of 41
7/31/2019 Chapter 4 Latest
1/41
Chapter 4:Electronic Filter SynthesisTopics: Filter background, definition and application
The properties of an ideal filter
Filter classification Filter design
Basic and characteristic filter
7/31/2019 Chapter 4 Latest
2/41
An introduction to filter
A filter is1. a circuit designed to pass signals with desired frequencies and
reject the others.
2. important to attenuate unwanted frequencies noise due to thebackground noise or nonlinear characteristic of some
electronic devices3. suitably used to
(a) block high pitches that cannot be efficiently broadcast.
(b) remove high frequencies which causes image blurring (vialow pass filter)
(c) remove low frequencies, sharpen and enhance edge (viahigh pass filter)
(d) In biomedical applications for instance; to filter out theEEG signals, to detect abnormal signal in ECG analysis, todistinguish the heart beat sound between mother and foetus.
7/31/2019 Chapter 4 Latest
3/41
Properties of an ideal filter
Must have identical gain at all frequencies
in its pass band
Should have zero gain outside its pass
band
Tolerable frequencies response curve
around +/-3dB
7/31/2019 Chapter 4 Latest
4/41
Ideal low-pass filter
Ideal high-pass filter
Ideal band-pass filter
Ideal notch filter
Common types of filters
7/31/2019 Chapter 4 Latest
5/41
Properties of fundamental filters
Low pass filter
o Pass only signals with frequencies lower than fc and
attenuate the rest
High pass filter
Pass only signals with frequencies greater than fc and
attenuate the rest
Band pass filter
Pass only signals with frequencies between f1
and f2
Band stop filter
attenuate only signals with frequencies between f1 and f2
7/31/2019 Chapter 4 Latest
6/41
Filter classification
Passive filterconsists only ofpassive elements(R,L,C)
Active filterconsists ofactive element (transistor, op-
amp) in addition to passive elements
Passive filter Active filter
7/31/2019 Chapter 4 Latest
7/41
Comparing Active and Passive filters
Aspect Passive filter Active filter
Gain Cannot generate gain >1 and is
lossy
Provide amplification and voltage gain
but need external sources
cost Expensive (because of the use
of inductors)
Cheap
Complexity Simple topology More complex topology
Flexibility - Isolation of each stage of filters from
source and load impedance effects can
be achieved by the use of buffer
amplifiers
Practicality Difficult to implement at low
frequencies (f
7/31/2019 Chapter 4 Latest
8/41
Designing a filter circuit
Parameters to be considered in designinga filter
1. Quality factor
-The higher the Q factor, the more (frequency)selective a circuit is, producing a high gain at thecircuit end
2. Damping ratio
- Determine the filter response characteristic3. RC pole
- A filter pole consists of a RC circuit. Each poledoubles the roll-off rate.
7/31/2019 Chapter 4 Latest
9/41
Passive Low-pass filters circuit(using RC network)
First order Low-pass filter
Second order Low-pass filter
Cut-off frequency:
Cut-off frequency:
7/31/2019 Chapter 4 Latest
10/41
Second order Low-pass filter has higher rate of roll-offin which itattenuates higher frequency more quickly.
Application where filter with high roll-off is desired:
1. to prevent crosstalk between adjacent channels on telephone FDM
(frequency division multiplexing) systems
2. Improve EEG signals
The frequency response of First and second order
low pass filter
7/31/2019 Chapter 4 Latest
11/41
Passive High-pass filter circuit(using RC network)
First order High-pass filter
Second order High-pass filter
Cut-off frequency:
Cut-off frequency:
7/31/2019 Chapter 4 Latest
12/41
Identify the types of filter (1) Given that we have a circuit as follow;
We know that
Using the voltage division, high fleads to lowXc hence low voltagedrop across capacitor, where Vout:
the otherwise is true iffis low.
