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0 0 0 0 0 0 0 0 1 }{CSTYLE "" -1 344 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 1 }{CSTYLE "" -1 345 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 1 } {CSTYLE "" -1 346 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 1 }{CSTYLE "" -1 347 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 1 }{PSTYLE "Normal" -1 0 1 {CSTYLE "" -1 -1 "" 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 1 }0 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }} {SECT 0 {EXCHG {PARA 0 "" 0 "" {TEXT 256 34 "Signal generation by an R C circuit" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 50 "We consider the RC circuit with time constant tau=" }{TEXT 257 2 " RC" }{TEXT -1 83 " driven by a time-dependent external source signal o f triangular shape with period " }{TEXT 258 1 "T" }{TEXT -1 56 ". We w ish to calculate the voltage across the capacitor." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 189 "We begin with the proble m for a single cycle of the source voltage, and then generalize to obt ain the voltage for a finite number of cycles. We find the capacitor v oltage for the regime of " }{TEXT 283 1 "T" }{TEXT -1 90 " < tau, and \+ study the transition from the transient to the steady-state capacitor \+ voltage." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 31 "We proceed as in the worksheet " }{TEXT 19 11 "RLCcirc.mw s" }{TEXT -1 274 " to generate the time evolution of the current (or v oltage across the resistor) using Kirchhoff's law and the relationship s between voltage and current for the two elements. We need to modify \+ the rhs of Kirchhoff's loop rule to allow for a time-dependent electri c potential." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "restart;" }} }{EXCHG {PARA 0 "" 0 "" {TEXT -1 77 "We begin by defining the voltage \+ drops across the resistor and the capacitor." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 12 "V_R:=R*i(t);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 28 "V_C:=(1/C)*Int(i(s),s=0..t);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 68 "According to Kirchhoff's rule we equate the sum of the drops across " }{TEXT 325 1 "R" }{TEXT -1 5 " and " }{TEXT 324 1 "C" }{TEXT -1 46 " to the applied time-dependent signal voltage " }{TEXT 323 1 "U" }{TEXT -1 1 "(" }{TEXT 322 1 "t" }{TEXT -1 2 "):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 20 "KL_RC:=V_R+V_C=U(t);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 84 "We differentiate the relationship to obta in a differential equation for the current " }{TEXT 327 1 "i" }{TEXT -1 1 "(" }{TEXT 326 1 "t" }{TEXT -1 2 "):" }}}{EXCHG {PARA 0 "> " 0 " " {MPLTEXT 1 0 21 "KL_RC:=diff(KL_RC,t);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 52 "Note that in the case of a constant source voltage " } {TEXT 284 2 "U " }{TEXT -1 188 "the value of the applied voltage disap peared from the equation due to the differentiation. However, it ente red the solution through the initial condition. In the case of a time- dependent " }{TEXT 288 1 "U" }{TEXT -1 1 "(" }{TEXT 287 1 "t" }{TEXT -1 17 ") the derivative " }{TEXT 286 1 "U" }{TEXT -1 2 "'(" }{TEXT 285 1 "t" }{TEXT -1 47 ") appears in the differentiated Kirchhoff rule ." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 162 "We \+ need to define the external voltage as a function of time. We consider a triangle-shaped function. First we have a linear rise from zero to \+ a maximum value for" }}{PARA 0 "" 0 "" {TEXT -1 1 " " }{TEXT 328 2 "t \+ " }{TEXT -1 2 "< " }{TEXT 329 1 "T" }{TEXT -1 53 "/2 . We choose a slo pe of 1, i.e., the rise is 1 V/s:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 9 "U1:=t->t;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 41 "KL_RC1:=simplify(subs(U(t)=U1(t),KL_RC));" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 184 "Without simplification the rhs would not have been ev