BIOEN 302, Section 3: AC electronics
For this section, you will need to have very present the basics of complex number
calculus (see Section 1 for a brief overview) and EE215’s section on phasors.
1. Representation of sinusoids in the complex plane
For a magnitude (say, a voltage) W(t) varying sinusoidally with time, we can write:
W(t) = M cos(ωt + θ)
hence
W(t) = Re [M ej(ωt + θ)]
where M ej(ωt + θ) = W(t) is the complex-plane representation of W(t). W(t) can be
thought of as a rotating vector the real part of which represents a sinusoid.
2. Representation of sinusoids as phasors
The phasor representation of a time-varying sinusoid v(t) = Vm cos(ωt + θ) is exactly the
same as its complex representation, most often used in its exponential or polar form (for
convenience).
The nomenclature is slightly different:
V ≡ Vm e j(ωt +θ)
V ≡ Vm ∠θ
where Vm is the maximum value of the sinusoid, ωt is suppressed and θ is the phase angle
of the sinusoid. Thus, the phasor is a “transformed” version of a sinusoidal voltage or
current waveform and consists of the magnitude and phase angle information of the
sinusoid.
Equivalently, v(t) is then referred to as the representation of the phasor in the time
domain:
Phase lead is expressed by a positive angle θ in the phasor notation, while phase lag is
expressed by a negative angle θ in the phasor notation. See the diagrams below.
v( t ) = Vm cos(ωt + θ )
v( t ) = Re[V m e j(ωt + θ )]
Important: As we shall see, the phasor concept may be used ONLY when:
1) the circuit is linear,
2) the steady-state response is sought
3) all independent sources are sinusoidal and have the same frequency.
BIOEN 302, 3: AC electronics
1
Instructor: Albert Folch
BIOEN 302, Section 3: AC electronics
3. Operations with phasors
a) Addition of phasors: just like addition of vectors! (Please do NOT remember by ear!)
V = V ∠θ
V = V ∠θ
V + V = Re[V ] + Re[V ] + j {Im[V ] + Im[V ]}
V + V = V cos θ + V cos θ + j (V sin θ + V
0
0
0
1
0
1
V 0 + V1 =
θ
2
1
0
1
0
1
0
[(V
1
0
0
1
1
1
0
0
1
sin θ 1)
] ∠θ
1
2 2
cos θ 0 + V 1 cos θ 1) + (V 0 sin θ 0 + V 1 sin θ 1)
0
2


