ECE 211 · Lecture 23: Cutoff Frequencies of Second-Order Filters

§1 · One definition, applied three times

Every cutoff calculation starts from the same equation

The cutoff frequency $\omega_c$ is the frequency at which the magnitude of the frequency response falls to the passband gain divided by $\sqrt{2}$ . The passband gain is the maximum value that $|H(j\omega)|$ attains, so the definition reads

|H(j\omega_c)| = \frac{|H|_{\max}}{\sqrt{2}} \qquad\Longleftrightarrow\qquad |H(j\omega_c)|^2 = \frac{|H|_{\max}^2}{2}

The squared form on the right is the one to use in practice. Squaring removes every square root from the work, because the squared magnitude of a complex number is a sum of two real squares:

|x + jy|^2 = x^2 + y^2, \qquad \left|\frac{x_1 + jy_1}{x_2 + jy_2}\right|^2 = \frac{x_1^2 + y_1^2}{x_2^2 + y_2^2}

Lecture 22 established the standard form of the second-order filter, in which the numerator carries a single power of $s$ and the denominator carries the natural frequency $\omega_0$ and the damping ratio $\zeta$ :

H(s) = \frac{K s^N}{s^2 + 2\zeta\omega_0 s + \omega_0^2}, \qquad N = 0,\ 1,\ \text{or}\ 2

The two extreme gains follow by inspection, and they fix the filter type:

N	Transfer function	DC gain	Infinite-frequency gain	Type
0	K / (s² + 2ζω₀s + ω₀²)	K / ω₀²	0	low-pass
1	Ks / (s² + 2ζω₀s + ω₀²)	0	0	band-pass
2	Ks² / (s² + 2ζω₀s + ω₀²)	0	K	high-pass

The recipe, in five steps

The parameters are read off: $\omega_0^2$ is the constant term of the denominator, and $2\zeta\omega_0$ is the coefficient of $s$ .
The filter type is identified from $N$ , and the passband gain $|H|_{\max}$ is determined.
The substitution $s = j\omega_c$ is made, both sides are squared, and the fraction is cleared against $|H|_{\max}^2 / 2$ .
The result is a quadratic equation in $\omega_c^2$ , which is solved by the quadratic formula. The negative root is discarded, because $\omega_c^2$ cannot be negative.
The positive square root delivers $\omega_c$ in radians per second.

The passband gain is not always the DC gain

A caution is required at step 2. For $N = 0$ and $N = 2$ the passband gain equals the DC gain and the infinite-frequency gain respectively only when the response is monotonic, which happens when $\zeta \ge \tfrac{1}{\sqrt{2}} \approx 0.707$ . For smaller damping the response carries a peak that rises above both extremes, and that peak, not the DC value, is the passband gain. For $N = 1$ a peak is always present, and it always sits at $\omega_0$ .

The prototype in normalized coordinates. The horizontal axis is $\omega / \omega_0$ and the vertical axis is $|H(j\omega)|$ divided by the reference gain of its family: the DC gain $K/\omega_0^2$ for $N = 0$ , the quantity $K/\omega_0$ for $N = 1$ , and the infinite-frequency gain $K$ for $N = 2$ . The blue lines mark the half-power level and the resulting cutoff; the indigo dot marks the passband gain.

damping ratio ζ 0.5

ζ = 0.50, a low-pass response · passband gain = 1.155 × the reference gain · peak at ω/ω₀ = 0.71 · ω_c/ω₀ = 1.17

The DC gain is $K/\omega_0^2$ and the infinite-frequency gain is zero, so low frequencies are passed. The response is monotonically decreasing when $\zeta \ge 1/\sqrt{2}$ , in which case the passband gain is the DC gain. When $\zeta < 1/\sqrt{2}$ a peak of height $(K/\omega_0^2)\big/\big(2\zeta\sqrt{1 - \zeta^2}\big)$ appears at $\omega = \omega_0\sqrt{1 - 2\zeta^2}$ , and the passband gain is that peak.

