Fullscreen HTML5Med Res (??MB)HTML4

Circuit Surgery Regular clinic by Ian Bell Measuring the frequency response of a circuit using a PC sound card, part 4: analysing circuit frequency responses I n recent articles, we have been investigating how to measure frequency responses. This was motivated by the need to measure the response of a practical implementation of an example digital filter (from our series on DSP) without using advanced lab test equipment. We covered the general principles of frequency response measurement and the use of a “sound card” (audio interface) in a PC or laptop, together with free software called REW (Room EQ Wizard). Last month, we considered the signal conditioning that might be needed to interface the circuit under test to the sound card’s line inputs and outputs. We described part of a signal conditioning circuit built from op amps, the function of which was to get the right signal amplitude and DC levels in and out of the sound card and the microcontroller board implementing the digital filter. In the circuit description, we mentioned that it is common practice to place a low-value capacitor across the feedback resistor of an op amp amplifier (see Figs.1 & 2). This is primarily to improve stability (reduce the likelihood of unwanted oscillations), one potential cause of which is capacitive loading of the output. We will discuss this further in a future article. Low-pass filtering I also commented that this reduction of gain may help if there are unwanted high frequencies in the input, but more usually an RC low-pass filter and/or ferrite bead would be included in the input circuit to address this if needed. I was describing a non-inverting amplifier (Fig.2); in this case, the gain does not continuously decrease at higher frequencies, but levels off at unity – so the circuit is not very useful as a low-pass filter in most circumstances. Despite this, I considered including a formula for the cutoff frequency of the circuit with the capacitor shown in the article. I did not know it from memory, although I knew it was not simply 1 ÷ (2πRFCF). A quick online search gave me the wrong answer, and various links related to the wrong circuits or circumstances, so I did not include it in the previous articles. Later, I worked out the formula myself. This highlighted the fact that sometimes when dealing with circuits, it is useful to be able to derive formulae for their characteristics rather than assuming you can look them up. So this article will focus on how to go about such calculations. I was intrigued and concerned that Google’s AI search summary presented the wrong answer. I expect that this is because the circuit is not commonly used as a low-pass filter, so this is relatively little discussed. But this could be misleading to anyone forgetting the small print “AI responses may include mistakes”. A tale of four AIs I wondered whether other AI systems would provide better answers. I tried the full AI mode on the Google search site, Copilot on my PC, plus Perplexity and CF RF CF – RF Vout RI Vin + Vin – Vout RG + Fig.1: an op-amp-based inverting voltage amplifier with a feedback capacitor. 42 Fig.2: an op-amp-based non-inverting amplifier with a feedback capacitor. ChatGPT via their websites. I used the same prompt for each, which stated it was a non-inverting amplifier and that I wanted the -3dB frequency, specifically not the pole frequency (poles will be explained later in the article). I checked my formula and the ones from the AIs against LTspice simulations with idealised op amps. Only ChatGPT gave me a correct response, also providing a detailed derivation. Its final equation was in a similar form to the one I derived manually. Both Perplexity and Copilot produced a formula with the same general ‘look’ as the correct versions (a 1 ÷ 2πRC term multiplied by a square root term containing the resistor values or gain value), but these were not mathematically equivalent and did not produce the correct numerical values. Perplexity gave some explanation and source links. Copilot’s response was more minimal, so I asked about its sources, and it said its response was based on its internal understanding of analog circuit theory (which was unfortunately not quite good enough). While there were poor results from three out of the four AI systems I tried, they will probably improve in the future. Using AI to obtain circuit equations is potentially very useful but could also undermine the deeper understanding that can come from working through the maths yourself. This month we will go into detail on how to obtain the formula for the -3dB cutoff frequency of the circuit in Fig.2. This will make use of some techniques that are based on advanced maths, but in many situations only basic algebra and a little knowledge of complex numbers is required. The circuit itself may not be very useful as a filter, but it does provide a good example for discussing how to approach analysing a circuit. This is also a natural extension of our discussion so far on concepts related to circuit frequency response. We will discuss how to handle the frequency-dependence Practical Electronics | January | 2026 of capacitors and inductors in circuit calculations. We will also look at the concept of poles and zeros, which you will often see referred to when discussing issues such as op amp frequency response, and stability of circuits with feedback (eg, in technical