Time, frequency & scopes formulas

$N$ = input cycles counted in the gate time $T_g$. A 1 s gate reads to 1 Hz and a 10 s gate to 0.1 Hz.

$f_c$ = clock (time-base) frequency and $M$ = number of input periods the gate stays open for ($M=1$ for a single period). Averaging makes the resolution $1/(M{f}_c)$.

As a fraction of the reading. The error is ±1 count whatever the input.

Large at low frequencies. A 50 Hz input with a 1 s gate gives $N=50$, which is an error of ±2%.

Small at low frequencies. It grows with the input frequency.

The worst-case (guaranteed) bound as a fraction of the reading. In period and time-interval modes, add the trigger error.

The frequency at which the ±1-count error is the same in both modes, found by setting $1/(f{T}_g)=f/(M{f}_c)$. Below $f_{co}$, measure the period and invert it. Above it, measure frequency directly.

$L$ is measured from the centre of the plates to the screen. Assumes a uniform field between the plates (no fringing), a small deflection angle and a beam that clears the plates. $e/m$ cancels, so the deflection does not depend on it.

Convention used here: sensitivity is deflection per volt, and the deflection factor is volts per unit deflection (the V/div knob sets a deflection factor). Doubling $V_a$ halves the sensitivity.

$n_v$ and $n_h$ are numbers of divisions. $k_{probe}$ = probe attenuation (10 for a 10:1 probe, unless the scope already corrects for it). For a sine, $V_{rms}=V_{pp}/(2\sqrt{2})$.

$\Delta t$ is the delay between corresponding points and $T$ the common period. On a CRO, $\Delta t/T$ = (divisions of delay)/(divisions per cycle), so the time/div setting cancels.

The output switches between 0 and $V_H$. The positive-going zero crossing of the reference sets the flip-flop and the positive-going zero crossing of the other signal resets it, so the duty cycle is $\varphi /360^{\circ }$, where $\varphi$ is the lag of the other signal behind the reference. The reading is unambiguous over the full circle: a 30° lag reads 30° and a 30° lead reads 330°.

The output switches between 0 and $V_H$. Both inputs must be square waves of 50% duty cycle (sinusoids are squared first by comparators). The output is high whenever the inputs differ, which happens twice per cycle. For $180^{\circ }\le \varphi \le 360^{\circ }$ the output is $V_H(360^{\circ }-\varphi )/180^{\circ }$, so the detector cannot tell a lag of $\varphi$ from a lead of $\varphi$.

$f_y$ is on the Y (vertical) input and $f_x$ on the X (horizontal) input. $n_h$ = points touching a horizontal tangent; $n_v$ = points touching a vertical tangent. For an open figure whose ends are retraced, each free end counts as ½.

$y_0$ = the distance from the centre to the point where the ellipse cuts the Y axis (half the chord on the Y axis); $y_m$ = the maximum vertical deflection. The ratio of full chord to full height is the same. Centre the figure first, and assume the X and Y amplifiers add equal phase shifts. This gives the size of φ only, not whether Y leads or lags, and $\sin \varphi$ alone cannot separate φ from $180^{\circ }-\varphi$: the quadrants of the major axis decide.

$R_p$ and $C_p$ = the probe's series resistance and its trimmer capacitor; $R_s$ and $C_s$ = the scope's input resistance and total shunt capacitance (input plus cable). When compensated, the attenuation $R_s/(R_p+R_s)$ holds at every frequency.

Loading falls by the attenuation factor. For a 10:1 probe, the circuit sees ten times the resistance and one tenth of the capacitance of the bare scope input. With a source resistance $R_{src}\ll R_{in}$, $C_{in}$ forms a low-pass filter that adds a rise time of about $2.2R_{src}{C}_{in}$.

10%–90% rise time of a single-pole (first-order) response, from $t_r=2.2\tau$ and $BW=1/(2\pi \tau )$. Use 0.35 unless the question gives another constant.

Root-sum-square rule for cascaded stages with little or no overshoot. It is an approximation (exact for Gaussian responses). To find the true rise time of the signal, subtract the squares.

For a signal with no components above $f_{max}$. The inequality is strict: at exactly $2f$, every sample can fall on a zero crossing. In practice a real-time DSO samples well above $2f_{max}$ so that the reconstructed trace is accurate; a common rule of thumb is at least 2.5 samples per cycle of the highest frequency with $\sin x/x$ interpolation and about 10 with linear interpolation.

Any input above $f_s/2$ that is still inside the analogue bandwidth (with no anti-aliasing filter) is displayed at $f_a$. For example, a 1.02 MHz sine sampled at 1 MS/s appears as a 20 kHz sine.

For a 10-division screen filled by the record. With a fixed memory, a slower time base forces a lower sample rate, which is how aliasing appears at slow sweeps.

For an $n$-bit ADC, taking one step as $V_{FS}/2^n$ (some texts divide by $2^n-1$; the difference is negligible for large $n$). An 8-bit ADC gives 256 levels, so one step on a 10 V full-scale range is 39 mV.

The bin spacing of an $N$-point FFT, which covers 0 to $f_s/2$.

For an LO above the input (high-side LO). The image frequency also mixes down to $f_{IF}$; an input filter or a high first IF keeps it out.

Change in the displayed noise level when the RBW is changed from $RB{W}_1$ to $RB{W}_2$ (negative means lower). Displayed noise power is proportional to the RBW, so a 10× narrower RBW lowers the noise floor by 10 dB, while a sinusoid's displayed level stays the same, so small tones emerge from the noise. Two equal tones closer together than about one RBW merge into a single peak.

Levels are shown relative to 1 mW. Convert to voltage with the system impedance, usually 50 Ω.

$V_n$ = rms value of the nth harmonic; $L_n$ = its level in dB read off the display. This THD is relative to the fundamental unless it is stated as a percentage of the total rms.