aluated. This would have caused dsolve to fail in the lines below. The initial current vanishes, as there is no voltage applied at " }{TEXT 330 1 "t" }{TEXT -1 90 "=0, and we assume that the capacitor is unchar ged. Therefore, the initial condition reads:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 18 "IC1:=i(0)=U1(0)/R;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 33 "sol_c:=dsolve(\{KL_RC1,IC1\},i(t));" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 18 "i_RCc:=rhs(sol_c);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 266 "We need to know the voltage across the capacitor in order to be able to start the solution for the 'discharge' phase. By \+ 'discharge' we mean the second half of the pulse during which the sour ce voltage decreases. We obtain the capacitor voltage as the complemen t of " }{TEXT 260 1 "U" }{TEXT -1 26 "_R to the battery voltage " } {TEXT 261 1 "U" }{TEXT -1 1 "(" }{TEXT 290 1 "t" }{TEXT -1 25 "). We a lso introduce tau=" }{TEXT 259 2 "RC" }{TEXT -1 35 ", the time constan t of the circuit." }}{PARA 0 "" 0 "" {TEXT -1 75 "The voltage drop acr oss the resistor during the 'charge' phase is given as:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 27 "U_Rc:=subs(i(t)=i_RCc,V_R);" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 35 "U_Rc:=simplify(subs(R=tau/C, U_Rc));" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 139 "The voltage across th e capacitor is obtained as the complement between the voltage applied \+ to the circuit and the drop across the resistor:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 17 "U_Cc:=U1(t)-U_Rc;" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 95 "This function describes how the voltage builds up across \+ the capacitor from a value of zero at " }{TEXT 262 1 "t" }{TEXT -1 98 "=0 up to asymptotic times. We explore its time behaviour for a linear ly increasing source voltage " }{TEXT 281 1 "U" }{TEXT -1 1 "(" } {TEXT 280 1 "t" }{TEXT -1 2 ")." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 56 "plot([U1(t),subs(tau=1,U_Cc)],t=0..10,color=[blue,red]);" }}} {EXCHG {PARA 0 "" 0 "" {TEXT -1 189 "We see that the capacitor voltage is non-linear for small times, but approaches a linear regime with a \+ time delay as compared to the source voltage. We can explore the short -time behaviour:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 19 "taylor( U_Cc,t=0,3);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 134 "The quadratic de pendence on time, as well as the dependence of the coefficient on the \+ time constant of the RC circuit should be noted." }}}{EXCHG {PARA 0 " " 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 " " {TEXT -1 17 "Now we introduce " }{TEXT 263 8 "T, i.e.," }{TEXT -1 71 " the period of the triangle-wave signal that drives the RC circuit . At " }{TEXT 264 1 "t" }{TEXT -1 1 "=" }{TEXT 265 1 "T" }{TEXT -1 60 "/2 the wave stops growing, and the discharge regime begins. " }} {PARA 0 "" 0 "" {TEXT -1 201 "The so-called duty cycle for the driving signal was chosen to be 50%. An interesting extension of this workshe et would consist of introducing the ratio of charge to discharge regim es during the period " }{TEXT 337 1 "T" }{TEXT -1 41 " as a variable ( e.g., charge regime: 0 < " }{TEXT 336 1 "t" }{TEXT -1 3 " < " }{TEXT 335 1 "T" }{TEXT -1 26 "/3, and discharge regime: " }{TEXT 334 1 "T" } {TEXT -1 5 "/3 < " }{TEXT 333 1 "t" }{TEXT -1 3 " < " }{TEXT 331 2 "T \+ " }{TEXT -1 35 "corresponding to a 1/3 duty cycle)." }}{PARA 0 "" 0 " " {TEXT -1 69 "We denote the voltage to which the capacitor was charge d at the time " }{TEXT 339 1 "t" }{TEXT -1 1 "=" }{TEXT 338 1 "T" } {TEXT -1 9 "/2 as U0:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 33 "U_ Th:=simplify(subs(t=T/2,U_Cc));" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 85 "We formulate the solution for the discharge regime. We note that w ith the choice for " }{TEXT 306 1 "U" }{TEXT 