where;
f= oscillating frequency
C= capacitance
Fig. 1
7/31/2019 Chapter 4 Latest
13/41
Identify the types of filter (2) As a rule of thumb:
This shows that the circuit shown in Fig.1 is a low pass filter
Hi f => LowXc => Low VoutLowf=>HiXc =>Hi Vout
This is true only for thecircuit shown in Fig. 1
where;
7/31/2019 Chapter 4 Latest
14/41
Identify the types of filter (3)Alternatively, we can also identify the types of filter for the circuit
shown in Fig. 1 by writing its transfer function, H(S) which isgiven as
RCj
CjR
Cj
V
VSH
in
out
1
1
1
1
)(
Substituting the value of =0 and = into the transfer function gives
01
1)0(
H
01
1)(
H
Showing that it is a low pass filter
7/31/2019 Chapter 4 Latest
15/41
Passive band-pass and band-stop filters using RC network
Band-pass filter via
cascaded RC circuits
Band-stop filter via
Twin T
7/31/2019 Chapter 4 Latest
16/41
Passive band-pass filter using RCnetwork
Band-pass filter via cascaded RC circuits
1. An ideal bandpass filter would have a completely flat passband(e.g. with no
gain/attenuation throughout) and would completely attenuate all frequencies
outside the passband.2. The bandwidth of the filter is simply the difference between the upper and lower
cutoff frequencies, while the centre frequency is given by;
Frequency response of a Band-pass filter
where
L is the lower -3dB cut-off frequency point
H is the upper -3dB cut-off frequency point
7/31/2019 Chapter 4 Latest
17/41
Band-stop filter via Twin T
The twin T provides a large degree of rejection at a particular frequency,
e.g. to filter out unwanted noise at 50 or 60 Hz that may be entering a circuit.
The response provided by the filter consists of a low level of attenuation away
from the notch frequency. As signals move closer to the notch frequency, the
level of attenuation rises, giving the typical notch filter response.
The notch frequency, fnotch, is given by
Passive band-stop filter using RCnetwork
7/31/2019 Chapter 4 Latest
18/41
Example of second order RLCpassive filters:
RLC low pass filter
RLC band pass filter
7/31/2019 Chapter 4 Latest
19/41
ExerciseIdentify the types of filter for the following circuit and give your
reason why.
7/31/2019 Chapter 4 Latest
20/41
Solution
At low frequency, lets assumefapproaching 0 (f> 0), XL appeared as a
short while XC an open, so that the circuit becomes
There is no current passing through the circuit hence no signal appear at
the output at low frequency
The reactance of capacitor , XC, and inductor , XL, in the circuit is given by
7/31/2019 Chapter 4 Latest
21/41
Whereas at high frequency, lets assumefapproaching infinity (f> ), XL
appeared as an open while XC as a short, so that the circuit becomes
Again, there is no current passing through the circuit hence no signal
appear at the output at high frequency
However, at the resonance frequency, the circuit appeared as
The reason is because:Minimum impedance at resonant
frequency for L-C connected in series
Maximum impedance at resonant
frequency for L-C connected in parallel
7/31/2019 Chapter 4 Latest
22/41
At the resonance frequency,fr, there is voltage drop across parallel L-C
branch, hence Vo is nonzero.
The above derivations show that this circuit is a Band-pass filter and itsfrequency response can be sketched as
7/31/2019 Chapter 4 Latest
23/41
ExampleBased on the filter circuit below
1. Find the transfer function.
2. Given that |H()| = , calculate the cutoff frequency. Use L = 4.5mH, C =
25F and R = 100k.
sCRsL
sCR
v
vsH
i1
1)( 0
Solution:Using the voltage division rule, H(s) is given by:
sRC
R
sCR
sCR
sCR
1/1
/1
where
7/31/2019 Chapter 4 Latest
24/41
RsLRLCs
R
sRCRsL
sRCRsH
2)1/(
)1/()(
hence
RLCLjR
RjH
2)(
Since s =j, so
Solution for (b)
The magnitude of H(j) is given by
210
|)(|222
2
LRLCR
RH
Squaring the |H()|, |H()|2 reduces the equation to
2
2)1(2
R
LLC cc
7/31/2019 Chapter 4 Latest
25/41
2
3
3
263
10100
)105.4()]1025)(105.4(1[(2
x
xxx cc
Substitute the values into the equation gives:
Solve for the quadratic equation ofc:
a
acbb
2
42
)1025.1(2
)11025.14()1025.2()1025.2(14
14277
x
xxxxx
c =
Therefore,
c= 4657 rad/s @ 4.657 krad/s
7/31/2019 Chapter 4 Latest
26/41
First order Active filter circuits(using RC circuit)
Active low-pass filter
Active high-pass filter
7/31/2019 Chapter 4 Latest
27/41
Second order Active filter circuits(using RC circuit)
Active band-pass filter
7/31/2019 Chapter 4 Latest
28/41
Again,
Given that we have an active filter as follows;
Iff> ,Xc> 0, the circuit becomes:
The capacitor acts as a short, the amount of current passed through R2
becomes negligible so the gain of the amplifier goes to zero
Identify the types of Active filter(part1)
Fig.2
7/31/2019 Chapter 4 Latest
29/41
Iff> 0,Xc
> , the circuit becomes:
The capacitor acts as an open, the circuit will act as an amplifier with gain
-R1/R2
*This shows the circuit is a low pass filter.