+
sin
sin
= arctan  V 0 θ 0 V 1 θ 1 
V 0 cos θ 0 + V 1 cos θ 1 
b) Multiplication of phasors: just like exponentials (NOT vectors)!
V = V ∠θ
V = V ∠θ
V ⋅ V = V ⋅V ∠(θ + θ )
0
0
0
0
1
1
0
1
1
0
1
1
c) Division of phasors: just like exponentials (NOT vectors)!
V = V ∠( − )
θ θ
V V
0
0
1
1
0
1
Exercise 1:
Given V0=120∠30° and V1=120∠120°, give V0 + V1 and V0/V1
Answer:
V0 + V1=169.7∠75°
V0/V1=1∠-90°
Example 2:
Reduce the equation
v(t) = 10 cos(ωt) + 5 sin(ωt + 60o) + 5 cos(ωt + 90o)
to an equation of the form
v(t) = Vm cos(ωt + θ)
This operation is greatly simplified by using complex (phasor) notation. Converting all
sinusoids to cosines, we get
v(t) = 10 cos(ωt) + 5 cos(ωt + 60o – 90o) + 5 cos(ωt + 90o)
All cosines are in turn the real part of complex numbers:
o
o
v(t) = 10 Re[ejωt] + 5 Re[ej(ωt–30 )] + 5 Re[ej(ωt+90 )]
BIOEN 302, 3: AC electronics
2
Instructor: Albert Folch
2
BIOEN 302, Section 3: AC electronics
o
o
v(t) = Re[(10 + 5 e–j30 + 5 ej90 ) ejωt]
In polar form, we have
v(t) = Re[(10∠0o + 5∠– 30o + 5∠90o) ejωt]
Now, we can combine the complex numbers as
10∠0o + 5∠– 30o + 5∠90o = 10 + 4.33 – 2.5j + 5j = 14.33 + 2.5j
= 14.54∠9.9o = 14.54 ej9.90
o
Thus we obtain
o
o
v(t) = Re[(14.54 ej9.90 ) ejωt] = Re [14.54 ej(ωt+9.90 )] = 14.54 cos(ωt + 9.90o)
Answer: Vm = 14.54 and θ = 9.90o
Example 3:
Exercise 4:
Reduce the following expressions using phasors.
BIOEN 302, 3: AC electronics
3
Instructor: Albert Folch
BIOEN 302, Section 3: AC electronics
Exercise 5.
Answer:
Exercise 6.
Exercise 7.
Summarizing, when performing operations with phasors, rather than remembering the
formulas, think of phasors as if they were arrows described by a length and an angle
when you are multiplying or dividing, and think of them as arrows with two rectangular
coordinates when you are adding or subtracting – it’s as easy as that!
4. Complex impedance
a) Inductance
Suppose the sinusoidal current going through an inductance is given by
iL(t) = Imax sin(ωt+θ) ; in other words, iL(t) = Imax cos(ωt+θ-π/2)
In complex notation:
iL(t) = Re[Imax ej(ωt+θ-π/2)]
BIOEN 302, 3: AC electronics
4
Instructor: Albert Folch
BIOEN 302, Section 3: AC electronics
The voltage across an inductance is given by vL(t) = L diL(t)/dt, hence
vL(t) = Re[jωLImax ej(ωt+θ-π/2)] = Re[ωLImax ejπ/2 ej(ωt+θ-π/2)] = ωLImax Re[ej(ωt+θ)]
or
vL(t) = ωLImax cos(ωt+θ)
In phasor notation, we have that the phasors for current and voltage are
IL = Im ∠(θ – 90o)
VL = ωLIm ∠θ
[Discuss: is this equivalent to writing θ+90o for VL and θ for IL?] It is the current, and not
the voltage, that carries an extra 90o when put into phasor notation, because otherwise it
does not satisfy iL(t) = Re [IL]. We subtract 90o to keep the sign of the real part constant.
VL = (ωL∠90o) × (Im ∠(θ – 90o)) = (ωL∠90o) × IL
VL = jωL IL Æ Remember this !!!!
We refer to the term (ωL∠90o) = jωL ≡ ZL as the impedance of the inductance.
Summarizing,
Note: Current lags voltage by 90o in a pure inductance, as illustrated below
BIOEN 302, 3: AC electronics
5
Instructor: Albert Folch
BIOEN 302, Section 3: AC electronics
b) Capacitance
Suppose the voltage across a capacitor is given by
vC(t) = Vmax sin(ωt+θ) ; in other words, vC(t) = Vmax sin(ωt+θ-90o)
In complex notation:
vC(t) = Re[Vmax ej(ωt+θ-π/2)]
The current charging this capacitor is given by iC(t) = C dvC(t)/dt, hence
iC(t) = Re[jωCVmax ej(ωt+θ-π/2)] = ωCVmax Re [ejπ/2ej(ωt+θ-π/2)] = ωCVmax Re[ej(ωt+θ)]
or
iC(t) = ωCVmax cos(ωt+θ)
In phasor notation, we have that the phasors for current and voltage are
VC = Vmax ∠(θ – 90o)
IC = ωCVmax ∠θ
We refer to the term (1/(ωC))∠-90o = 1/jωC ≡ ZL as the impedance of the capacitance.
VC = IC / jωC
IC = jωC VC
Æ Remember this !!!!
o
ZC = 1/(jωC) = 1/(ωC) ∠ -90
Summarizing:
Note: Current leads voltage by 90o in a pure capacitance:
BIOEN 302, 3: AC electronics
6
Instructor: Albert Folch
BIOEN 302, Section 3: AC electronics
c) Resistance
v(t) = R i(t)
VR = R IR
IR = VR / R
ZR = R ∠ 0o
Æ You already knew this …
Summarizing:
Note: VR and IR are in phase in a pure resistance:
BIOEN 302, 3: AC electronics
7
Instructor: Albert Folch
BIOEN 302, Section 3: AC electronics
Overall summary:
1) Imprint in your memory that ZC = 1/jωC and ZL = jωL because you will be using
it a lot.
2) Appreciate the simplification provided by the phasor notation:
