ECE 3336
Introduction to Circuits & Electronics
Note Set #12
Frequency Response
More About Filters
Spring 2013,
TUE&TH 5:30-7:00 pm
Dr. Wanda Wosik
1
Outline
Frequency Response and System Concept
We will cover the following topics:
• Second-Order Filters
• Transfer Function Calculation
• Bode Plots for Those Filters
2
Bode Plots
We will repeat what we know about Bode plots from the first order filters:
Bode Plots will be used to simplify graphical representation of the
magnitude and phase of the transfer functions. We will use them instead of
plots such as shown in the figure below for high-pass filters and obtained
from calculation of |H(j)| and H(j).
0.707
Phase
45°
Magnitude
0
0
Bode Plots will give us approximation of the transfer function in the whole
range of frequency changes. Accuracy will be acceptably even at the
breakpoint frequencies.
Bode Plots will give us straight line approximations both for the magnitude
and phase of the Transfer Function. That will allow us to easily predict its
frequency dependence.
3
Derivation of Bode Plots
Now we will use approximations for the magnitude and the phase
instead of calculating step by step the transfer characteristics. They
will be accurate!
If we use decibels (dB) defined as
A0
Ai
= 20log10
dB
A0
Ai
We will find the magnitude of the transfer function
H( jw ) =
1
1
=
Ð - tan-1(w /w 0 )
1+ jwRC
1+ (w /w 0 ) 2
which will be expressed in decibels |H(j)dB|
H ( jw ) dB =
V0 ( jw )
1
= 20 log10
Vi ( jw ) dB
1+ (w / w 0 )2
4
Plotting Bode Plots of the Transfer Function
H( jw ) =
1
1
=
Ð - tan-1(w /w 0 )
1+ jwRC
1+ (w /w 0 ) 2
We will plot the MAGNITUDE at selected multiples of 0. Then we will plot the PHASE
For <<0
At =0
|H(j)|dB=-20log10(1)=0 dB
|H(j)|dB=-20log10√(1+1)=-3 dB
w
)
w0
H ( jw ) dB = -20 log10 (
For >> 0
=100
=1000
=10000
|H(j)|dB=-20log(10)=-20 dB
|H(j)|dB=-20log(100)=-40 dB
The 0 affects
the phase only
locally: within
two decades
only
V0 ( jw )
= -tan-1 (w /w 0 )
Vi ( jw )
=0/10
=0
H(j-tan-1(1/10)=0°
H(j-tan-1(1)=-45°
=100
H(j j-tan-1(10)=-90°
0°
etc.
The influence
of 0 is seen in
|H(j)| for all
subsequent
frequencies.
Ð
-20dB/dec
0
-45°
-45°/dec
0
also known as
3dB frequency
-90°
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Bode Plots of Higher-Order Filters
•
•
•
In determining the transfer function the most important step is to find the
breakpoint (cutoff, half-power, 3dB=all synonyms) frequency. The influence of
the breakpoint frequency is seen in a decrease (or increase) of the amplitude
response (magnitude of |H(j)|) everywhere beyond 0 but is local around 0 by
±1 decade only for the phase.
In the higher-order frequency response functions there may be many cutoff
frequencies. These frequencies may affect the magnitude by ±20 dB/dec and
the phase by ±45°/dec. However, in most cases there will be significantly higher
slops than ±20 dB/dec and ±45°/dec in the corresponding Bode plots. Such
filters will allow for better selectivity in signal detection or attenuation.
The frequency response function of higher order filters can be found from
cascaded transfer functions:
H( jw) = H1( jw)H2 ( jw)H3 ( jw)
As earlier we will find the amplitude response that we want to measured in dB:
H( jw) dB = H1 ( jw) dB + H2 ( jw) dB + H3 ( jw) dB
And the phase response.
ÐH( jw) = ÐH1( jw) + ÐH2 ( jw) + ÐH3 ( jw)
Such a format allows us to use Bode plots to look at the frequency dependences
of the transfer function.
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Bode Plots of Higher-Order Filters
Cascaded Transfer Function
EXAMPLE: A circuit has the frequency response function
jw + 5
H( jw ) =
( jw + 10)( jw + 100)
We have to find the cutoff frequencies, calculate the amplitude response and the phase
response. We will do it by factoring each term to obtain a familiar term: j0+1.
K
H( jw ) =
1
5( jw /5 + 1)
0.005( jw /5 + 1)
=
10( jw /10 + 1)100( jw /100 +1) ( jw /10 +1)( jw /100 +1)
Now we calculate the amplitude response
H ( jw ) = 0.005 dB
2
3
jw
jw
jw
+
+1 +1 +1
5
10
100 dB
dB
dB
And the phase response
jw
jw
jw
ÐH ( jw ) = Ð0.005 + Ð( + 1) - Ð( + 1)- Ð(
+ 1)
5
10
100
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Bode Plots for our Example Higher-Order Filter
jw
jw
jw
H ( jw ) = 0.005 dB +
+1 +1 +1
5
10
100 dB
dB
dB
5
10
5 100
100
10
jw
jw
jw
ÐH ( jw ) = Ð0.005 + Ð( + 1) - Ð( + 1)- Ð(
+ 1)
5
10
100
5
10 100
5
100
10
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Bode Plots of Another Example of a Higher-Order Filter
The new transfer function is now as follows:
We factor the function to the standard form:
10-3 ( jw ) 2 + 0.1jw
H( jw ) =
[1/9 ×10 4 ]( jw ) 2 + (3,030 /90,000) jw + 1
H ( jw ) =
K ( jw /w1 + 1) ××××( jw /w m + 1)
( jw /w m+1 + 1) ××××( jw /w n + 1)
In order to find the cutoff frequencies. They are: 100, 30, and 3,000 rad/s.
H ( jw ) =
0.1 jw ( jw /100 + 1)
( jw / 30 + 1)( jw / 3,000 + 1)
There is the j term in the numerator. There is no breakpoint frequency associated with this
term. So we have to find a specific point on the amplitude response
w
jw dB = 20 log10 . For the phase Ðjw = p / 2 we have always 90° since  does not
1
affect the phase here. The complex number
for this term j has only an imaginary part.
Amplitude response
Phase response
30
30
100
30
100
3,000
final
3,000
final
3,000
100
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Bandpass Filters
Bandpass filters allow for passing selective ranges of frequency. The
difference from the first order filters is that in the bandpass filters there are
lower and upper limits of passing frequencies.
|H(j)|
Such filters are usually
made using both storage
elements: the capacitor
and inductor
However, they can also be
made as RC filters
|H(j)|
band
band
pass
rejection

