FREQUENCY RESPONSE &
RESONANT CIRCUITS
Filters, frequency response, time domain
connection, bode plots, resonant circuits.
OUTLINE AND TOPICS
Reading
1. Boylestad Ch 21.1-21.11
2. Boylestad Ch 20.1-20.8
Low-pass filters
 High-pass filters
 Other filters
 Resonance (Ch 20)
 Ideal op-amps and active filters
 Decibels & log scales
 Linear systems and transfer functions
 Bode plots

FILTERS
FILTERS
 Any
combination of passive (R, L, and C)
and/or active (transistors or operational
amplifiers) elements designed to select or
reject a band of frequencies is called a
filter.
 In communication systems, filters are
used to pass those frequencies containing
the desired information and to reject the
remaining frequencies.
FILTERS

In general, there are two classifications of filters:
Passive filters-gain always<1
 Active filters-gain can be >1

Gain=Vo/Vi
Can also be represented
By a differential eqn.
Vi
Circuit or
system
T(jω)=T(s)
Vo
Gain works for DC always
For AC, only for amplitude/phase of sine/cosine
FILTERS
Couple of definitions:
Impedance of a circuit element is the
AC analog of resistance. Only works
for sines and cosines!
Z = R + jX
Z-impedance
R-real part of impedance-resistance
X-imag part of impedance-REACTANCE
Resistance is always positive
Reactance can be positive or negative
Positive reactance-inductive
Negative reactance-capacitive
FIG. 21.7 Defining the
four broad categories of
FILTERS
Impedance of a circuit element is the
AC analog of resistance. Only works
for sines and cosines! WHY?
capacitor :
dv
dt
if input and output are sine or cosine,
we can represent them as
i=C
v = v0 e jwt
i = i0 e jwt
e jwt = cos(w t) + j sin(w t)
cos(Why?
w t) = 1 / 2[e jwt + e- jwt ]
sin(w t) = 1 / 2 j[e jwt - e- jwt ]
Then, the “resistance-like” ratio
of the amplitudes of current and
voltage
FIG. 21.7 Defining the
four broad categories of
ZC =
v0
1
1
=
= j()
i0 jwC
wC
R-C LOW-PASS FILTER
FIG. 21.8 Lowpass filter.
ZC = 1 / jwC = j(-1 / wC)
ZC (w = 0) = - j¥
ZC (w ® ¥) = 0
FIG. 21.9 R-C lowpass filter at low
frequencies.
R-C LOW-PASS FILTER
FIG. 21.10 R-C lowpass filter at high
frequencies.
Applying voltage divider
Vo
ZC
gain =
=
Vi ZC + R
ZC =
1
= jXC
jwC
Zc-impedance
Xc-reactance
FIG. 21.11 Vo versus frequency
for a low-pass R-C filter.
We must take the magnitude
To get the size of the gain. This
magnitude gives rise to the square
root we see in the textbook.
We take the “argument” or angle
To get the phase of the gain.
R-C LOW-PASS FILTER
Links to help with sketching sines and
cosines
http://www.youtube.com/watch?v=ijTIraykUk&feature=relmfu
http://www.youtube.com/watch?feature=e
ndscreen&v=RzZyyIu9IvA&NR=1
FIG. 21.12 Normalized plot of
Fig. 21.11.
Length is
amplitude
gain
At a given frequency, gain
is a complex number that
can be drawn as
Imaginary/”reactive”
θ-phase change
Real
R-C LOW-PASS FILTER-PHASE
FIG. 21.13 Angle by which Vo
leads Vi.
R-C LOW-PASS FILTER
FIG. 21.14 Angle by which
Vo lags Vi.
R-C LOW-PASS FILTER
FIG. 21.15 Low-pass RL filter.
FIG. 21.16
Example 21.5.
R-C LOW-PASS FILTER
Remember,
w (rad / s) = 2p f (Hz)
FIG. 21.17 Frequency response for the low-pass R-C
network in Fig. 21.16.
R-C LOW-PASS FILTER
FIG. 21.18 Normalized plot of
Fig. 21.17.
R-C HIGH-PASS FILTER
FIG. 21.19 Highpass filter.
R-C HIGH-PASS FILTER
FIG. 21.20 R-C high-pass
filter at very high frequencies.
Vo
R
gain =
=
Vi ZC + R
jw RC
=
jw RC +1
FIG. 21.21 R-C highpass filter at f = 0 Hz.
jw RC
w RC
| gain |=|
|=
jw RC +1
1+ (w RC)2
As w=0, |gain|=0
As winf, |gain|1
R-C HIGH-PASS FILTER
FIG. 21.22 Vo versus frequency for a highpass R-C filter.
R-C HIGH-PASS FILTER
FIG. 21.23 Normalized plot of
Fig. 21.22.
R-C HIGH-PASS FILTER
FIG. 21.24 Phase-angle response for the highpass R-C filter.
Same as Boylestad result
jw RC
Ðgain = Ð
jw RC +1
= Ðjw RC - Ð( jw RC +1)
= 90° - arctan(wRC)
R
1 / jwC + R
= ÐR - Ð(1 / jw C + R)
Ðgain = Ð
= 0° - arctan(-(1 / wC) / R)
= arctan(XC / R)
R-L HIGH-PASS FILTER
Z L = jw L
Vo
ZL
gain =
=
Vi Z L + R
jw L
jw (L / R)
=
=
jw L + R jw (L / R) +1
What is L/R?
w=0, |gain|=0
winf., |gain|1
FIG. 21.25 High-pass RL filter.
What kind of filter is this?
R-L HIGH-PASS FILTER
FIG. 21.26 Normalized plots for a low-pass and a high-pass filter
using the same elements.
R-L HIGH-PASS FILTER
FIG. 21.27 Phase plots for a low-pass and a high-pass filter using
the same elements.
PASS-BAND FILTERS
FIG. 21.28 Series resonant passband filter.
CR gives you high pass, LR gives you low pass.
RLC CIRCUITS-RESONANCE!
 The
resonant electrical circuit must have
both inductance and capacitance.
 In addition, resistance will always be
present due either to the lack of ideal
elements or to the control offered on the
shape of the resonance curve.
 When resonance occurs due to the
application of the proper frequency ( fr),
the energy absorbed by one reactive
element is the same as that released by
another reactive element within the
system.
SERIES RESONANT CIRCUIT
A resonant circuit (series or parallel) must have
an inductive and a capacitive element.
 A resistive element is always present due to the
internal resistance of the source (Rs), the
internal resistance of the inductor (Rl), and any
added resistance to control the shape of the
response curve (Rdesign).

