filter with a function generator. A two-channel oscilloscope reads actual input and output using high electrical impedance. Sweep the frequency, record the voltage amplitude ratio, and locate $1/\sqrt{2}$. A control compares a cable with a $10\times$ oscilloscope probe on a resistive voltage divider, isolating measurement capacitance before using the probe for the filter. A slow square wave then lets discharge timing estimate the time constant independently. Agreement supports the RC low-pass frequency response; source output impedance, probe loading, and instrument signal bandwidth limit it. These are prescribed measurements, with no measured dataset supplied here.^[ref]

The two-channel setup can also test the predicted phase difference directly: for positive frequency $f$, measure the output's time lag $\Delta t$ between corresponding features of the two sinusoids. For this single-stage model the phase lies between $-\pi /2$ and $0$, so choose the lag within a quarter cycle and calculate $\varphi =-2\pi f\Delta t$. At the cutoff frequency the predicted lag is one eighth of a cycle. This tests the timing part of the frequency response as well as the voltage ratio.

A spark gap can rapidly discharge a charged capacitor through an inductor. After excitation, the stored energy runs down through Joule heating, radiation and other losses. A key held down can permit repeated discharges: each discharge starts a damped train of rapid oscillations. The spark repetition frequency, the electrical oscillation frequency within each train, and the rate of keyed symbols are different quantities. Repetition does not make each train a maintained sinusoid.

An original lumped model replaces the conducting spark gap and other losses by a constant total electrical resistance $R>0$ in an isolated series RLC circuit. For electric charge $q$ on the capacitor, the Kirchhoff cycle law givesFor $Q>1/2$, its characteristic roots are $-\alpha \pm \mathrm{i}{\omega }_d$, with ${\omega }_d=\sqrt{{\omega }_0^2-{\alpha }^2}$, proving the stated damped form. The electric current $\dot{q}$ has the same exponential decay scale. Its wave envelope falls by $\exp (-2\pi \alpha /{\omega }_d)$ per cycle; the energy envelope falls twice as fast in the exponent. A maintained sinusoid instead requires continuing input power balancing losses.

With $L=5\mathrm{mH}$, $C=2\mathrm{nF}$ and $R=100\Omega$, the model gives $f_0={\omega }_0/(2\pi )=50.329\mathrm{kHz}$, Q factor $Q=15.811$, amplitude time constant $1/\alpha =100\mu \mathrm{s}$, and damped frequency ${\omega }_d/(2\pi )=50.304\mathrm{kHz}$. These are chosen component values. A real spark gap switches nonlinearly, an antenna radiates, and coupling exchanges energy between circuits; one constant-$R$ free decay does not model the complete transmitter. Rapid damping also broadens the frequency spectrum, which helps explain why short trains make sharp receiver selection difficult.

In spatial sampling of a wave, a detector records a wave at positions $x_n=np$ separated by pitch $p>0$. Angular wavenumbers differing by $2\pi m/p$, with integers $m$, give identical samples of a complex Fourier mode:This is spatial aliasing. To distinguish arbitrary sinusoids within a prescribed band, require $|k|<\pi /p$, equivalently $|{\nu }_x|<1/(2p)$, the Nyquist spatial frequency. Thus $p<{\lambda }_{\min }/2$ for the shortest allowed wavelength. At equality a sine can vanish at every sample, so phase information can be lost. The Nyquist–Shannon sampling theorem states the precise reconstruction conditions for a bandlimited function; a finite scan introduces further limitations.

The synthetic two-tone snapshot iswhere both ${\nu }_j$ count cycles per metre. The component wavelengths are $0.50\mathrm{m}$ and $0.125\mathrm{m}$. With $p=0.05\mathrm{m}$, both lie below the Nyquist spatial frequency of $10$ cycles per metre. With $p=0.10\mathrm{m}$, the $8$-cycle component aliases to the $2$-cycle component: at every $x_n=n/10$, $\cos (2\pi \cdot 8x_n)=\cos (2\pi \cdot 2x_n)$. The sampled sum is therefore indistinguishable from $1.5\cos (4\pi x)$, although the continuous curves differ between samples. Drawing lines between sample points cannot resolve this ambiguity.

