EDITOR: B. SOMANATHAN
NAIR
1. INTRODUCTION
Filters
are a very important class of circuits in communication engineering. There are
mainly four types of filters. They are low-pass, high-pass, band-pass, band-
reject filters, respectively. Consider Fig. 1, which shows a passive RC circuit. We prove that this network
can act as a low-pass filter.
From Fig. 1, we find
Vi = IR + I(1/jωC)
(1)
Vo = I(1/jωC)
(2)
The
transfer gain of the circuit is now obtained using (1) and (2) as
H(ω) = Vo/ Vi =(1/jωC)/
[R + (1/jωC)]
= 1/(1+ jωCR)
=
1/[1+ j(ω/ωc)] (3)
where
ωc = 1/CR is the cut-off frequency. Equation
(3) may be expressed in the form
│H(ω) │2 = 1/[1+ (ω/ωc)2]
∟-tan-1 (ω/ωc) (4)
Equation
(4) represents the transfer gain of a first-order passive low-pass filter.
Consider (4). We observe that the
magnitude H(ω) will be almost constant at 1 upto ω < ωc. At ω = ωc,
we find that the magnitude drops to 0.707. Above this value, the gain gets
reduced drastically at the rate of ‒20 dB/decade. This means that the gain gets
reduced by 20 dB when the frequency gets multiplied by a factor of 10. For
example, if gain at 1 kHz is 0 dB, at 10 kHz, it reduces to ‒20 dB, at 10 kHz,
it reduces to ‒40 dB, and so on. Thus the circuit permits all frequencies less
than ωc to pass through it
and prevents frequencies higher than ωc
from passing through it. Hence we call this circuit as a first-order low-pass
filter, as there is only one RC
section in it. It may be noted that the cut-off of the single-section RC filter is not sharp. To get sharper
cut-off, we have to increase the number of RC
sections. Thus, a two-section RC
filter, called as second-order filter, has a gain drop at the rate of ‒40
dB/decade. Similarly, a third-order filter has three RC sections with a gain drop off of ‒60 dB/decade. In general, with an n-section RC filter, we have a gain drop off of ‒20n dB/decade. Equation (4) may now be modified as
│H(ω)
│2 = 1/[1+ (ω/ωc)2n]1/2
(5)
Equation
(5) represents the transfer gain of an nth-order
Butterworth filter.
Figure 2 shows the frequency-response characteristics
of first-order, second-order, and third-order Butterworth low-pass filters.
Butterworth filters have the following features:
·
The pass band
can be seen to be flat upto ω = ωc and thereafter drops off
uniformly in the stop (attenuation) band.
·
The higher the
order of the filter, the sharper the drop off.
·
The expression
for the transfer gain of the nth-order
filter shows that the filter can be designed by knowing the values of the order
of the filter n.
2. SECOND-ORDER RLC LOW-PASS FILTER
It
is customary to design and construct filters of first and second orders. It is
also customary to construct filters of higher order by cascading suitable
number of first- and second-order filters. For example, a third-order filter is
obtained by cascading a second-order with a first-order filter. Similarly, a
sixth-order filter is obtained by cascading three second-order filters.
Figure 3 shows a second-order
low-pass RLC filter. The transfer
gain (function) of this filter can be obtained as follows:
From Fig. 3, we find
Vi = I(R+jωL+1/jωC)
(6)
And
Vi = I/jωC (7)
The
transfer function, therefore, is
H(ω)
= Vo/ Vi =(1/jωC)/
[R +jωL+(1/jωC)] (8)
Now,
we use Laplace transform in (8) by writing s = ω.
Performing this operation yields
H(s)
=1/sC)/ [R +sL+ (1/sC)]
=1/(sCR+s2LC +1)
= (1/lLC)/[s2+(R/L)s+ (1/LC)] (9)
Equation
(9) is expressed in a more convenient form by choosing 1/LC = ωc2
(cut-off frequency) and R/L = 2δωc
(attenuation factor). Substituting these new terms in (9) yields
H(s)
= ωc2/[s2+ 2δωcs+ ωc2] (10)
Equation
(10) is a design equation.
Example 1: Design a second-order RLC LPF for a cut-off frequency of 1 kHz. Assume damping factor to be
0.5.
Solution: From (10), we obtain
1/LC
= ωc2 (11)
R/L = 2δωc (12)
Substituting
for ωc from (10) into (11)
yields
R/L
= 2δ/√LC (13)
Manipulating (13), we find
R = 2δ√(L/C)
(14)
Since
fc = 1 kHz, we find
ωc2 = (2πx1000)2
= 1/LC (15)
Assume
the value of capacitor
C = 0.01 μF
(16)
we get

The
value of R can now be computed as
R = 2δ√(L/C)
= 2x0.5x√2.53/0.01x10‒6 = 15.9 kΩ (18)
Equations (16), (17), and (18) give the
desired values of the filter components.
In passive filter design, the values of the
components are chosen from their availability in the market. Usually, we first
choose C as it is difficult to get
capacitors of all values. However, we an get inductors of desired values by
winding them to specifications. Resistors can be obtained by using trimmer
potentiometers.
In the forthcoming blogs, we shall discuss
the design of analog and digital Butterworth filters in detail.
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