ECE 211 · Lecture 22: Filters

§1 · What a filter is

A filter is a transfer function chosen for its magnitude response

A filter is a circuit whose transfer function is designed so that some bands of frequency reach the output and others do not. Only one block is involved: an input is applied, a transfer function acts on it, and an output is produced. Everything that follows is a statement about $|H(j\omega)|$ , the magnitude of that transfer function on the imaginary axis.

The filter as a single block. The transfer function $H(s)$ relates the input to the output; the frequency response $H(j\omega)$ is the same function evaluated at $s = j\omega$ , as established in Lecture 19.

The four descriptors

Four quantities describe a filter completely enough for the purposes of this course, and every worked example on this page reports all four.

Type: LP, HP, or BP Order: how many energy storage devices Passband gain: the maximum value Cutoff frequency: ω_c

The order is counted as the number of energy storage devices, that is, the number of capacitors and inductors. A higher order produces a response closer to the ideal, because the transition between the passband and the stopband is made steeper; the price is a larger circuit.

The half-power definition of the cutoff

The cutoff frequency $\omega_c$ is the frequency at which the magnitude has fallen to $1/\sqrt{2}$ of its maximum value:

|H(j\omega_c)| = \frac{\text{max}}{\sqrt{2}}

Because power is proportional to the square of an amplitude, a factor of $1/\sqrt{2}$ in amplitude is a factor of $1/2$ in power, which is why $\omega_c$ is also called the half-power frequency. In the decibel measure of Lecture 20 the same point sits $3.01$ dB below the maximum.

The ideal and the realizable

The ideal filter passes its band with a constant gain and rejects everything else completely, which draws as a rectangle. No finite circuit produces a rectangle; a realizable response falls off smoothly, and the cutoff frequency is the agreed marker on that smooth fall.

The ideal response in blue and a first-order or second-order realizable response in orange, both normalized to a passband gain of 1. The dashed level is $\text{max}/\sqrt{2}$ , and the marked frequencies are the cutoffs. Select a type:

The low-pass filter passes frequencies below $\omega_c$ and attenuates those above it. The maximum is reached at DC, so the passband gain is $|H(j0)|$ , and the response falls monotonically from there. The single cutoff frequency is the frequency at which the fall reaches $\text{max}/\sqrt{2}$ .

The high-pass filter attenuates frequencies below $\omega_c$ and passes those above it. The maximum is approached as $\omega \to \infty$ , so the passband gain is $|H(j\infty)|$ , and the response rises monotonically toward it. The cutoff frequency is again the half-power point.

The band-pass filter passes a band between two cutoff frequencies, $\omega_{c1}$ and $\omega_{c2}$ , and attenuates both outside them. The maximum is reached at an interior frequency $\omega_0$ , and the half-power level is therefore crossed twice, once on the way up and once on the way down. Two cutoff frequencies are reported rather than one.

§2 · First-order filters

One pole, two options

A first-order filter contains one energy storage device, so its transfer function has one pole. With the pole written at $s = -\alpha$ , only two numerators are available if the transfer function is to remain proper: the numerator may carry $s^1$ or $s^0$ . Those two options are the whole of the first-order case.

H(s) = \frac{Ks}{s + \alpha} \qquad\Longrightarrow\qquad H(j\omega) = \frac{K \cdot j\omega}{j\omega + \alpha}

The DC gain vanishes and the infinite-frequency gain is $K$ :

|H(j0)| = 0, \qquad |H(j\infty)| = K

A response that is zero at DC and $K$ at high frequency is a high-pass filter, and its passband gain is $K$ .

H(s) = \frac{K}{s + \alpha} \qquad\Longrightarrow\qquad H(j\omega) = \frac{K}{j\omega + \alpha}

The DC gain is $K/\alpha$ and the infinite-frequency gain vanishes:

|H(j0)| = \frac{K}{\alpha}, \qquad |H(j\infty)| = 0

A response that is $K/\alpha$ at DC and zero at high frequency is a low-pass filter, and its passband gain is $K/\alpha$ .

The cutoff frequency is the pole

The half-power condition is applied to each option in turn. Squaring both sides removes the square root of the magnitude, and in both cases the constant $K$ cancels, which is the reason the answer does not depend on it.