Both extreme gains vanish, so only a band of intermediate frequencies is passed. The denominator of $|H(j\omega)|$ is smallest at $\omega = \omega_0$ for every damping ratio, so the passband gain is $K/(2\zeta\omega_0)$ and it is always reached at $\omega_0$ . Two half-power frequencies therefore exist, one on each side of the peak, and the band between them narrows as $\zeta$ decreases.

The DC gain is zero and the infinite-frequency gain is $K$ , so high frequencies are passed. The substitution $\omega \to \omega_0^2/\omega$ carries the low-pass magnitude into the high-pass magnitude, so the high-pass response is the mirror image of the low-pass response about $\omega_0$ : it peaks under exactly the same condition $\zeta < 1/\sqrt{2}$ , at $\omega = \omega_0 \big/ \sqrt{1 - 2\zeta^2}$ .

§2 · Worked example 1

A low-pass filter with a cutoff at 5.33 rad/s

The problem

Find the cutoff frequency of

H(s) = \frac{100}{s^2 + 80s + 400}

The transfer function is second order, because the denominator is a quadratic, and the numerator carries no power of $s$ .

Read the parameters from the denominator

\omega_0^2 = 400 \;\Longrightarrow\; \omega_0 = 20\ \text{rad/s}, \qquad 2\zeta\omega_0 = 80 \;\Longrightarrow\; 2\zeta(20) = 80 \;\Longrightarrow\; \zeta = 2

With $N = 0$ the filter is low-pass. Its DC gain is $100/400 = \tfrac14$ , and its infinite-frequency gain is zero.

The passband gain, and the level to solve for

Because $\zeta = 2$ exceeds $1/\sqrt{2}$ , the response is monotonically decreasing and the passband gain is the DC gain, $\tfrac14$ . The half-power level is therefore

\frac{|H|_{\max}}{\sqrt{2}} = \frac{1}{4\sqrt{2}} \approx 0.177

Substitute, square, and clear the fraction

\left|\tm{lp.cv}{\frac{100}{-\omega_c^2 + j80\omega_c + 400}}\right|^2 = \left(\tm{lp.hp}{\frac{1}{4\sqrt{2}}}\right)^2 \;\Longrightarrow\; \frac{10{,}000}{(400 - \omega_c^2)^2 + (80\omega_c)^2} = \frac{1}{32}

Cross-multiplying and expanding the square gives

320{,}000 = 160{,}000 - 800\omega_c^2 + \omega_c^4 + 6400\omega_c^2 \;\Longrightarrow\; \omega_c^4 + 5600\,\omega_c^2 - 160{,}000 = 0

Solve the quadratic in $\omega_c^2$

\omega_c^2 = \frac{-5600 \pm \sqrt{5600^2 + 4(160{,}000)}}{2} = \frac{-5600 \pm 4000\sqrt{2}}{2}

The lower sign produces a negative number, which is discarded because $\omega_c^2$ is a square. The upper sign gives $\omega_c^2 = 2000\sqrt{2} - 2800 = 28.43$ , so

\boxed{\ \omega_c = 5.33\ \text{rad/s}\ }

Check: evaluate the response at the answer

At $\omega = 5.3317$ rad/s the denominator of the frequency response is $400 - 28.43 = 371.57$ in its real part and $80(5.3317) = 426.54$ in its imaginary part, so

|H(j\omega_c)| = \frac{100}{\sqrt{371.57^2 + 426.54^2}} = \frac{100}{565.69} = 0.1768

and $\tfrac14 / \sqrt{2} = 0.1768$ ✓. The board notes round the answer to 5.3 rad/s; the unrounded value is 5.332 rad/s.

§3 · Worked example 2

A high-pass filter with a cutoff at 58.33 rad/s

The problem

Find the cutoff frequency of

H(s) = \frac{5s^2}{s^2 + 60s + 100}

The numerator is $Ks^N$ with $K = 5$ and $N = 2$ , so the prototype table applies directly.

Read the parameters, and the two extreme gains

\omega_0^2 = 100 \;\Longrightarrow\; \omega_0 = 10\ \text{rad/s}, \qquad 2\zeta(10) = 60 \;\Longrightarrow\; \zeta = 3

The DC gain is zero and the infinite-frequency gain is $K = 5$ , so the filter is high-pass.