documents from semiconductor manufacturers). Circuit overview Having a general or intuitive understanding of a circuit’s behaviour before attempting a mathematical analysis is helpful. The equations should fit within the general understanding. For example, if you know that increasing a particular parameter (component value etc) should reduce some other parameter (gain, frequency point etc), the structure of the equation should correspond with this. The capacitor in the circuits in Figs.1 & 2 causes the circuit to act as a low-pass filter because the capacitor’s reactance (effective resistance) reduces as the signal frequency increases. At low frequencies, the capacitor is effectively an open circuit and has little impact on the gain. As the frequency increases, the capacitor’s reactance magnitude becomes comparable with the feedback resistor value, reducing the total parallel impedance. At still higher frequencies, the capacitor dominates the parallel impedance, which tends towards zero as the frequency increases to infinity. Last month, we discussed the differences between the inverting and non-inverting amplifier configurations. The impact of a feedback capacitor on the amplifier’s frequency response is another area where these circuits differ. For the inverting amplifier, the gain without the capacitor is -RF ÷ RI, which is also the DC gain with the capacitor (A0). At higher frequencies, as CF bypasses RF, reducing the effective overall value of the feedback resistance, the gain of the circuit with the capacitor reduces and tends towards zero (assuming all components are ideal). For the non-inverting amplifier, the gain without the capacitor (and DC gain with it) is 1 + RF ÷ RG. At higher frequencies, as CF bypasses RF, the gain with the capacitor reduces and tends towards unity (the RF ÷ RG term approaches zero, leaving just the 1 term). The inverting amplifier with a feedback capacitor has a standard first-order low-pass response, whereas the non-­ inverting amplifier exhibits a shelving filter response. A shelving filter provides different gains at low and high frequencies, but with flat gain-versus-frequency on both sides of the transition region, rather than the gain falling towards zero at the high or low end. Practical Electronics | January | 2026 The sinc correction filter discussed in the November 2024 issue has a shelving response. Shelving filters are also common in audio equalisation applications. The circuits in Figs.1 & 2 are not commonly used as low-pass filters because, with the addition of an extra capacitor and resistor, you can create second-order (faster roll-off) filters using topologies such as Sallen-Key and multiple feedback. Complex numbers recap In the October 2025 issue, we discussed frequency response in general, and briefly discussed complex number representation. Complex numbers are often used in the representation of frequency-­dependent circuits, signals and components. As a reminder, we need to deal with the fact that signals have both amplitude and phase; components and circuits may influence both aspects of a signal. Therefore, a single number is insufficient to represent, for example, the input-output relationship (gain or transfer function) of a circuit, except in simple cases, such as circuits using only resistors. We can represent amplitude A and phase Ф directly as A∠Ф (see Fig.3), but this does not lead to the most straightforward mathematical analysis of circuits. Therefore, we often use complex numbers where the value of the signal in Fig.3 is x + jy (j is √-1, often written as i by mathematicians). In x + jy, x is the ‘real’ part and jy is the ‘imaginary’ part (since √-1 does not exist as a real number). It is well known that if we apply a sinewave voltage to a capacitor or inductor, the current is 90° phase shifted with respect to the voltage. Thus, the currentvoltage relationship of capacitors and inductors and circuits built from them requires the use of either an amplitudephase or complex number approach. Reactance and impedance For resistors, the current (I) to voltage (V ) relationship is Ohm’s law, VR = RIR, Imaginary where all the values are real numbers. Capacitors and inductors have a similar relationship, with reactance (X) rather than resistance. For example, VC = XCIC for a capacitor. More generally, for circuits with a mixture of components, we use impedance (Z), so V = ZI. Reactance and impedance involve complex numbers. The mathematics underlying circuit analysis with complex numbers will not be familiar to all our readers. However, in some common cases, it is possible to use complex impedances for analysing circuits, along with well-known rules such as Ohm’s law, series and parallel component combinations and the potential divider formula, without a deep knowledge of advanced maths. The following is an informal ‘hand-­ waving’ explanation of the complex reactance of capacitors and inductors. As frequency increases, the reactance of a capacitor decreases – it is inversely proportional to frequency (proportional to 1/f ). Also, at a given frequency, increasing the capacitance decreases the reactance; again, it is inversely proportional. We can calculate it as 1 ÷ 2πfC with frequency in hertz (f ) or 1 ÷ ωC with frequency in radians per second (ω), but we also need to consider the phase shift. The 90° phase shift means that the voltage-to-current relationship for the capacitor must be represented