SERIES RESONANT CIRCUIT
At resonance, ZC=-ZL, or jXC=j(-XL) reactance
Therefore, reactances cancel!
Only resistive components remain
C
XC = -1/ wC XC = X L
XL = w L
w r =1 / LC = 2p fr
FIG. 21.28 Series resonant passband filter.
L
fr, resonance
ω
PASS-BAND FILTERS
FIG. 21.29 Parallel resonant passband filter.
PASS-BAND FILTERS
FIG. 21.30 Series resonant pass-band filter
for Example 21.7.
PASS-BAND FILTERS
FIG. 21.31 Pass-band response for the
network.
PASS-BAND FILTERS
FIG. 21.32 Normalized plots for the pass-band
filter in Fig. 21.30.
SELECTIVITY
Q is the “sharpness” or
selectivity of the resonance
We define it as:
Q=
fr (location)
BW (width)
Q –Quality factor
Q high, narrow BW, sharp,
High quality resonance.
Q small, wide BW
Poor quality resonance
FIG. 20.15 Effect of R, L,
and C on the selectivity curve
for the series resonant
circuit.
SELECTIVITY
FIG. 20.16 Approximate series resonance
curve for Qs ≥ 10.
PASS-BAND FILTERS
T(s) = T( jw )
This is a common shorthand notation
T(s) is the ‘transfer’ function, or the ‘gain’ function
Describes how voltage is ‘transferred’ from in to out
T1(s)
FIG. 21.33 Passband filter.
T2(s)
Total Gain=T(s)=T1(s)xT2(s)
PASS-BAND FILTERS
FIG. 21.34 Pass-band
characteristics.
PASS-BAND FILTERS
FIG. 21.35 Passband filter.
FIG. 21.36 Pass-band
characteristics for the filter in
Fig. 21.35.
PASS-BAND FILTERS
FIG. 21.37 Network of Fig. 21.35 at f =
994.72 kHz.
BAND-REJECT FILTERS