A transfer function packages the response of a linear time-invariant system into a rule for turning a plant input into a plant output. Instead of solving a differential equation separately for every signal, take a Laplace transform, multiply by $G(s)$, then transform back. This also makes it possible to compare slow and fast oscillations through the continuous-time frequency response.

Here consider a continuous-time, causal, finite-dimensional linear time-invariant system with one plant input $u(t)$ and one plant output $y(t)$. With every initial state set to zero, its transfer function is defined byThe ratio $G(s)=Y(s)/U(s)$ is used where $U(s)\ne 0$ in a common Laplace convergence region; $G$ describes the system, independently of that particular plant input. Zero initial state matters: stored motion or another nonzero initial condition can contribute to the plant output even when the plant input vanishes. ^[ref]

The first-order low-pass filter below shows the same transfer function predicting both a step and a sustained sinusoid. Its pole gives a decay rate, its value at zero gives the eventual gain, and its values along the imaginary axis give the continuous-time frequency response. These descriptions make a useful bridge between a physical relaxation model and control theory calculations.

An original teaching model exposes the roles of input loading, output loading and the supply. Choose a small-signal model with positive input electrical resistance $r_i$, transconductance $g_m>0$, and output electrical resistance $r_o>0$. Its controlled output current source draws $g_m{v}_i$ from the output node toward the reference node. This chosen sign gives an inverting voltage gain; it is not the historical contact circuit's sign convention.

Assume a fixed operating point, negligible reverse coupling, a frequency range where coupling capacitors pass the signal, and negligible other capacitances and transit delays. A collector resistor $R_D$ connects the output to a steady ideal supply $V_{DD}$; an external electrical load $R_L$ is coupled through a capacitor, so it draws no direct current. The input source has series electrical resistance $R_s$. All voltages below are small departures from the operating point. An ideal fixed supply has zero incremental voltage and therefore becomes the reference node in the small-signal model.

Define $R_O=R_D\parallel r_o$ and $R_P=R_O\parallel R_L$, where $a\parallel b=ab/(a+b)$. Input voltage division and output Kirchhoff's current law giveConsequently the port voltage gain, overall source-to-output voltage gain, and load current gain magnitudes areThis current gain uses load electric current, not all the controlled-source electric current. It is also a different port definition from the historical emitter-to-collector current gain. Larger $R_s$ attenuates the input; smaller $R_L$ attenuates the output.

For a sinusoid input signal, use root mean square voltages. The power gain from accepted input signal power $P_i=v_i^2/r_i$ to useful load power $P_L=v_o^2/R_L$ isUnlike $|A_i|$, this tends to zero as either $R_L\to 0$ or $R_L\to \infty$. Differentiation givesThus matching the output electrical load maximizes delivered power in this unilateral resistive model. For fixed source voltage, input loading multiplies $P_L$ by $[r_i/(R_s+r_i){]}^2$; the accepted-input power gain above excludes the loss in $R_s$.

Choose $r_i=R_s=1\mathrm{k\Omega }$, $g_m=2\mathrm{mS}$, $r_o=20\mathrm{k\Omega }$, and $R_D=2\mathrm{k\Omega }$. Then $R_O=20/11\mathrm{k\Omega }$. With $R_L=2\mathrm{k\Omega }$ and $v_s=20\mathrm{mV}$ root mean square, the input is $10\mathrm{mV}$ and the inverted output has magnitude $19.0476\mathrm{mV}$ root mean square. Hence $P_i=0.100\mu \mathrm{W}$ and $P_L=0.181406\mu \mathrm{W}$: $G_P=1.81406$. These are calculated values of the chosen model, not historical measurements.

With the same parameters but $R_L=100\Omega$, $|A_i|=1.89573$ while $G_P=0.359381<1$. At $R_L=10\mathrm{k\Omega }$, $|A_v|=3.07692$ while $G_P=0.946746<1$. Neither current gain nor voltage gain alone establishes power gain.

