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An acoustic wavelength measurement makes an invisible spatial pattern readable with a ruler. The authored guide Interesting and inexpensive experiments for high school physics by Joe Wolfe places a loudspeaker near a pipe's open end, drives it with an oscillator, and slides a microphone attached to a metre rule through the pipe. An oscilloscope or computer displays the microphone signal. Its oscillation amplitude rises and falls with position; adjacent minima locate pressure nodes of a standing wave a distance $d=\lambda /2$ apart. A cap gives a pressure antinode at the closed end; an open end has a pressure node just outside it. These are pressure patterns, distinct from displacement patterns. The guide advises wavelengths longer than the pipe diameter, sufficient pipe length to locate both a node and an antinode, and frequency spectrum analysis to distinguish the intended tone from other sounds.^[ref]

The inference follows from superposition of two oppositely directed acoustic plane waves. In an ideal uniform, lossless pipe, taking equal pressure amplitudes $P$ and a rigid cap at $x=0$ givesThe time-varying pressure perturbation $p^{\prime }$ has local oscillation amplitude $2|P\cos (kx)|$. Its zeros are separated by $\pi /|k|=\lambda /2$. Unequal incident and reflected amplitudes leave nonzero minima, but their spacing remains $\lambda /2$ in this ideal model. A microphone detects that pressure variation through a changing electrical signal, rather than watching a wave crest move. Record the tone's frequency too, then infer phase speed $v_p=2fd$.

For proposed controls, keep the drive frequency and amplitude fixed during a scan, verify the frequency on the recorded signal, and repeat the scan in both directions. Fit several minimum positions $x_j=a+jd$ using linear regression to reduce dependence on one pair of positions. Check the ruler scale and use one consistent reference point on the microphone. Compare with the source switched off to locate background measurement noise; use frequency spectrum analysis to check for additional tones, and reduce drive if the trace ceases to resemble a sinusoid. Record air temperature and avoid strong air motion. These are suggested controls, not additional procedures or observations attributed to the guide.

The pipe must support approximately plane waves over the measured region: transverse structure at shorter wavelengths, the microphone's finite size, and disturbance by the ruler can compromise the interpretation. Weak reflection gives shallow minima, while damping and other reflections can shift the extrema of the simple two-wave model. Measure interior spacings rather than assuming the first minimum is exactly at the pipe lip. A short scan may not include enough minima to estimate $d$ reliably, and drift during sequential scanning can imitate spatial variation.

For an explicitly hypothetical calculation, $f=1000\mathrm{Hz}$ and $d=0.1715\mathrm{m}$ give $\lambda =0.3430\mathrm{m}$, $K\simeq 18.3183\mathrm{rad}{\mathrm{m}}^{-1}$, and $v_p=343.0\mathrm{m}{\mathrm{s}}^{-1}$. No such measurements are claimed here. If independent small measurement errors in $f$ and fitted $d$ have standard deviations ${\sigma }_f$ and ${\sigma }_d$, first-order error propagation gives ${\sigma }_{v_p}/v_p\simeq \sqrt{({\sigma }_f/f{)}^2+({\sigma }_d/d{)}^2}$. A systematic scale error or changing air conditions requires separate assessment.

For integers $m,q,n$, adding $qN$ to a frequency index changes the complex exponential by $e^{2\pi iqn}=1$. Thus different continuous Fourier modes become identical discrete Fourier modes on the sampling grid. This is aliasing, caused by sampling before any discrete Fourier transform is computed.

For $N=8$, compare the continuous functions $\cos (2\pi mt)$ and $\sin (2\pi mt)$ with $m=1$ and $m=9$, sampled at $t=n/8$, $n=0,\dots ,7$. With $a=\sqrt{2}/2$, both cosines give exactly the first row and both sines give exactly the second:The nine-cycle sine agrees in sign as well as magnitude with the one-cycle sine: the frequency indices differ by $8$, rather than by a sign reversal. Consequently, either cosine produces the same discrete Fourier transform, $X_1=X_7=4$ with all other coefficients zero; either sine produces $X_1=-4i$, $X_7=4i$, with all other coefficients zero.