Since the characteristics of a band-reject filter
(also called stop-band or notch filter) are the
inverse of the pattern obtained for the band-pass
filter, a band-reject filter can be designed by
simply applying Kirchhoff’s voltage law to each
circuit.
BAND-REJECT FILTERS
FIG. 21.38 Demonstrating how an applied signal of fixed
magnitude can be broken down into a pass-band and band-reject
response curve.
BAND-REJECT FILTERS
FIG. 21.39 Band-reject filter using a series
resonant circuit.
BAND-REJECT FILTERS
FIG. 21.40 Band-reject filter using a parallel
resonant network.
BAND-REJECT FILTERS
FIG. 21.41 Bandreject filter.
BAND-REJECT FILTERS
FIG. 21.42 Band-reject
characteristics.
OPERATIONAL AMPLIFIERS
Active filters
AMPLIFIERS GIVE GAIN
Simple amp-1 input and 1 output
Gain, A=Vout/Vin
EXAMPLE
If the amplifier above gives an output voltage of
1000V with an input voltage of 50V, what is the
gain?
IDEAL OPERATIONAL-AMPLIFIER
(OP
-AMP)
http://www.youtube.com/watch?v=TQB1VlLBgJE
Inputs draw no current-infinite input impedace
Vout=A(Vplus-Vminus) A-open loop gain.
A is ideally infinity-How is this useful?
Output can provide as much voltage/current as needed-zero output
impedance
NEGATIVE FEEDBACK
Negative feedback (NF) tries to reduce the
difference
with NF, Vplus=Vminus ALWAYS
summing point constraints
virtual ground.
INVERTING AMPLIFIER
Input goes into Vminus input-INVERTING input
Gain, Ainv=-R2/R1, gain is negative because
inverting
INVERTING AMPLIFIER
Vplus=Vminus
Inputs draw no current
NON-INVERTING AMPLIFIER
Input goes into Vplus input-NON-INVERTING
input
Gain, Ainv=1+R2/R1, gain is positive
UNITY GAIN BUFFER
Gain is 1 i.e. Vin=Vout
Used to isolate one side from the other
REAL OP-AMPS
http://www.national.com/mpf/LM/LM324.html#Overview
Single LM741
Quad LM324
Output voltage determined by rails (power supply)
Open loop gain not infinity
Inputs draw small amount of current-nA’s or less
BANDPASS
FILTER AMPLIFIER
f1=0.3Hz, f2=10Hz
High freq., cap is short, low freq., cap is open
FREQUENCY<F1
all caps are open.
What is the gain?
F1<FREQUENCY<F2
C1 is short. C2 is open.
What is the gain?-midband gain.
FREQUENCY>F2
All caps are shorts
What is the gain?
FILTER OP-AMP
What is T(s)?
sC1R2
T (s) = (sC2 R2 +1)(sC1R1 +1)
FILTER OP-AMP
zero at s=0
poles at 1/R1C1 and 1/R2C2
What happens at the zero? At the poles?
DECIBELS & BODE PLOTS
The key to amplifiers and control systems.
INTRODUCTION