Identify the types of Active filter(part2)
7/31/2019 Chapter 4 Latest
30/41
Identify the types of Active filter(part3)In the case of Fig.3
This circuit will attenuate low frequencies ( 1/R1C1), but will pass intermediate frequencies with a gainof -R1/R2. So, it is a band-pass filter.
Fig.3
E l
7/31/2019 Chapter 4 Latest
31/41
ExampleFigure below shows a filter circuit
1. Prove that the transfer function is given as:
RCj
RCj
R
RH
f
11)(
1
2. Identify the type of filter
7/31/2019 Chapter 4 Latest
32/41
According to the golden rule of op-amp: V+ =V- , therefore op-amp appeared as
an open circuit
It can be seen from the circuit that voltage across Node A gives V+, while
voltage drop across Node B gives V- , and we know that
Using voltage division rule to give V+ and V- as
Eq. 1
Eq. 2
Eq. 3
7/31/2019 Chapter 4 Latest
33/41
Substitute Eq.2 and Eq.3 into Eq. 1 produces
Rearranging the equation gives
The AOL (open loop gain) of an op-amp is always approaching infinity, hence
the transfer function of this circuit can be reduced to
RCj
RCj
R
R
V
VsH
i
f
i
o
1
1)(
7/31/2019 Chapter 4 Latest
34/41
0)0(1
)0(1)0(
RCj
RCj
R
RH
i
f
1)(1
)(1)(
RCjRCj
RRH
i
f
Solution (Q2):
Substituting the value of = 0 and = into the equation
The above shows that this circuit is a high pass filter.
7/31/2019 Chapter 4 Latest
35/41
Characteristic filter: Butterworth filter(Part I)
The third-order low-pass design shown above is a simple
example of Butterworth filter
7/31/2019 Chapter 4 Latest
36/41
1. The distinctive characteristic of this filter is the maximally flatfrequency response which is desired for various applications.
2. The roll-off rate for this filter starts with 20n dB/decade. The
higher number of n, the flatter the response will be - hence the
filter is approaching to an ideal frequency curve
Characteristic filter: Butterworth filter(Part II)
7/31/2019 Chapter 4 Latest
37/41
Characteristic filter: Chebyshev filter I
1. This filter is well-known by the existence of ripple in the pass band and
rapidly increasing attenuation in the transition band.
2. Standard roll-off rate for Chebyshev filter is 40dB/decade which meansthe response will be steeper at the edge- hence it has a sharper cutoff as
compared to Butterworth filter.
3. This filter belongs to active filter group and which has irregular pass
band response.
4. The higher number of N the faster the roll off rate but also results in an
increase in the amount of ripple in the pass band
7/31/2019 Chapter 4 Latest
38/41
Characteristic filter: Chebyshev filter II
1. The response is monotone at the pass band but equiripple
response at the stop band.
2. This filter is not commonly in used as it gives inconsistent
gain due to fluctuations.
7/31/2019 Chapter 4 Latest
39/41
Sallen Key (Two-Pole) Low Pass Filter
-
+
+V
-V
R1
Rf1
Rf2
C1
vin
vout
C2
R2
Low Pass Filter
1. To obtain an nth order filter, n/2SK circuits should be cascaded
(assume K=1)
2. As K increased from 0 to 3, the
transfer function displays more
peaking3. The circuit becomes unstable
when K>3
where,
R1 = R2
C1 = C2
Rf1 = (K-2)Rf2
7/31/2019 Chapter 4 Latest
40/41
Sallen Key (Two-Pole) High Pass Filter
-
+
+V
-V
R1
Rf1
Rf2
C2
vin
vout
R2
C1
High Pass Filter
1. Using op amp, the SK is nottruly a high-pass filter,
because the gain of the op
amp eventually falls off.
2. However, the frequencies at
which the op amp gain is
fairly high, the circuit
behaves as a high-pass filter.
3. Since the HP SK circuit is
equivalent as the low-pass,
the empirical values for K
would be still valid in thiscase also.
where,
C1 = C2
Rf1 = (K-2)Rf2
7/31/2019 Chapter 4 Latest
41/41
Peaking clearly be seen in
the stopband
The typical Sallen Key frequency response: