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Inside Collection (Textbook):

Textbook by: Carlos E. Davila. E-mail the author

# Second-Order Filters

Module by: Carlos E. Davila. E-mail the author

Summary: Summarizes second order lowpass, bandpass, and highpass fitlers.

## Second-Order Lowpass Filters

The second-order lowpass filter has system function

H ( s ) = Ω n 2 s 2 + 2 ζ Ω n s + Ω n 2 H ( s ) = Ω n 2 s 2 + 2 ζ Ω n s + Ω n 2
(1)

where ΩnΩn is the undamped natural frequency, and ζζ is the damping ratio. The undamped natural frequency and damping ratio are properties of the physical devices used to implement the second-order filter (capacitors, inductors, and resistors, in the case of an electrical circuit). It so happens that the damping ratio satisfies ζ0ζ0. Using the formula for the roots of a quadratic polynomial, the two poles of H(s)H(s) are easily found to be

s 1 = - ζ Ω n + Ω n ζ 2 - 1 s 2 = - ζ Ω n - Ω n ζ 2 - 1 s 1 = - ζ Ω n + Ω n ζ 2 - 1 s 2 = - ζ Ω n - Ω n ζ 2 - 1
(2)

There are three possible sets of poles that are categorized as follows:

1. Overdamped: the poles are real and distinct. This occurs if ζ>1ζ>1. In this case the impulse response is given by:
h(t)=Ωn2ζ2-1es1tu(t)-Ωn2ζ2-1es2tu(t)h(t)=Ωn2ζ2-1es1tu(t)-Ωn2ζ2-1es2tu(t)
(3)
2. Critically damped: corresponds to ζ=1ζ=1. The two poles are repeated with,
s1=s2=-ζΩn=-Ωns1=s2=-ζΩn=-Ωn
(4)
and the impulse response is given by
h(t)=Ωn2te-ζΩntu(t)h(t)=Ωn2te-ζΩntu(t)
(5)
3. Underdamped: corresponds to 0ζ<10ζ<1, giving a pair of complex conjugate poles. In this case, the impulse response is given by
h(t)=Ωn1-ζ2e-ζΩntsin1-ζ2Ωntu(t)h(t)=Ωn1-ζ2e-ζΩntsin1-ζ2Ωntu(t)
(6)
Note that in the underdamped case, the magnitude of the poles is |s1|=|s2|=Ωn|s1|=|s2|=Ωn, and when ζ=0ζ=0, the two poles are on the imaginary axis which corresponds to an impulse response that is a pure sinusoid and the system is unstable.

A root locus diagram shows the paths that the poles of H(s)H(s) would take as the damping ratio is decreased from some number greater than 1 down to 0, and is shown in Figure 1.

The frequency response of the second-order lowpass filter can be found using the substitution H(jΩ)=H(s)|s=jΩH(jΩ)=H(s)|s=jΩ, giving

H j Ω = Ω n 2 Ω n 2 - Ω 2 + j 2 ζ Ω n Ω H j Ω = Ω n 2 Ω n 2 - Ω 2 + j 2 ζ Ω n Ω
(7)

The frequency response magnitude is shown in Figure 2 for Ωn=10Ωn=10 and several values of ζζ. Note that for the underdamped case, there is a resonance or peak that would be expected to maximize at the undamped natural frequency ΩnΩn, when ζ=0ζ=0.

We also observe that for the critically damped case, since

H ( s ) = Ω n 2 s + Ω n 2 H ( s ) = Ω n 2 s + Ω n 2
(8)

setting s=jΩs=jΩ gives HΩn=12HΩn=12. Moreover, it is clear that for Ω<ΩnΩ<Ωn, HΩn>12HΩn>12 and for Ω>ΩnΩ>Ωn, HΩn<12HΩn<12. These ideas will be used later when we discuss Bode plots.

## Second-Order Filter Implementation

Second order filters can be implemented using either passive or active circuit elements. We will consider the series RLC circuit shown in Figure 3. A lowpass filter results by taking the filter output to be the voltage across the capacitor. Using voltage division, the system function is easily found to be

H l p ( s ) = V l p ( s ) V i ( s ) = 1 L C s 2 + R L s + 1 L C H l p ( s ) = V l p ( s ) V i ( s ) = 1 L C s 2 + R L s + 1 L C
(9)

Comparing Equation 9 with Equation 1 we find that the undamped natural frequency is Ωn=1LCΩn=1LC, while 2ζΩn=RL2ζΩn=RL, giving the attenuation coefficient:

α ζ Ω n = R 2 L α ζ Ω n = R 2 L
(10)

A second-order highpass filter is implemented by taking the output of the filter to be the inductor voltage in the series RLC circuit. The resulting system function is

H h p ( s ) = V h p ( s ) V i ( s ) = s 2 s 2 + R L s + 1 L C H h p ( s ) = V h p ( s ) V i ( s ) = s 2 s 2 + R L s + 1 L C
(11)

So in terms of the undamped natural frequency and damping ratio, the second-order highpass filter is given by

H h p ( s ) = s 2 s 2 + 2 ζ Ω n s + Ω n 2 H h p ( s ) = s 2 s 2 + 2 ζ Ω n s + Ω n 2
(12)

The graph of the frequency response magnitude of this filter is shown in Figure 4 for several damping ratios and Ωn=10Ωn=10.