Element
Differential equation
Phasor notation
Impedance
L
v = L di/dt
V = jωL I
jωL
C
i = Cdv/dt
I = jωC V
1/jωC
In other words, except for the fact that we use complex arithmetic, sinusoidal
steady-state analysis of RLC circuits is virtually the same as the analysis of
resistive circuits – no RLC differential equations! (We pay a price: the phasor
notation applies only to steady-state sinusoidal signals.)
3) Rules of conversion from one notation to another:
W = A+jB Æ W = W∠θ where W = (A2+B2)½ and θ = arctan (B/A)
W = W∠θ Æ W = A+jB, where A = Wcosθ and B = Wsinθ
Exercise 8:
Exercise 9:
BIOEN 302, 3: AC electronics
8
Instructor: Albert Folch
BIOEN 302, Section 3: AC electronics
Exercise 10:
Exercise 11:
BIOEN 302, 3: AC electronics
9
Instructor: Albert Folch
BIOEN 302, Section 3: AC electronics
5. Circuits with multiple impedances
We now know:
1) The expressions for the impedances (i.e. phasor notations) of R, L, and C
2) The rules for adding R, L, and C in series vs. parallel in non-phasor notation
What about the rules for adding impedances?
Importantly, impedances add just like resistors (albeit in the complex plane): in series,
they sum to yield the total impedance; in parallel, their inverses (also called admittances,
denoted by Y) add to yield the inverse of the total impedance.
Note: In Z ≡ R + jX, R is called resistance and X reactance (both in ohms)
In Y ≡ 1/Z = G + jB, G is called conductance and B susceptance
(both in siemens)
Important: G ≠ 1/R !!!
It is straightforward to see that both Kirchhoff’s laws hold for phasors.
Example 12: A series RLC circuit
The exercise will consist of finding the
currents and voltages given the following
circuit:
The given expression for the source voltage
vs(t) tells us that the peak voltage is 100 V,
the angular frequency is 500, and the phase
angle is 30o. The phasor for the voltage
source is thus
Important: a disadvantage of the polar notation is that the angular frequency is implicit –
it’s up to you to remember its value!
The complex impedances of the inductance and the capacitance are, respectively,
The circuit in phasor/complex notation is shown below:
BIOEN 302, 3: AC electronics
10
Instructor: Albert Folch
BIOEN 302, Section 3: AC electronics
Note that all three elements are in series, therefore we can find the equivalent impedance
of the circuit by adding the impedances of the elements: Zeq = R + ZL + Zc. Substituting
values, we obtain Zeq = 100 + 150j – 50j = 100 + 100j. Converting to polar form,
Now we can find the phasor current by dividing the phasor voltage by the equivalent
impedance:
As a function of time, the current is:
Therefore, we can express the phasors for the voltage drops across R, L and C as:
As a function of time, the voltages are:
vR(t) = 70.7 cos (500t – 15o)
vL(t) = 106.1 cos (500t + 75o)
vC(t) = 35.4 cos (500t – 105o)
Exercise 13: Consider the circuit below:
BIOEN 302, 3: AC electronics
11
Instructor: Albert Folch
BIOEN 302, Section 3: AC electronics
i1(t)
i3(t)
i2(t)
Find the values and units of Vs, ZR, ZC, ZR, ZL1, and ZL2. Find an expression for the total
impedance, for vo(t), for the currents through each branch of the circuit (call them i1(t),
i2(t), and i3(t)), and for the voltages across each circuit element (call them vR(t), vC(t),
vL1(t), and vL2(t)).
Exercise 14. Suppose that the input signal for the circuit shown below is
Exercise 15.
BIOEN 302, 3: AC electronics
12
Instructor: Albert Folch
BIOEN 302, Section 3: AC electronics
Exercise 16.
Exercise 17.
BIOEN 302, 3: AC electronics
13
Instructor: Albert Folch
BIOEN 302, Section 3: AC electronics
Exercise 18.
Exercise 19.
BIOEN 302, 3: AC electronics
14
Instructor: Albert Folch
BIOEN 302, Section 3: AC electronics
6. Low-pass and high-pass filters
Until now we have only talked generically about how we can analyze circuits with a pure
(single frequency) sinusoidal input – so we didn’t have to worry about the response of the
system to other frequencies. Let’s take a closer look at the equations for impedance and
try to think about what will happen when the frequency of the input is varied.
L
Consider the following simple circuit:
I
R
V
Note that the current I and the voltage V1 are related as follows:
V
V
R − jω L
=V 2
I= =
2
Z R + jω L
R + (ωL )
I=
V
[R
2
]
1
2 2
+ (ωL )
ωL 