as a purely imaginary number (the real and imaginary axes are at 90° in Fig.3). The fact that voltage lags current in a capacitor means that in VC = XCIC, the value of XC is purely imaginary and negative, giving XC = -j/ωC. Multiplying by -j/-j gives XC = -1/-jωC = 1/jωC, which is the form you often see it written in. Similar formula manipulations for an inductor (where voltage leads current and impedance is proportional to inductance and frequency) gives ZL = jωL. Transforming problems Fundamentally, the instantaneous current (i) in a capacitor (C) is proportional to the rate of change of voltage (v) across it. We can write this as: i=C jy j = √(–1) A Here, dv/dt is the rate ) change of v B ( s of H (time s )= (t) – we are in the with respect to A ( s ) for an inrealm of calculus. Similarly, ductor, we have v 1+ = L(di/dt). ( 1/628 )The s resistor s )=time domain equation is is simplerH–( its 1+ ( 1/ 4712 ) s Ohm’s law, v = Ri. For circuits containing ( − 628 )2 628 resistors, inductors can use f =and capacitors, = we=100 Hzthese 2π three component equations, ( 2 π )2characteristic along with circuit theory rules, to develop ) soutput 1+ (input 1/6.28 × 107to equations relating signals H ( s )= signals. The results time-domain dif( 1/628 )s 1+are ferential equations. Unfortunately, such √ Φ x Real Fig.3: the relationship between amplitude, phase and complex numbers. dv dt v out ( s )= √ Xc 1/ sC v ( s )= v (s) R+ X c in R+1/ sC in 43 v out ( s ) 1 i=C equations are difficult to directly manipulate and solve. To overcome this difficulty, we use mathematical transforms. Transforms convert problems between different domains (the way in which signals are represented), for example, the time domain or the frequency domain. This can greatly simplify things; for example, requiring solutions to algebraic rather than differential equations. The most commonly used transforms in electronics are the Fourier and Laplace transforms for continuous signals, and the Z transform for sampled signals. Others, such as the wavelet transform, are also used in specific circumstances where they provide advantages. Developing the theory, proofs and properties of transforms from scratch takes a significant amount of advanced maths, but this has already been done. So, in straightforward cases, we do not have to directly apply the transform maths. We can use the known results, like the component reactances (1/jωC and jωL) discussed above, together with ‘resistorbased’ circuit theory. Of course, we need some knowledge of manipulating complex numbers. Often, this just involves finding the magnitude of an impedance, or a gain-magnitude for a frequency response calculation (remember it is the gain magnitude that is plotted in a Bode frequency response graph). The magnitude of a complex number, x + jy, is √x² + y². This follows from applying Pythagoras’ theorem to find A in Fig.3. When finding the magnitude of a division of two complex numbers, say M ÷ N, we find the individual magnitudes of M and N first, then perform the division. For purely capacitive and inductive reactance, we only have an imaginary part (eg, jy), so the magnitude is √y² = y. For example, this is 1 ÷ ωC or 1 ÷ 2πfC for a capacitor, which is well-known as the formula for the ‘resistance’ of a capacitor. It might look like we just removed j to get 1 ÷ 2πfC, but if we have a combination of resistance and reactance, this does not work. We have to use the full root-sumsquared calculation to get the impedance magnitude – as we did last month when we looked at the cutoff frequency of the AC-coupled inverting amplifier. Enter the s-domain The Fourier transform represents signals in the frequency domain (f or ω), based on the fact that signals can be represented by a sum of sinusoids. The Laplace transform uses the s-domain, which is a form of complex-number frequency. The Laplace transform is a generalisation of the Fourier transform, which can handle signals that the Fourier transform can’t (when you delve into the transform 44 maths), so it is more widely applicable. In the s-domain, the capacitor and inductor reactances are 1/sC and sL, respectively. The s-domain is able to handle a wide range of signals and situations, but if we limit ourselves to the steady-state, singlefrequency behaviour of linear circuits, we can use the substitution s = jω = j2πf. We discussed steady-state conditions previously in our introduction to frequency response concepts (October 2025). Typically (for steady state), we can perform the algebra of our circuit analysis using s-domain component reactances, make the substitution for s in the result, then find the magnitude to get the real numbers we need. We could do all the calculations using jω or j2πf instead of s; you will find examples of both approaches in technical documentation. A common approach to circuit analysis is to use the s-domain component reactances to obtain the transfer function H(s) in the form of a division of two dv polynomials in i=C s, with the denominator dt that is: A(s) and numerator B(s); H ( s )= B (s) A (s) A polynomial of1+ a variable ( 1/628 ) s(eg, s) is an expression only the variable H (involving s )= 1+ ( 1/integer 4712 ) exponents s and its non-negative (powers), which are 2 multiplied by coeffi( − 628 ) positive 628 cientsf = (dimensionless or negative = =100 Hz 2 2 π numbers) and summed, eg, s² + 2s + 5. (2 π ) √ √ 1+ ( 1/6.28 × 10 ) s Poles and zeros H ( s )=of s that cause B(s) and A(s) The values 1+ ( 1/628 ) s 7 to be zero indicate important frequencies in theXcircuit’s transfer function, 1/ sC c v out ( s )= v shape v (s) in ( s ) = of the frequendetermining R+ Xthe R+1/ sC in c cy response. When the numerator B(s) is zero, the s-domain v out ( s ) transfer 1 function (gain) value and H ( s )is=zero, =this is given the v ina(zero. s ) 1+sRC obvious name of When denominator A(s) of the s-­domain 1 1 is 1zero, the trans1 transfer function = value + + +… R 1 R 2 infinity R3 R P approaches fer function – this is called a pole (a pole sticking up; a high R F ( 1/ s C F ) R values). X CF point in the Z F = s =F 0 in = transfer function Putting the R F + X CF R + ( 1/ s C F ) gives the DC gain. ForF example, this Gain (dB) Z F= DC gain R 1+sR F C F F Pole Gain falling by 20db per decade R G + R F / ( 1+sR F CF) H ( s )= RG dv dt B (s) )= one pole and one H ( shas transfer function A (s) zero: 1+ ( dv 1/628 ) s H ( s )=i=C dt 1+ ( 1/ 4712 ) s B ( s ) 1 + (s ÷ 628) The zero is found H ( s)2)=using ( − 628 628 √ = 0 (numerator equals s ÷ 628 f= = A zero), ( s =100 ) so Hz π substitute s = = -1, hence )2 If2we √ (s2=π-628. ( 1/628 ) s To find jω = j2πf, we get f1+ -628 ÷ (j2π). 7 H ( s )= ( = )s 1/6.28 × 10 the frequency,1+ we calculate: H ( s )= 1+ ( 1/ 4712 ) s ( 1/628 ) s 1+ 2 ( − 628 ) 628 √ f= X = =100 1/ sC Hz v out ( s )= √ ( 2c π )v2 in ( s )2=π v (s) R+ X c R+1/ sC in There is a zero(at 100Hz. The 1+ 1/6.28 × 107 ) spole is found H using ( s )=1 +vsout÷( 4712 s ) = 01which, fol) s 750Hz. 1+ ( 1/628 ( s )= calculation, = gives lowing aHsimilar v in0)( sis) 1. 1+sRC The DC gain (s = Xc 1/ sC Remember that ( s )= (s) v out v ins( sis)=a complex v(two1 1 1 1 R+ X R+1/ sC in to dimensional) related = cvalue + that + is+… R 2 frequency R3 R PbutRis1 not frequency as we (s) v out 1 experience it. An s-domain transfer funcH ( sR)=X = 1/ s C R ( ) tion going to does not mean the F F Finfinity CF 1+sRC v in ( = s) Z F =magnitude real gain of the circuit is R F + X CF R F + ( 1/ s C F ) infinite. Having 1 1that 1are at negative 1 poles = not + mean + negative +… values ofRs does real R RR P 1 2 F R3 frequencies.Z F = 1+sR of C F s-domain The abstract nature 1/ s C ) RF F (the R F X CF and advanced maths=behind it canF be offZ F= RisX R it+ ( 1+sR sCCFcircuit G+ F /R ( 1/Ffor putting, but aCFR useful tool F+ F) ) H ( s )=F analysis calculations. Furthermore, poles RG R F insights into and zeros provide important Z F = and circuit stability. frequency response R +1+sR R F +sR R C F CFF G F H ( s )= G above At frequencies the pole frequenR G +sR F R G C F cy, gain decreases rate of C 20dB per R G + RatF /a( 1+sR F F) decade than at frequencies below ( s )= H more R +R 1 the pole. rate-off z = ThisG is a Fchange =R G in the A 0 πR F with R G C frequency, 2 πR F C not change of2gain the R + R +sR R C G F so if F the G gain F was absolute rate of change, H ( s )=R + R / R +sR C ( G20dB decreasing by decade R G +sR F )per F F GR G C FF below ( s )= H a pole, it would decrease 40dB per 1+sR F Cby F R G it. + RF decade above 1 f = = A 0 -90°. z A pole the2 phase by 2also πR Fshifts R GAC0 +sR πR F CFFC ( ) Hthe s= Fig.4 shows frequency response char1+sR FC acteristics of aRcircuit with aFsingle pole. ( G + R F ) / R G +sR F C F Similarly, for a zero, the gain will H ( s )= 2 2 AC0 F more per Pof 0 +1+sR F increase at √a Arate 20dB = 2 decade above√the frequency com√ 2C 1+APzero 0 +sR F adds F pared to below it. A zero +90° of H ( s )= 2 the frequency phase shift. Fig.5 shows 1+sR C A F F P 2= 2 0 of a circuit with response characteristics A20 − 2 a single zero.√ A 2 + P A 0 = 0 Gain (dB) 2 f A 1√ 1+ P A 0√ 2 f − 3 dB = = by p2 0 ∙ increasing 2 2 π R F C Gain 2 √ A0 − 2 20db √ AA20per−decade P2= 2 0 A0 − 2 Zero f A A0 1 = p2 0 ∙ 2 f 2 π RF C Z √ A0 − 2 √ A0 − 2 Phase shift Log frequency / f DC gain R + R +sR F R G C F Log frequency / f H ( s )= G fP F Phase shift R G +sR F R G C F 0° f z= –45° RG + R F 1 = A 2 πR F R G C 2 πR F C 0 –90° f − 3 dB = +90° +45° 0° CF ( R G + R Fand ) / Rphase G +sR F Fig.4: the frequency response H ( s )= of a circuit with a single 1+sRpole. F CF H ( s )= A 0 +sR F C F 1+sR F C F Fig.5: the frequency and phase response of a circuit with a single zero. Practical Electronics | January | 2026 i=C H ( s )= dt B (s) A (s) LTspice Laplace 1+ ( 1/628 ) s LTspiceHcan ( s )=directly implement Laplace functions 1+ using controlled and ( 1/ 4712 )s behavioural sources. Fig.6 shows a 2 ( − 628 ) voltage voltage-­controlled 628 source (VCVS, f = = =100 Hz 2 implementing E SPICE element) a simi2π (2 π ) lar example to the one above: √ √ H ( s )= 1+ ( 1/6.28 × 107 ) s 1+ ( 1/628 ) s Here, theXpole is at 100Hz, the zero 1/ sC c )= v outat( s10MHz = gain is 1.vVery is andv the in ( s )DC in ( s ) R+ Xand R+1/ sC were c different pole zero frequencies chosen for this example so that the effect v out ( s ) response 1 of each on is well H (the s )=frequency = separated and clear 1+sRC ) see. v in ( sto The VCVS is configured using Laplace=<func(s)> 1 1 1as the source 1 = +The keyword + +…Laplace “value” (its gain). R P R1 R2 R3 is followed by a function (func) in terms of s. The above 1/ s C Fformat R XequationRinF (LTspice ) is Laplace=(1+s/6.28e07)/(1+s/628). Z F = F CF = R + X R + 1/ s C ( theF ) simF the CFresults F from Fig.7 shows ulation in Fig.6. We can see that the R F magnitude rebreakpoints in the