Not surprisingly, a second-order bandpass filter results by taking the output of the filter be the resistor voltage in the series RLC circuit. The system function is then

H b p ( s ) = V b p ( s ) V i ( s ) = R L s s 2 + R L s + 1 L C H b p ( s ) = V b p ( s ) V i ( s ) = R L s s 2 + R L s + 1 L C
(13)

This can be expressed in terms of ζζ and ΩnΩn as

H b p ( s ) = 2 ζ Ω n s s 2 + 2 ζ Ω n s + Ω n 2 H b p ( s ) = 2 ζ Ω n s s 2 + 2 ζ Ω n s + Ω n 2
(14)

## Frequency Response of Second-Order Bandpass Filter

Setting s=jΩs=jΩ in Equation 14 gives the frequency response of the second-order bandpass filter

H b p ( j Ω ) = j 2 ζ Ω n Ω Ω n 2 - Ω 2 + j 2 ζ Ω n Ω H b p ( j Ω ) = j 2 ζ Ω n Ω Ω n 2 - Ω 2 + j 2 ζ Ω n Ω
(15)

The magnitude of this frequency response is shown in Figure 5 where Ωn=10Ωn=10.

Evidently, the frequency response peaks at Ω=ΩnΩ=Ωn. To prove this, we can divide the numerator and denominator of Equation 15 by j2ζΩnΩj2ζΩnΩ, giving

H b p ( j Ω ) = 1 1 - j Ω n 2 - Ω 2 2 ζ Ω n Ω = 1 1 + j Ω 2 - Ω n 2 2 ζ Ω n Ω = 1 1 + j A ( Ω ) H b p ( j Ω ) = 1 1 - j Ω n 2 - Ω 2 2 ζ Ω n Ω = 1 1 + j Ω 2 - Ω n 2 2 ζ Ω n Ω = 1 1 + j A ( Ω )
(16)

where

A ( Ω ) = Ω 2 - Ω n 2 2 ζ Ω n Ω A ( Ω ) = Ω 2 - Ω n 2 2 ζ Ω n Ω
(17)

It is easy to check that A(Ωn)=0A(Ωn)=0 and A(Ω)>0,ΩΩnA(Ω)>0,ΩΩn. Therefore Hbp(jΩ)Hbp(jΩ) does in fact peak at Ω=ΩnΩ=Ωn.

The bandwidth of bandpass filters is typically measured as the difference between the two frequencies, Ω2Ω2 and Ω1Ω1, at which the gain has dropped by 3 decibels from its peak of 0 dB. A diagram of this is shown in Figure 6.

Formulas for Ω1Ω1 and Ω2Ω2 can be found by solving

A ( Ω ) = ± 1 A ( Ω ) = ± 1
(18)

since if A(Ω)=±1A(Ω)=±1, we get H(jΩ)=12H(jΩ)=12. Solving the quadratic equation A(Ω)=1A(Ω)=1 gives two solutions:

Ω ˜ 1 = ζ Ω n + Ω n 1 + ζ 2 Ω ˜ 1 = ζ Ω n + Ω n 1 + ζ 2
(19)
Ω ˜ 2 = ζ Ω n - Ω n 1 + ζ 2 Ω ˜ 2 = ζ Ω n - Ω n 1 + ζ 2
(20)

while solving A(Ω)=-1A(Ω)=-1 gives two additional solutions

Ω ˜ 3 = - ζ Ω n + Ω n 1 + ζ 2 Ω ˜ 3 = - ζ Ω n + Ω n 1 + ζ 2
(21)
Ω ˜ 4 = - ζ Ω n - Ω n 1 + ζ 2 Ω ˜ 4 = - ζ Ω n - Ω n 1 + ζ 2
(22)

Since Ωn1+ζ2>ζΩnΩn1+ζ2>ζΩn, we pick the two positive solutions as the “corner” frequencies:

Ω 1 = - ζ Ω n + Ω n 1 + ζ 2 Ω 1 = - ζ Ω n + Ω n 1 + ζ 2
(23)
Ω 2 = ζ Ω n + Ω n 1 + ζ 2 Ω 2 = ζ Ω n + Ω n 1 + ζ 2
(24)

The two negative frequencies are just the negatives of the two positive frequencies (recall that H(jΩ)H(jΩ) has even symmetry). So the bandwidth is BW=Ω2-Ω1=2ζΩnBW=Ω2-Ω1=2ζΩn. It can also be readily verified that Ωn=Ω1Ω2Ωn=Ω1Ω2, that is, the center frequency of the bandpass filter is the geometric mean of the two corner frequencies. The qualify factor, QoQo is defined as

Q o = Ω n B W Q o = Ω n B W
(25)

and is a measure of the narrowness of the filter bandwidth, with respect to its center frequency. It can be seen that Qo=12ζQo=12ζ.

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