The unit decibel (dB), defined by a logarithmic
expression, is used throughout the industry to
define levels of audio, voltage gain, energy, field
strength, and so on.
Vo
Po
gain(dB) = 20 log10 (| |) = 20 log10 (| T(s) |) =10 log10 ( )
Vi
Pi
Disambiguate now:
Vo/Vi-refers to |T(s)| i.e. ratio of amplitudes
T(s) is the full “gain” equation before taking the absolute value
INTRODUCTION
LOGARITHMS
 Basic


Relationships
Let us first examine the relationship between
the variables of the logarithmic function.
The mathematical expression:
INTRODUCTION
LOGARITHMS
 Some

Areas of Application
The following are some of the most common
applications of the logarithmic function:
1. The response of a system can be plotted for a range
of values that may otherwise be impossible or
unwieldy with a linear scale.
 2. Levels of power, voltage, and the like can be
compared without dealing with very large or very
small numbers that often cloud the true impact of the
difference in magnitudes.
 3. A number of systems respond to outside stimuli in
a nonlinear logarithmic manner.
 4. The response of a cascaded or compound system
can be rapidly determined using logarithms if the
gain of each stage is known on a logarithmic basis.

INTRODUCTION
LOGARITHMS
FIG. 21.1 Semilog
graph paper.
INTRODUCTION
LOGARITHMS
FIG. 21.2 Frequency
log scale.
INTRODUCTION
LOGARITHMS
FIG. 21.3 Finding a value
on a log plot.
FIG. 21.4
Example 21.1.
PROPERTIES OF LOGARITHMS

There are a few characteristics of logarithms that
should be emphasized:





The common or natural logarithm of the number 1
is 0
The log of any number less than 1 is a negative
number
The log of the product of two numbers is the sum of
the logs of the numbers
The log of the quotient of two numbers is the log of
the numerator minus the log of the denominator
The log of a number taken to a power is equal to
the product of the power and the log of the number
PROPERTIES OF LOGARITHMS
CALCULATOR FUNCTIONS
 Using
the TI-89 calculator, the common
logarithm of a number is determined by
first selecting the CATALOG key and
then scrolling to find the common
logarithm function.
 The time involved in scrolling through the
options can be reduced by first selecting
the key with the first letter of the desired
function—in this case, L, as shown below,
to find the common logarithm of the
number 80.
DECIBELS
Power Gain
 Voltage Gain
 Human Auditory Response

DECIBELS
TABLE 21.1
DECIBELS
TABLE 21.2 Typical sound levels and their
decibel levels.
DECIBELS
FIG. 21.5 LRAD (Long Range Acoustic
Device) 1000X. (Courtesy of the American
Technology Corporation.)
DECIBELS
INSTRUMENTATION
FIG. 21.6 Defining the relationship between a dB scale
referenced to 1 mW, 600Ω and a 3 V rms voltage scale.
LINEAR SYSTEMS
RLC circuits, op-amps are linear circuit elements
i.e. a differential equation can describe them.
 You can add solutions at a given ω



i.e. if exp(jωt) and exp(-jωt) are solutions,
exp(jωt)+exp(-jωt)=2cos(ωt) is a solution.
t
t
t
t
LINEAR SYSTEMS
Any voltage signal can be represented by a sum of sinusoidal voltage signalsFourier/Laplace theorems
If s=jω, the input voltage is represented by:
V0exp(jωt)= V0exp(st)
Allows us to use algebra instead of differential eqns.
RLC circuit, for example.
t
t
t
t
BODE PLOTS
There is a technique for sketching the frequency
response of such factors as filters, amplifiers, and
systems on a decibel scale that can save a great
deal of time and effort and provide an excellent
way to compare decibel levels at different
frequencies.
 The curves obtained for the magnitude
and/or phase angle versus frequency are
called Bode plots (Fig. 21.44). Through the
use of straight-line segments called
idealized Bode plots, the frequency response
of a system can be found efficiently and
accurately.