For the direct current load line in the figure, choose $V_{DD}=6\mathrm{V}$, $I_Q=1\mathrm{mA}$ and $V_Q=4\mathrm{V}$, consistent with $V_Q=V_{DD}-R_D{I}_Q$. Locally extend the model as $I=I_Q+g_m{v}_i+(V-V_Q)/r_o$. With the external electrical load isolated at direct current, its intersection with $I=(V_{DD}-V)/R_D$ is $V=V_Q-g_m{R}_O{v}_i$. Keeping $|v_o|$ small compared with the available voltage headroom is essential; extrapolation to arbitrarily large signals would be unphysical.

At the chosen operating point the collector supply delivers $V_{DD}{I}_Q=6\mathrm{mW}$: $2\mathrm{mW}$ is dissipated in $R_D$ and $4\mathrm{mW}$ in the output device. Under a small signal, that energy flow is redistributed to provide load power. The incremental controlled current source hides the supply's steady energy flow; it does not create energy. A real complete amplifier also needs its input biasing network and has further losses. This calculation establishes neither its total efficiency nor a universal law for point-contact transistors.

A linear time-invariant circuit sends each sinusoidal input to an output at the same frequency, multiplied by its complex transfer function. Two inputs therefore give the sum of two phase-shifted sinusoids. If their output amplitudes are equal, the sum factors into a carrier at the mean frequency and a signed cosine envelope at half the frequency difference. The envelope magnitude repeats at the full difference frequency. Unequal amplitudes give a nonzero minimum; a nonlinear detector is needed to generate a separate difference-frequency signal.

For the toy models here, visual contrast means brightness relative to a positive uniform background $L_0$: $C(x,y)=(L(x,y)-L_0)/L_0$. Positive values mean brighter than background, negative values darker, and zero unchanged. This convention separates spatial structure from the background level. A sinusoid $C=a\cos (\cdots )$ with $0\le a\le 1$ keeps $L=L_0(1+C)$ nonnegative. Different experimental conventions for visual contrast require their own definitions; the toy values below do not specify a historical stimulus intensity.

The wavelength $\lambda$ is the spatial period of a monochromatic wave, measured at a fixed time along the wavevector. For a plane wave with a sinusoid as its profile, it is the distance between consecutive wave crests, or between any consecutive points with the same wave phase modulo $2\pi$:Here $\mathbf{k}$ uses angular wavenumber, $f$ is ordinary frequency, and $v_p$ is phase speed. A distance between arbitrary zero crossings need not be a wavelength: consecutive zeros of a sinusoid are only $\lambda /2$ apart, with opposite slopes. In a standing wave, consecutive nodes of a standing wave are also $\lambda /2$ apart; $\lambda$ refers to the constituent travelling waves, even though the nodes of a standing wave do not travel.

Across a stationary interface between stationary linear media, the incident, reflected and transmitted monochromatic waves have the same frequency. The boundary conditions must hold at every time: terms oscillating with different angular frequencies cannot cancel throughout the entire time record. Each medium's dispersion relation then determines its own angular wavenumber and wavelength, soFor electromagnetic radiation in a transparent medium with refractive index $n$, $v_p=c_0/n$ and $\lambda ={\lambda }_0/n$, where $c_0$ is the vacuum speed of light. A synthetic example with ${\lambda }_0=600\mathrm{nm}$ and $n=1.5$ gives $\lambda =400\mathrm{nm}$ at the same frequency. This conservation argument assumes time-independent linear material response and a stationary interface in the chosen frame. A moving interface can produce a Doppler effect; time-dependent or nonlinear response can change or generate frequencies.^[ref]

A broad frequency spectrum generally supplies many component wavelengths, rather than one wavelength for an entire pulse. Even two tones have both their individual wavelengths and a different overall repetition distance: the $2$- and $8$-cycle-per-metre sum in spatial sampling of a wave repeats after $0.50\mathrm{m}$, but contains the finer $0.125\mathrm{m}$ structure. A narrow wave packet can have an approximate carrier wavelength; the wave envelope propagates approximately at group velocity, determined by the slope of the dispersion relation. Substituting that group velocity into $\lambda =v_p/f$ generally gives the wrong carrier wavelength. Where the medium varies slowly, local wavelength gives a more useful description.