An observed low frequency therefore does not establish that physical high frequencies are absent. To exclude such alternatives, the Nyquist–Shannon sampling theorem requires a specified class of bandlimited functions. For arbitrary pure sinusoids with frequencies at most $B$, choose sampling rate $f_s>2B$; equality needs care because $\sin (2\pi (f_s/2)t)$ vanishes at every $t=n/f_s$. The finite-energy L2 space version of the Nyquist–Shannon sampling theorem permits equality under its own hypotheses. In either case, exact continuous reconstruction generally requires the entire infinite sample sequence; eight observed samples alone do not supply it. Demanet derives the grid aliasing condition as a difference by an integer multiple of the sampling angular frequency.^[ref]

In the single-sinusoid AM voltage model, $m$ is the peak fractional excursion of the signed multiplying factor from its mean value $1$. At $m=0$ there is only the carrier wave; at $m=0.5$ its multiplying factor ranges from $0.5$ to $1.5$; at $m=1$ it just touches zero. For $0\le m\le 1$, the wave envelope extrema areThe last equality follows by adding and subtracting the first two. An oscilloscope estimate uses local carrier peaks near the message extrema and needs enough carrier cycles to approximate those bounds. Consistent local peak-to-peak amplitude readings give the same ratio because their common factor of two cancels. Mixing a peak reading with a peak-to-peak reading does not.

For $A=2\mathrm{V}$ and $m=0.5$, the wave envelope ranges from $1$ to $3\mathrm{V}$, giving $(3-1)/(3+1)=0.5$. These are model values. At $m=1.4$, the signed factor has minimum $-0.4$, but the nonnegative wave envelope reaches zero at its sign changes; the extrema formula would misleadingly report $1$. It cannot measure the true depth of overmodulation from the folded envelope alone.

Apply the product-to-sum formula to the voltage model:The carrier wave component has peak voltage amplitude $A$; each AM sideband has peak voltage amplitude $Am/2$. These are temporal frequency spectrum components, unlike the spatial sidebands used in pattern stability. The separation $\Omega >0$ makes the three positive angular frequencies distinct, and $\Omega <{\omega }_c$ keeps the lower one positive. The span between sidebands is $2\Omega$, or $2f_m$ in hertz when $f_m=\Omega /(2\pi )$; this is the single-tone span, not a universal signal bandwidth for arbitrary messages.

A one-sided cosine-amplitude display assigns $A$ to the carrier line. A two-sided complex-exponential expansion instead assigns $A/2$ at each of $\pm {\omega }_c$, and $Am/4$ at each of the four sideband locations. The signal is identical; confusing the conventions creates a factor-of-two error. An ideal infinite-duration sinusoid gives a spectral line. A finite recording and finite analyser resolution broaden the displayed peak.

For a voltage applied across an ideal resistor of electrical resistance $R>0$, a component of peak voltage $B$ has root mean square voltage $B/\sqrt{2}$ and mean power $B^2/(2R)$ by Ohm's law. ThusThe proof of additivity is to average $s(t{)}^2/R$ over a long time: cross-products of distinct frequencies have zero mean. Consequently the whole signal's root mean square voltage is $A\sqrt{(1+m^2/2)/2}$, not simply its maximum wave envelope divided by $\sqrt{2}$. The carrier-component $A/\sqrt{2}$ remains unchanged as $m$ varies.

With $A=2\mathrm{V}$, $R=50\Omega$ and $m=0.5$, the carrier supplies $40\mathrm{mW}$ and each $0.5\mathrm{V}$ peak sideband supplies $2.5\mathrm{mW}$, giving $45\mathrm{mW}$ total. At $m=1.4$ each sideband is $1.4\mathrm{V}$ peak and supplies $19.6\mathrm{mW}$, giving $79.2\mathrm{mW}$ total. Neither case changes the carrier-component power.

The fraction of mean power in the message-bearing sidebands is $m^2/(2+m^2)$. It is $1/9$ at $m=0.5$ and $1/3$ at $m=1$. Conventional AM spends the remaining power on a carrier that enables an inexpensive envelope detector. Increasing $m$ beyond $1$ increases this fraction in the linear model but breaks ordinary envelope recovery. Removing the carrier gives suppressed-carrier amplitude modulation and changes the receiver requirement.

