{VERSION 2 3 "IBM INTEL NT" "2.3" } {USTYLETAB {CSTYLE "Maple Input" -1 0 "Courier" 0 1 255 0 0 1 0 1 0 0 1 0 0 0 0 }{PSTYLE "Normal" -1 0 1 {CSTYLE "" -1 -1 "Times" 1 12 0 0 0 1 2 2 2 0 0 0 0 0 0 }0 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }{PSTYLE "R3 Fo nt 0" -1 256 1 {CSTYLE "" -1 -1 "Helvetica" 1 12 255 0 0 1 2 1 2 0 0 0 0 0 0 }0 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }{PSTYLE "R3 Font 2" -1 257 1 {CSTYLE "" -1 -1 "Courier" 1 12 0 0 0 1 2 2 2 0 0 0 0 0 0 }0 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }} {SECT 0 {EXCHG {PARA 0 "" 0 "" {TEXT -1 79 "PHYSICS RESOURCE PACKETS P ROJECT File: FILTRC.MWS" }}{PARA 0 "" 0 "" {TEXT -1 110 "Rose-Hulman Institute of Technology \+ Authors: Perry Peters & Greg Williby" }}{PARA 0 " " 0 "" {TEXT -1 97 "http://www.rose-hulman.edu/~moloney \+ Software: Maple V Release 4" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 407 "R-C circuits are often u sed as simple frequency filters in electronic circuits. To see how thi s works, apply an input voltage of v(t) = A cos (wt) to a series resis tor-capacitor (the input voltage is applied at the resistor R and the \+ output voltage is taken at the point where the resistor R and capacito r C join together. One side of the capacitor is connected to the resis tor and other side is grounded.)." }}{PARA 0 "" 0 "" {TEXT -1 208 "V i s the voltage across the capacitor, and q is the charge magnitude on e ither plate, so q = VC, from the definition of capacitance. The time \+ derivative of this relation is : dq/dt = current = I = C dV/dt." }} {PARA 0 "" 0 "" {TEXT -1 80 "a) Show that the loop equation for volt age can be written v(t) = RC dV/dt + V." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 77 "b) Use dsolve to obtain a solution t o the second of these equations for V(t)." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 287 "c) Assume R= 10 k ohms, C = 0.12 microfarads, A = 0.350 V, and w=6328 rad/s. Plot the voltage across \+ the capacitor vs. time. If the voltage amplitude across the capacit or is close to A, then the circuit is 'passing' most of the input volt age through to the output at this frequency." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 153 "d) Plot the ratio of the ampli tude of the voltage across the capacitor to the input voltage amplitud e A from frequencies from 10 rad/sec to 2000 rad/sec." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 89 "e) Based on this plot, would you say this filter is a 'high-pass' or a 'low-pass' filter?" } }{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 8 "Solution " }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "restart;" }}}{EXCHG {PARA 0 "> \+ " 0 "" {MPLTEXT 1 0 12 "with(plots):" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 37 "Tell Maple that C and R are pos itive." }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 12 "assume(C>0);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 12 "assume(R>0);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 6 "Part A" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 43 "Current through the capacitor at the output" }} {PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "i:=C*diff(Vout(t),t);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 22 "Voltage acoss resistor" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 15 "eq1:=Vr(t)=i*R;" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 42 "Write Vr(t) in terms of Vin(t) and Vout(t)" }}{PARA 0 "> " 0 " " {MPLTEXT 1 0 36 "eq2:=subs(Vr(t)=Vin(t)-Vout(t),eq1);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 61 "Substitute the input voltages given in th e problem for Vin(t)" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 36 "DE:=subs(Vi n(t)=A*cos(omega*t),eq2);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 6 "Part \+ B" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 70 "Solv e the differential equations for Vout(t) using initial conditions." }} {PARA 0 "> " 0 "" {MPLTEXT 1 0 36 "sol:=dsolve(\{DE,Vout(0)=0\},Vout(t ));" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 51 "Set Vout_t equal to the ri ght side of the equation." }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 17 "Vout_t :=rhs(sol);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 6 "Part C" }}{PARA 0 " " 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 39 "Substitute in valu es for the constants." }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 67 "Vout_tsub: =subs(\{R=10000,C=0.12*10^(-6),A=0.35,omega=6238\},Vout_t);" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 53 "plot(Vout_tsub,t=0..0.005,ti tle=`Vout_tsub vs time`);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 6 "Part \+ D" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 61 "Vout_tsub2:=subs(\{R=10000,C=1 2/100*10^(-6),A=35/100\},Vout_t);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 68 "Find the steady state response by eliminating the exponential term s." }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 49 "Vss:=simplify(subs(exp(-2500/ 3*t)=0,Vout_tsub2));" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 26 "Max :=maximize(Vss,t,0..1);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 76 "The am plitude of Vss is the term in Max with the radical in the denominator. " }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 13 "Aout:=Max[2];" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 67 "plot(Aout/0.35,omega=0..2000,title=`Gain \+ of the circuit vs omega`);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 6 "Part F" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 45 "The circuit is acting like a low pass filter." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 38 "Now solve the problem in the S \+ domain." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 34 "Tell Maple that omega is positive." }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 16 "assume(omega>0);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }} {PARA 0 "" 0 "" {TEXT -1 33 "Use voltage division to find Vout" }} {PARA 0 "> " 0 "" {MPLTEXT 1 0 46 "VoutS:=(35/100)/(C*I*omega)/(R+1/(C *I*omega));" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 33 "Find the amplitude of the output." }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 18 "AoutS:=abs(VoutS );" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 83 "Find the amplitude of the o utput using the values for R and C given in the problem." }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 59 "AoutSsub:=simplify(subs(\{R=10000,C=12/100*10 ^(-6)\},AoutS));" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 67 "Compair this \+ answer to the solution from the differential equation." }}{PARA 0 "> \+ " 0 "" {MPLTEXT 1 0 14 "Aout-AoutSsub;" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 23 "Solutions are the same." }}}}{MARK "22 0 0" 0 }{VIEWOPTS 1 1 0 1 1 1803 }