The passband gain, and the level to solve for

Because $\zeta = 3$ exceeds $1/\sqrt{2}$ , the response climbs monotonically toward its asymptote and the passband gain is 5. The half-power level is $5/\sqrt{2} \approx 3.54$ .

Substitute, square, and clear the fraction

\left|\tm{hf.cv}{\frac{-5\omega_c^2}{-\omega_c^2 + j60\omega_c + 100}}\right|^2 = \left(\tm{hf.hp}{\frac{5}{\sqrt{2}}}\right)^2 \;\Longrightarrow\; \frac{25\,\omega_c^4}{(100 - \omega_c^2)^2 + (60\omega_c)^2} = \frac{25}{2}

The factor of 25 cancels. Cross-multiplying,

2\omega_c^4 = 10{,}000 - 200\omega_c^2 + \omega_c^4 + 3600\omega_c^2 \;\Longrightarrow\; \omega_c^4 - 3400\,\omega_c^2 - 10{,}000 = 0

Solve the quadratic in $\omega_c^2$

\omega_c^2 = \frac{3400 \pm \sqrt{3400^2 + 4(10{,}000)}}{2} = 1700 \pm 100\sqrt{290}

The lower sign gives $-2.94$ , which is discarded. The upper sign gives $\omega_c^2 = 3402.9$ , so

\boxed{\ \omega_c = 58.33\ \text{rad/s}\ }

Where the high-pass differs from the low-pass

At $s = j\omega_c$ the numerator becomes $-5\omega_c^2$ , and its squared magnitude, $25\omega_c^4$ , sits on the left of the equation. Clearing the fraction therefore cancels most of the $\omega_c^4$ contributed by the denominator, which is why the coefficient of $\omega_c^2$ comes out negative here and positive in Section 2, and why the admissible root lands above $\omega_0$ rather than below it. The minus sign on $-5\omega_c^2$ is squared away in the magnitude, but it must be kept whenever a phase is also wanted.

§4 · Worked example 3

A numerator that is not a prototype

The transfer function of this example carries a zero at $s = -18$ rather than at the origin, so the table of Section 1 does not apply. The definition still does, and it is applied in exactly the same order.

The problem

Find the cutoff frequency of

H(s) = \frac{s + 18}{s^2 + 12s + 36}

The denominator is still second order, so $\omega_0$ and $\zeta$ may still be read off; the numerator, however, is neither $K$ , nor $Ks$ , nor $Ks^2$ .

Read the parameters, and the two extreme gains

\omega_0^2 = 36 \;\Longrightarrow\; \omega_0 = 6\ \text{rad/s}, \qquad 2\zeta(6) = 12 \;\Longrightarrow\; \zeta = 1

The DC gain is $18/36 = \tfrac12$ and the infinite-frequency gain is zero, because the denominator grows as $\omega^2$ while the numerator grows only as $\omega$ . The behavior is therefore low-pass.

The passband gain follows from a factored denominator

Because $\zeta = 1$ , the denominator is a perfect square, $s^2 + 12s + 36 = (s + 6)^2$ , and the squared magnitude simplifies:

|H(j\omega)|^2 = \frac{\tm{mx.cv}{18^2 + \omega^2}}{\left(36 - \omega^2\right)^2 + 144\omega^2} = \frac{324 + \omega^2}{\left(\omega^2 + 36\right)^2}

This quantity decreases at every frequency, as its derivative with respect to $\omega^2$ , namely $-(\omega^2 + 612)/(\omega^2 + 36)^3$ , confirms, so the passband gain is the DC gain, $\tfrac12$ , and the half-power level is $1/(2\sqrt{2}) \approx 0.354$ .

