Gary Hecht
RLC Low-Pass Filter Transfer Function
Consider the second-order RLC low-pass filter circuit below:
Vo
=
1
I • ──────
j2πfC
Vo
───
Vs
=
Gain
=
=
Vs
1
─────────────────── • ─────
1
j2πfC
R + j2πfL + ─────
j2πfC
1
─────────────────────────────
j2πfRC + (j2πfL)(j2πfC) + 1
Note that the denominator in the above transfer function is not in standard Bode form and,
furthermore, it may not be immediately obvious how to manipulate it into standard Bode form. The
equation below is the same as the equation just above except that the denominator has been
reorganized so as to make it clear that the denominator is a quadratic equation in terms of j2πf.
Gain
=
1
──────────────────────────────
(j2πf)(j2πf)LC + (j2πf)RC + 1
=
1
────────────────────────
(j2πf)2LC + (j2πf)RC + 1
As such, the denominator can be viewed as follows:
(j2πf)2LC + (j2πf)RC + 1
=
LCx2 + RCx + 1
where x = j2πf
The well-known quadratic-root equation can be used to determine the “roots” of a quadratic equation
as documented below:
If ax2 + bx + c = 0 then the values of x for which this is true (known
as the roots r1,2) are given by:
r1,2
=
-b ± (b2 – 4ac)1/2
─────────────────
2a
However, in our situation we do not desire to set the quadratic equation to zero – instead it is our
desire to factor the quadratic equation. But note that the quadratic-root equation can be used to
factor a quadratic equation if the quadratic equation is first divided by the coefficient of the squared
term (e.g., ‘a’). Symbolically this is illustrated below:
2
[ax2
+ bx + c]/a
=
x2 + (b/a)x + c/a
and then
x2 + (b/a)x + c/a
=
(x - r1)•(x – r2)
where r1,2 are given by the
quadratic-root equation (using
a’ = 1, b’ = b/a, and c’ = c/a)
Before illustrating the actual factoring process for the gain equation derived earlier, examine below
how the denominator of our gain equation, after applying the process just above, will be almost in
standard Bode form:
[(j2πf)2LC
+ (j2πf)RC + 1]/(LC) = (j2πf)2 + (j2πf)R/L + 1/(LC)
= (j2πf - r1)•(j2πf – r2)
The process of using the quadratic-root equation for factoring the denominator of the gain equation
previously derived will now be illustrated:
Gain
=
1
1/(LC)
──────────────────────── • ──────
1/(LC)
(j2πf)2LC + (j2πf)RC + 1
Gain
=
1/(LC)
────────────────────────────
(j2πf)2 + (j2πf)R/L + 1/(LC)
Gain
=
1/(LC)
───────────────────────
(j2πf - r1)•(j2πf – r2)
◄—— view as x2 + (R/L)x + 1/(LC)
and also as (x - r1)•(x – r2)
where r1,r2 are the roots
of the quadratic equation
where
r1,2
=
-b ± (b2 – 4ac)1/2
─────────────────
2a
with
a = 1,
b = R/L, and
c = 1/(LC)
Specifically:
r1,2
=
[-R/L ± (R /L
2
2
– 4/(LC)
) ]/2
1/2
The roots r1,2 will have the following characteristics:
► If R2/L2 – 4/(LC) > 0 then the roots r1,2 will both be real and
negative in value (they will be negative since –R/L ± (a value less
than R/L) = a negative value)
► If R2/L2 – 4/(LC) = 0 then the roots r1,2 will be equal, real and
negative in value (r1,2 = -R/(2L))
► If R2/L2 – 4/(LC) < 0 then the roots r1,2 will both be complex (due to
the square root of a negative number)
3
The last version of the gain equation will now be manipulated into standard Bode form:
Gain
=
1/(LC)
1/[(-r1)(-r2)]
─────────────────────── • ───────────────
1/[(-r1)(-r2)]
(j2πf - r1)•(j2πf – r2)
Gain
=
1/(LCr1r2)
────────────────────────────────
(j2πf/(-r1) + 1)•(j2πf/(-r2) + 1)
Gain
=
1/(LCr1r2)
────────────────────────────────
(1 + j2πf/(-r1))•(1 + j2πf/(-r2))
(and note that r1 and r2 will both be negative in value making the terms
-r1 and -r2 both positive (unless the roots are complex))
The general form of the Bode gain (magnitude) graph will now be made for each of the three
situations in regards to the roots of the denominator's quadratic equation.
► If the roots r1 and r2 are real and unequal in value (and they will be negative for our quadratic
equation) this will result in two unequal break-point frequencies in the denominator of the gain
equation:
Gain
-20db/decade
1 = 0db
-40db/decade
fBP1
fBP2
freq.
fBP1 = R/(4πL) – (R2/L2 – 4/(LC))1/2/(4π)
fBP2 = R/(4πL) + (R2/L2 – 4/(LC))1/2/(4π)
► If the roots r1 and r2 are real and equal in value (and they will be negative for our quadratic
equation) this will result in two equal break-point frequencies in the denominator of the gain
equation:
Gain
-40db/decade
1 = 0db
fBP
fBP = R/(4πL)
freq.
4
► If the roots r1 and r2 are complex this will result in a gain greater than 1 in the vicinity of a
double break-point frequency:
Gain
-40db/decade
1 = 0db
freq.
fBP
Although it seems counterintuitive that a passive circuit can have a gain greater than 1 – it is indeed
possible if the circuit contains capacitance and inductance as well as sufficiently low resistance. To
further illustrate this, we will begin by reexamining an earlier version of the transfer function for the
RLC low-pass filter circuit and viewing it from a slightly different perspective than before:
Gain
=
1
──────────────────────────────
(j2πf)(j2πf)LC + (j2πf)RC + 1
=
1
────────────────────────
-4π2f2LC + (j2πf)RC + 1
Note that j•j = -1 and this substitution is used in the first term of the denominator to make a
negative value (with no j). Doing this does not lead to transforming the transfer function into
standard Bode form – but it will allow us to more easily see why a gain greater than 1 can occur.
With this substitution we can see that there will be a frequency where the -4π2f2LC term will be
equal to -1 which, when added to the +1 term in the denominator, equals to 0 leaving only the
(j2πf)RC term in the denominator. Furthermore, if the value of R is sufficiently low, then the
(j2πf)RC term in the denominator can have a magnitude that is less than 1 making the gain
greater than 1 (because 1 divided by a number that is less than 1 will result in a value that is greater
than 1).
The preceding discussion noted that there is a frequency when the -4π2f2LC term in the
denominator will be equal to -1 which, when added to the +1 term in the denominator, equals to 0
leaving only the (j2πf)RC term in the denominator. Note that there will actually be a small range
of frequencies where the -4π2f2LC term will be near the value -1 thereby creating a relatively
small range of frequencies where the gain can be greater than 1 (as illustrated in the above
representative Bode gain graph).
Finally, note that the following changes can be made to the original second-order RLC low-pass filter
circuit (see page 1) resulting in the outcomes noted below:
► The inductor (L) and capacitor (C) can have their positions swapped
(i.e., the voltage across the inductor then creates Vo) resulting in a
second-order high-pass filter
► The resistor (R) and capacitor (C) can have their positions swapped
(i.e., the voltage across the resistor then creates Vo) resulting in a
second-order bandpass filter