Transcript Document
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
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Bode Plots
We will repeat what we know about Bode plots from the first order filters: Bode Plots magnitude will be used to and phase simplify graphical representation of the of the transfer functions. We will use them instead plots such as shown in the figure below for high-pass filters and obtained from calculation of |H(j )| and H(j ).
of 0.707
0 Magnitude 0 45 ° Phase 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 .
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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 A
0
A i dB
= 20log 10
A
0
A i We will find the magnitude of the transfer function H
(
j
w ) = 1 + 1
j
w
RC
= 1 + ( w / w 0 ) 2 Ð tan 1 ( w / w 0 ) which will be expressed in decibels
|H(j
) dB | H
(
j
w )
dB
=
V
0 (
j
V
i
(
j
w w ) )
dB
= 20 log 10 1 1 + ( w / w 0 ) 2 4
Plotting Bode Plots of the Transfer Function
H
(
j
w ) = 1 + 1
j
w
RC
= 1 + 1 ( w / w 0 ) 2 Ð tan 1 ( w / w 0 ) We will plot the MAGNITUDE at
selected multiples of
0
. For
<<
0 |H(j
)| dB =-20log 10 (1)=0 dB At
=
0 |H(j
)| dB =-20log 10 √(1+1)=-3 dB For
>>
0
=10
0 |H(j
)| dB =-20log(10)=-20 dB
=100
0
=1000
0 H
(
j |H(j
)| dB =-20log(100)=-40 dB
etc .
w )
dB
= 20 log 10 ( w w 0 )
0
Ð °
V V
0
i
=10
0
Then we will plot the PHASE (
=
0 /10
=
0
(
j j
w w ) ) = tan H(j 1 ( w / w 0 ) H(j -tan -1 (1/10)=0 ° H(j -tan -1 (1)=-45 ° j -tan -1 (10)=-90 ° 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
0 also known as 3dB frequency -45
°
0 -45
°
/dec -90
° 5
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 ±1 decade only for the phase . by 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
(
j
w ) =
H
1 (
j
w )
H
2 (
j
w )
H
3 (
j
w ) As earlier we will find the amplitude response
H
(
j
w )
dB
=
H
1 (
j
w )
dB
+ that we want to measured in dB:
H
2 (
j
w )
dB
+
H
3 (
j
w )
dB
phase response .
Ð
H
(
j
w ) =Ð
H
1 (
j
w ) +Ð
H
2 (
j
w ) +Ð
H
3 (
j
w ) Such a format allows us to use Bode plots to look at the frequency dependences of the transfer function. 6
Bode Plots of Higher-Order Filters Cascaded Transfer Function
EXAMPLE: A circuit has the frequency response function
H
(
j
w ) = (
j
w
j
w + 10)( +
j
5 w + 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:
H
(
j
w ) = 10(
j
w /10 5(
j
w /5 + 1)100( + 1)
j
w /100 + 1) Now we calculate the amplitude response = (
j
w K 0.005(
j
w /10 + 1)(
j
2 w 1
j
/5 + 1) /100 3
0 +1.
+ 1)
H
(
j
w
)
=
0.005
dB
+
j
w +
1 5
dB
-
j
w
10
+
1
dB
-
j
w
100
+
1
dB
And the phase response Ð
H
(
j
w ) = Ð 0.005
+Ð (
j
w 5 + 1) -Ð (
j
w 10 + 1) -Ð (
j
w 100 + 1) 7
H
(
j
w
Bode Plots for our Example Higher-Order Filter )
=
0.005
dB
+
j
w
5
+
1
dB
-
j
w
10
+
1
dB
-
j
w
100
+
1
dB
5 10 100 5 10 100 Ð
H
(
j
w ) = Ð 0.005
+Ð (
j
w 5 + 1) -Ð (
j
w 10 + 1) -Ð (
j
w 100 + 1) 5 10 100 5 10 100 8
Bode Plots of Another Example of a Higher-Order Filter The new transfer function
H
(
j
w ) = 10 3 (
j
[1/9 × 10 4 ](
j
w ) 2 is now as follows: w ) 2 + 0.1
j
w + (3,030 /90,000)
j
w + 1 We factor the function
H
(
j
w ) =
K
(
j
(
j
w / w 1 w / w
m
+ 1 to the standard form: + 1) ×××× (
j
+ 1) ×××× (
j
w / w
m
+ 1) w / w
n
+ 1) In order to find the cutoff frequencies . They are: 100, 30, and 3,000 rad/s.
H
(
j
w ) = (
j
w 0.1
/ 30
j
w + (
j
1)( w
j
/100 + 1) w / 3,000 + 1) There is the
j
term in the numerator
. There is no breakpoint frequency associated with this
j
w
dB
= 20 log 10 w . 1 For the phase Ð
j
Amplitude response w = p / 2 we have always 90 ° since does not affect the phase here. The complex number for this term j has only an imaginary part. Phase response 30 3,000 final 100 30 3,000 final 30 100 3,000 100 9
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
|H(j
)| band pass
|H(j
)| rejection
passing
Resonance filters rejection
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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 11
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
and phase
of such filters using Bode plots .
plot the
magnitude
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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 V 0 (j ) as a function of V i (j ) . As earlier, we will use the voltage divider
V
0 (
j
w ) =
V
i
(
j
w )
R
+ 1/
R j
w
C
+
j
w
L
Which gives us a quadratic equation:
H
(
j
w ) =
V
0 (
j
V
i
(
j
w ) w ) =
R
+
j j
w
CR
w
CR
+ (
j
w ) 2
LC
H
(
j
w ) =
V
0 (
j
V
i
(
j
w w ) ) = (1 +
j
w
A j
w / w 1 )(1 +
j
w / 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
H
(
H
(
j
w ) =
j
w
V
i
)
V
0 ( ( =
j j
V
0 (
V
i
(
j j
w ) w ) w w ) ) = = (1 + (1 +
j
w /
j
w w 1 )
jA
w / w 1 )(1 + 2 (1 + The transfer function must be
j
w
j
w / w 2 ) represented by its magnitude and phase so we rewrite
H(j
)
to show |H(j )| and H(j .
A
w Ð 90 / w 2 ) 2 Ð [tan 1 ( w / w 1 ) + tan 1 ( w / w 2 )] = = (1 +
A
w
j
w / w 1 ) 2 (1 +
j
w / w 2 ) 2 Ð [ 90 tan 1 ( w / w 1 ) tan 1 ( w / w 2 )] Phase So that we can find the magnitude phase and 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 14
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: w )
H
(
j
w ) =
V
0 (
V
i
(
j j
w ) = (
j
w ) 2
j LC
w
CR
+
j
w
CR
+ 1 = (
j
w / w
n
(2 z ) 2 / w
n
) + (2 z
j
w / w
n
)
j
w + 1 = = (
j
w / (1/
Q
w
n
) w
n
) 2 + (1/
Q j
w w
n
)
j
w + 1 The role the quality factor Q plays in the Where: w
n Q
= 1 2 z = 1 =
LC
1 w
n RC
Is a natural or resonant frequency = 1
R L C
Is a quality factor
B
= w
n Q
Amplitude
B
= w 2 w 1 z = 1 2
Q
=
R
2
C L
Is a damping ratio Q It is known as Half-power Bandwidth (1/√2) Half-power Frequencies 1 and 2 1 2 Q Phase 15
Summary
• Periodic signals are represented by Fourier series increasing frequency and decreasing amplitude.
where harmonics have • 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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