ECE 3336 Introduction to Circuits

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ECE 3336 Introduction to Circuits Electronics
Lecture Set 12 Frequency Response More About
Filters
Fall 2007, TUETH 400-530 pm Dr. Wanda Wosik
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Outline Frequency Response and System Concept

• We will cover the following topics
• Second-Order Filters
• Transfer Function Calculation
• Bode Plots for Those Filters

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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?)
H(j?)
band
Such filters are usually made using both storage
elements the capacitor and inductor However,
they can also be made as RC filters
band
rejection
?
pass
?
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
Z1/j?C, while the inductive impedance increases
with frequency Zj?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 will study transfer functions of bandpass and
resonant filters using the same techniques that
we used for the first order filters. We will also
plot the magnitude and phase of such filters
using Bode plots.
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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.
Bandpass Filters - Transfer Functions
To find H(j?) we have to find V0(j?) as a
function of Vi(j?). As earlier, we will use the
voltage divider
Which gives us a quadratic equation
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!
The A is a constant that does no depend on ?
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Plotting the Transfer Function
The transfer function must be represented by its
magnitude and phase so we rewrite H(j?) to show
H(j?) and ?H(j??.
So that we can find the magnitude and phase
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
Where
Is a natural or resonant frequency
Is a quality factor
The role the quality factor Q plays in the
bandwidth B (notice it affects the selectivity)
Is a damping ratio
Amplitude
Q
It is known as Half-power Bandwidth (1/v2)
defined by Half-power Frequencies ?1 and ?2
Q
Phase
?2
?1
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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?). 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.
0.707
Phase
Magnitude
45
?0
?0
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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
We will find the magnitude of the transfer
function
which will be expressed in decibels H(j?)dB
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Plotting Bode Plots of the Transfer Function
We will plot the MAGNITUDE at selected multiples
of ?0.
Then we will plot the PHASE
For ?ltlt?0 H(j?)dB-20log10(1)0 dB At
??0 H(j?)dB-20log10v(11)-3 dB
??0/10
?H(j???-tan-1(1/10)0
For ?gtgt ?0
??0
?H(j???-tan-1(1)-45
?10?0 H(j?)dB-20log(10)-20 dB
?10?0
?H(j? j???-tan-1(10)-90
H(j?)dB-20log(100)-40 dB
?100?0
0
etc.
?1000?0
The influence of ?0 is seen in H(j?) for all
subsequent frequencies. The ?0 affects the
phase only locally within two decades only
-20dB/dec
-45/dec
-45
?0
?0
-90
also known as 3dB frequency
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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) 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

As earlier we will find the amplitude response
that we want to measured in dB
And the phase response.
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
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???01.
?1
K
?3
?2
Now we calculate the amplitude response And
the phase response
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Bode Plots for our Example Higher-Order Filter
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5
100
5
100
10
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
In order to find the cutoff frequencies. They
are 100, 30, and 3,000 rad/s.
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
.
For the phase
we have always 90 since ? does not affect the
phase here. The complex number for this term j?
has only an imaginary part.
Amplitude response
Phase response
final
3,000
30
30
3,000
100
final
100
???30
???100
???3,000
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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.