gain Z F= sponse (solid green trace) at 100Hz 1+sR F Coccur F and 10MHz, corresponding to the pole and zero frequencies. plot Falso / ( 1+sR C F )shows R G + R FThe H ( s )response = the phase (dotted green trace and right-hand axis). R G As the frequency increases past the R + R +sR R G C F depole, the )= Gof theF gainFresponse H ( sslope creases by 20dBRper decade 0dB G +sR F R G Cfrom F per decade (horizontal) to -20dB per decade. TheRpole the 1 phase shift G + Rcauses F f z = by -90°, = A0 to change the phase 2 πR F R Gwith C -45° 2 πRof FC change occurring at the pole frequency. The specific(change to -90° R G + R Fhere +sR 0° ) / R Gfrom F CF ( s )= over a range of frequencies takesHplace 1+sR FC F above and below the pole frequency. As frequency increases past the zero, A 0 +sR F C F increasthe slope H of (the response s )=gain 1+sRfrom es by 20dB per decade F C F -20dB per decade back to 0dB per decade (horizontal again). The the phase A A 20 +zero P 2 causes = 0with +45° of shift to change by 2+90°, √at2 the zero fre1+ P phase change occurring quency. The specific change from -90° 2 2takesA 0 back to 0° again place over a range P= 2 of frequencies. A0 − 2 The DC (or low-frequency) gain is ×1 (0dB), which is seenfin Fig.7. 1 the gain atA1Hz 0 p A0 fFig.8 =using ∙ the − 3 dB =shows a simulation 2 2 2 π R C A 0 −zero 2 frequency A0 − 2 circuit in Fig.6F with the changed to 500Hz. The closeness of the pole and zero means that the full effect of the pole (-90° phase and -20dB per decade slope) does not occur before the zero reverses the situation. Fig.6: implementing a Laplace transfer function using a controlled source in LTspice. Fig.7: the simulation results from Fig.6. √ √ √ Fig.8: the simulation results from Fig.6 with closer pole and zero frequencies. √ RC circuits The discussion so far has been somewhat abstract, so we will look at a simple circuit example – an RC low-pass filter, shown in Fig.9 along with a Laplace version of its transfer function implemented using a behavioural voltage source (B1). The cutoff frequency is well known for Practical Electronics | January | 2026 Fig.9: an LTspice simulation for the Laplace representation of an RC low-pass filter. 45 i=C dt H ( s )= B (s) A (s) this circuit; f = 1 ÷ 2πRC, which with the 1+ ( 1/628 s × 1000 × values used ÷ (2 ×) π )= is 1dv H ( shere 159×10-9) = 1.0kHz. i=C 1+ ( 1/ 4712 ) s dt divider formed The circuit is a potential 2 ( ) − 628 by a resistor (R) and628 capacitor (C). In f= = B ( s )=100 Hz (2s resistor’s )= the s-domain,Hthe 2 π( s ) resistance is (2 π ) A simply R and the capacitor’s reactance 7 is Xc = 1 ÷ sC,1+ as (1+ previously )s ( 1/628 ) sdiscussed. 1/6.28 × 10 ( ) ( ) H s = H spotential = Using the divider formula, the ( ) ( ) 1+ 1/ 4712 s 1+ 1/628 s output is: √ √ X c )2 628 1/ sC ( s )= =100 Hzv ( s ) v out ( sf )= =√ ( − 628 v = R+( 2Xπc )2 in 2 πR+1/ sC in √ Note that we write vin(s) because we are 7 ( s ) and v out 1this ( 1/6.28 ) sapplies workingHin( sthe s-domain )=1+ = × 10 ( ) H s = to the voltages as as the component 1+sRC ( ) v inwell s 1+ ( 1/628 ) s values. Multiplying the top and bottom X c 1 v1in to get by sC and1moving sC expres11/ the )= gain, )= v out ( sfor v+ v (s) = H(s), +the +… in ( sin sion polynomial R+1/ sC in R2 R RR+ c 1 P XR 3 form, we obtain: R Xv out ( s ) R F ( 1/1 s C F ) H ( s )F= CF Z F= == R F + XvCF + ( 1/ s C F ) in ( s ) R F1+sRC This response has no zeros (there is no 1 1 1 pole at 1 + 1 s in the numerator). = + ItR+Fhas a+… R 3 If we substiR P Z FR=1 1+sR sRC = 0, meaning s =R-1/RC. 2 F CF tute s = jω = j2πf and find the magnitude: C F )) R FRXGCF F ( 1/ s C + R F2 / (R1+sR F F ZHF = ( sR )= +√X(−1= ) R + 1/ 1 sC ) CF f =F = FG ( F R √ ( 2 πRC )2 2 πRC R F F RG C F R Gthe + Rsame F +sR ThisHfollows pattern as the ( s )=Z F = R1+sR R(s FFC FG C numerical example above = -628). We G +sR F are back to the well-known -3dB cut-off +circuit: R F / ( 1+sR R Rthe +GR frequency of f 1= F1C÷F )2πRC. Hz =( s )= G F = f A0 The pole 2 frequency frequency R2G-3dB πR F R G Cand πR F C are the same here, but in general, you cannot assume case.CCF RR Gthis ++RRFis /the R G +sR F)+sR F RG F F )=( Gvalue ( s( )s= HHnegative The of s is important in R G1+sR +sR FFRCGFC F wider use of s-domain analysis, but here it disappears when we take the magniR G + R FA 0 +sR F C 1F tude ftoz = find frequency. A0 H (the s )=pole = 2 πR F R G1+sR C 2FπR C FF C ( R G 2+ R F2) / R G +sR F C F H ( s )= √ A 0 + P = A 0 2 2F √F C √ 1+ P1+sR A 0 +sR C A 20 F F H ( s )= 2 P =1+sR C A 20 − 2F F √1 A + P =A A f A = ∙ √ 22 = A − 2 2 π R√ 1+ C P√ A − √ 2 2 0 f − 3 dB 0 0 2 F P2= f − 3 dB = A 2 0 2 0 p 0 2 0 2 0 A −2 f A A0 1 = p2 0 ∙ 2 2 π RF C √ A0 − 2 √ A0 − 2 Fig.10: the simulation results from Fig.9. 