ECE 201 Section 4 Lecture 6 Prof Peter Bermel January 23 2012 HW 2 Overview Returning today Grades were very strong Average 95 Standard deviation 6 Passive sign convention biggest problem 2A 1 23 2012 10 V ECE 201 4 Prof Bermel Recap from Friday Series resistors currents equal Parallel resistors voltages equal Series parallel circuits Analyzed iteratively 1 23 2012 ECE 201 4 Prof Bermel Dependent Sources Can use current or voltage to control output current or voltage Control type Voltage Output type 1 23 2012 Current Voltage Current VCVS V vx CCVS V IR VCCS I gV CCCS I Ix ECE 201 4 Prof Bermel Dependent Sources Example What is the output voltage and current gain and total power dissipated 4 I1 V1 20 V 12 I V1 5 1 23 2012 ECE 201 4 Prof Bermel 2 I2 4 Dependent Sources Solution 4 I1 V1 20 V 12 Io V1 5 2 I2 4 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 W 1 23 2012 ECE 201 4 Prof Bermel Voltage Sources Ideal voltage source V V Vs I Vs I Non ideal voltage source R 1 23 2012 V Vs IR V I Vs I Note R is generally small ECE 201 4 Prof Bermel Ideal Current Sources Ideal current source V Is I Is I Non ideal current source Is I Is V R R I 1 23 2012 V Note R is generally large ECE 201 4 Prof Bermel Nodal 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 nodes 1 23 2012 ECE 201 4 Prof Bermel Nodal Analysis Example For these 5 resistors with a voltage source solve for the voltages and currents everywhere G1 I1 A G4 B I3 I4 G3 C I2 G2 I5 G5 Io Io V a 1 23 2012 ECE 201 4 Prof Bermel Nodal Analysis Example G1 Using KCL Using Ohm s Law I1 A C I5 G5 Io Io ECE 201 4 Prof Bermel I4 G3 G2 1 23 2012 I3 I2 G4 B V a Nodal Analysis Example G1 Grouping by voltages I1 A 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 Bermel G4 B I3 I4 G3 C I2 G2 I5 G5 Io Io V a Nodal Analysis Example Rewriting as matrix Use matrix inversion formula 1 det Note det Now check 1 1 0 0 1 det 1 23 2012 ECE 201 4 Prof Bermel Nodal Analysis Example Using matrix inversion formula we obtain 1 If 0 2 0 4 0 5 0 1 0 7 5 then 1 1 6 0 5 0 2 2 524 5 2 039 1 03 0 5 0 8 0 4 And 0 495 1 185 0 243 0 252 1 428 Finally check KCL is obeyed 1 23 2012 ECE 201 4 Prof Bermel Nodal 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 techniques 1 23 2012 ECE 201 4 Prof Bermel Software Licensed software Download MATLAB from https engineering purdue edu PULS ptViewVers ion 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 Bermel Homework 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 63 1 23 2012 ECE 201 4 Prof Bermel
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