∠ − arctan

R 

Importantly, the function of the circuit depends on what part of the circuit we consider
the “output”:
•
If the output Vo ≡ VR = IR, then Vo varies with ω in the same way as I does.
VR =
RV
[R
2
]
1
2 2
+ (ωL )
ωL 

∠ − arctan

R 

If you plot the magnitude of the voltage gain, Gv ≡ Vo / V, you will see that this is
a “low-pass” filter:
As you can see, the response is far from “ideal”.
BIOEN 302, 3: AC electronics
15
Instructor: Albert Folch
BIOEN 302, Section 3: AC electronics
An approximate idea of what the filter response is like can be obtained by looking
at the limits when ω Æ 0 and Æ ∞:
V1
∠0 °
R
What happens when ω = R/L?
V
I = 1 ∠ − 45°
2R
ω
lim 0 I =
ω
lim ∞ I = 0∠ − 90°
The frequency ω = R/L is therefore referred to as the “half-power frequency” (or
also “break frequency” for reasons we will see later).
•
If the output Vo ≡ VL = IjωR, then I lags Vo by 90o.
VL =
jωL
V=
R + jωL
or also VL =
ω 2 L2 V
[R
1
1− j
R
ωL
2
+ (ωL )
V=
]
1
2 2
ωL 

∠ 90 o − arctan

R 

V
R 

∠ arctan

ωL 

 R 
1+ 

 ωL 
2
If you plot the magnitude, you will see that this is a “high-pass” filter:
Exercise 20. Repeat the previous exercise but with an RC circuit instead of RL. Consider
also both cases, Vo ≡ VC (left) and Vo ≡ VR (left).
BIOEN 302, 3: AC electronics
16
Instructor: Albert Folch
BIOEN 302, Section 3: AC electronics
What type of filter are these? Sketch a plot for the voltage gain Gv ≡ Vo / V1 in both
cases.
7. The Transfer Function
We have started talking about voltage gain, Gv ≡ Voutput / Vinput. The voltage gain is just
one of several “transfer functions”, which are defined as the ratio of the output phasor (of
interest) to the input phasor (defined by the source). They are expressed usually as a
function of frequency, H(f), where f = ω/2π is measured in Hertz.
Let’s find the transfer function of the following circuit:
ZT = R +
1
jω C
The phasor current is the input voltage Vin divided by the complex impedance ZT of the
circuit:
Since the output is defined across the capacitor, the phasor for the output voltage is the
product of the phasor current and the impedance of the capacitance, 1/jωC:
Substituting for I, we have
The transfer function is thus:
If is useful to define the parameter fB,
Then, the transfer function can be written as:
We see that the transfer function is a complex quantity having a magnitude and a phase:
BIOEN 302, 3: AC electronics
17
Instructor: Albert Folch
BIOEN 302, Section 3: AC electronics
Plots of the magnitude and phase are shown below:
This is a low-pass filter. Also, fB is the half-power frequency and the frequency at which
the output lag is half of 90o.
It is important to note that the low-pass filter built with R and L (VL as the output, see
above) would have the same transfer function, even though the definition of fB would be
different. This makes the concept of fB all the more useful – it characterizes the filter,
regardless of how it is physically built.
The same applies to the two high-pass filters: 1) series RL circuit (output taken across L),
and 2) series RC circuit (output taken across R).
8. Bode Plots
Most of the times, it is easier to visualize the ranges of H(f) and of f in a logarithmic
scale, in what is called a Bode plot.
In a Bode plot, we choose to represent 20 log H(f) as a function of log ω instead of
G(ω) as a function of ω. In this representation, a change in 20 dB (decibels) represents
a change in an order of magnitude in  H(f). A change in √2 represents 3 dB.
Let’s see an example, taking again the transfer function of the low-pass filter:
In decibels,
Thus,
Using the properties of the logarithm, we obtain
BIOEN 302, 3: AC electronics
18
Instructor: Albert Folch
BIOEN 302, Section 3: AC electronics
Since the logarithm of 1 is zero and the logarithm of x½ is ½ log(x),
The magnitude Bode plot for the first-order low-pass filter thus looks like:
Note the following:
•
For very small f, f << fB , H(f)dB ≈ 0 (a horizontal line)
•
For very large f, f >> fB , H(f)dB ≈ – 20 log (f / fB) (a straight line in
the log(f) scale that intersects the frequency axis at fB)
•
The high- and low-frequency asymptotes appear to intersect (or “break”)
at fB, hence the name “break frequency”.
A table of useful values for the f >> fB approximation is shown below:
Often in the context of decibels and Bode plots, you will hear the terms “decade” (a
frequency range spanning one order of magnitude, i.e. a factor of 10) and “octave”
(spanning a factor of 2). In the logarithmic scale:
BIOEN 302, 3: AC electronics
19
Instructor: Albert Folch
BIOEN 302, Section 3: AC electronics
An important feature of the Bode plots and their decibel representation is that a product
of two transfer functions “appears” as a sum of the two:
This is important when one tries to figure out the transfer function corresponding to a
given, more complex Bode plot with multiple breaks. For example, a band-pass filter can
be straightforwardly seen as the addition of the Bode plots of the high-pass and the lowpass filters, but in reality the two filters can be assembled in two stages (e.g. the input of
the high-pass stage is the output of the low-pass stage, so the total transfer function is the
product of the transfer functions of each stage).
Exercise 21.
Exercise 22.
Exercise 23.
Exercise 24.
BIOEN 302, 3: AC electronics
20
Instructor: Albert Folch
BIOEN 302, Section 3: AC electronics
Example 25.
Example 26.
BIOEN 302, 3: AC electronics
21
Instructor: Albert Folch
BIOEN 302, Section 3: AC electronics
BIOEN 302, 3: AC electronics
22
Instructor: Albert Folch
BIOEN 302, Section 3: AC electronics
Example 27.
BIOEN 302, 3: AC electronics
23
Instructor: Albert Folch
BIOEN 302, Section 3: AC electronics
9. Resonance in RLC circuits
Suppose we connect a series RLC circuit to a voltage supply as indicated. The impedance
of the circuit is:
BIOEN 302, 3: AC electronics
24
Instructor: Albert Folch
BIOEN 302, Section 3: AC electronics
Note we can also write
2
1 
1 
1 


