47 4747 47392010-05-13 00:43:32 / rev b667c9e4c1f1+2.3 Low-pass filtersThe next example is an analysis that originated in the study of circuits(Section 2.3.1). After those ontological bonds are snipped – once thesubject is “considered independently of its original associations” – thecore idea (the abstraction) will be useful in understanding diverse naturalphenomena including temperature fluctuations (Section 2.3.2).2.3.1 RC circuitsRCVinVoutLinear circuits are composed of resistors, ca-pacitors, and inductors. Resistors are the onlytime-independent circuit element. To get time-dependent behavior – in other words, to get anyinteresting behavior – requires inductors or ca-pacitors. Here, as one of the simplest and mostwidely applicable circuits, we will analyze the behavior of an RC circuit.The input signal is the voltage V0, a function of time t. The input signalpasses through the RC system and produces the output signal V1(t). Thedifferential equation that describes the relation between V0and V1is(from 8.02)dV1dt+V1RC=V0RC. (2.11)This equation contains R and C only as the product RC. Therefore, itdoesn’t matter what R and C individually are; only their product RCmatters. Let’s make an abstraction and define a quantity τ as τ ≡ RC.This time constant has a physical meaning. To see what it is, give thesystem the simplest nontrivial input: V0, the input voltage, has been zerosince forever; it suddenly becomes a constant V at t = 0; and it remainsat that value forever (t > 0). What is the output voltage V1? Until t = 0,the output is also zero. By inspection, you can check that the solution fort ≥ 0 isV1= V1 −e−t/τ. (2.12)In other words, the output voltage exponentially approaches the inputvoltage. The rate of approach is determined by the time constant τ. Inparticular, after one time constant, the gap between the output and input48 4848 48402010-05-13 00:43:32 / rev b667c9e4c1f1+voltages shrinks by a factor of e. Alternatively, if the rate of approachremained its initial value, in one time constant the output would matchthe input (dotted line).tV0inputτThe actual inputs provided by the world are more complex than a stepfunction. But many interesting real-world inputs are oscillatory (and itturns out that any input can be constructed by adding oscillatory inputs).So let’s analyze the effect of an oscillatory input V0(t) = Aeiωt, where A isa (possibly complex) constant called the amplitude, and ω is the angularfrequency of the oscillations. That complex-exponential notation reallymeans that the voltage is the real part of Aeiωt, but the ‘real part’ notationgets distracting if it is repeated in every equation, so traditionally it isomitted.The RC system is linear – it is described by a linear differential equation– so the output will also oscillate with the same frequency ω. There-fore, write the output in the form Beiωt, where B is a (possibly complex)constant. Then substitute V0and V1into the differential equationdV1dt+V1RC=V0RC. (2.13)After removing a common factor of eiωt, the result isBiω +Bτ=Aτ, (2.14)orB =A1 + iωτ. (2.15)This equation – a so-called transfer function – contains many generalizablepoints. First, ωτ is a dimensionless quantity. Second, when ωτ is smalland is therefore negligible compared to the 1 in the denominator, thenB ≈ A. In other words, the output almost exactly tracks the input.Third, when ωτ is large, then the 1 in the denominator is negligible, so49 4949 49412010-05-13 00:43:32 / rev b667c9e4c1f1+B ≈Aiωτ. (2.16)In this limit, the output variation (the amplitude B) is shrunk by a factor ofωτ in comparison to the input variation (the amplitude A). Furthermore,because of the i in the denominator, the output oscillations are delayedby 90◦relative to the input oscillations (where 360◦is a full period).Why 90◦? In the complex plane, dividing by i is equivalent to rotatingclockwise by 90◦. As an example of this delay, if ωτ  1 and the inputvoltage oscillates with a period of 4 hr, then the output voltage peaksroughly 1 hr after the input peaks. Here is an example with ωτ = 4:tVinputIn summary, this circuit allows low-frequency inputs to pass through tothe output almost unchanged, and it attenuates high-frequency inputs.It is called a low-pass filter: It passes low frequencies and blocks highfrequencies. The idea of a low-pass filter, now that we have abstracted itaway from its origin in circuit analysis, has many applications.2.3.2 Temperature fluctuationsThe abstraction of a low-pass filter resulting from the solutions to the RCdifferential equation are transferable. The RC circuit is, it turns out, amodel for heat flow; therefore, heat flow, which is everywhere, can beunderstood by using low-pass filters. As an example, I often prepare acup of tea but forget to drink it while it is hot. Slowly it cools toward roomtemperature and therefore becomes undrinkable. If I neglect the cup forstill longer – often it spends the night in the microwave, where I forgot it– it warms and cools with the room (for example, it will cool at night asthe house cools). A simple model of its heating and cooling is that heatflows in and out through the walls of the mug: the so-called thermalresistance. The heat is stored in the water and mug, which form a heatreservoir: the so-called thermal capacitance. Resistance and capacitanceare transferable abstractions.50 5050 50422010-05-13 00:43:32 / rev b667c9e4c1f1+If Rtis the thermal resistance and Ctis the thermal capacitance, theirproduct RtCtis, by analogy with the RC circuit, a thermal time constant τ.To measure it, heat up a mug of tea and watch how the temperature fallstoward room temperature. The time for the temperature gap to fall by afactor of e is the time constant τ. In my extensive experience of neglectingcups of tea, in 0.5 hr an enjoyably hot cup of tea becomes lukewarm. Togive concrete temperatures to it, ‘enjoyably warm’ is perhaps 130◦F, roomtemperature is 70◦F, and lukewarm is perhaps 85◦F. The temperature gapbetween the tea and the room started at 60◦F and fell to 15◦F – a factorof 4 decrease. It might have required 0.3 hr to have fallen by a factor of e(roughly 2.72). This time is the time constant.How does the teacup respond to daily temperature variations? In thissystem, the input signal is the room’s temperature; it varies with a fre-quency of f = 1 day−1. The output signal is the tea’s temperature.