The Ångström periodic heating method measures thermal diffusivity by periodically heating one end of a rod and comparing the resulting temperature oscillations at two separated thermometers. The farther signal has a smaller amplitude and a later wave phase. Their combination measures how quickly the material redistributes heat, without requiring the absolute heater power. The method bears the name of Anders Jonas Ångström, the physicist who became a professor at Uppsala in 1858.^[ref]

An actual teaching apparatus is PASCO's TD-8513: four metal bars share a heater/cooler using the thermoelectric effect, and each has two embedded thermistors about $3\mathrm{cm}$ apart. Foam covers reduce environmental heat transfer. Its manual instructs students to measure sensor spacing and connect the thermistors through a temperature interface to a computer. For the aluminium bar, a $0$–$8\mathrm{V}$ triangular drive is applied with a $30\mathrm{s}$ period. Recording begins after the amplitudes stabilize; two cycles provide amplitudes and the delay between corresponding peaks. Repeated runs at $40$ and $50\mathrm{s}$ test consistency. The manual converts the inferred thermal diffusivity to thermal conductivity using mass density and specific heat capacity.^[ref]

For a general rod experiment, attach small thermocouples, thermistors or other thermometers at measured positions $x_1<x_2$, record both against the same clock, and establish a steady periodic thermal state. A sinusoid is especially convenient. For a switched or triangular drive, isolate the same Fourier series component at both sensors: unrelated peaks in distorted waveforms need not measure that component's phase difference. Subtract each sensor's mean temperature; a steady spatial background does not belong in the oscillating amplitude ratio. Small excursions help keep thermal conductivity, mass density and specific heat capacity constant.

The appeal is that the rod supplies two complementary clocks: decreasing amplitude measures spatial damping, while increasing wave phase lag measures delay. Unlike laser flash analysis, which follows a transient after one pulse, this method repeatedly samples a steady periodic thermal state. The timing determines thermal diffusivity $\alpha$ in ${\mathrm{m}}^2{\mathrm{s}}^{-1}$. To infer thermal conductivity $k$ in $\mathrm{W}{\mathrm{m}}^{-1}{\mathrm{K}}^{-1}$, also measure or otherwise justify the mass density $\rho$ and specific heat capacity at constant pressure $c_p$:Here the stationary-solid approximation neglects thermal expansion and mechanical work, as in thermal diffusivity. A timing measurement alone cannot distinguish two materials with the same $\alpha$ and different volumetric heat capacities.

The capillary length measures when gravity and surface tension are comparably important in shaping a liquid interface. For a liquid of uniform mass density $\rho$, beneath a gas whose mass density is negligible, with uniform gravitational acceleration $g>0$ and constant surface tension $\gamma >0$,A deformation of height $a$ and horizontal length scale $L$ has hydrostatic pressure variation of order $\rho ga$ and small-slope capillary pressure of order $\gamma a/L^2$. Equating these gives $L\sim {\ell }_c$: surface tension smooths small features, whereas gravity matters for broad ones. For two resting fluids, replace $\rho$ by their positive mass density difference $\Delta \rho$ in this static balance. This change alone does not give the two-fluid wave dynamics, because both fluids can have inertia. ^[ref]

With the stipulated illustrative parameters $\gamma =0.072\mathrm{N}{\mathrm{m}}^{-1}$, $\rho =1000\mathrm{kg}{\mathrm{m}}^{-3}$, and $g=9.81\mathrm{m}{\mathrm{s}}^{-2}$, the capillary length is ${\ell }_c=2.71\mathrm{mm}$. These parameters define the ideal calculations below; they are not a measurement of a particular sample.

For a sinusoid with wavenumber $k>0$, the relevant horizontal scale is $L=1/k$, while its wavelength is $\lambda =2\pi /k$. Thus the Bond number for this scale and the reciprocal restoring-term ratio areEquality is at $k_c=1/{\ell }_c$, hence ${\lambda }_c=2\pi {\ell }_c=17.02\mathrm{mm}$ for these parameters. Using $L=\lambda$ instead produces a different numerical Bond number; state the length convention. The capillary length itself is not the equality wavelength. This distinction helps choose whether a measured ripple should be modeled as a surface gravity wave, a nearly pure capillary wave, or a gravity-capillary wave. ^[ref]

A fluctuation in the time at which a signal is sampled produces an amplitude error proportional to its local time derivative. For a sinusoid, the characteristic fractional error grows with the product of frequency and timing uncertainty. Sampling-time uncertainty is distinct from uniform quantization error.