309 1 "1" }{TEXT -1 1 "( " }{TEXT 307 1 "t" }{TEXT -1 2 ")=" }{TEXT 308 1 "t" }{TEXT -1 1 "," } {TEXT 332 1 " " }{TEXT -1 39 "and a given choice of the period (e.g. \+ " }{TEXT 310 1 "T" }{TEXT -1 71 "=1) we imply a scale for the voltage \+ axis, since the source voltage at " }{TEXT 341 1 "t" }{TEXT -1 1 "=" } {TEXT 340 1 "T" }{TEXT -1 36 "/2 will rise to some maximum value (" } {TEXT 312 1 "U" }{TEXT -1 13 "_max=1/2 for " }{TEXT 311 1 "T" }{TEXT -1 28 "=1). The source voltage for " }{TEXT 292 1 "T" }{TEXT -1 5 "/2 \+ < " }{TEXT 291 1 "t" }{TEXT -1 3 " < " }{TEXT 289 1 "T" }{TEXT -1 13 " is given as:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 11 "U2:=t->T-t ;" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 45 "Kirchhoff's law is now formu lated as follows:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 41 "KL_RC2 :=simplify(subs(U(t)=U2(t),KL_RC));" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 102 "The initial condition on the current is given by the amount th at flows at the end of the charge cycle:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 40 "IC2:=i(T/2)=simplify(subs(t=T/2,i_RCc));" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 33 "sol_d:=dsolve(\{KL_RC2,IC2\},i(t)); " }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 44 "We combine the exponentials i n the solution:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 45 "i_RCd:=c ombine(subs(C=tau/R,rhs(sol_d)),exp);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 103 "The voltage drop across the resistor is calculated in or der to obtain the voltage across the capacitor:" }}}{EXCHG {PARA 0 "> \+ " 0 "" {MPLTEXT 1 0 37 "U_Rd:=simplify(subs(i(t)=i_RCd,V_R));" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 27 "U_Cd:=simplify(U2(t)-U_Rd); " }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 77 "The voltage at t=T will be ne eded to continue the solution in the next cycle." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 20 "U_T:=subs(t=T,U_Cd);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 82 "Now we have the voltage across the capacitor for the c harge regime valid up until " }{TEXT 269 1 "t" }{TEXT -1 1 "=" }{TEXT 268 1 "T" }{TEXT -1 62 "/2, as well as for the discharge regime voltag e valid between " }{TEXT 267 1 "T" }{TEXT -1 7 "/2 and " }{TEXT 266 1 "T" }{TEXT -1 69 ". We need to piece them together, and to display the m for one period " }{TEXT 270 1 "T" }{TEXT -1 86 " as a function of t ime with a choice of the parameter tau. The voltage scale for the " } {TEXT 271 1 "y" }{TEXT -1 16 " axis is set by " }{TEXT 300 1 "U" } {TEXT 301 1 "1" }{TEXT -1 1 "(" }{TEXT 305 1 "t" }{TEXT -1 6 ") and " }{TEXT 303 1 "U" }{TEXT 302 1 "2" }{TEXT -1 1 "(" }{TEXT 304 1 "t" } {TEXT -1 17 ") once we choose " }{TEXT 272 1 "T" }{TEXT -1 36 "=1 to s et a scale for the time axis." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {PARA 0 "" 0 "" {TEXT -1 148 "We reassign the previously defined volta ges across the capacitor at the end of the charge and discharge cycles to refer specifically to the case of " }{TEXT 342 1 "T" }{TEXT -1 3 " =1:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 31 "U_Cc:=simplify(subs( T=1,U_Cc));" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 31 "U_Cd:=simpli fy(subs(T=1,U_Cd));" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 127 "It is of \+ interest to observe the results of the charge&discharge cycles for the case where the time constant of the RC circuit " }{TEXT 19 3 "tau" } {TEXT -1 28 " is shorter than the period " }{TEXT 343 1 "T" }{TEXT -1 46 " of the driving signal. With the scale set by " }{TEXT 274 1 "T" } {TEXT -1 15 " we can choose " }{TEXT 19 3 "tau" }{TEXT -1 4 " << " } {TEXT 273 1 "T" }{TEXT -1 4 " as " }{TEXT 19 3 "tau" }{TEXT -1 5 " <<1 :" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 50 "P1c:=plot(subs(tau=1/1 0,U_Cc),t=0..1/2,color=red):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 