46 The syntax for implementing the Laplace function using an LTspice behavioural source is a little different from the controlled sources such as the VCVS used above. If the Laplace function definition is included in the source definition, it is applied to what would usually be the source output, as defined by the source expression, V=<expression>. In this case, the expression is just V=V(in), so the source is applying the voltage on node in to the Laplace function, which is what we need here. The Laplace statement is written after the source expression, separated by a space, and is the H(s) function for the RC circuit from above with the specific R and C values in it: Laplace=1/ (1+1e03*1.59e-07*s). The results from the simulation shown in Fig.10 confirm that the RC circuit and Laplace function have the same frequency response. This example illustrates that with the s-domain reactances, basic circuit theory, and a little knowledge of complex numbers we can obtain and use the Laplace domain transfer function without having to do any difficult calculations, such as applying the transform maths directly to a circuit equation. It’s worth emphasising that a pole or zero will always indicate a breakpoint in the frequency response, but this will not necessarily be a 3dB point. The effect of poles and zeros can overlap, which leads to more complex relationships between the poles and zeros and gain at these frequencies. In fact, if there is a pole and a zero at exactly the same frequency, the two breakpoints are present, but their effects cancel one another and no change in gain or phase will occur. Poles and zeros are fundamental in terms of defining the shape of the frequency response. A 3dB point dv is an arbitrary i=C definition of a cut(but commonly used) dt off frequency. It is not used universally. For example, a Chebyshev B ( s ) filter’s cutoff H ( s )= defined as the frefrequency is commonly A (s) dv quency after the last passband peak where i=C the response has 1+ dropped by dt ( dv 1/628 ) sthe ripple H ( sis)=not i=Cnecessarily at -3dB. level, which 1+ ( 1/ dt4712 (s) ) s B H ( s )= 2 dv Non-inverting amplifier A ( s ) response B √ ( − 628 ( s) )= 628 H i=C Forf = the non-inverting amplifier, =100 Hz the dt 2 A 2 π( s ) and ) s (also DC 1+ ( dv 1/628 ( ) 2 π gain without the capacitor √ H ( s )=i=C B ( s ) dt4712 gain with theH capacitor) is A ))s0s7=) 1 + (RF 1+ 1/628 ( 1+ )=((1/ × 10 HG (+s )Z1+ =F)s(÷1/6.28 ÷ RG) =H(R R . With the scapaci( ) A s G ( s )= 1+ 2 ( 1/ ( sreplace ) ) s R with Bto4712 ( −we ) 628 tor in place, need 628 √ ( ) 1/628 s Hz 1+ H ( s )= F f= = A (Z =100 ) sRF and CF, ( ) s 2 ( 1/628 the parallel impedance ) of 2 1+ F 2π )π=) ) 628 H√( − (X2s628 √ 1/ )sC which is the cparallel combination of (RF) =( 1/ =100 v out ( sf )= = v21+ =4712 s ) ss XHzv.inSo, 1+ in ( s( )1/628 2 π and the capacitor’s reactance, ( ) 2 π R+ √ sX)1+ H = ) s CF c (21/6.28R+1/ × 107sC 1/ 4712 ) s7 ( s( − )1=628 H √is the gain + (Z1+ )F ÷(R 628 G). f= = =100 ()1/628 ) s Hz (1+ 1/6.28 × 10 s The parallel 1+ resistance 2out ( v s P) 1of )multiple 2 2 π(R )(= HH√( s(√s− ) 2 π ) ) = = 628 628 resistors and so ) sC s is 1+ f = (RX1, cR2v,2inR =100 Hzgiven 1/on) 1+sRC (3= s())1/628 7 ( s )= ( by well-known formula: v outthe v s = v (s) 2 π in ( ) 2 π ( ) 1+ 1/6.28 × 10 √)X=Xc c R+1/ sCs in 1/ sC (R+ H s 1i=C ( 1s( dv )1/628 v out ( s )= 1 v1+ = 1 ) s 7 v in ( s ) + R+1/ (+in1/6.28 )s × +… 10 sC R+=X1+ c dt ( sP)= Rv1out (Rs )2 R 3 1 HR H ( sX)= = 1/) sC s 1+ ( 1/628 c For ( s )two ( this )=( s1+sRC v out = resistors, vRinP( s=) )becomes vvvout inin( ss) B 1 1/ sC R R+ X R+1/ sC R X ) ( ) HR s = ( s1X)F+= = F F c (CF (R1R1) ÷H(R ). As well as resistors, = c vv2in ((= )1/ sC v ( of ss))A=( s1+sRC ( Zs )F= v out can we use impedances in 1 1Laplace 1 ( 1/ ssC 1R Fthe + X R + C F ) in s ) CF F R+ X R+1/ = + + +… cv out ( ) s 1 the capacitor and inductors R11 1+ R1(21/628 HR1(Ps()s= = R13 )ins these for)= mula, so H we= get: 1+sRC + +… ( ) v s R in ) +F4712 v out 1+(R(s1/ 1) s 1 2= R3 )F=FXR=CF HR(PsRZ 1/ R ( FC F s C F ) 1+sR F 1+sRC s1) 1 Z F =√1( − 628 1v in)2( = += 628 +… + XCFCF R+RFF+=100 1/ 1/ ssCCFF)) f = RRRF= FX R R 2/2( 1+sR R(3( CHz 12+ R= Z F =1P( 2RπG1both ) π F F 1 1 ) Multiplying the numerator R s CFF ) and )=F=+ X CF H ( s√ + RR+F + ( 1/ +… denominator sC , we get: F R R R R R PRZ by F 2 RG (31/ s7C ) X=1 1/6.28 F F Z FH=( s )=F F1+CF(1+sR =R FF C×F10 ) s RRFZ+R X R + 1/ s C ( R F( 1/628 +sR CFFF)) CF =+ RFF (F1/R ) sGC 1+ F FXGCF1+sR ( ) C s = ZH = = F F F +GR+sR R GR ( 1+sR R(G1/CFsC F /R F) F+ CF R XF c+ X sCFCZFF) into ( R ) this Putting expression FF 1/for ( sH ) =s = ( ) v out v s = v (s) ZRF = R G asF C G +inRwritten F / ( 1+sR the gain equation, R+1/ sCFG) +inZF) C F1 (R RX c R1+sR )= H ( sR+ G+ F RF =RF G A ÷ RG,fwe z = get: ZR F= R F +sR 2 πR R+1+sR 2 πR CC F 0 G FR Fv GC FG F C F 1F C F ) H ( s )= R G +outR( sF )/ ( 1+sR s )=R GR+GR+sR =F RFGRCGFC F ( s ()= HH ( sFF))/+sR v+inR /(R1+sR RG1+sRC H ( s )=( R G+ G +sRC F C)F R R G F F F R +sR HH( (ss)= G F RG C F ) = R + R Multiplying the numerator 1 C G both F1+sR F F R G1F R G CAF 0 and 1G + R1F= f z= 1 R +sR denominator 1++CsR we get: +CπR F, +… πR R H (Rs2)= R=Gby + R F GF R F2R 1 FCC 1 G 2 F R3C +sR A +sR f z = P RRR = F GRF FCA 0 G F G F ( sF)= πR R+GRC0F +sR 2 πR FC C )= H ( s2H RFGRC +sR ( RX+GR+ R1+sR F ) /R F F FC +sR 1/ s C R ( R G F G F F 1 F) H ( s )= FG CFF Zf zF= = ( R + R1+sR == A C +sR F G F ) / RFG This form of equation allows RπR +FAthe X2CF +πR sFCCFF0)us to ( 1/ GC FC H ( s )2= P 2 R 2FA RF√Gand +RR 0+ 01 that we get F1+sRNote find the poles zeros. f = == FFCCFF A 0 AC02of +sR the DCz gain (the gain non-­ inverting 2+sR 2HπR R 1+ PF )R/ the F CF +GR R2√GπR CF = (ZRs√F)= F G 1+sR C amplifier without C0F+sR ) asFexpected with H ( s )= A C F F F F 1+sR C 2 H (( R s )G=+ R1+sR FF C FF s = 0. / 0R +sR F )A GC F CF 2 1+sR FR C = 0, at H ( sis)=a √pole PA 2= There atPR22G +FAsR 0++ 0CF G 1+sR C A − 2 R / 1+sR R A +sR F F ( 0 G F = F FF C F ) 0 R whichHs( s=H -R C 22 G terms cancel). (√s√FA )1+ =2(the )= 20the pole fre+ P P 0 Using s = jω = j2πf gives