The continuous-time frequency response is the frequency response of a continuous-time linear time-invariant system. From zero initial state, a BIBO stable system driven by a sustained sinusoid $u(t)=A\cos (\omega t)$ approaches the steady responseAny decaying initial or switching transient is separate. The modulus $|G(i\omega )|$ is the amplitude gain; the complex argument is the shift in oscillation phase. Thus evaluating a transfer function at $s=i\omega$ measures both how much of an oscillation passes through and how far it lags.

For the first-order low-pass filter, with $K>0$, direct complex number arithmetic givesThe phase uses the inverse tangent branch between $-\pi /2$ and $\pi /2$. Low angular frequencies have gain close to $K$ and little lag. High angular frequencies have gain approximately $K/(\omega \tau )$ and lag approaching $\pi /2$. This is why the model is a first-order low-pass filter: it suppresses fast variation compared with slow variation. Rowell relates frequency response gain and phase to poles and then introduces Bode plots in MIT 2.004, lecture 32, sections 1 and 2.

At the corner angular frequency,For $K=2$ and $\tau =3\mathrm{s}$, ${\omega }_c=1/3\mathrm{rad}{\mathrm{s}}^{-1}$, the gain is $\sqrt{2}\approx 1.4142$, and the lag is $45$ degrees. The angular frequency $\omega$ is measured in radians per second; ordinary frequency $f$ is measured in hertz (cycles per second), with $\omega =2\pi f$. Consequently the same corner is $f_c=1/(2\pi \tau )\approx 0.0531\mathrm{Hz}$, not $1/3\mathrm{Hz}$.

MIT's demonstration places two 120-turn coils face to face, each with a capacitor. A swept sinusoid drives one circuit; an oscilloscope observes the other circuit's capacitor electric potential difference. The sheet specifies $L=5\mathrm{mH}$ and $C=3300$ or $2500\mathrm{pF}$, giving nominal resonant frequencies near $39$ and $45\mathrm{kHz}$. It presents a procedure, not a measured dataset.^[ref]

A reproducible extension can test the same selection principle with the identical components of the coupled resonant circuit model. Prepare two separate series loops, measure their actual inductances, capacitances and coil electrical resistances, and include generator Thévenin resistance in the driven loop's total $R_1$. Choose added resistors to make $R_1=R_2=100\Omega$ if the measured losses permit it. Include a known $R_s=50\Omega$ sense resistor within $R_2$. Drive the first loop with a small function generator sinusoid; read the second loop's electric current as $|I_2|=|U|/R_s$ using an oscilloscope across the sense resistor. Here $U$ denotes the sense reading, distinct from the source electric potential difference $V$ in the model.

Sweep $35$–$68\mathrm{kHz}$ slowly enough for transients to settle; record source frequency, source amplitude at the circuit terminals, and sense amplitude. Repeat at fixed coil spacing after changing one capacitor, then at fixed components after changing spacing. Turn the drive off to establish the background; rotate the receiving coil toward a right angle to test inductive coupling; restore its orientation and repeat to check drift. These are proposed controls, not observations reported by MIT. Keep probe loading small and account for it if not negligible. A grounded oscilloscope can electrically join the loops if attached to both: use an appropriately isolated differential input for any simultaneous primary measurement, or measure the primary separately.

Recovering deliberately chosen long and short generator bursts on the receiver trace demonstrates symbol timing. The essential inference is narrower than a long-distance radio test: nearby coils primarily exchange energy through mutual inductance in the near and far field regime dominated by the near field. Coil spacing, detuning and background controls distinguish this coupling from a direct wire connection or incidental pickup. They do not by themselves demonstrate reception of propagating radio waves in the radiation zone.

Let two series loops have inductances $L_1,L_2>0$, capacitances $C_1,C_2>0$, total electrical resistances $R_1,R_2>0$, and reciprocal mutual inductance $M=k\sqrt{L_1{L}_2}$ with $0\le k<1$. For a steady sinusoid of angular frequency $\omega >0$, choose complex peak amplitudes with time convention $e^{\mathrm{i}\omega t}$. An ideal source of peak electric potential difference $V$ drives loop 1; loop 2 has no independent source. ThenEliminating $I_1$ proves the response formula. Reversing one coil changes the sign of $M$ and the relative wave phase, but leaves this amplitude response unchanged. The measured sense ratio is $|R_s{I}_2/V|$, with $R_s$ included in $R_2$, not added a second time.