Substitute, square, and clear the fraction

\frac{18^2 + \omega_c^2}{(36 - \omega_c^2)^2 + 12^2\omega_c^2} = \left(\tm{mx.hp}{\frac{1}{2\sqrt{2}}}\right)^2 = \frac{1}{8} \;\Longrightarrow\; 8\left(324 + \omega_c^2\right) = 1296 - 72\omega_c^2 + \omega_c^4 + 144\omega_c^2

\omega_c^4 + 64\,\omega_c^2 - 1296 = 0

Solve the quadratic in $\omega_c^2$

\omega_c^2 = \frac{-64 \pm \sqrt{64^2 + 4(1296)}}{2} = \frac{-64 \pm \sqrt{9280}}{2}

The lower sign is negative and is discarded; the upper sign gives $\omega_c^2 = 16.17$ , so

\boxed{\ \omega_c = 4.02\ \text{rad/s}\ }

Check: evaluate the response at the answer

At $\omega_c^2 = 16.17$ ,

|H(j\omega_c)| = \frac{\sqrt{324 + 16.17}}{16.17 + 36} = \frac{18.44}{52.17} = 0.3536

and $\tfrac12 / \sqrt{2} = 0.3536$ ✓. The board notes round the answer to 4 rad/s; the unrounded value is 4.021 rad/s.

§5 · The same three answers in closed form

Normalizing the frequency removes the numbers from the algebra

The three examples repeat one calculation with different constants, so the calculation is worth performing once in symbols. Let $u = \omega/\omega_0$ and $x = u^2$ . For the prototype with $N = 0$ , dividing the squared magnitude by the square of the DC gain gives

\frac{|H(j\omega)|^2}{\left(K/\omega_0^2\right)^2} = \frac{1}{(1 - x)^2 + 4\zeta^2 x}

Setting this equal to half of the squared passband gain produces a quadratic in $x$ , whose positive root is the cutoff. Two cases arise, according to whether the response peaks:

\boxed{\ \omega_c = \omega_0\sqrt{a + \sqrt{a^2 + 1}}\ }\quad \left(\zeta \ge \tfrac{1}{\sqrt{2}}\right), \qquad \boxed{\ \omega_c = \omega_0\sqrt{a + 2\zeta\sqrt{1 - \zeta^2}}\ }\quad \left(\zeta < \tfrac{1}{\sqrt{2}}\right), \qquad a = 1 - 2\zeta^2

The high-pass case needs no separate derivation. The substitution $\omega \to \omega_0^2/\omega$ exchanges the two magnitudes, so at equal damping the two cutoffs satisfy $\omega_{c,\text{LP}}\,\omega_{c,\text{HP}} = \omega_0^2$ , and the high-pass answer is obtained by inverting the low-pass one. The band-pass case factors exactly:

\omega_{c1},\ \omega_{c2} = \omega_0\left(\sqrt{1 + \zeta^2} \mp \zeta\right), \qquad \omega_{c2} - \omega_{c1} = 2\zeta\omega_0, \qquad \omega_{c1}\,\omega_{c2} = \omega_0^2

The bandwidth of a second-order band-pass filter is therefore exactly the coefficient of $s$ in its denominator, and $\omega_0$ is the geometric mean of the two cutoffs rather than the arithmetic mean.

Filter	ω₀	ζ	Closed form	ω_c (rad/s)
§2 low-pass	20	2	20 √(−7 + √50)	5.33
§3 high-pass	10	3	10 / √(−17 + √290)	58.33
Lecture 22 band-pass	100	1	100 (√2 ∓ 1)	41.42 and 241.42

The closed form is a check, not a substitute

The examination question is the derivation, not the formula. Section 4 is the reminder: its numerator is $s + 18$ , the normalization above does not apply to it, and only the definition $|H(j\omega_c)|^2 = |H|_{\max}^2/2$ produces the answer. The closed forms are useful for confirming an answer quickly once the quadratic has been solved by hand.

§6 · Your turn

Practice: a low-pass filter that peaks

The examples of this lecture were all monotonic, because every damping ratio exceeded $1/\sqrt{2}$ . The problem below is not, so step 2 of the recipe carries the whole difficulty. The problem should be attempted on paper before the hints are opened, in order.

H(s) = \frac{200}{s^2 + 10s + 100}

Find the filter type, the natural frequency, the damping ratio, the passband gain, and the cutoff frequency.

