Passive Filters
 A filter is a circuit that is designed to pass signals with desired frequencies and reject or attenuate others.
 A filter is a passive filter if it consists of only passive elements R , L , and C .
 A lowpass filter passes low frequencies and stops high frequencies.
 A highpass filter passes high frequencies and rejects low .
 A bandpass filter passes frequencies within a frequency band and blocks or attenuates frequencies outside the band.
 A bandstop (band reject) filter passes frequencies outside a frequency band and blocks or attenuates frequencies within the band.
1

Filter Responses
Ideal Filter Classification
Low pass f filter
High pass f filter
Band pass f filter
Band reject f filter
2

Passive (1 st order) Lowpass Filters
RL circuit
(  
R
 
 j
R L
  R L

1
1
Plot |H(j  )| versus  :
Ideal
Actual
H j  
1
2
(  ) max

R
L
H
LPF
( j  
1
1 j c
3

Passive (1 st order) Lowpass Filters
RC circuit
(  
R
1 ( 
 
)

1 ( j C ) 1
Plot |H(j  )| versus  :
1
Ideal
Actual
H j  
1
2
(  ) max

1
RC
H
LPF
( j  
1
1 j c
4

Passive (1 st order) Highpass Filters
(  
R j L

1
 c

R
L

1 j   c j c
 c

1
RC
(  
R
R
 

1 ( j C ) 1

1 j   c j c
5

Example
Effects of Loading on Filter Performance
H
)
 sL
R sL


Ks
/
R
, K
L
 1
H
H

)

Z eq , Z eq
 ||
L eq
)

 s
R s
 sLR
L sL R
L

 sLR
L sL R
L
R
L

R
L

R
L

L


L

 sLR
L
( 
L
)  sLR
L
Ks
 
,
K 
R
L
R R
L
 1;
) )
 
)
6

Example cont`d
Let R=R
L
= 500  and L =5.3mH
Then f c
(unloaded) = 15kHz and f c
(loaded) = f c`
= 7.5 kHz
Effects of Loading on Filter Performance
 Changes the cutoff frequency
 Changes the gain in the passband
For R
L
= R
7

Passive 2 nd order Lowpass Filter

1 sC
  1 sC

1 

(1 LC ) s 2  ( )  (1 LC )

 2 o s 2 s 2 o
 H
LPF
1

1 LC
8

Passive Bandpass Filters
Series RLC circuit
H ( s ) 

R  sL
R
 1 s 2 sC
 ( R
( R
L )
L s
)
 s
( 1

RsC
LC )


RsC s 2 LC s 2
 1

 s
 s   o
2
 H
BPF
( s )
B
1 LC
R L
9

Passive Bandpass Filters
Parallel RLC circuit

Z eq , Z eq
 sL ||1 sC 
L C sL  1 sC eq


R sL
L C
 1 sC
 L C sL  1 sC

 s s 2   s   2 o

L C


( 1 sC ) L C s 2
(1 )
 (1 )  (1 LC )
 H
BPF
( ) 1
B    1
LC
RC
10

Passive Bandpass Filters
Frequency response characteristics:
Only 2 of the 5 characteristics of a bandpass filter can be specified independently. The others are related as follows:
    c 2
B c
Q c 1
11

Example
Design a bandpass filter with a center frequency of 2 kHz and a bandwidth of 500 Hz. Use a
250  resistor.
B 
1
RC
 C 
1
 R

1
2 500Hz 250  o
1
LC
 L
 1.27 F

1
2 C

 4.97 mH
Note: Q
 3
 4
500
The Q is so low in order to achieve the specified bandwidth.
1
12

Passive (Bandstop) Bandreject Filters
Series RLC circuit
 sL 
R sL
1
 sC
1 sC


 1
 1
 s 2  ( s 2  (1 LC )
)  (1 LC )
 s 2 s 2   2 o
    2 o
 H
BRF
( )
1 LC
B    R L
13

Passive (Bandstop) Bandreject Filters
Parallel RLC circuit

R
, Z eq
 sL ||1 sC 
L C sL  1 sC eq

R 
R
L C sL  1 sC

 s 2   2 o s 2 s 2 o
(
(
 1
 1 sC
 H
BRF
( )
) sC

)
L C
 s 2 s 2  (1 LC )
 (1 )  (1 LC )
1
B    1
LC
RC
14

Example 1
Determine what type of filter is shown. Calculate the corner or cutoff frequency.
Take R = 2 k  , L = 2 H, and C = 2 μ F.
15

