Filters & waveforms formulas

R in series from the input to the + terminal, C from the + terminal to ground; $R_1$ (− terminal to ground) and $R_f$ (output to − terminal) set the passband gain and do not affect $f_c$. $|H(jf)|=A_0/\sqrt{1+(f/f_c{)}^2}$ and $\angle H=-{\tan }^{-1}(f/f_c)$. With $R_f=0$ (voltage follower), $A_0=1$.

R and C swap places: C in series to the + terminal, R from the + terminal to ground. $|H(jf)|=A_0(f/f_c)/\sqrt{1+(f/f_c{)}^2}$, phase $90^{\circ }-{\tan }^{-1}(f/f_c)$. A real op-amp's finite gain-bandwidth product eventually turns every active high-pass into a band-pass; the ideal model ignores this.

$R_1$ from the input to the − terminal, $C_f$ in parallel with $R_f$ in the feedback path, + terminal grounded. Only the feedback pair sets $f_c$. The passband phase is $180^{\circ }$, so the total phase is $180^{\circ }-{\tan }^{-1}(f/f_c)$.

$R_1$ and $C_1$ in series from the input to the − terminal, $R_f$ alone in the feedback path. Only the input pair sets $f_c$; the high-frequency gain is $-R_f/R_1$.

K is the passband gain: at DC for the low-pass, at high frequency for the high-pass, at ${\omega }_0$ for the band-pass, and in both outer bands for the notch. $\zeta =1/(2Q)$. At $\omega ={\omega }_0$: $|H_{LP}|=|H_{HP}|=KQ$, $|H_{BP}|=K$ and $H_N=0$.

$C_1$ is the capacitor to the output, $C_2$ the capacitor to ground; DC gain K. Dividing through by $R_1{R}_2{C}_1{C}_2$ gives $Q=\sqrt{R_1{R}_2{C}_1{C}_2}/\left[(R_1+R_2)C_2+R_1{C}_1(1-K)\right]$.

With equal resistors, Q is set by the capacitor ratio alone: Butterworth ($Q=1/\sqrt{2}$) needs $C_1=2C_2$, the feedback capacitor twice the grounded one. Equal R and equal C at unity gain give only $Q=0.5$.

Holds for both the low-pass and the high-pass Sallen-Key; the gain alone sets Q. Stable only for $K<3$: at $K=3$ the poles reach the $j\omega$ axis and the circuit oscillates at ${\omega }_0$, and for $K>3$ they are in the right half-plane. Butterworth: $K=3-\sqrt{2}=1.586$, i.e. $R_f=0.586R_a$. At $\omega ={\omega }_0$ the gain is $KQ=K/(3-K)$.

Low-pass of order n (for a high-pass, replace $\omega /{\omega }_c$ by ${\omega }_c/\omega$). Maximally flat at $\omega =0$ (the first $2n-1$ derivatives of $|H{|}^2$ vanish there), no ripple, exactly −3 dB at ${\omega }_c$ for every n, asymptotic roll-off $20n$ dB/decade, and phase $-n\times 45^{\circ }$ at ${\omega }_c$. The poles lie equally spaced on a circle of radius ${\omega }_c$ in the left half-plane. Other standard approximations make different trade-offs: Chebyshev accepts ripple in the passband for a sharper transition at the same order, and Bessel accepts a gentler transition for a nearly constant group delay; all three roll off at $20n$ dB/decade asymptotically.

$A_p$ = the largest attenuation (dB) allowed at the passband edge ${\omega }_p$; $A_s$ = the least attenuation (dB) required at the stopband edge ${\omega }_s$ (low-pass: ${\omega }_s>{\omega }_p$; high-pass: use ${\omega }_p/{\omega }_s$). Round n UP to an integer. If ${\omega }_p$ is the −3 dB cut-off, $10^{0.1A_p}-1\approx 1$. Example: at least 40 dB at $2{\omega }_c$ needs $n\ge {\log }_{10}(9999)/(2{\log }_{10}2)=6.64$, so $n=7$.

$f_L$ and $f_H$ are the −3 dB frequencies of the overall response. Exact for any second-order band-pass ($f_L{f}_H=f_0^2$ and $f_H-f_L=f_0/Q$): the centre is the geometric, not the arithmetic, mean of the cut-offs. For a cascade of first-order stages, the stage cut-offs $f_1$ and $f_2$ are not the overall $f_L$ and $f_H$ unless $f_2\gg f_1$ (see above), although the gain always peaks at $\sqrt{f_1{f}_2}$.