For identical loops at their uncoupled resonant frequency ${\omega }_0=1/\sqrt{LC}$,Writing $x={\omega }_{0}M/R$ makes this $x/[R(1+x^2)]$. Its derivative has the sign of $1-x^2$, so response at this fixed frequency is largest when $x=1$, equivalently $k=1/Q$. Too little coupling transfers little excitation; excessive coupling reduces the response at the original centre frequency. This is a fixed-frequency optimum, not a universal optimum over every possible receiver and drive frequency.

For the chosen values, $k=1/Q=0.063246$ gives $M=0.31623\mathrm{mH}$. At source peak $V=0.100\mathrm{V}$, the secondary peak electric current is $0.500\mathrm{mA}$ and the sense peak electric potential difference is $25.0\mathrm{mV}$. The primary peak electric current is also $0.500\mathrm{mA}$, so source mean power is $25.0\mu \mathrm{W}$: each loop dissipates $12.5\mu \mathrm{W}$, checking conservation of energy. Here $V$ is the equivalent ideal source behind $R_1$; a generator's terminal reading must be converted using its measured Thévenin resistance.

In the lossless identical-loop limit, the normal modes have equal or opposite electric currents, making their effective inductances $L+M$ or $L-M$. ThusAt $k=0.20$ these are $45.944$ and $56.270\mathrm{kHz}$. Finite electrical resistance broadens and shifts the driven peaks, so the exact response formula, rather than these lossless values, generates the plot. Changing a capacitor moves the uncoupled tuning through $f_0\propto C^{-1/2}$: increasing it by $21\%$ lowers $f_0$ by the factor $1/1.1$. The model explains why tuning and coupling must be adjusted together; it supplies no transatlantic propagation or detector-threshold prediction.

For the constant-coefficient model $x^{\prime }(t)=-kx(t-\tau )$ with $k>0$ and $\tau \ge 0$, the zero equilibrium point is asymptotically stable if and only if $k\tau <\pi /2$. For positive $\tau$, perturbations are measured by the supremum norm of the initial history of delayed feedback; in the stable regime, every continuous history produces a solution tending to zero. At equality, nondecaying sinusoids prevent asymptotic stability. For $k\tau >\pi /2$, growing oscillatory solutions exist. These statements concern this ideal linear time-invariant system, rather than an arbitrary controlled plant. The corresponding linear equation and its threshold are developed in Cain and Reynolds, section 6.4, pages 174–177.

Substitute the exponential function $x(t)=e^{st}$, allowing a complex number $s$. The characteristic equation isFor positive feedback delay this is not a characteristic polynomial: it has infinitely many roots. A root's real part determines growth or decay of its exponential mode, and its imaginary part determines the angular frequency. To find the first boundary, set $s=i\omega$ with $\omega >0$ and use Euler's formula:The smallest positive crossing is $\omega =k$ and $\omega \tau =\pi /2$. Thus at ${\tau }_c=\pi /(2k)$ the complex conjugate roots are $s=\pm ik$. In fact, with the matching initial history of delayed feedback,solves the delay differential equation exactly: a quarter-cycle feedback delay turns the past displacement into precisely the current derivative. Its cycle lasts $2\pi /k=4{\tau }_c$. The amplitude is arbitrary; this linear model does not select a finite-amplitude oscillation.

The roots cross toward growth as feedback delay increases. Differentiating the characteristic equation at fixed $k$ givesThis calculation identifies the direction of the first destabilization. The full threshold also requires ruling out earlier unstable roots: if $s=\alpha +i\beta$ with $\alpha \ge 0$ and $\beta >0$, the characteristic equation implies $\cos (\beta \tau )\le 0$ and $0<\beta \le k$, hence $k\tau \ge \beta \tau \ge \pi /2$. A real nonnegative root is impossible because $s+k{e}^{-s\tau }>0$ there. This supplies the geometric reason the threshold is a quarter-cycle rather than an arbitrary fitting constant.

