252 Practical MATLAB® Applications for Engineers
Solving for E 1 and E 2 by hand yieldsE 1
50 5
03
10 5 5
5315 5
60 50 2626 23 ∠
∠j
j
jj
jjj
j. ..1986
E 2
10 5 50 0
50
10 5 5
5325
60 50 4152j
j
jj
jjj
j ∠
. ∠85 236.
The voltages E 1 and E 2 are verifi ed by using MATLAB as follows:>>Y = [10 – 5j – 5j:5j j + 3]
Y =
10.0000 – 5.0000i 0 + 5.0000i
0 + 5.0000i 3.0000 + 1.0000i
>>l = [5:0]
l =
5
0
>>node _ voltage = inv(Y)*l
node _ voltage =
0.2414 + 0.1034i
0.0345 – 0.4138i
>>magnitude _ voltages = abs(node _ voltage)
magnitude _ voltages =
0.2626
0.4152
>>phase _ voltages = angle(node _ voltage)*180/pi
phase _ voltages =
23.1986
–85.2364The instantaneous voltages are then given bye 1 (t) = 0.2626 cos(10t + 23.1986°) Ve 2 (t) = 0.4152 cos(10t − 85.2364°) VR.3.59 The Thevenin’s and Norton’s theorems developed for DC discussed in Chapter 2
can be extended to include the AC case. Recall that the Thevenin’s theorem states
that any linear network with a load connected to terminals aa′ can be replaced by a
voltage source called the Thevenin’s voltage VTH in series with an impedance ZTH,
where the VTH is the open circuit voltage measured or calculated across terminals
aa′ (by removing the load), and ZTH is the impedance across terminals aa′ when all
sources are set to zero (with no load).
Recall that the Norton’s theorem states that any linear network can be replaced
by a current source IN, which is the short circuit current across aa′, and the
Thevenin’s impedance ZTH, connected in parallel.
R.3.60 Thevenin’s as well as Norton’s equivalent circuits can be evaluated if all the network
sources have the same frequency, and they are expressed as either sines or cosines.