The input drives both op-amp inputs: through $R_1$ to the − terminal (with $R_f=R_1$ from the output to the − terminal), and through R to the + terminal, with C from the + terminal to ground. Then $v_{+}=v_i/(1+sRC)$ and $v_o=2v_{+}-v_i$. The gain is 1 at every frequency; the phase goes from $0^{\circ }$ at DC through $-90^{\circ }$ at $\omega =1/RC$ towards $-180^{\circ }$. Swapping R and C gives $H(s)=(sRC-1)/(1+sRC)$, phase $180^{\circ }-2{\tan }^{-1}(\omega RC)$.

Ideal op-amp, no feedback: the virtual-short rule $v_{+}=v_{-}$ does NOT apply. If the output is limited by a series resistor followed by back-to-back Zener diodes from the limited output node to ground, the levels become $\pm (V_Z+V_D)$ (constant-drop diode model, $V_D\approx 0.7$ V).

Non-inverting comparator, input $V_m\sin \omega t$, reference $V_R$ with $|V_R|<V_m$; D is the fraction of each period with the output at $+V_{sat}$, and ${\sin }^{-1}$ is in radians. For the inverting comparator the fraction is $\frac{1}{2}+\frac{1}{\pi }{\sin }^{-1}(V_R/V_m)$. $V_R=0$ gives exactly one half.

Input to the − terminal; $R_1$ from the + terminal to the reference $V_R$ (ground if $V_R=0$), $R_2$ from the output to the + terminal. By superposition $v_{+}=(R_1{v}_o+R_2{V}_R)/(R_1+R_2)$ with $v_o=\pm V_{sat}$. With $V_R=0$: $V_{UT}=+\beta V_{sat}$, $V_{LT}=-\beta V_{sat}$, $\beta =R_1/(R_1+R_2)$. The output goes LOW when $v_{in}$ rises above $V_{UT}$ and HIGH when it falls below $V_{LT}$. The loop is centred at $R_2{V}_R/(R_1+R_2)$.

Input through $R_1$ to the + terminal, $R_2$ from the output to the + terminal, reference $V_R$ on the − terminal (ground if $V_R=0$). Switching happens when $v_{+}=(R_2{v}_{in}+R_1{v}_o)/(R_1+R_2)$ reaches $V_R$. With $V_R=0$: $V_{UT}=+(R_1/R_2)V_{sat}$ and $V_{LT}=-(R_1/R_2)V_{sat}$. The output goes HIGH when $v_{in}$ rises above $V_{UT}$ and LOW when it falls below $V_{LT}$.

Time for a capacitor charging or discharging through R, from $V_{initial}$ toward the asymptote $V_{final}$, to reach $V_{switch}$ (which must lie between the two). Every period and pulse-width formula in this section and the 555 section is this formula with specific values.

An inverting Schmitt trigger ($R_1$ from the + terminal to ground, $R_2$ from the output to the + terminal) with R from the output to the − terminal and C from the − terminal to ground. With symmetric $\pm V_{sat}$, each half-period the capacitor charges from $\mp \beta V_{sat}$ toward $\pm V_{sat}$ until it reaches $\pm \beta V_{sat}$, so the output is a square wave with 50% duty cycle and T does not depend on $V_{sat}$. The capacitor voltage is a chain of exponential segments, not a true triangle. $R_1=R_2$ gives $T=2RC\ln 3=2.197RC$; $R_1=0.86R_2$ gives $T\approx 2RC$.

The astable above with a diode across C (anode at the − terminal) that clamps $v_C$ at $+V_D$ (constant-drop model, $V_D\approx 0.7$ V) while $v_o=+V_{sat}$; this stable state needs $\beta V_{sat}>V_D$. A negative trigger pulse at the + terminal sends $v_o$ to $-V_{sat}$; C then charges from $+V_D$ toward $-V_{sat}$, and the circuit flips back when $v_C$ reaches $-\beta V_{sat}$. With $\beta =0.5$ and $V_D$ neglected, $T=RC\ln 2=0.693RC$. After the pulse, C must recover to $+V_D$ before the next trigger.

Schmitt trigger: $R_1$ from the triangle output to the + terminal, $R_2$ from the square output to the + terminal, − terminal grounded. Integrator: input resistor R, feedback capacitor C. The triangle swings between $\pm V_p$ (peak-to-peak $2V_p$; needs $R_1<R_2$) and the square between $\pm V_{sat}$. From $T/2=2V_p/(V_{sat}/RC)$; f is independent of $V_{sat}$ because the ramp rate and the thresholds both scale with it.