The tradeoff is concrete. Without feedback delay, increasing $k$ shortens the exponential decay time constant $1/k$. With known $\tau >0$, the same proportional control must satisfy $k<\pi /(2\tau )$. For $\tau =0.2\mathrm{s}$, the threshold is $k\approx 7.854{\mathrm{s}}^{-1}$: $k=5{\mathrm{s}}^{-1}$ is stable and $k=8{\mathrm{s}}^{-1}$ is unstable. A design close to the boundary leaves little room for additional feedback delay; reducing $k$ or reducing measurement latency increases that room. The zero-delay time constant is only a comparison scale, not the decay time of the delayed response. Frank discusses the gain–delay compromise in Control Theory Tutorial, section 13.4 and figure 13.3.

A diode is a two-terminal component with an asymmetric current-voltage characteristic: reversing its terminal electric potential difference need not reverse its electric current with the same magnitude. Name the terminals diode anode $A$ and diode cathode $K$. Define $V_D={\varphi }_A-{\varphi }_K$ using their electric potentials, and take positive electric current $I_D$ from $A$ to $K$. The bar in the diode symbol marks $K$. Positive $V_D$ is the forward direction; negative $V_D$ is the reverse direction.^[ref]

The asymmetry makes a rectifier possible: a sinusoid can drive electric current through a resistor on one half of each cycle. An ordinary resistor obeying Ohm's law instead reverses its electric current when the applied electric potential difference reverses. The useful idea is directional selection; the ideal diode model captures it, while the Shockley diode equation describes a smooth exponential function rather than an abrupt switch.

Electrical impedance is the ratio of terminal electric potential difference and electric current phasors at one specified positive angular frequency $\omega$. For an electrical circuit modeled as a linear time-invariant system, use the sinusoidal steady state after any decaying transient has settled. With the passive sign convention, positive electric current enters the terminal assigned positive electric potential difference. Define peak complex amplitudes byThe physical $v(t)$ and $i(t)$ are real; the complex numbers encode their amplitudes and wave phases. The unit of electrical impedance is the ohm, $1\Omega =1\mathrm{V}/\mathrm{A}$. Both phasors may instead use root mean square magnitudes, with the same ratio $Z$. A nonlinear diode can generate additional frequencies; a single $Z(\omega )$ then does not describe the whole response.

Write the electrical impedance in Cartesian and polar forms asHere $R$ is the resistive real part, $X$ the signed electrical reactance, and $\varphi$ the voltage-minus-current phase difference, defined modulo a full turn. The Pythagorean theorem gives the modulus from a right triangle in the complex plane. The complex argument gives the rotation from the current phasor to the voltage phasor. Thus $V_{\mathrm{rms}}=|Z|I_{\mathrm{rms}}$, while negative $\varphi$ means electric current leads terminal electric potential difference. For $R>0$, $\varphi =\arctan (X/R)$ lies between $-\pi /2$ and $\pi /2$; outside that case, use the quadrant of the complex number rather than a bare inverse tangent. The geometric magnitude and phase difference are the two pieces of the AC relation.^[ref]

Multiplication by $Z$ acts on phasors, not on instantaneous physical signals. For example, when $i(t)=I_0\cos \omega t$,The second term is in phase quadrature with electric current; it cannot be represented by pointwise multiplication $v(t)=Zi(t)$ when $X\ne 0$. This is the useful compression: one complex multiplication replaces separate amplitude and phase difference calculations, while the real part restores the physical signal.

An ideal resistor, inductor and capacitor have $Z_R=R$, $Z_L=\mathrm{i}\omega L$ and $Z_C=1/(\mathrm{i}\omega C)$. Their electrical impedances add in a series RLC circuit, giving $Z=R+\mathrm{i}(\omega L-1/(\omega C))$. The current amplitude is $V_0/|Z|$, so cancellation of the imaginary parts explains the resonant series RLC response. The component-law derivation and opposite complex exponential convention are in reactance phase convention.

