University of California, San Diego
Spring 2016
J. Connelly
ECE 45 Discussion 1 Notes
Topics
• Steady State Analysis of RLC Circuits
• Impedance
Phasor Representation of Sinusoidal Functions:
• Phasors are used to represent sinusoidal functions and allow for easier representation of linear
(resistor, capacitor, and inductor) circuits with sinusoidal voltages and currents.
• Represent A cos(ωt + φ) as A ejφ .
• We do not include the frequency ω in the representation, but it is implicit.
• Differentiation and Integration:
f (t) = A cos(ω t + φ) ←→ A ejφ = F
df (t)
←→ jω F
dt
Z t
1
f (τ ) dτ ←→
F
jω
−∞
Note: We can only use phasor representation when our function (input to circuit) is sinusoidal!
Impedance:
When the inputs to our circuit are sinusoidal, we can represent resistors, capacitors, and inductors as
generalized components called impedances. This allow for simple Voltage/Current relationships.
vR (t) = iR (t) R
←→ VR = IR R
Z t
1
IC
vC (t) =
iC (τ )dτ ←→ VC =
C −∞
jωC
diL (t)
vL (t) = L
←→ VL = jω L IL
dt
We can lump together the terms to end up with a general expression: V = I Z so
ZR = R,
ZC =
Please report any typos/errors to j2connelly@uscd.edu
1
,
jωC
ZL = jωL.
Steady State Analysis:
Because the impedance equation (V = I Z) has the same structure as Ohm’s Law (v = i R), we can
use circuit analysis techniques from DC circuit analysis such as:
• Parallel/Series Combinations
• KCL and KVL Analysis
• Source Transformations
• Voltage/Current Dividers
• Thevenin and Norton Equivalence
Examples:
1. Represent the following sinusoidal function as phasors as a single complex number in rectangular
form:
2 cos(4π t + π/4) − 3 sin(4π t − π/3).
Both terms have ω = 4π so we can write the function in phasor form:
!
√
h √
i
√
j
3
j
1 √
1
−3 −
−
2 2+3 3+j 2 2+3
=
2 ejπ/4 − 3 e−j5π/6 = 2 √ + √
2
2
2
2
2
2. Assume ω = 2π and represent the following phasor in sinusoidal form:
=
√1
2
√1
2
+j
√1
2
e−jπ/6
1
e−jπ/6 .
1+j
√
√
1 e−jπ/6
2 −j5π/12
2
=
e
←→
cos(2πt − 5π/12)
=√
jπ/4
2
2
2 e
3. For what frequencies is the circuit component below purely resistive? (i.e. Zef f = X + j 0 = X)
What is the effective resistance at each frequency?
R1
R2
Z R1
L
Z R2
C
ZL
ZC
Where R1 =
Zef f (ω) = ZC //ZR1 + ZR2
√
8 Ω, R2 =
√
2 Ω, C = 1/4 F , L = 1 H
+ ZL = ... = R2 +
R1
(R1 Cω)2 + 1
+ j ωL −
R12 Cω
(R1 Cω)2 + 1
The impedance is purely resistive if the imaginary portion of Zef f (ω) equals 0:
1 2
R12 Cω
2
2
3 2
= ω R1 LC + ω(L − R1 C) = ω
ω −1
0 = ωL −
(R1 Cω)2 + 1
2
√
√
∴ ω0 = 0, ω1 = 2, ω2 = − 2
and so the effective resistance at each frequency is:
√
Zef f (0) = R2 + R1 = 3 2 Ω
√
Zef f (± 2) = R2 +
√
R1
=
2
2Ω
2R12 C 2 + 1
4. Find i1 (t)
R2
i(t)
C
R1
i1(t)
v(t) +−
L
v(t) = cos(2t) V
i(t) = sin(2t) A
R1 = 2 Ω
C = 1/2 F
L = 1/2 H
R2 = 1 Ω
Since the current and voltage sources are both of the same frequency, we can represent the voltages and currents as phasors and the resistors, capacitor, and inductor as impedances:
VA
ZR
2
ZR
1
I
ZC
I2
I1
V
ZL
+
−
V =1
I = −j
ZR1 = 2
ZC = −j
ZL = j
ZR2 = 1
By Ohm’s Law, we have:
VA
V − VA
, I2 =
ZR1 + ZL
ZR2 + ZC
and by KCL we have: I + I2 = I1 . By substituting in I1 and I2 expressions, we have
I1 =
I+
VA
V − VA
=
ZR2 + ZC
ZR1 + ZL
Solving for VA gives us:
I+
VA = (ZR2 + ZC ) (ZR1 + ZL )
V
ZR2 +ZC
ZR2 + ZC + ZL + ZR1
.