R from $V_{CC}$ to pins 6 and 7, C from there to ground, trigger on pin 2. In the stable state the output is low and C is held discharged. A trigger below $\frac{1}{3}{V}_{CC}$ sets the output high; C charges from 0 toward $V_{CC}$, and the pulse ends when $v_C=\frac{2}{3}{V}_{CC}$. The trigger must return above $\frac{1}{3}{V}_{CC}$ before T ends (a narrow trigger pulse); in the basic circuit, further triggers during the pulse do not extend it. T does not depend on $V_{CC}$.

$R_A$ from $V_{CC}$ to pin 7, $R_B$ from pin 7 to pins 6 and 2, C from pins 6 and 2 to ground. C charges through $R_A+R_B$ and discharges through $R_B$ alone, between $\frac{1}{3}{V}_{CC}$ and $\frac{2}{3}{V}_{CC}$ (a ratio of 2 each way, hence $\ln 2$). With D defined as high time over period, D is always above one half and approaches it only when $R_B\gg R_A$. $R_A$ must not be zero, or pin 7 shorts $V_{CC}$ to ground when the discharge transistor turns on.

A diode across $R_B$ (anode at pin 7) bypasses $R_B$ while C charges, so C charges through $R_A$ only and discharges through $R_B$ only. Ideal-diode result; with a real diode drop, $t_H$ comes out somewhat longer. $R_A=R_B$ gives a 50% duty cycle, and duty cycles below 50% become possible.

Pin 5 driven to $V_{CON}<V_{CC}$ moves the thresholds to $V_{CON}$ and $V_{CON}/2$. $t_L$ is unaffected (the discharge still halves $v_C$), while raising $V_{CON}$ lengthens $t_H$ and lowers f. This is how a 555 becomes a voltage-controlled oscillator and, in monostable mode, a pulse-width modulator. $V_{CON}=\frac{2}{3}{V}_{CC}$ recovers the standard formulas.

$K_v$ in Hz/V, $K_o$ in (rad/s)/V. Linear model. Check which unit a question uses before multiplying.

Multiplier of gain $K_m$: inputs $V_1\sin (\omega t+{\theta }_i)$ and $V_2\cos (\omega t+{\theta }_o)$ average to $\frac{K_m{V}_1{V}_2}{2}\sin ({\theta }_i-{\theta }_o)$, so $K_d=K_m{V}_1{V}_2/2$ V/rad ($V_1{V}_2/2$ for unit multiplier gain). XOR with logic levels 0 and $V_{DD}$: with 50% duty-cycle square waves a phase difference $\varphi$ apart, the output is high for a fraction $\varphi /\pi$ of the time, so $K_d=V_{DD}/\pi$ V/rad; ${\theta }_e=\varphi -\pi /2$, measured from the midpoint output $V_{DD}/2$.

Locked loop, linear PD model ${\bar{v}}_d=K_d{\theta }_e$, loop-filter DC gain $F(0)$, so $v_c=F(0)K_d{\theta }_e$. $K_V$ (units 1/s) is the DC loop gain; the $2\pi$ cancels if both the offset and the VCO gain are in Hz. A larger $K_V$ gives a smaller phase error for the same offset. With a sinusoidal PD, the second relation gives $\sin {\theta }_e$ instead of ${\theta }_e$.

${\bar{v}}_{d,max}$ is the largest PD output measured from its zero point: $K_d$ for the multiplier (at ${\theta }_e=\pm \pi /2$), and $K_d\pi /2$ for the XOR, whose characteristic is linear over $|{\theta }_e|\le \pi /2$. A loop that is already locked stays locked while the input moves up to this far from ${\omega }_{fr}$, i.e. as far as the largest available control voltage can pull the VCO (assuming the VCO itself can go that far). It depends on the filter's DC gain, not its bandwidth. Some questions quote the total width ($2\Delta f_L$) instead of the one-sided value; read the wording.

One-sided ranges; loop filter $F(s)=1/(1+s\tau )$ with $F(0)=1$; an approximation valid when $\Delta {\omega }_C\tau \gg 1$. Starting unlocked, the PD output is a beat at the difference frequency $\Delta \omega$, which the filter attenuates by about $1/(\Delta \omega \tau )$; the loop pulls in only if the attenuated beat still swings the VCO far enough, which gives $\Delta {\omega }_C\approx \Delta {\omega }_L/(\Delta {\omega }_C\tau )$. The capture range never exceeds the lock range. A narrower filter (larger $\tau$) narrows the capture range and leaves the lock range unchanged.