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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

where Ω n is the undamped natural frequency , and ζ 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 ζ 0 . Using the formula for the roots of a quadratic polynomial, the two poles of 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

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

  1. Overdamped: the poles are real and distinct. This occurs if ζ > 1 . In this case the impulse response is given by:
    h ( t ) = Ω n 2 ζ 2 - 1 e s 1 t u ( t ) - Ω n 2 ζ 2 - 1 e s 2 t u ( t )
  2. Critically damped: corresponds to ζ = 1 . The two poles are repeated with,
    s 1 = s 2 = - ζ Ω n = - Ω n
    and the impulse response is given by
    h ( t ) = Ω n 2 t e - ζ Ω n t u ( t )
  3. Underdamped: corresponds to 0 ζ < 1 , giving a pair of complex conjugate poles. In this case, the impulse response is given by
    h ( t ) = Ω n 1 - ζ 2 e - ζ Ω n t sin 1 - ζ 2 Ω n t u ( t )
    Note that in the underdamped case, the magnitude of the poles is | s 1 | = | s 2 | = Ω n , and when ζ = 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 ) would take as the damping ratio is decreased from some number greater than 1 down to 0, and is shown in [link] .

Root locus of second-order lowpass filter having Ω n = 10 .

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

H j Ω = Ω n 2 Ω n 2 - Ω 2 + j 2 ζ Ω n Ω

The frequency response magnitude is shown in [link] for Ω n = 10 and several values of ζ . Note that for the underdamped case, there is a resonance or peak that would be expected to maximize at the undamped natural frequency Ω n , when ζ = 0 .

Frequency response of second-order lowpass filter with Ω n = 10 and several values of ζ .

We also observe that for the critically damped case, since

H ( s ) = Ω n 2 s + Ω n 2

setting s = j Ω gives H Ω n = 1 2 . Moreover, it is clear that for Ω < Ω n , H Ω n > 1 2 and for Ω > Ω n , H Ω n < 1 2 . These ideas will be used later when we discuss Bode plots.

Second-order filter implementation

Series RLC circuit.

Second order filters can be implemented using either passive or active circuit elements. We will consider the series RLC circuit shown in [link] . 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

Comparing [link] with [link] we find that the undamped natural frequency is Ω n = 1 L C , while 2 ζ Ω n = R L , giving the attenuation coefficient:

α ζ Ω n = R 2 L

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

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

The graph of the frequency response magnitude of this filter is shown in [link] for several damping ratios and Ω n = 10 .

Frequency response of second-order highpass filter with Ω n = 10 and several values of ζ .

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

This can be expressed in terms of ζ and Ω n as

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

Frequency response of second-order bandpass filter

Setting s = j Ω in [link] 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

The magnitude of this frequency response is shown in [link] where Ω n = 10 .

Frequency response of second-order bandpass filter with Ω n = 10 and several values of ζ .

Evidently, the frequency response peaks at Ω = Ω n . To prove this, we can divide the numerator and denominator of [link] by j 2 ζ Ω n Ω , 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 )


A ( Ω ) = Ω 2 - Ω n 2 2 ζ Ω n Ω

It is easy to check that A ( Ω n ) = 0 and A ( Ω ) > 0 , Ω Ω n . Therefore H b p ( j Ω ) does in fact peak at Ω = Ω n .

The bandwidth of bandpass filters is typically measured as the difference between the two frequencies, Ω 2 and Ω 1 , at which the gain has dropped by 3 decibels from its peak of 0 dB. A diagram of this is shown in [link] .

Definition of bandpass filter bandwidth, B W = Ω 2 - Ω 1 .

Formulas for Ω 1 and Ω 2 can be found by solving

A ( Ω ) = ± 1

since if A ( Ω ) = ± 1 , we get H ( j Ω ) = 1 2 . Solving the quadratic equation A ( Ω ) = 1 gives two solutions:

Ω ˜ 1 = ζ Ω n + Ω n 1 + ζ 2
Ω ˜ 2 = ζ Ω n - Ω n 1 + ζ 2

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

Ω ˜ 3 = - ζ Ω n + Ω n 1 + ζ 2
Ω ˜ 4 = - ζ Ω n - Ω n 1 + ζ 2

Since Ω n 1 + ζ 2 > ζ Ω n , we pick the two positive solutions as the “corner” frequencies:

Ω 1 = - ζ Ω n + Ω n 1 + ζ 2
Ω 2 = ζ Ω n + Ω n 1 + ζ 2

The two negative frequencies are just the negatives of the two positive frequencies (recall that H ( j Ω ) has even symmetry). So the bandwidth is B W = Ω 2 - Ω 1 = 2 ζ Ω n . It can also be readily verified that Ω n = Ω 1 Ω 2 , that is, the center frequency of the bandpass filter is the geometric mean of the two corner frequencies. The qualify factor , Q o is defined as

Q o = Ω n B W

and is a measure of the narrowness of the filter bandwidth, with respect to its center frequency. It can be seen that Q o = 1 2 ζ .

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Source:  OpenStax, Signals, systems, and society. OpenStax CNX. Oct 07, 2012 Download for free at http://cnx.org/content/col10965/1.15
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