TRANSFER FUNCTION
Transfer function T(s), or H(s) describes how the
output is affected by the input.
 i.e. T(s)=Vo/Vi
 s=jω, so ZC=1/sC and ZL=sL
 The ‘s’ notation is convenient shorthand, but is
also important in the context of Laplace
Transforms, which you will see later in the class.
 Transfer because it describes how voltage is
“transferred” from the input to output.

LINEAR SYSTEMS
T(s) has zeros when T(s)=0
T(s) has poles when T(s)=infinity
POLES & ZEROS
Write T(s) in the form A(s)/B(s), where A and B
do not have any fractions in them. They should
look like factored polynomials.
 All transfer functions have poles and zeros.
 Zeros are when T(s)=0 i.e. A(s)=0
 Poles are when 1/T(s)=0 or T(s)=∞ i.e. B(s)=0
 These contribute very distinct behaviors to the
frequency response of a system. They tell us that
these are the critical frequencies in the system.

TRANSFER FUNCTION FOR LOW-PASS

Again we go to our good friend, the low-pass
filter.
0.16uF
FIG. 21.16
Example 21.5.

Now, we will redo this in the language of
“transfer function”
LOW PASS FILTER TRANSFER FUNCTION
Vo
ZC
T (s) =
=
Vi ZC + R
1 / sC
1
=
=
1 / sC + R 1+ sRC
0.16uF
Zeros: NONE
FIG. 21.16
3
Poles: s=-1/RC6.28x10 rad/s or 1kHz Example 21.5.
(We can ignore the minus signs)
Why?
T(s)1 as w0, so gain is 1 (0dB) and phase is 0
T(s)-j/wC as winf., so gain0 and phase-90deg
BODE PLOTS
There is a technique for sketching the frequency
response of such factors as filters, amplifiers, and
systems on a decibel scale that can save a great
deal of time and effort and provide an excellent
way to compare decibel levels at different
frequencies.
 The curves obtained for the magnitude
and/or phase angle versus frequency are
called Bode plots (Fig. 21.44). Through the
use of straight-line segments called
idealized Bode plots, the frequency response
of a system can be found efficiently and
accurately.

WHAT DOES A BODE PLOT TELL YOU?
Only works for sine/cosine
 If input is , cos(wt) output becomes


|T(s)|cos(wt+phase(T(s)))
Vi=Cos(wt)
Circuit or
system
T(jω)=T(s)
Vo=|T(s)|cos(wt+phase(T(s)))
BODE PLOTS
Once you have the transfer function, there is a
clear recipe that lets you sketch the frequency
response by hand very accurately.
 There is a surprisingly easy way to do this.







List the poles and zeros from low to high.
Go to w=0 and w=infinity to get start/end points.
At each pole/zero, apply the 20dB/dec add/subtract
At each pole/zero apply the 90deg phase shift
These contributions are additive.
Because of the simplicity, there are some errors,
but only small ones near the critical cutoff
frequencies.
LOW-PASS RESPONSE-MAGNITUDE |T(S)|
FIG. 21.52 Bode plot for the high-frequency region of a
low-pass R-C filter.
LOW PASS RESPONSE ANGLE(T(S))
MORE RULES

Phase starts at 0deg UNLESS
-sign, starts at +/-180 deg.
 Pole/zero at 0 (see next slide).


Gain starts with slope of 0dB/dec UNLESS
Pole/zero at 0 (see next slide)
 (Remember, the starting value of gain could be any
value and must be determined from the transfer
function)

SPECIAL CASE POLES AND ZEROS AT W=0

Zero
Pole
Zero@0rad/s
Gain Magnitude
+20dB/dec
-20dB/dec
+20dB/dec
Pole@0rad/s
-20dB/dec
Phase
+45deg at zero, +90 deg thru zero.
-45deg at pole, -90 deg thru pole.
Phase starts at +90 deg (in addition to
minus signs which give 180 deg)
Phase starts at -90 deg (in addition to
minus signs which give 180 deg)
We will now work some examples that have been
posted
‘REVERSE’ BODE PLOTS

If I give you the Bode plot, and tell you what the
input is, can you sketch the output?