Sinusoidal Steady State Analysis
Huseyin Bilgekul
Eeng224 Circuit Theory II
Department of Electrical and Electronic Engineering
Eastern Mediterranean University
Chapter Objectives: Apply previously learn circuit techniques to sinusoidal steady-state
analysis.
Learn how to apply nodal and mesh analysis in the frequency domain.
Learn how to apply superposition, Thevenin’s and Norton’s theorems
in the frequency domain.
Transform a voltage source in series with an impedance to a current source in
parallel with an impedance for simplification or vice versa.
Source Transformation
Source Transformation
If we transform the current source to a voltage source, we obtain the circuit shown in Fig. (a).
Practice Problem 10.4: Calculate the current Io
Source Transformation Practice Problem 10.4: Calculate the current Io
Thevenin Equivalent Circuit Thévenin’s theorem, as stated for sinusoidal AC circuits, is changed only to
include the term impedance instead of resistance.
Any two-terminal linear ac network can be replaced with an equivalent circuit
consisting of a voltage source and an impedance in series. VTh is the Open circuit voltage between the terminals a-b.
ZTh is the impedance seen from the terminals when the independent sources are
set to zero.
Norton Equivalent Circuit The linear circuit is replaced by a current source in parallel with an impedance.
IN is the Short circuit current flowing between the terminals a-b when the
terminals are short circuited.
Thevenin and Norton equivalents are related by:
Th N N Th NV Z I Z Z
Thevenin Equivalent Circuit P.P.10.8 Thevenin Equivalent At terminals a-b
Thevenin Equivalent Circuit P.P.10.9 Thevenin and Norton Equivalent
for Circuits with Dependent Sources
To find Vth , consider the circuit in Fig. (a).
Thevenin Equivalent Circuit P.P.10.9 Thevenin and Norton Equivalent for Circuits with Dependent Sources
Thevenin Equivalent Circuit P.P.10.9 Thevenin and Norton Equivalent for Circuits with Dependent Sources
Thevenin Equivalent Circuit P.P.10.9 Thevenin and Norton Equivalent for Circuits with Dependent Sources
Since there is a dependent source, we can find the impedance by inserting a voltage source
and calculating the current supplied by the source from the terminals a-b.
OP Amp AC Circuits Practice Problem 10.11: Calculate vo and current io
OP Amp AC Circuits Practice Problem 10.11: Calculate vo and current io
Capacitance multiplier: The circuit acts as an equivalent capacitance Ceq
2
1
11i
i eq
i eq
V RZ C C
I j C R
OP Amp Capacitance Multiplier Circuit
( )1i o
i i o
V VI j C V V
j C
0 2
0
1 2 1
0 0ii
V V RV V
R R R
2 2
1 1
Substituting, (1 ) or (1 )ii i
i
IR RI j C V j C
R V R
Oscillators
An oscillator is a circuit that produces an AC waveform as output when
powered by a DC input (The OP AMP circuit needs DC to operate).
A circuit will oscillate if the following criteria (BARKHAUSEN) is satisfied.
The overall gain of the oscillator must be unity or greater.
The overall phase shift from the input to ouput and back to input must be
zero.
Oscillators An oscillator is a circuit that produces an AC waveform as output when powered by a
DC input (The OP AMP circuit needs DC to operate).
OUTPUT
+ INPUT
- INPUT
Phase shift circuit to
produce 180 degree
shift
Produce overall gain
greater than 1
Assignment to be Submitted
Construct the PSpice schemmatic of the oscillator shown Prob. 10.91 from the
textbook which is also shown above.
Display the oscilloscope AC waveforms of V2 and Vo to show the phase
relationship.
Submit the printout of your circuit schemmatic and the oscilloscope waveforms of
V2 and Vo as shown in the next page for a similar circuit.
Do you obtain the required phase shift and the oscillation frequency? If not it will
not oscillate to produce a pure sine wave.
Submission date 21 March 2007.
The analytic solution is given in the next page to help your simulation.
Vo V2
Assignment (Analytic Solution)
Chapter 10, Solution 91.
voltage at the noninverting terminal of the op amp 2V
oV output voltage of the op amp
110 p o sk R R j L
j C
Z Z
2 2
2
) )( ( 1
p o o
o s p o oo
R CR
j C R RR R j L
C
j LC
ZV V
V Z Z V
For this to be purely real,
2
-3 -9
1 1 11 0
2 2 (0.4 10 )(2 10180kHz Osc. Fr
)eq.o o oLC f
LC LC
o
o
oo
oo
o
2
RR
R
)RR(C
CR
V
VAt oscillation,
This must be compensated for by
2
8040
11 5 4
20k
5
o ov o
o
RR R
R R
VA
V
Similar Oscillator as the Assignment
OP Amp AC Circuits Practice Problem 10.11: Calculate vo and current io
The frequency domain equivalent circuit.