Example 1 cont’d
R = 2 k  , L = 2 H, C = 2 μ F
Note:
Q p
 R
1
LC

C
L
1
 6
 500 rad/s
3
2 10  6
2
 2
16

Example 2
For the circuit shown, obtain the transfer function V o( ω )/ V i( ω ) . Identify the type of filter the circuit represents and determine the corner frequency. Take R
1
= 100 
= R
2
, L = 2 mH.
This is a highpass filter.
17

Example 2 cont’d
18

Example 3
If this bandstop filter is to reject a 200-Hz sinusoid while passing other frequencies, calculate the values of L and C .
Take R = 150  and the bandwidth as 100 Hz.
19

Example 4
Design this bandpass filter with a lower cutoff frequency of 20.1 kHz and an upper cutoff frequency of 20.3 kHz. Take R = 20 k  . Calculate
L , C , and Q .
 Too large !!
20

Take R = 20  .
Example 4 cont’d
Q
L


B o  101

R
B

20 
C 
1
 o
QR

 
 15.916 mH
1
  
 3.9 nF
21

Example 5
Use a 500 nF capacitor to design this band-reject filter. The filter has a center frequency of 4 kHz and a quality factor of 50.
1
LC
Q   o
RC
 L 
1
 2
0
C
 R 
Q
 o
C



1
   3
 2
 
50
      9
 9
 3.166 mH
 3.98 k 
22

Active Filters
Passive Filter Disadvantages:

Maximum gain in the pass band is 1 – no amplification in the pass band!

Adding a load to a passive filter changes the filter’s characteristics, like the cutoff frequency.

Passive band pass and band reject filters require the use of inductors, which are large, costly, and can introduce stray electromagnetic field effects.

No obvious method to make the filter characteristic more ideal.
23

Active Filters
Active filters
 Based on circuits with op amps, which are active devices, since they require an external power supply (  V
CC
).
 Eliminate all of the disadvantages of passive filters:
 Can create a pass band gain > 1;
 Can add loads without changing the filter characteristics;
 All four types of filters can be created using op amps, resistors, and capacitors – no inductors;
 Can cascade simple (first- and second-order) filters to create higher order filters that are more nearly ideal.
24

First-Order Active Lowpass Filter

Find the transfer function:

  o i
 
Z f  
Z i

 
R
1
1 
2


1
  sC
R
1

|| R
2

R
2
R R R R sC
 

R R s 
1
1

1 R C
R C

 s
K  c
  c
1
K  
R
2
R
1
25

Example
Design an active low pass filter with a gain of -5 and a cutoff frequency of 1 kHz. Use a 10 nF capacitor.
1
R
2

1
 9
  15.916 k 
R
1

R
5
2 
15,916
5
 3.184 k 
26

First-Order Active Highpass Filter

Find the transfer function:
H ( s )



V o
( s )
V i
( s )
 
R
1
R
2 sC sC  1
Z
Z i
 f

 s
 

R
2
1
R
2
R
1
R
1

 s
1
R
1
C sC
For K = -1
1
K  
R
2
R
1
27

Example 1
Design an active high pass filter with a gain of -10 and a cutoff frequency of 500 rad/s. Use a 0.1  F capacitor.
1
 500 rad/s
R
1

1
R
2
 10 R
1
 200k 
 20k 
28

Example 2
Design a first-order highpass op amp filter whose transfer function is given by
    4000

Use capacitances of any value. But you only have 10 k  resistors available
- no inductors!
The transfer function of an inverting op amp is
Thus,
 
1 s
 
 3 
 3
  s
 
 

3

 3
1
 9 s
 3 
 
 3
Therefore, f
( ) R 10k  i
( )  
1 sC where R  10k  and C  25nF s i f
7
 

3

 3
1
 8 s
29

First-Order Active Bandpass Filter
+ + +
=
30

First-Order Active Bandpass Filter
Gives  c2
Gives  c1
Gives K
31

First-Order Active Bandpass Filter
H
BPF
( )  H
LPF
( )

 s
 c 2
  c 2
 s
 s
  c 1

 R
R
 ( R f
R i
)  c 2 s i f s
2

HPF
( ) 
Inv
( )
( c 1 c 2
) s    c 2
 s
2 s
2
0 if    c 2
So this cascading technique only produces broadband filters.
32

Example
Design an active band pass filter with a passband gain of 2 and cutoff frequencies of 500 Hz and 50 kHz. Use a 200 nF capacitor.
Lowpass filter:
R
L

1
 15.9

Highpass filter:
R
H

1
 1.592 k 
Gain:
R f
R i
 500 
33

First-Order Active Bandstop Filter
Gives  c1 Gives K
Gives  c2
LPF
Summing
Amp.
HPF
34

First-Order Active Bandstop Filter
Find the transfer function:
H
BRF
( s )  H
Inv
( s )[ H
LPF
( s )  H
HPF
( s )]

 R f
R i

 s


 c 1
 c 1
 s
 s
  c 2




( R f s 2 
R i
) [ s 2
(  c 1

 2
 c 2
 c 1 s
) s 
  c 1

 c 1
 c 2 c 2
]
K s 2
(
 s 2
 s
 

2
0

)
2
0 if  c 1
  c 2
So again, this technique only produces broadband filters.
35

Example
Design a notch filter between 20 krad/s and 100 krad/s with a gain of
K = 5. You have only 10 k  resistors available.
Lowpass filter:
C
L

1
20krad/s 10k 
 5nF
Highpass filter:
C
H

1
100krad/s 10k 
 1nF
Gain:
R f
R i
 5  R i
 10 k 
R f
5 10 k 
36

Example cont’d
Note: Due to the fact that the lowpass and highpass filters used in this example go down with only 20 dB/dec, attenuation in the stopband might be limited.
37

v i
Second-Order Op-Amp Filters
Bandpass Filter
H v o j 
 
1  j  2
R
1
 R
3
R
1
 R
3
C
C  ( j  ) 2
R
1
 R
3
C 2
Exercise: Try to derive this transfer function.

0

1 R
1
 R
3 , Q 
2
1


1
2

0
, B 

0 
2
Q R C
, H   K 
R
2
2 R
1
Advantage: The bandwidth, thus Q factor, is entirely controlled by R
2
.
38

Example 2 nd -order Bandpass Filter
Design for 3kHz, K =2, Q =10:
39

v i
H
Bandstop Filter
H
 k

1 ( j  ) 2 R C 2

1  j  2(2  )  ( j  ) 2 R C 2 v o 1 2
H
 k

1   2 R C 2


 1   2 R C 2

 2
   2(2  )
 2
H 


 
1
RC



 0    dB
40

Scaling
•
There are two ways of scaling a circuit: magnitude or impedance scaling , and frequency scaling .
•
Both are useful in scaling responses and circuit elements to values within the practical ranges.
•
Magnitude scaling leaves the frequency response of a circuit unaltered.
•
Frequency scaling shifts the frequency response up or down the frequency spectrum.
41

Magnitude Scaling
 Magnitude scaling is the process of increasing all impedances in a network by a factor, the frequency response remaining unchanged .
Z
Z
Z
R
L
C
 
 
  k Z k Z k Z
R
L
C

1
R   k R
 

 k m

1
R   k R
L   k L
C  
C k m
 The primed variables are the new values and the unprimed variables are the old values. Consider the series or parallel RLC circuit. We now have
 o
 
1

1
 m
 m


1
LC
  o
42

Frequency Scaling
 Frequency scaling is the process of shifting the frequency response of a network up or down the frequency axis while leaving the impedance the same .
   k f

R   R
 
L    j k f
1
 
C 

1
R   R
L   /
C   / f f
 Again, if we consider the series or parallel RLC circuit, for the resonant frequency
 o
 
1


/ f
1

/ f

 k f
LC
 k f
 o
B   k B
43

Magnitude and Frequency Scaling
 If a circuit is scaled in magnitude and frequency at the same time, then
R   k R
L   k m k f
L
C  
C
   k f
 B   k B
44

Example 1
Redesign the active low pass filter to use a 200 pF capacitor.
Use magnitude scaling:
C  
C k m k m

C

C 
10nF
200pF
 50
R
1
 
R
2
 
50(3183.8 ) 159.19 k 

45

Example 2
Redesign the active low pass filter, which was designed for 1 kHz, so that the cutoff frequency is 7.5 kHz, using a 250 pF capacitor.
Use frequency scaling first, then magnitude scaling:
 c
C  
 k
C
 

C   250 pF; k f k m

  c 
 c

C

7500
1000

 7.5
10nF
(7.5)(250pF)
 5.3
R
1
 R
2
 5 R
1
 84.9k

46

Example 3
The fourth-order Butterworth lowpass filter is designed such that the cutoff frequency is ω c
10-k  resistors.
= 1 rad/s. Scale the circuit for a cutoff frequency of 50 kHz using
R   k R L   k m k f
L
C  
C
   k f

Frequency scaling: k f



' c 
  
1
3
Magnitude scaling: k m

R

R
 3
1
 10 4
10 5
47

mH
48

Example 4
The third-order Butterworth filter is normalized to ω c
= 1. Scale the circuit to a cutoff frequency of 10 kHz. Use 15 nF capacitors.
k f

C ' 


' c 
 
1
4
C
   4
 k m

C f

1
15 10
 9    4

10
4
3 
R '  k R
L '  k m k f
L


10 4
3 
1   1.061k

10 4 2H
   4
 33.77 mH
C
49