 L
2
Z = R + j ωL −
 = R +  ωL −
 ∠ arctan ω −
ωC 
ωC 
 R ωCR 


which has a minimum in magnitude when ωL = 1/ωC, i.e. at ω0 = (LC)–½.
Most important, at this frequency the phase becomes zero. By definition, the frequency at
which the output (either current or voltage, or both) is in phase (“in resonance”) with the
input (a condition termed “resonance”) is called the resonant frequency, which is often
(but not always) characterized by yielding a maximal output. The circuits themselves are
said to be in resonance at this frequency. Remember this value because it will come back
again and again!
For a parallel RLC circuit we would write the total admittance (i.e. the inverse of the total
impedance) of the circuit as:
1
1
Y( jω) = + jωC +
R
jωL
where Y(jω) ≡ 1/Z(jω). It is straightforward to see that a plot of the magnitude of
Y(jω),Y(jω), for the parallel RLC circuit is essentially the same as the series-RLC
Z(jω) impedance plot, with a minimum at ω = ω0 = (LC)–½.
Summarizing for the series as well as parallel RLC circuits:
•
ω0 = (LC)–½
•
At resonance (by definition) Z(ω0) = R∠0o (no imaginary component, i.e. no
phase either)
•
At low frequencies,Zis dominated by the capacitive term for the series circuit
and by the inductive term for the parallel circuit
•
At high frequencies,Zis dominated by the inductive term for the series circuit
and by the capacitive term for the parallel circuit
As a general rule (for any circuit),
BIOEN 302, 3: AC electronics
25
Instructor: Albert Folch
BIOEN 302, Section 3: AC electronics
•
Resonance conditions are defined as those that satisfy Im(Z) or Im(Y) = 0, not
necessarily ω0 = (LC)–½
•
At resonance current and voltage are by definition in phase because the
impedance won’t change their phase. (Recall the phase is introduced by the
imaginary component.)
•
Example of a more general case: consider a circuit of the type RL//C (R in series
with L, C in parallel with them), then
1
R
ωL