An actual teaching experiment turns the phasor definition into readings. Leipzig University's laboratory handout Measurements with an oscilloscope, experiment E 3e, uses a function generator, a two-channel oscilloscope, a known resistor and a capacitor. Figure 5 places the capacitor above the resistor, with a common return below the resistor. Channel 1 measures the applied terminal electric potential difference $v_1$; channel 2 measures $v_R$. The handout asks for peak-to-peak values and relative timing, then comparison of measured and calculated electrical impedances with $R=3\mathrm{k\Omega }$, $C=0.47\mu \mathrm{F}$ at $100$ and $300\mathrm{Hz}$. It also asks students to check the oscilloscope time base against a digitally counted generator frequency. These are apparatus and procedural specifications, not reported measured results.^[ref]

For this arrangement, Ohm's law gives $i(t)=v_R(t)/R$ if probe loading and the sensing resistor's electrical reactance are negligible. Record both voltage magnitudes with the same convention: peak, peak-to-peak, or root mean square. After the transient settles, compare matched rising zero crossings $t_1$ of $v_1$ and $t_R$ of $v_R$, with period $T=1/f$. ThenThe recovered electrical impedance includes the known resistor. For a separate unknown component in series with that resistor, subtract $R$ from the inferred total electrical impedance. A current crossing earlier than the applied voltage crossing gives $\Delta t<0$, hence negative $\varphi$. For the stipulated series RC impedance example at $1000\mathrm{Hz}$, this predicts $\Delta t=-0.160717\mathrm{ms}$; it is a calculated timing, not a lab observation. Timing supplies the complex argument that an amplitude ratio alone cannot recover.

As additional measurement controls, first feed the same sinusoid to both channels to check relative delay and polarity; then replace the reactive chain with a known resistor, expecting nearly zero phase difference. Keep both probe returns at the common return shown in figure 5. A reversed voltage polarity adds half a turn to its inferred wave phase. Match crossings with the same slope and choose $-T/2<\Delta t\le T/2$; adjacent cycles otherwise change $\varphi$ by a full turn. A sweep continuous in frequency helps track that branch. These controls test the signed inference rather than assuming that the scope's delay readout has the desired sign.

Measure the loaded terminal electric potential difference on channel 1 at every frequency. The generator's output impedance can change the applied magnitude; its front-panel setting is not the measured terminal value. The generator's internal electrical impedance is outside the terminal ratio above, while the known sensing resistor is inside it. Probe loading, finite timing resolution, component tolerances and distorted waveforms limit the inference. A repeat at lower drive amplitude tests whether one linear electrical impedance still fits. Electrical reactance bench measurement develops these loading and reference-direction issues further.

For a message consisting of a single sinusoid with frequency $f_m>0$, the FM modulation index is $\beta =\Delta f/f_m$. It is dimensionless and equals the peak wave phase excursion in radians in the chosen sine-phase convention. Keeping $\Delta f=5\mathrm{Hz}$ while changing $f_m$ from $1$ to $2\mathrm{Hz}$ reduces $\beta$ from $5$ to $2.5$: the same frequency deviation acts for a shorter portion of each message cycle. A general message containing many frequencies does not have a single $f_m$ with this meaning.

A Gabor filter has a kernel formed by multiplying a Gaussian function by a sinusoid or a complex oscillation. In two-dimensional image processing, its orientation and spatial frequency select local image structure; evaluating the kernel's translated dot product with an image is spatial filtering. The real even example under signed spatial filter model of a simple cell uses a cosine.

There is experimental motivation for this family without making every simple cell an exact Gabor filter. Jones and Palmer fitted two-dimensional Gabor filters to measured spatial and spectral response profiles of simple cells in cat cortex and compared independently estimated parameters as a test of spatial linearity. Their reported residuals supported the approximation in that dataset. The chosen parameters and orientation tuning curve in this article are not extracted from their measurements. ^[ref]

A half-wave rectifier with one diode and a resistor load passes one half of each input cycle.^[ref]

For an original ideal example, let an ideal voltage source supply the sinusoid $v_S(t)=V_p\sin (\omega t)$, with $V_p=5\mathrm{V}$ and angular frequency $\omega >0$. Connect its positive reference terminal to the diode anode, the diode cathode to the upper terminal of a load resistor with electrical resistance $R_L=1\mathrm{k\Omega }$, and the lower terminal to the source return. Let $v_L$ be the upper-minus-lower load electric potential difference, and $i_L$ the downward load electric current. Assume the ideal diode model, zero source electrical resistance, and no filter capacitor, inductor or switching delay.

