ECE 201, Section 4Lecture 6Prof. Peter BermelJanuary 23, 2012HW #2 Overview• Returning today• Grades were very strong– Average: 95– Standard deviation: 6• Passive sign convention biggest problem1/23/2012 ECE 201-4, Prof. Bermel2 A10 V+-Recap from Friday• Series resistors: == / ; currents equal• Parallel resistors == /; voltages equal• Series-parallel circuits– Analyzed iteratively1/23/2012 ECE 201-4, Prof. BermelDependent Sources• Can use current or voltage to control output current or voltage1/23/2012 ECE 201-4, Prof. BermelVoltage CurrentVoltage VCVSV=µvxCCVSV=IRCurrent VCCS I=gVCCCSI=βIxControl typeOutput type+-+-Dependent Sources Example• What is the output voltage and current, gain, and total power dissipated?1/23/2012 ECE 201-4, Prof. BermelI=V1/54 Ω+-12 Ω+ -20 VV1I1I22 Ω4 ΩDependent Sources Solution• Voltage division V1=15V, Io=3A, I1=2A, I2=1A• Gain g=I2R2/(20V)=0.2 (a 7 dB attenuator)• Power dissipated=400/16+4*3=37 W1/23/2012 ECE 201-4, Prof. BermelIo=V1/54 Ω+-12 Ω+ -20 VV1I1I22 Ω4 ΩVoltage Sources• Ideal voltage source:• Non-ideal voltage source:1/23/2012 ECE 201-4, Prof. BermelVIV=VsVIV=Vs-IRVsRI+-+-VsI++--Note: R is generally smallIdeal Current Sources• Ideal current source• Non-ideal current source:1/23/2012 ECE 201-4, Prof. BermelVII=IsVII=Is-V/RIs+-RIs+-Note: R is generally largeNodal Analysis• General linear circuits aren’t simple combination of series and parallel circuits• Instead, must apply KCL and Ohm’s law to solve for voltage at all unknown nodes1/23/2012 ECE 201-4, Prof. BermelNodal Analysis Example• For these 5 resistors with a voltage source, solve for the voltages and currents everywhere:1/23/2012 ECE 201-4, Prof. BermelG4G2G3G1BCG5Va+ -I3I1I4I2I5IoAIoNodal Analysis Example• Using KCL:= += += += +• Using Ohm’s Law:−= −+= −+−−+(−) = +1/23/2012 ECE 201-4, Prof. BermelG4G2G3G1BCG5Va+ -I3I1I4I2I5IoAIoNodal Analysis Example• Grouping by voltages:++−= −+(++)= ++(+)= (+)• Third equation is sum of first two and can be eliminated; good cross-check• Leaves 2 equations for 2 unknowns, and 1/23/2012 ECE 201-4, Prof. BermelG4G2G3G1BCG5Va+ -I3I1I4I2I5IoAIoNodal Analysis Example• Rewriting as matrix:++−−++=• Use matrix inversion formula:= =1det() −− • Note det = −. Now check:=1det() −− =1 00 11/23/2012 ECE 201-4, Prof. BermelNodal Analysis Example• Using matrix inversion formula, we obtain:=1++++−++++• If %= (0.2,0.4,0.5,0.1,0.7), = 5, then:=11.031.6 0.50.5 0.80.20.45 =2.5242.039And %= (0.495,1.185,0.243,0.252,1.428)Finally, check KCL is obeyed1/23/2012 ECE 201-4, Prof. BermelNodal Analysis• Should always be able to solve problems with 2 unknowns using matrix inversion formula• What about more than 2 unknowns?– Adjoint method (calculate cofactor matrix, take transpose, divide by determinant)– Software techniques1/23/2012 ECE 201-4, Prof. BermelSoftware• Licensed software:– Download MATLAB from https://engineering.purdue.edu/PULS/ptViewVersion?version_id=18643• Only appears to work from on campus• You may need an ECN account first• Free software:– Download Octave from http://octave.sourceforge.net/1/23/2012 ECE 201-4, Prof. BermelHomework• HW #4 solution posting this morning• HW #5 due today by 4 pm at EE 325B• HW #6 due Wednesday: DeCarlo & Lin, Chapter 2:– Problem 39– Problem 46– Problem 631/23/2012 ECE 201-4, Prof.
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