51 "P1d:=plot(subs(tau=1/10,U_Cd),t=1/2..1,color=blue):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 12 "with(plots):" }}}{EXCHG {PARA 0 "> \+ " 0 "" {MPLTEXT 1 0 17 "display(P1c,P1d);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 193 "We see that the capacitor voltage nearly follows the sou rce voltage with some rounding off occuring. The voltage to which the \+ capacitor charges at the end of the rise in the source voltage is:" } }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 31 "evalf(subs(T=1,tau=1/10,U_ Th));" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 53 "The value at the end of \+ the cycle can be read off as:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 30 "evalf(subs(T=1,tau=1/10,U_T));" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 80 "For comparable time constants of the RC circuit and the square \+ wave signal, tau=" }{TEXT 313 2 "T " }{TEXT -1 10 "we obtain:" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 47 "P2c:=plot(subs(tau=1,U_Cc),t =0..1/2,color=red):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 48 "P2d: =plot(subs(tau=1,U_Cd),t=1/2..1,color=blue):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 17 "display(P2c,P2d);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 46 "The voltage to which the capacitor charges at " }{TEXT 293 1 "T" }{TEXT -1 15 "/2 is given as:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 28 "evalf(subs(T=1,tau=1,U_Th));" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 39 "while at the end of the period we have:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 27 "evalf(subs(T=1,tau=1,U_T));" }}} {EXCHG {PARA 0 "" 0 "" {TEXT -1 24 "We observe that as tau, " }{TEXT 295 4 "i.e." }{TEXT -1 90 ", the time constant of the RC circuit appro aches the time constant of the source voltage, " }{TEXT 296 1 "T" } {TEXT -1 324 " , the capacitor voltage does not follow the source volt age. In fact, the capacitor charged during the 'discharge' phase to a \+ higher voltage! Further below we will demonstrate that this is so only during the transient regime when the capacitor is charged to a low vo ltage as compared to the maximum value (set by a scale of " }{TEXT 314 1 "U" }{TEXT -1 33 "_max=1/2 in the source voltage " }{TEXT 298 1 "U" }{TEXT -1 1 "(" }{TEXT 297 1 "t" }{TEXT -1 3 "))." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 51 "Finally we g o to the other extreme given by tau >> " }{TEXT 299 1 "T" }{TEXT -1 1 ":" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 48 "P3c:=plot(subs(tau=10 ,U_Cc),t=0..1/2,color=red):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 49 "P3d:=plot(subs(tau=10,U_Cd),t=1/2..1,color=blue):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 17 "display(P3c,P3d);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 179 "The answer is clearly not periodic after a sin gle cycle, we would need to piece together several charge/discharge se gments to see whether a steady-state solution will be achieved." }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 195 " The signal is strongly attenuated. The RC circuit acts as a low-pass f ilter: the voltage across the capacitor reflects the triangular-wave i nput to the circuit for low frequencies (time constant " }{TEXT 275 1 "T" }{TEXT -1 233 " large compared to tau) with a rounding effect on t he edges of the pulse. It suppresses throughput at high frequencies wh ile distorting the shape considerably. The details of this behaviour a re shown in the solution for many periods " }{TEXT 294 1 "T" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 47 "The voltage to which the capacitor charges at " }{TEXT 316 1 "T" }{TEXT -1 7 "/2 and " }{TEXT 315 1 "T" }{TEXT -1 17 " respec tively is:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 29 "evalf(subs(T= 1,tau=10,U_Th));" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 28 "evalf(s