R √FAC F = A1+sR A G0C, 02+sR F C F f p A0 1 quency as f = 1 ÷ 2πR is the = ∙P 2F √ 2which f − 3 dB = Hp( s√)1+ = 2 C 2 2A 2+ 1+sR 2 0 2 π inverting R C same as the amplifier. R R +sR R C F F F AG0= + P√FA 0 −AF20 G√ AF 0 − 2 P H (iss )a=√zero There atA2R2AF= + sRFRGC = −2+2R2GG C 2R 20 0F√R 2 G +sR 1+ P √ P = 0, which gives √ Aus:0 + PA 2=− 2A 0 F R1 + R FP 2 0AA2 0√ 21 f p A 0 f − 3 dB = A0 2 ∙ =0 f z= = G√P1+ 2 22ππR R C= 20−πR 2 F C√fAp20A−02 2AA 1 FF R √ GC 0 2 f − 3 dB = ∙ A 0A−02 2 = 2 2 π RP = FC 2 √C A 20 − 2 √F )2A/−0R−2G +sR R G + RA ( F FA Practical Electronics 0|AJanuary f| p 2026 1 0 ( s )= f − 3H ∙1+sR2 0F C = dB = 2 π R F C √ AA0 − 2 F √fA 20A− 2 1 0 f = = p 0 ∙ i=C √ A +P = A 2 0 dv dt B (s) 628 √ ( − 628 ) =frequency Finally, f = the -3dB =100 Hz 2 dv 2 π we have to find To find the point, π )i=C √ ( 2-3dB dtthe magnitude of the frequency at which 7 ( 1/6.28 10 ) s gain the gain is 3dB1+ below the × passband H ( s )= B (s) (the DC gain in ( s1+ )=case). H this ( 1/628As) sdiscussed A ( s )the gain is relast month, this is when X 1/we sC need to duced by a factor of √2, so c )s ( s( dv )1/628 v out (the s )=frequency v1+ =) at which v in ( s ) in(f find H R+( sX)= R+1/ sC|H(s)| -3dB i=C c 1+ ( 1/ dt4712 ) s = A0 ÷ √2. The algebra to solve (and similar v out 2 ( s ) this ) 1 – there B=( s‘messy’ ) )=quite −)= 628 628 √ ( H s problems) could be ( H s f= =100 Hz v (= s )A )track of, and are many√terms 2 π( s1+sRC ( 2 π )2into keep it could take quite a few steps. The in1/628 s7 1 1+1(are 1 ) difficult 1 dividual H operations or × +… 10 ) s (= + not s )1+ = (+1/6.28 ( ) H s = R R R R ( ) 1+ 1/ 4712 s advanced P(add, 1subtract, and 2 3multiply 1+ ( 1/628 ) s square root to manipulate the equations), 2 ( ) − 628 628 1/sC s Cand R √ R X (1/ but itf is easy to make a mistake, F F ) not X F CF c =s )= =100 Hz = obvious v out ( Zs )= v v in ( s ) 2 in (= F= necessarily what the most 2 π + ( 1/ ssC C F effec√R(F2X+πcX) CF R FR+1/ ) R+ 2 tive approach is. 7 It is possible1+ manipulate the ( 1/6.28 ) s equa( s )R F × 10 vtoout 1 ( ) H s = tion in different to get to the result. )=F =ways H ( sZ = )s 1+ C Fthe ( s()1/628 v in1+sR F 1+sRC Of course, when you see finished working presented nicely in Xc 1/an sCarticle or R G1v+inR( F1s /)= v out ( s )you = 1don’t (if1+sR )v in ( s ) FC book, a1took aFcouple + + +… R+1/ sC )==X c know H ( sR+ of attempts Don’t R 1 a slick R 2 Rapproach. R R Pto get G 3 worry if you try similar calculations and v out ( s ) 1 things don’t go Rsmoothly. RXGtotally + R F +sR RsGC CFF) H ( sR)F= = F (F1/ CF ZH =( s )=to approach v in ( = s ) 1+sRC One this particular Fway Cs C F ) R + XR +sR R F +R(G1/ example is Fto seeCFGthat the DCFgain for1 1 1 1 mula can beR= found within the expression +F 1+… R G + R+ for H(s). the R=1 F R= Rnumerator R Pwe f z =If A and Z Fdivide 2 3 2 πRby C 0 1+sR CπR F RR GGC denominator , weF2get: F F R F ( 1/ s C F ) R X + R= / R G +sR Z F = (FR GGCF F FC F )/ ( 1+sR FC F )F R HH( (ss)= )=F + X CF R F + ( 1/ s C F ) 1+sR C R GF F RF A +sR C ZR F= G + R0F +sRFF RFG C F 1+sR F C F )=( s )= H ( sH R G1+sR +sR FFRCGFC F We can alsoRmake ( 1+sR F C Fsubsti) G 2+ R Fa2/ temporary R√GsR + A 01 P indicates AR +F P Hof( sP)= 0 tution = R C, where F G = f z= A R = 0 πR√F1+ R GPC2 2G√πR this is the2term associated 2with F C the pole and the parallel and capacitor. R +resistor R F +sR F RG C F We doHthis theF number ( s )=(just R GG2+toRreduce / 20R G +sR C F )A C F ItF also H ( s )we = have F R G out. P R=G +sR of terms to2 write 1+sR A 0equation − 2F C F comprishelps us see that the R G + RtoF the DC 1gain and the es terms f z =related = F C F fA 0A A 0 +sR 1sF)= p 0 2 πR R G∙this, C Awe 20 πRcan pole. Having F Cwrite the f − 3 dB = H ( done = 1+sR 2 F C=F A ÷ 2√2 as: -3dB gain2equation π R F C √|H(s)| A 0 − 2 √0 A 0 − 2 ( R G 2+ R F2) / R G +sR F C F H ( s )= √ A 0 + P A 1+sR = F C0 F 2 √ 1+ P √ 2 A +sR F C F We needHto P (make P ( s )solve = 0 this A 20 for 2 1+sR F C F This takes the subject of P the =equation). A 20both − 2 sides, cross a few steps: square 2 2 A 0cancel terms, + P multiply, then and √ Agather 0 = A 0 2 in Pf2 pare arranging so 1that all2 terms √ = A 0 on 1+ ∙P f − 3side. dB = We √ one 2 2 π Rget: C F √ A20 − 2 √ A 20 − 2 A P2= 2 0 A0 − 2 f p A0 A0 1 Practical f = Electronics∙ | January = | 2026 2 π RF C √ A −2 √ A −2 2 0 0 √ 1+ P2 √ 2 ( s )=of the amplifier. In H gain A0 is the DC A ( s )amplifier does comparison, the inverting not have a zero in transfer ( 1/628 ) s function 1+its ( s )= – its gainHdecreases continuously with ( 1/ 4712 )s 1+pole frequency past the frequency. − 3 dB 2 2 0 2 A 0 original terms 2 We can now P reinsert = 2 the in P, make the s = j2πf and A 0 −substitution 2 make f the subject: f − 3 dB = f A A0 1 = p2 0 ∙ 2 2 π RF C √ A0 − 2 √ A0 − 2 We could also substitute the resistors back into the A0 terms, but this form of the equation is more useful for understanding the circuit as it more directly shows the relationship between the -3dB point, the pole frequency and the DC circuit gain. What the formulae tell you When you obtain an equation for a circuit, you can look at the relationship between terms to gain insights into the characteristics of the circuit. Earlier, we noted that the inverting and non-inverting circuits differed in their pole frequencies. The inverting amplifier analysis was straightforward, and non-inverting case far less so. The + 1 term in the