\left|\frac{jK\omega_c}{j\omega_c + \alpha}\right|^2 = \left(\frac{K}{\sqrt{2}}\right)^2 \;\Longrightarrow\; \frac{K^2 \omega_c^2}{\alpha^2 + \omega_c^2} = \frac{K^2}{2}

\frac{\omega_c^2}{\alpha^2 + \omega_c^2} = \frac{1}{2} \;\Longrightarrow\; 2\omega_c^2 = \alpha^2 + \omega_c^2 \;\Longrightarrow\; \boxed{\omega_c = \alpha}

\left|\frac{K}{j\omega_c + \alpha}\right|^2 = \left(\frac{K/\alpha}{\sqrt{2}}\right)^2 \;\Longrightarrow\; \frac{K^2}{\alpha^2 + \omega_c^2} = \frac{K^2}{2\alpha^2}

2\alpha^2 = \alpha^2 + \omega_c^2 \;\Longrightarrow\; \alpha^2 = \omega_c^2 \;\Longrightarrow\; \boxed{\alpha = \omega_c}

The same result is obtained in both cases: the cutoff frequency of a first-order filter is the magnitude of its pole. The numerator selects the type and the passband gain; the denominator alone fixes the cutoff.

Option 1, high-pass. The response crosses max/√2 at ω_c = α.

Option 2, low-pass. The response crosses max/√2 at the same ω_c = α.

α, the pole (rad/s) 4

ω_c = α = 4.00 rad/s f_c = ω_c2π = 0.64 Hz

Both plots are divided by their own passband gain, so that the one feature they are meant to display, the crossing at $\omega_c = \alpha$ , is comparable whatever $K$ happens to be. Moving the slider moves the pole, and the crossing follows it in both plots at once.

§3 · Worked example A: a first-order transfer function

$H(s) = \dfrac{4s + 1}{s + 2}$

The problem, and the order

The magnitude response of

H(s) = \frac{4s + 1}{s + 2} \qquad\Longrightarrow\qquad H(j\omega) = \frac{4 \cdot j\omega + 1}{j\omega + 2}

is to be classified, and its cutoff frequency is to be found. The transfer function has one pole, so this is a first-order filter.

The DC gain

Setting $\omega = 0$ leaves only the constant terms:

H(j0) = \frac{1}{2}

The response does not vanish at DC, so this is not the textbook high-pass of Section 2. The value $1/2$ is the floor of the response, not its maximum.

The infinite-frequency gain, which is the maximum

As $\omega \to \infty$ the two constant terms become negligible against the terms in $\omega$ :

H(j\infty) = \frac{4 \cdot j\omega}{j\omega} = 4

The magnitude rises monotonically from $1/2$ to $4$ , so the response is high-pass in character and the maximum value is $4$ . The passband gain is therefore $4$ , and the half-power construction is applied to that number, not to any other.

The half-power level

|H(j\omega_c)| = \frac{\text{max}}{\sqrt{2}} = \frac{4}{\sqrt{2}} \approx 2.83

The dashed level is drawn at $2.83$ . Because the response rises from $1/2$ and $1/2 < 2.83 < 4$ , the level is crossed exactly once, and one cutoff frequency exists.

Solving the half-power condition

\left|\frac{j4\omega_c + 1}{j\omega_c + 2}\right|^2 = \left(\frac{4}{\sqrt{2}}\right)^2 \;\Longrightarrow\; \tm{e1.mag}{\frac{16\omega_c^2 + 1}{\omega_c^2 + 4}} = 8

16\omega_c^2 + 1 = 8\omega_c^2 + 32 \;\Longrightarrow\; 8\omega_c^2 = 31 \;\Longrightarrow\; \boxed{\omega_c = \sqrt{\tfrac{31}{8}} \approx 1.97\ \text{rad/s}}

Type: high-pass Order: 1 Passband gain: 4 ω_c = 1.97 rad/s

The cutoff is measured from the maximum, not from unity

The half-power level is $\text{max}/\sqrt{2}$ , and the maximum here is $4$ , so the level is $2.83$ and not $0.707$ . A response whose floor is nonzero, as this one is, is a shelving response: it does not reject the stopband completely, it merely attenuates it by the ratio $4 : 1/2$ , that is, by a factor of eight.