ubs(T=1,tau=10,U_T));" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 4 "The " } {TEXT 282 19 "long-time behaviour" }{TEXT -1 25 " of the capacitor vol tage" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 447 " The solution to a first-order equation depends on a single integration constant. It is the actual value of the current at the end of the pr evious time segment that determines the further time evolution. The so lution to the problem is continuous and differentiable in this case as there are no discontinuities in the source voltage (the derivative of the source voltage is discontinuous). Given that the source voltage d oes not change abruptly at " }{TEXT 317 1 "t" }{TEXT -1 1 "=" }{TEXT 318 1 "T" }{TEXT -1 194 "/2 we should assume as the correct initial co ndition for the discharge cycle that the final value of the current in the charge cycle represents the correct initial value for the dischar ge cycle." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 80 "Now we are interested in piecing together a solution over several periods, e.g. " }{TEXT 19 7 "nmax=20" }{TEXT -1 9 " of them. " }}{PARA 0 "" 0 "" {TEXT -1 97 "We follow a simple strategy and let M aple do the work. We integrate the equation for the current " }{TEXT 320 1 "i" }{TEXT -1 1 "(" }{TEXT 319 1 "t" }{TEXT -1 183 ") as derived from the Kirchhoff rules for the charge and discharge regimes respect ively, and calculate the voltages as before. We pick the case where th e period of the driving signal " }{TEXT 344 1 "T" }{TEXT -1 48 " equal s the time constant tau of the RC circuit." }}}{EXCHG {PARA 0 "> " 0 " " {MPLTEXT 1 0 30 "i0:=0: T:=1: tau:=1; nmax:=20;" }}}{EXCHG {PARA 0 " > " 0 "" {MPLTEXT 1 0 24 "for j from 1 to nmax do:" }}{PARA 0 "> " 0 " " {MPLTEXT 1 0 12 "tj:=(j-1)*T;" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 37 " solc:=dsolve(\{KL_RC1,i(tj)=i0\},i(t));" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 31 "U_Rc:=subs(i(t)=rhs(solc),V_R);" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 35 "U_Rc:=simplify(subs(R=tau/C,U_Rc));" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 20 "U_Cc:=U1(t-tj)-U_Rc;" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 15 "U_Ccj[j]:=U_Cc;" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 46 "P_c[j]:=pl ot(U_Ccj[j],t=tj..tj+T/2,color=red):" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 44 "i0:=evalf(subs(t=tj+T/2,R=tau/C,rhs(solc)));" }}{PARA 0 "> " 0 " " {MPLTEXT 1 0 41 "sold:=dsolve(\{KL_RC2,i(tj+T/2)=i0\},i(t));" }} {PARA 0 "> " 0 "" {MPLTEXT 1 0 31 "U_Rd:=subs(i(t)=rhs(sold),V_R);" }} {PARA 0 "> " 0 "" {MPLTEXT 1 0 35 "U_Rd:=simplify(subs(R=tau/C,U_Rd)); " }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 20 "U_Cd:=U2(t-tj)-U_Rd;" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 15 "U_Cdj[j]:=U_Cd;" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 49 "P_d[j]:=plot(U_Cdj[j],t=tj+T/2..tj+T,color=blue):" }} {PARA 0 "> " 0 "" {MPLTEXT 1 0 42 "i0:=evalf(subs(t=tj+T,R=tau/C,rhs(s old)));" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 92 "print(`U_CTh: `,evalf(su bs(t=tj+T/2,U_Ccj[j]),3),` U_CT: `,evalf(subs(t=tj+T,U_Cdj[j]),3));" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 3 "od:" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 75 "We can display the sequences of graphs for the charge and discharge cycles:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 53 "displ ay(seq(P_c[j],j=1..nmax),seq(P_d[j],j=1..nmax));" }}}{EXCHG {PARA 0 " " 0 "" {TEXT -1 276 "We observe that after a few cycles of the square- wave signal a steady-state regime is reached if tau=1. For larger tau \+ it takes more cycles to reach the steady state. The steady-state regim e is reached as the oscillations become symmetric with respect to the \+ midpoint voltage " }{TEXT 276 1 "U" }{TEXT -1 5 "=1/4." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 195 "Various properties \+ of this solution can be exploited in practice: we mentioned the low-pa ss filtering property already; another property is the fact that under the right circumstances, i.e., for " }{TEXT 277 1 "T" }{TEXT -1 159 " << tau, we can consider the circuit to be an approximate integrator. \+ We learned that it turns a square-wave signal into an approximately tr iangular shape (in " }{TEXT 19 14 "RLCcircuit.mws" }{TEXT -1 324 "). T his was found to be correct in the sense that an exponential function \+ is approximated for small arguments by a straight line. Here we explor ed the response to a triangular-shape source signal and found that the response is a sequence of parabolic pieces that look alomost like a s ine function in the steady-state regime." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 311 "Thus, one can use this circuit in function generators to generate interesting shapes. One can feed the \+ near-sine wave signal coming from one RC circuit to drive a second one . To understand the response of the second RC circuit one then needs t o solve a differential equation with a time-dependent driving force " }{TEXT 279 1 "U" }{TEXT -1 1 "(" }{TEXT 278 1 "t" }{TEXT -1 128 "). Th is worksheet can be used for many further explorations in this respect , as the user has control over the input voltages U1(" }{TEXT 346 1 "t " }{TEXT -1 9 ") and U2(" }{TEXT 345 1 "t" }{TEXT -1 2 ")." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 68 "We explore the be haviour of the solution in the steady-state regime:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 10 "U_Ccj[20];" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 56 "P1:=plot(U_Ccj[20],t=19..19.5,color=green): display(P 1);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 214 "We calculate the precise \+ location of the minimum to be able to compare the solution with a sine shape. The sine needs to be phase shifted for comparison, and we can \+ use the location of the minimum for this purpose." }}}{EXCHG {PARA 0 " > " 0 "" {MPLTEXT 1 0 36 "t_min:=solve(diff(U_Ccj[20],t)=0,t);" }}} {EXCHG {PARA 0 "" 0 "" {TEXT -1 167 "We also need the amplitude precis ely. We assume (observe from the graph) that we have symmetrical oscil lations about U_avg=0.25, and obtain for the expected amplitude:" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 39 "A:=evalf(0.25-subs(t=t_min,U _Ccj[20]));" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 137 "The capacitor vol tage is nearly periodic in the steady-state regime. We can compare it \+ to a sine function with the expected period (2 Pi/" }{TEXT 321 1 "T" } {TEXT -1 57 "), the expected average voltage of 0.25 and an amplitude \+ " }{TEXT 347 1 "A" }{TEXT -1 62 " and phase shift (related to t_min) d erived from the solution:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 86 "P2:=plot(0.25-A*sin((2*Pi/T)*(t-t_min)+Pi/2),t=19..19.5,color=maro on): display(P1,P2);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 660 "This jus tifies the use of a single RC integrator circuit as an approximate sin e-wave generator. Better generators are based on the property that dif ferential equations can be modelled by coupling several elements with \+ feedback (two integrators can model a second-order differential equati on), ie., true sine solutions can be obtained in electronics. The inte resting message from this worksheet is (apart from explaining the imag es generated often on oscilloscopes and probably rarely understood) th at piecing together exponential solutions can result in a nearly perio dic response. This is the result of mathematics that leads to a limit \+ cycle in the solution." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 135 "On modern digital oscilloscopes it is relatively st raightforward to record the transient (turn-on) behaviour by single-sh ot triggering." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}} {MARK "0 0 0" 34 }{VIEWOPTS 1 1 0 1 1 1803 1 1 1 1 }{PAGENUMBERS 0 1 2 33 1 1 }