|H(j)|

|H(j)|
passing

rejection

Resonance filters
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RC Filters for Bandpass Applications
• Bandstop filters can be also made using RC as
shown below. We will NOT work out these examples
in class.
• Bandpass filters can be also made using RC as
shown below
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Bandpass and Resonant Filters
Bandpass filters will most frequently include all three elements: a capacitor, inductor,
and resistor. The main reason these filters operate as bandpass (or rejection) is that
capacitors have their impedance Z decreasing with frequency Z=1/jC, while the
inductive impedance increases with frequency Z=jL. We can expect therefore that
there will be a range (band) of frequency when both these elements will pass (or
stop) the signals.
Examples of such filters are shown below.
We can study transfer functions of bandpass and resonant filters using the same
techniques that we used for the first order filters. We can also plot the magnitude
and phase of such filters using Bode plots.
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Bandpass Filters - Transfer Functions
To monitor the frequency dependence of a filter we need to derive the Transfer
Function H(j). We will use the same approach as we used for the low-pass
and high-pass filters.
We start with the first example.
To find H(j) we have to find V0(j) as a
function of Vi(j).
As earlier, we will use the voltage divider
R
V0 ( jw ) = Vi ( jw )
R + 1/ jwC + jwL
Which gives us a quadratic equation:
H( jw ) =
V0 ( jw )
jwCR
=
Vi ( jw ) R + jwCR + ( jw ) 2 LC
H( jw ) =
V0 ( jw )
jwA
=
Vi ( jw ) (1+ jw /w1)(1+ jw /w 2 )
The “A” is a constant that does no depend on 
Two roots of this quadratic
equation define two
frequencies 1 and 2 that
determine the passband or
bandwidth B= 2- 1
1 and 2 are determined by
the circuit parameters!
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Plotting the Transfer Function
V ( jw )
jAw
H( jw ) = 0
=
Vi ( jw ) (1 + jw /w1 )(1+ jw /w 2 )
The transfer function must be
represented by its magnitude
and phase so we rewrite H(j)
to show |H(j)| and H(j.
V0 ( jw )
AwÐ90
H( jw ) =
=
=
2
2
-1
-1
Vi ( jw )
(1 + jw /w1 ) (1 + jw /w 2 ) Ð[tan (w /w1 ) + tan (w /w 2 )]
=
Aw
(1 + jw /w1 ) (1 + jw /w 2 )
2
-1
2
Ð[ 90 - tan (w /w1 ) - tan
Phase
-1
So that we can
(w /w 2 )] find the
magnitude and
phase
Magnitude
If we plot both plots we notice that they seem to be composed of high- and low pass filters
as
highpass
filter
as
lowpass
filter
as
highpass
filter
as
lowpass
filter
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Parameters of the Transfer Function
We will introduce parameters that will help us to easily interpret Bode plots, which describe filter
operation. We start with the Transfer Function:
H( jw ) =
=
V0 ( jw )
(2z /w n ) jw
jwCR
=
=
=
2
2
Vi ( jw ) ( jw ) LC + jwCR + 1 ( jw /w n ) + (2z /w n ) jw + 1
(1/Qw n ) jw
( jw /w n ) 2 + (1/Qw n ) jw + 1
Where:
The role the quality factor Q plays in the
bandwidth B (notice: it affects the selectivity)
B=
wn
Q
It is known as
Half-power
Bandwidth
(1/√2)
defined by
Half-power
Frequencies
1 and 2
wn =
1
LC
Is a natural or resonant frequency
Q=
1
1
1 L
=
=
2z w n RC R C
Is a quality factor
z=
1 R C
=
2Q 2 L
Is a damping ratio
Amplitude
B = w 2 - w1
Q
Q
1
2
Phase
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Summary
• Periodic signals are represented by Fourier series where harmonics have
increasing frequency and decreasing amplitude.
• The frequency response of linear circuits gives the important information
about the distortion of signals caused by amplitude and phase changes
in the circuit.
• This information is used to design Filters, which are tuned to specific
frequencies for attenuation.
• First and higher order filters are analyzed using the Transfer Function.
We can predict the amplitude response and phase response as a
function of frequency. We represent these responses using Bode plots.
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