The Kirchhoff cycle law gives $v_S=V_D+v_L$, and Ohm's law gives $v_L=R_L{i}_L$. If $v_S>0$, the blocking assumption would give $V_D=v_S>0$, contradicting the ideal diode model, so the diode conducts with $V_D=0$. If $v_S<0$, conduction would give $i_L=v_S/R_L<0$, also forbidden, so the diode blocks. At the crossings both values vanish. ThusThe peak load electric current is $5\mathrm{mA}$. The output is unidirectional but still varies in time, reaching zero throughout every negative half-cycle. Over a period of a function $2\pi /\omega$, write the wave phase as $\theta =\omega t$. Its average value of a function and root mean square follow from two integrals:Consequently the average load power is $v_{L,\mathrm{rms}}^2/R_L=6.25\mathrm{mW}$, not ${\overline{v}}_L^2/R_L$. The ideal diode has zero loss; the resistor converts supplied energy into Joule heating. The same switching argument predicts $v_L=\min (0,v_S)$ when the diode is reversed. Directional selection changes the waveform and its average value of a function; it does not by itself produce a constant output.

In his 2000 lecture, Jack St. Clair Kilby describes an August 1958 comparison circuit made from separate silicon elements. Resistors were cut and etched; capacitors came from diffused, metallized semiconductor wafers. It tested whether semiconductor elements could perform all the required functions, but was still a discrete circuit. For the September prototype he used already diffused germanium material with contacts, cut into bars, and hand-applied wax masks to define a transistor and a distributed resistor-capacitor region. Applying supply power produced an oscillation.^[ref]

An inspected eyewitness interview describes the germanium on a microscope slide with protruding wires; changing its connections changed the oscilloscope display from a straight line to a sinusoid. This supplied a qualitative connected/disconnected comparison, not a calibrated record of amplitude, loading or repeatability.^[ref]

The inference was feasibility: integrated semiconductor elements could cooperate as an operating electrical circuit. The hand-wired germanium prototype did not yet supply the deposited integrated circuit interconnects of Robert Noyce's silicon design. Nor does a working prototype establish manufacturing yield or long-term reliability.

A lock-in amplifier measures the component of an input signal synchronized with a reference at a chosen frequency. It multiplies the input by a reference sinusoid, then uses a low-pass filter to retain the slowly varying product. Two references in phase quadrature recover both amplitude and phase difference. A shared timing source keeps the relative wave phase stable: this is coherent detection.^[ref]

The useful trick is to move the wanted frequency to zero before averaging. An electronic amplifier alone enlarges both the wanted signal and unwanted noise; a lock-in amplifier instead selects a component by its timing. A photodiode's electric current can first be converted to an electric potential difference, so the same calculation can measure weak modulated light. The chopped-light lock-in experiment makes this selection tangible.

Consider this original constructed input electric potential difference, with $t$ in seconds and $f_0=1\mathrm{Hz}$:The unwanted component has three times the reference frequency and more than twice the wanted peak amplitude. Nevertheless, the lock-in quadratures over $T=4/f_0=4\mathrm{s}$ are exactlyso $R=0.4\mathrm{V}$ and $atan2(Y,X)=\pi /3=60^{\circ }$. This is a peak amplitude calculation, not an SR830 display prediction.

To see why the larger component vanishes, multiply $B\cos (3\omega t+\psi )$ by each reference:Each sinusoid has zero average value of a function over a whole reference cycle. More generally, a component $B\cos (n\omega t+\psi )$ with integer $n\ge 2$ gives products at $(n-1)\omega$ and $(n+1)\omega$, whose integrals vanish over any integer number of reference cycles. This is the orthogonality behind selecting a Fourier coefficient with sinusoidal references. A nonsinusoidal reference can contain additional frequencies and need not have the same rejection property.

An unwanted component at the reference frequency survives. If $u(t)=B\cos (\omega t+\psi )$ is added to the input, its coordinate vector adds to the wanted one:For example, adding $B=0.1\mathrm{V}$ with $\psi =0$ to the wanted component gives $X=0.3\mathrm{V}$, $Y=0.2\sqrt{3}\mathrm{V}$, $R=\sqrt{0.21}\mathrm{V}\approx 0.458258\mathrm{V}$, and phase difference $atan2(0.2\sqrt{3},0.3)\approx 49.1066^{\circ }$. Longer averaging cannot separate these two synchronized contributions.