Hint 1: the parameters

The constant term gives $\omega_0^2 = 100$ , so $\omega_0 = 10$ rad/s, and the coefficient of $s$ gives $2\zeta(10) = 10$ , so $\zeta = 0.5$ . With $N = 0$ the filter is low-pass, and its DC gain is $200/100 = 2$ .

Hint 2: the DC gain is not the passband gain

Since $\zeta = 0.5 < 1/\sqrt{2}$ , the response peaks. Section 1 gives the peak frequency $\omega_0\sqrt{1 - 2\zeta^2}$ and the peak height $(K/\omega_0^2)\big/\big(2\zeta\sqrt{1 - \zeta^2}\big)$ . Evaluate both before writing the cutoff equation, and use the peak height, not the DC gain, on the right-hand side.

Hint 3: the equation to solve

With $|H|_{\max} = 4/\sqrt{3}$ , the equation to clear is

\frac{200^2}{(100 - \omega_c^2)^2 + (10\omega_c)^2} = \frac{1}{2}\left(\frac{4}{\sqrt{3}}\right)^2 = \frac{8}{3}

Solution

Parameters. $\omega_0 = 10$ rad/s, $\zeta = 0.5$ , $N = 0$ : a second-order low-pass filter with a DC gain of 2 and an infinite-frequency gain of zero.

Passband gain. Because $\zeta < 1/\sqrt{2}$ the response peaks, at

\omega_p = \omega_0\sqrt{1 - 2\zeta^2} = 10\sqrt{0.5} = 7.07\ \text{rad/s}, \qquad |H|_{\max} = \frac{2}{2\zeta\sqrt{1 - \zeta^2}} = \frac{2}{\sqrt{3}/2} = \frac{4}{\sqrt{3}} = 2.309

The peak stands 1.155 times above the DC gain, which is less than $\sqrt{2}$ , so direct current still lies inside the passband.

Cutoff. Clearing the fraction in Hint 3, with $u = \omega_c^2$ ,

15{,}000 = 10{,}000 - 200u + u^2 + 100u \;\Longrightarrow\; u^2 - 100u - 5000 = 0 \;\Longrightarrow\; u = \frac{100 + 100\sqrt{3}}{2} = 136.6

\boxed{\ \omega_c = \sqrt{50\left(1 + \sqrt{3}\right)} = 11.69\ \text{rad/s}\ }

Check. At $\omega_c^2 = 136.6$ the squared magnitude is $40{,}000 / \left[(100 - 136.6)^2 + 100(136.6)\right] = 40{,}000/15{,}000 = 2.667$ , so $|H(j\omega_c)| = 1.633$ , and $|H|_{\max}/\sqrt{2} = 2.309/1.414 = 1.633$ ✓. The closed form of Section 5 agrees: with $a = 1 - 2(0.25) = 0.5$ , $\omega_c = 10\sqrt{0.5 + 2(0.5)\sqrt{0.75}} = 10\sqrt{1.366} = 11.69$ rad/s.

Note that the cutoff exceeds $\omega_0$ , and that it would be badly underestimated if the DC gain had been used as the passband gain: that error would place the half-power level at 1.414 instead of 1.633, and the answer at 12.72 rad/s.

Every cutoff calculation starts from the same equation

The passband gain is not always the DC gain

A low-pass filter with a cutoff at 5.33 rad/s

The problem

Read the parameters from the denominator

The passband gain, and the level to solve for

Substitute, square, and clear the fraction

Solve the quadratic in ωc2\omega_c^2ωc2​

A high-pass filter with a cutoff at 58.33 rad/s

The problem

Read the parameters, and the two extreme gains

The passband gain, and the level to solve for

Substitute, square, and clear the fraction

Solve the quadratic in ωc2\omega_c^2ωc2​

A numerator that is not a prototype

The problem

Read the parameters, and the two extreme gains

The passband gain follows from a factored denominator

Substitute, square, and clear the fraction

Solve the quadratic in ωc2\omega_c^2ωc2​

Normalizing the frequency removes the numbers from the algebra

Practice: a low-pass filter that peaks

Solve the quadratic in $\omega_c^2$

Solve the quadratic in $\omega_c^2$

Solve the quadratic in $\omega_c^2$