Y = jωC +
= 2
+ j ωC − 2
2 2
2 2 
R + jωL R + ω L
R +ω L 

The frequency that satisfies Im(Y)=0 is
1 R2
−
LC L2
… which, incidentally, is not a maximum or minimum of Y. In other words, in
general the output is not necessarily maximum at resonance. “Resonance” is a
term more related with the phase of the output than with its magnitude.
ω0 =
Example 28. Consider the voltage drop
across the resistor in a series RLC circuit:
In phasor notation, the input voltage Vs
divided by the total impedance Z gives us
the current going through the circuit, I =
Vs/Z. The voltage drop across the resistor R
will thus be Vo = I × R = (R/Z) × Vs. Hence
we obtain:
[Discuss: what is the expression for the voltage across the inductance? And across the
capacitance?] Rearranging,
and substituting values,
The plots of the magnitude and phase of Vo look as follows. Note the resonance peak in
Vo at low frequencies.
BIOEN 302, 3: AC electronics
26
Instructor: Albert Folch
BIOEN 302, Section 3: AC electronics
Because of their steep slopes, magnitude and phase plots are usually shown with
frequency on a log axis:
Let’s take a closer look at the phases for each particular phasor in the circuit. In the case
of the series RLC circuit, we consider and plot the phasors Vc, VL and VR versus I
because I is what is conserved along the circuit. (Here V1 represents Vs in the above
circuit.) [Discuss the following plots.]
Note what happens at resonance: the voltage across the resistor and the current through
the resistor are in phase. That is precisely because Z(ω0) = R, hence there is no change in
phase when we compute I = Vs/Z, and both ZL = ZC = 0 at resonance. [Discuss the
implications on energy storage.]
BIOEN 302, 3: AC electronics
27
Instructor: Albert Folch
BIOEN 302, Section 3: AC electronics
10. Filters, Bandwith, and Quality factor
Let’s take a closer look at the expression for the gain in the series RLC circuit as defined
above (i.e. output voltage is defined as the voltage across the resistor):
Vo
=
Vs
R
1 

R + j ωL −

ωC 

=
R
1
=
 ω ωL
 ω ω0 
ω0 
 1 + jQ

−
−
R + j 0
ω
ω
ω
ω
ω
C
0
0


 0

Thus,
G≡
Vo
=
Vs
1
, with Q ≡
ω0 L
1
(recall ω0 = (LC)-½ )
=
R
ω0 RC
 ω ω0 

1 + jQ
−
 ω0 ω 
Writing the phasor G in terms of magnitude and phase, G = M(ω) ejφ(ω), we have:
1
M (ω) =
2
ω0 
2 ω

−
1 + Q 
 ω0 ω 
and
  ω ω0  
 
φ(ω) = − tan −1  Q
−

ω
ω
0



An approximate log graph of these two functions is
shown here:
We define bandwidth (BW in the graph) as the difference between the two half-power
frequencies or, equivalently (since power is the square of magnitude), as the range of
frequencies that satisfy M(ω) > 1/√2. Looking at the above relationship for M(ω), that
condition is satisfied when
2
 ω ω0 
 < 2
1 + Q 
−
 ω0 ω 
2
which in turn is satisfied when
BIOEN 302, 3: AC electronics
28
Instructor: Albert Folch
BIOEN 302, Section 3: AC electronics
 ω ω0 
 > 1
−
Q
 ω0 ω 
i.e. ω2 – ω0ω/Q – ω02 > 0
or when
or else
 ω ω0 
 < −1
−
Q
 ω0 ω 
ω2 + ω0ω/Q – ω02 < 0
Each equation has 2 roots, but only one is positive; discarding these “negative
frequencies” (no physical meaning), we only have one value per equation. We call the
lower value ωLO and the larger one ωHI. In other words, M(ω) > 1/√2 in the interval ωLO
< ω < ωHI, where
ω LO
 1
≡ ω 0 −
+
 2Q

2

 1 

 + 1

 2Q 

and
ω HI
 1
≡ ω0 
+
 2Q

2

 1 

 + 1

 2Q 

The difference between ωHI and ωLO gives us (by definition) the bandwidth:
BW ≡ ωHI – ωLO = ω0/Q
i.e. BW = R/L for the series RLC circuit.
The latter is generally used as the general definition of Q, i.e. Q ≡ ω0/BW.
Hence circuits with high Q have narrow bandwidths. In addition, ω02 = ωLO × ωHI, i.e. the
resonance frequency is the geometric mean of the two half-power frequencies.
Bode plots of the transfer function vs. frequency for various values of Q are shown
below. As we can see, the series RLC circuit (with output taken across the resistor) is a
(second-order) bandpass filter.
Exercise 29.
Show that for a series RLC circuit the ratio of impedance at resonance, Z0 ≡ Z(ω0), to
impedance at any radian frequency, Z, is:
Z0
Y
=
=
Z Y0
BIOEN 302, 3: AC electronics
29
1
 ω ω0 

