Kevin D. Donohue, University of Kentucky 1Voltage-Controlled Switches, and Designing ResponsesKevin D. Donohue, University of Kentucky 2Determine the pulse response for io(t) the circuit below with input vs(t) = 40v (t), where pulse duration T is 2 ms:Show:0.1 mF10 Ω40 Ωio(t)vs(t)()( )[]amps)m2())m20(125exp(8.08.0)()0125exp(8.08.0)(pulse0−−−−−−−= tuttutti[] [ ]amps)m2(1825.12)()1250exp(8.0)m2()(8.0)(pulse−−−−−−= tututtututio0 1 2 3 4 5 6 7 800.10.20.30.40.50.60.70.8Pulse ResponsemsmaKevin D. Donohue, University of Kentucky 3Do a SPICE Simulation to determine the pulse response for io(t) the previous circuit with input vs(t) = 40v(t), where pulse duration T is 2 ms:Ø To solve you must do a transient analysis, define the voltage source as piece-wise linear, and describe its transient properties. V0R 10 R0 40 C0.0001VAmtranex-Transient-14-TableTIME I(VAM)(s) (Amp)+0.000e+000 +0.000e+000+10.000n +99.999n+10.840n +109.104n::::::+7.285m +972.534u+7.445m +795.710u+7.605m +651.035u+7.765m +532.665u+7.925m +435.817u+8.000m +396.630uI(VAM)Time (s)tranex-Transient-14(Amp)+0.000e+000+500.000m+0.000e+000 +2.000m +4.000m +6.000m +8.000mKevin D. Donohue, University of Kentucky 4Use voltage controlled switches in SPICE to determine the pulse response for io(t) in the previous circuit with input vs(t) = 40v(t), where pulse duration T is 2 ms:Ø To solve you must set up separate circuits for the controlling voltages defined as piece-wise linear voltage to achieve switching action.V40R040C0.0001R 10 VAmSwitchV1SwitchV0V1 for SwitchV1Time Voltage01.00001 -1.00199 -1.00201 1.008 1V0 for SwitchV0Time Voltage0-1.00001 1.00199 1.00201 -1.008 -1IVmV00 V10IVm0For switch settings, you should give each a unique name and indicate thecontrolling source (V1 or V0) or avoltmeter name. You can also modifythe closed and open resistances of theswitch, which are default 1 ohm and1G ohm, respectively. When controlvoltage is less than 0, switch is closed.When control voltage is greater than 0,switch is open.Kevin D. Donohue, University of Kentucky 5Compare the results from response using a pulsed voltage source to that of the result using switches (Closed resistance = 1 Ohm, Open resistance = 1G Ohm:I(VAM)Time (s)tranexsw-Transient-17(Amp)+0.000e+000+200.000m+400.000m+600.000m+0.000e+000 +2.000m +4.000m +6.000m +8.000mTime (s)tranex-Transient-14(Amp)+0.000e+000+500.000m+0.000e+000 +2.000m +4.000m +6.000m +8.000m735mAPulse SourceExample706mASwitch ExampleKevin D. Donohue, University of Kentucky 6The circuit below represents a channel for sending digital data in the form of unit pulses transmitted from vp. A pulse with amplitude -1 V is a binary 0 and a 1 V pulse is a binary 1. The receiver has an infinite input impedance and is connected at vr. Find R0to make the system critically damped. For the critically damped case using a detector that requires the voltage to be greater than 0.95 volts for a binary 1 (below -0.95 volts for binary 0) for at least 50% of the pulse duration, what is the smallest pulse duration that should be transmitted over this channel? R0R=2Ω L=10µHC=.1µF+vr-vpKevin D. Donohue, University of Kentucky 7Characteristic Equation for System:0102=+÷øöçèæ++LCsLRRsRoots:242002,1LCLRRLRRs−÷øöçèæ+±÷øöçèæ+−=Ω−Ω=∴ 22,180RWhile a negative resistance can be achieve with electronics and a power supply, it will be more expensive than a simple positive resistance so chooseΩ=180RKevin D. Donohue, University of Kentucky 8For R0= 18Ω, solution becomes:ïîïíì−+−+−−+−+=Interval 0Binary For )10exp()10exp(1Interval 1Binary For )10exp()10exp(162616261ttBtBttAtAvrAssume a worse case transition of -1 V to 1 V, where steady-state had effectively been reached before the transition. Therefore initial conditions become vc(0+) = -1 and iL(0+) = 0: )10exp()10(2)10exp(21666tttvr−−−−=Kevin D. Donohue, University of Kentucky 9Can write a Matlab program to plot result and numerically find pulse width:Since waveform must be above.95 volts for 50% of the pulseduration, the smallest pulse thatshould be used is 11.14 µs.This would correspond to a bitrate of 8.97 kbits/sec. Thiscould carry 0.064 channels ofstereo CD quality audio, or1.4 telephone quality voice signals. Plot of vrin -1 to 1 transition0 1 2 3 4 5 6 7 8 9 10-1-0.8-0.6-0.4-0.200.20.40.60.81microsecondsvoltsvr exceeds .95 Vat t=5.57 microsecondsKevin D. Donohue, University of Kentucky 10% This program will find the time point where the waveform vr = .95% Since time constant is 1/1e6 = 1 microsecond, create a time axis forabout 10 time constants:t = 10*(1/1e6)*[0:9999]/10000; % Create 10000 points over 10 time constantsvr = 1 - 2*exp(-1e6*t) - 2e6*t.*exp(-1e6*t); % Create function pointsA Matlab program is referred to as a script (type “help script” at the Matlab prompt for more information). This is a text file that ends with a *.m extension (also called an mfile). Program comments are preceded by the percent symbol %.Kevin D. Donohue, University of Kentucky 11% Plot waveform, divide axis by 1e-6 to put units in microsecondsplot(t/1e-6,vr)xlabel('microseconds') % Label x-axisylabel('volts') % Label y-axis% Plot and plot a dashed line at .95 voltshold on % Keep current plot in figure box and plot over it with …plot(t/1e-6,.95*ones(size(t)), 'k--')% in above, k means black, and -– means broken line (see help plot)% Loop to find points where vr exceed .95:k=1; % Initialize array indexkend = length(t); % Find total number of points in array% While loop, increment k until vr equals or exceeds .95while (vr(k)<.95)k=k+1;endKevin D. Donohue, University of Kentucky 12tint = t(k) % Find time point from indexplot([tint, tint]/1e-6, [-1 1], 'k--') % Plot vertical linehold