Circuit Simulation
10-11-2016, 04:32 AM (This post was last modified: 10-21-2016 08:20 PM by toshk.)
Post: #1
 toshk Member Posts: 195 Joined: Feb 2015
Circuit Simulation
eNet(M1)
M1 has 6 columns
M1=[[busfrom] [busto] [Resistor] [Independent Voltage] [Independent Current] [Opamp] ]
M1 for linedata: column1=busFrom, column2=busTo, column3=Z, column4=Voltage, column5=Curr.Source, column6=Oamp
BusFrom To BusTo will follow current path
Current flows through the -ve terminal of the battery

"\left(\begin{array}{cccccc}\n1 & 2 & 10 & 0 & 0 & 0 \\\n1 & 1 & 0 & 0 & 5 & 0 \\\n2 & 3 & 40 & 0 & 0 & 0 \\\n1 & 3 & 20 & 0 & 0 & 0 \\\n2 & 2 & 50 & 0 & 0 & 0 \\\n3 & 3 & 0 & 0 & 2 & 0\n\end{array}\right) "

node1 to node2 has 10_Ohms (or node2 to node1 has 10_Ohms )
node1 to node1 has 5_Amps (to node1 to ground, 5 because current flow into Node1)
node2 to node3 has 40_ohms
node1 to node3 has 20_Ohms
node2 to node2 has 50_Ohms (node2 to ground)
node3 to node3 has 2amps (to node3 to ground, 2 because current flow into Node3)

eNet(M1)
results:
"\left(\begin{array}{c}\n404.285714286 \\\n350 \\\n412.857142857\n\end{array}\right) "

interpretative of the the results: type: nkll

CAS programming you can enter variables for elements:
node1 to node2 has r1 (or node2 to node1 has 10_Ohms )
node1 to node1 has Ig1 (to node1 to ground, 5 because current flow into Node1)
node2 to node3 has r2
node1 to node3 has r3
node2 to node2 has r4 (node2 to ground)
node3 to node3 has Ig3 (to node3 to ground, 2 because current flow into Node3)

For Capacitor impedance: 1/s*c1 into resistor column3; (c1, c2.... not c) qualify the variable c
For Inductor impedance: s*l1 into resistor column3
Opamp: [1 2 0 0 0 4] (positive node1 to 1, -ve node2 to 2) 0 in column5 (for ideal Opamp) and node4 for the feedback path;

"\left(\begin{array}{cccccc}
1 & 1 & 0 & 0 & 4 & 0 \\
1 & 1 & 500 & 0 & 0 & 0 \\
1 & 2 & 450 & 0 & 0 & 0 \\
2 & 2 & 1500 & 0 & 0 & 0 \\
2 & 3 & 600 & 0 & 0 & 0 \\
3 & 3 & 550 & 0 & 0 & 0 \\
3 & 3 & 0 & 0 & 5 & 0
\end{array}\right) "

OR

"\left(\begin{array}{cccccc}
1 & 1 & 500 & 0 & 4 & 0 \\
1 & 2 & 450 & 0 & 0 & 0 \\
2 & 2 & 1500 & 0 & 0 & 0 \\
2 & 3 & 600 & 0 & 0 & 0 \\
3 & 3 & 550 & 0 & 0 & 0 \\
3 & 3 & 0 & 0 & 5 & 0
\end{array}\right) "

"\left(\begin{array}{cccccc}
1 & 2 & 4 & 0 & 0 & 0 \\
2 & 3 & 0 & 10 & 0 & 0 \\
1 & 3 & 8 & 0 & 0 & 0 \\
1 & 1 & 0 & 20 & 0 & 0 \\
3 & 3 & 12 & 0 & 0 & 0 \\
2 & 2 & 16 & 0 & 0 & 0
\end{array}\right) "

row2: can be written as
[3 2 0 -10 0 0 ]

"\left(\begin{array}{cccccc}
1 & 1 & 0 & 1 & 0 & 0 \\
1 & 2 & 6000 & 0 & 0 & 0 \\
2 & 2 & 9000 & 0 & 0 & 0 \\
2 & 3 & 18000 & 0 & 0 & 0 \\
3 & 4 & 54000 & 0 & 0 & 0 \\
4 & 4 & 45000 & 0 & 0 & 0 \\
3 & 3 & 0 & 0 & 0 & 4
\end{array}\right) "
last row: Opamp input

Attached File(s)
10-11-2016, 12:06 PM (This post was last modified: 10-11-2016 12:06 PM by compsystems.)
Post: #2
 compsystems Senior Member Posts: 1,374 Joined: Dec 2013
RE: Circuit Simulation

Code:
 #cas examples():= BEGIN .... END; #end
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