We can substitute this value of VA into our expression for I1 from Ohm’s Law:
I1 =
(ZR2 + ZC ) I + V
−j (1 − j) + 1
−j
1
=
=
= e−jπ/2
ZR2 + ZC + ZL + ZR1
1−j+j+2
3
3
converting back to the time domain gives us:
i1 (t) =
1
1
cos(2t − π/2) = sin(2t).
3
3
5. Let v(t) = 2 cos(3t + π/6) V , i(t) = cos(3t − π/6) A, R = 2 Ω, C = 1/6 F , and L = 1 H.
Determine an equivalent circuit as a voltage source in series with a resistor and an inductor.
L
R
Reff
L eff
+
v(t) +−
+
vth(t) +−
i(t) vo(t)
C
vo(t)
−
−
Since the voltage and current sources are both sinusoidal of the same frequency, we can represent
the circuit using phasors and impedances:
ZR
V
+
−
ZL
ZC I
+
V = 2 ejπ/6
Vo
−
I = e−jπ/6
ZR = 2
ZC = −2j
ZL = 3j
To solve for Zth , set voltage sources to 0V (short) and current sources to 0A (open).
ZR
ZL
ZC
Zth
−4j
2j
ZR ZC
= 3j +
= 3j −
Zth = ZR //ZC + ZL = ZL +
ZR + ZC
2 − 2j
1−j
1+j
1+j
= 3j −
−2 + 2j
2
= 1 + 2j
To solve for Vth , leave the output open and solve for Vth :
IR
V
IL
ZR
ZL
+
I C ZC
I
Vth
+
−
V = 2 ejπ/6
I = e−jπ/6
ZR = 2
ZC = −2j
ZL = 3j
−
By Ohm’s Law:
IR = (Vth − V )/ZR
IC = Vth /ZC
IL = 0
By KCL:
IR + IC + I + IL = 0
Substituting the expressions for IR , IC , and IL into the KCL equation yields:
Vth
Vth − V
+
+I =0
ZR
ZC
Solving for Vth gives us:
Vth =
=
V
−I
ZR
1
+ Z1C
ZR
V
−4j
=
− I (ZC //ZR ) = (ejπ/6 − e−jπ/6 )
ZR
2 − 2j
!!
√
√
√
3 j
3 j
+ −
−
(1 − j) = j (1 − j) = 1 + j = 2 ejπ/4
2
2
2
2
We could have instead solved for the short-circuit current Isc by connecting the terminals (shorting) at the output as follows:
In general, we only need to calculate two
out of the three of Vth , Zth , and Isc . The
R
L
third quantity follows from the equation
Vth = Isc Zth . In our case:
Z
V
+
−
Z
ZC
I
Isc
Isc =
Vth
1+j
(1 + j)(1 − 2j)
3−j
=
=
=
Zth
1 + 2j
5
5
We know Vth =
we have
√ jπ/4
2e
and Zth = 1 + 2j = Ref f + jωLef f , so converting to the time domain
Ref f = Re {Zth } = 1 Ω
Im{Zth }
= 2/3 H
ω
√
vth (t) = 2 cos(3t + π/4)
Lef f =
Alternatively, we could a series of use source transformations and series and parallel combinations to solve for Zth and Vth ,
ZR
V
+
−
ZC
ZL
+
I
Vo
−
Z L + (Z R// Z C)
+
(Z R// Z C) (V_
− I)
R
+
−
Vo
−
Zth = ZL + ZR //ZC = ... = 1 + 2j
Vth = (V /ZR − I) (ZR //ZC ) = ... =
√
2ejπ/4