1 + jQ
−
ω0 ω 
ω L
where Q ≡ 0
Instructor: Albert Folch
R
BIOEN 302, Section 3: AC electronics
Show also that for a parallel RLC circuit the expression is the same provided we invert
the definition of Q:
Vs
R
L
Z0 Y
=
=
Z Y0
1
 ω ω0 

1 + jQ
−
ω0 ω 
R
where Q ≡
ω0 L
C
Summarizing:
•
The expression of Q for a series RLC circuit is the inverse of the expression of Q
for a parallel RLC circuit. For example, Q decreases with increasing R in the
series circuit, but it increases with increasing R in the parallel circuit.
Exercise 30.
Exercise 31.
Exercise 32.
10. Other filters using RLC series circuits
We have only been considering the series RLC circuit with the output defined as the
voltage across the resistor. What would happen if we chose the voltage across the
capacitor VC (or across the inductance VL) as the output of the circuit, such as depicted
below?
Output ≡ Capacitor
BIOEN 302, 3: AC electronics
30
Instructor: Albert Folch
BIOEN 302, Section 3: AC electronics
We have:
I = Vs/Z
,
VL = jωLI
and
VC = I /(jωC)
where Z = R + jωL + 1/(jωC)
NOTE that the outputs VC and VL can NEVER be in phase with the input Vs, but the
current at the output (L, C or R) CAN be in phase with Vs. In such cases, we still speak of
resonance when the current at the output is in phase with the input voltage/current. In L
or C, we can’t have current and voltage in phase!
Thus,
VC =
1
jω C
R + jω L +
1
jωC
Vs =
Vs
1 − ω LC + jωCR
2
Note that I = IC is in phase with Vs for ω0 = (LC)-½ , but note also that at this frequency
the phase of VC is -90o (with respect to Vs).
The transfer function H(f) = Vout/Vin = Vc / Vs, can be written as:
with Qs = ω0L/R = 1/(ω0RC) and f0 = ω0/(2π) = (2π)-1 (LC)-½
The magnitude of the transfer function is plotted below. This is a second-order low-pass
filter circuit, similar to the RC (“first-order”) one, compared here side-by-side:
BIOEN 302, 3: AC electronics
31
Instructor: Albert Folch
BIOEN 302, Section 3: AC electronics
Output ≡ Inductor
When the output of a series RLC circuit is defined as the voltage across the inductor, we
have:
VL =
jω L
R + jωL +
1
jω C
Vs =
Vs
1−
1
R
−j
ωL
ω LC
2
The magnitude of the transfer function is plotted below:
where with Qs = ω0L/R = 1/(ω0RC) and f0 = ω0/(2π) = (2π)-1 (LC)-½
Therefore, this is a (second-order) high-pass filter.
Exercise 33.
Suppose we need a filter that passes components higher in frequency than 1 kHz and
rejects components lower than 1 kHz. As we know, the series-RCL second-order circuit
configuration constitutes a high-pass filter when the output is taken across the inductor.
Determine the values or R and C, keeping in mind that we are interested in a filter that
has a transfer function that is approximately constant in the pass band (see plot above).
Solution: C = 0.507 µF, R = 314.1 Ω.
Now, let’s see what happens at resonance at both L and C.
At resonance, since Z(ω0) = R, we have
I(ω0) = Vs/R and
VR = Vs
Does that mean there is no voltage drop across L or C? No!! We have
VL(ω0) = jω0L×I(ω0) and
BIOEN 302, 3: AC electronics
V(ω0) = I(ω0)/(jω0C)
32
Instructor: Albert Folch
BIOEN 302, Section 3: AC electronics
Substituting I(ω0), we get
VL(ω0) = jω0L×Vs/R = j Q Vs
and
VC(ω0) = Vs /(jω0CR) = – j Q Vs
In other words, at resonance the voltage across the capacitor and across the inductor have
the same magnitude, the gain is Q, and they are out of phase by 180o, i.e. they add to 0.
However, QVs is not the maximum value of the magnitudes of VL and VC. For example,
VC =
1
jω C
R + jω L +
1
jωC
Vs =
Vs
1 − ω LC + jωCR
2
It is straightforward to show that the frequency that satisfies
d VC
dω
= 0 is
2
ω max =
ω 02
1
1 R
1
2
−   = ω0 −
= ω0 1 −
≠ ω0
2
LC 2  L 
2Q
2Q 2
At this frequency, we have
 Vc 
Q