non-­ inverting amplifier gain (1 + RF ÷ RG) is responsible for this. A commonly used approach to using formulae to understand circuits is to look at what happens when key terms in the expression corresponding to characteristics of the circuit (or signal) become very large, or very small, or take on specific values that have a significant impact on the way the equation works. We are looking for situations where a formula can be simplified – where values calculated by the formula are dominated by some terms and others become almost irrelevant. We are also interested in situations where the equation can’t be solved or produces (or tends towards) specific values like one, zero or infinity. We will look at some situations where these things happen in the formula for the -3dB frequency of the non-inverting amplifier, together with numerical examples to help make sense of it . Later we will show some related simulation results. At high DC gains, the – 2 term in A02 – 2 has little influence on the value of the denominator; for example, for A0 = 50, the denominator is √502 – 2 = 49.98. If we ignore the 2, the denominator becomes √A2 = A = 50. The difference is small enough (compared to normal tolerances in component values etc) to justify simplifying the equation under these conditions. If the denominator is simplified to A0, it cancels with the A0 in the numerator and the formula for the cutoff frequency reduces to f-3dB = fp = 1 ÷ 2πRFC, which is the same as for the non-inverting amplifier. At lower values of A0, we cannot make this simplification; for example, for A0 = 3 we have 2.65, which gives f-3dB = 1.13fp and hence a 13% error in -3dB frequency if we assume the pole and cutoff frequency are the same. The term √A2 – 2 tells us that A02 = 2 is significant. At this point, the denominator becomes zero, so the -3dB frequency is at infinity. When A 02 is less than 2, we have the square root of a negative value. This formula is based on the gain magnitude, which is a real number; we are not in the realm of complex numbers, so the square root of a negative value indicates that a solution does not exist. This means that when A02 < 2, there is no -3dB frequency. At exactly A02 = 2, the gain is A0 = √2 = 1.414 (3dB). Remember that the minimum gain of the non-inverting amplifier is 1 (0dB), so at A0 = √2, the maximum gain drop (going from DC to high frequency) is from √2 to 1, which is a factor of exactly 1/√2 (0.7071), that is, exactly a 3dB drop. But that will only happen at an infinitely high frequency. For DC gains lower than A0 = √2, the drop in gain from A0 to 1 as frequency increases will be by a factor of less than 1/√2 – there is insufficient gain ‘headroom’ to accommodate a 3dB drop, so there is no -3dB frequency. For example, at a DC gain of A0 = 1.2 (1.58 dB), the gain can only drop by 1.58dB. At low gains, the circuit still has a pole and a zero at the frequencies given by expressions derived above. For example, for A 0 = 1.2, with R F = 200Ω, RG = 1kΩ and CF = 796nF, we have a pole at 1kHz and a zero at 1.2kHz. As the gain reduces, the pole and zero get closer together. 5-year collections 2019-2023 £49.95 2018-2022 £49.95 2017-2021 £49.95 2016-2020 £44.95 Purchase and download at: www.electronpublishing.com 47 Fig.11: an LTspice schematic simulating the non-inverting amplifier circuit with various feedback capacitor values. LTspice examples Fig.11 shows an LTspice schematic for simulating four copies of the inverting amplifier with different gains but the same pole frequency of fp = 1kHz. The circuits were configured by setting all the RG resistors (R1, R3, R5 and R7) to 1kΩ and choosing a set of DC gains (A0 values of ×1.5, ×2.2, ×3.3 and ×5.0) that are approximately evenly spaced in decibel terms, at +3.5dB, +6.9dB, +10.4dB and +14.0dB, respectively. The op amps are the idealised Universal­OpAmp1 model from LTspice library, with the open loop gain and gain-bandwidth product increased from the defaults to more ideal values. The values of the RF resistors (R2, R4, R6 and R8) were calculated using the standard non-inverting amplifier gain formula. The capacitor values were calculated by rearranging the pole frequency formula for the circuit (as discussed above) to make the C the subject (C = 1 ÷ 2πfpRF). The results are shown in Fig.12, where we see that as the DC gain is reduced, the -3dB point moves further from the pole frequency. The DC gains are approximately evenly spaced, but the shift in -3dB frequency gets larger as the DC gain decreases. This is consistent with the discussion on the formula, which showed the -3dB frequency moving towards infinity as the DC gain gets close to +3dB. The -3dB frequency values shown in Fig.12 were measured from the simulation. They are very close to the frequencies calculated using the formula derived above. The largest difference is 2.979kHz from the simulation versus the 3kHz calculated value (a 0.7% difference). This is due to numerical inaccuracies in the simulation, errors in measuring from the graph and possibly slight nonPE idealities of the op amps. Simulation files Most Circuit Surgery columns feature the use of the free circuit simulation software LTSpice. It is used to support descriptions and analysis in Circuit Surgery. Fig.12: the results from the simulation in Fig.11. 48 The examples and files for this issue are available for download: https://pemag.au/link/ac9b Practical Electronics | January | 2026