V  =
1
 s  max
1−
4Q 2
Although ωmax ≠ ω0, for large Q (Q>>1), ωmax ≈ ω0 and the gain at ωmax is Vc/Vs ≈ Q.
Exercise 34.
Derive the transfer function for a series RLC circuit whose output is taken across L and C
combined (a “band-reject” filter):
BIOEN 302, 3: AC electronics
33
Instructor: Albert Folch
BIOEN 302, Section 3: AC electronics
Note, once more, that our choice of “output” determines what type of filter we have:
Gv =
R
Gv = (see Exercise 35 just above)
1 

R + j ωL −

ωC 

11. Resonance in the parallel RLC circuit
From what you know of phasors and resonance in RLC series circuit (and some examples
with parallel circuits above), you should be able to comfortably analyze the parallel RLC
circuit. Let’s take a quick look at the expression for the gain G = Vout/Iin in the parallel
RLC circuit:
It is straightforward to show that the magnitude of the gain is:
Exercise 35.
Find the total impedance Z = Vs/I, the resonance
frequency ω0, and the currents through each element, i.e.
IR, IC, and IL in the parallel RLC circuit below.
Compare (in arrow plots) the magnitudes and phases
of the resonance values IR(ω0), IC(ω0), and IL(ω0).
BIOEN 302, 3: AC electronics
34
I
IR
Vs
R
IL
L
IC
C
Instructor: Albert Folch
BIOEN 302, Section 3: AC electronics
Hint: We can now plot the phasors Ic, IL and IG versus Vs (≡ V1 below) because Vs is
what is conserved along the circuit. Similarly to what we did for the series RLC circuit,
we have:
Looking at the magnitude of the gain:
We note first that the magnitude of the gain has a maximum at the resonance frequency,
ω0 = (LC) –½ , and that at this frequency Gmax = R. (Note that the same conclusion is
reached by observing that by definition Z at resonance must have Im(Z) = 0, and when Z
is maximum, Vout and G are maximum.) If we plotted this curve, we would see that it
also has one peak (it looks very similar to the one for the series RLC circuit), and that
therefore the parallel RLC circuit acts as a bandpass filter.
To find the bandwidth of the circuit, we find the ωHI and ωLO that satisfy that the
magnitude of the gain is reduced by √2 (i.e. the power is reduced by 2):
Solving the equation and taking only the positive values of ω we obtain:
Thus the bandwidth (recall that in the series RLC circuit BW = R/L) is:
BW ≡ ωHI – ωLO = 1/RC
and the quality factor Q is:
Q ≡ ω0/BW = RC/(LC)½ = R (L/C) ½
BIOEN 302, 3: AC electronics
35
Instructor: Albert Folch
BIOEN 302, Section 3: AC electronics
which we already knew from our previous study of the parallel RLC circuit. [To visualize
the meaning of the Q factor, recall from your homework the values for the currents
through the capacitor and inductance.]
Using this value of Q, we can re-write the expressions for ωHI and ωLO as:
ω LO
 1
≡ ω 0 −
+
 2Q

2

 1 

 + 1

 2Q 

and
ω HI
 1
≡ ω0 
+
 2Q

2

 1 

 + 1

 2Q 

… which are the same expressions as the ones for the series RLC circuit! Hence the
usefulness of the Q factor concept.
Exercise 36.
Consider the parallel RLC circuit, with L = 10 mH, C = 100 µF, and R = 1 kΩ.
Determine a) ω0, b) ωHI, c) ωLO, d) BW, and e) Q.
Answer: a) ω0 = 1000 rad/s, b) ωHI = 1005 rad/s, c) ωLO = 995 rad/s, d) BW = 10 rad/s,
and e) Q = 100.
BIOEN 302, 3: AC electronics
36
Instructor: Albert Folch
BIOEN 302, Section 3: AC electronics
Exercise 37.
Exercise 38.
BIOEN 302, 3: AC electronics
37
Instructor: Albert Folch