Check: the value at the computed cutoff

Substituting $\omega_c^2 = 31/8$ back into the magnitude gives

|H(j\omega_c)|^2 = \frac{16 \cdot \tfrac{31}{8} + 1}{\tfrac{31}{8} + 4} = \frac{62 + 1}{\tfrac{63}{8}} = \frac{63 \cdot 8}{63} = 8 = \left(\frac{4}{\sqrt{2}}\right)^2\ \checkmark

The arithmetic is exact, which confirms both the algebra and the rounding to $1.97$ rad/s.

§4 · The first-order RC ladders

Two resistors and one capacitor, read at the two extremes

The two first-order options are realized by the same two components arranged in the two possible orders. The type of each ladder is settled without any algebra at all, by replacing the capacitor with its DC and its high-frequency limits.

The two limits of a capacitor

At DC the capacitor impedance $1/(j\omega C)$ is infinite, so the capacitor behaves as an open circuit. As $\omega \to \infty$ the same impedance vanishes, so the capacitor behaves as a short circuit. The inductor behaves in the opposite way, as established in Lecture 8.

The two ladders

In the left ladder the resistor is in series and the capacitor sits across the output. In the right ladder the capacitor is in series and the resistor sits across the output. Each is a voltage divider between two impedances, driven by $V_s$ and read at $V_o$ .

At DC the capacitor is an open circuit

With the capacitor open, no current flows in either ladder, so no voltage drops across the series element. In the left ladder the output therefore sits at the source, $V_o = V_s$ . In the right ladder the output is disconnected from the source and sits at $V_o = 0$ .

At infinite frequency the capacitor is a short circuit

With the capacitor shorted, the left ladder has its output terminals tied together, so $V_o = 0$ . In the right ladder the source is connected straight through to the resistor, so $V_o = V_s$ .

The two transfer functions

The two readings identify the types without algebra: the left ladder passes DC and rejects high frequency, so it is a low-pass filter; the right ladder does the opposite, so it is a high-pass filter. The voltage divider supplies the details:

H_{\text{LP}} = \frac{1/j\omega C}{R + 1/j\omega C} = \frac{1}{1 + j\omega RC} \qquad H_{\text{HP}} = \frac{R}{R + 1/j\omega C} = \frac{j\omega RC}{1 + j\omega RC}

Both denominators are the same, so both ladders have the same pole, and by the result of Section 2 both have the same cutoff frequency. Writing the low-pass form in the standard shape of Section 2 makes the pole explicit:

H_{\text{LP}}(s) = \frac{1}{1 + sRC} = \frac{1/RC}{s + 1/RC} \qquad\Longrightarrow\qquad K = \alpha = \frac{1}{RC} \;\Longrightarrow\; \boxed{\omega_c = \frac{1}{RC}}

The passband gains follow from the same reading: the low-pass ladder has $K/\alpha = 1$ and the high-pass ladder approaches $1$ , so neither ladder amplifies. A passive RC ladder can only attenuate, which is one reason the operational amplifier of Lecture 14 is added when gain is wanted as well as filtering.

§5 · Second-order filters

The series RLC and the standard form

A second-order filter contains two energy storage devices. The canonical example is the series RLC circuit, in which one inductor and one capacitor are placed in series with one resistor across a source. The impedance seen by the source is

Z = \frac{V_s}{I_s} = \tm{sr.r}{R} + \tm{sr.l}{j\omega L} + \tm{sr.c}{\frac{1}{j\omega C}}

and the frequency at which the inductive and capacitive parts cancel is the undamped natural frequency:

j\omega_0 L + \frac{1}{j\omega_0 C} = 0 \;\Longrightarrow\; \boxed{\omega_0 = \frac{1}{\sqrt{LC}}}

The series RLC circuit. One current flows through all three elements, so the three element voltages $V_R$ , $V_L$ , and $V_C$ are three different transfer functions of the same input. The choice of which one is called the output selects the type of the filter.

The standard form

Every second-order filter of this course is written in one standard form, in which the denominator is fixed and only the numerator changes:

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

The denominator carries the two parameters $\omega_0$ , the undamped natural frequency, and $\zeta$ , the damping ratio, both introduced in Lecture 10. The numerator power $N$ carries the type, and it does so by controlling the two extreme values.

N	Numerator	DC gain	∞ gain	Type and tap
0	K	K / ω₀²	0	LPF, across C
1	K s	0	0	BPF, across R
2	K s²	0	K	HPF, across L

The three rows are read from the same two limits used on the RC ladders. At DC the capacitor is an open and the inductor is a short, so the whole source voltage stands across the capacitor and none of it across the inductor. At infinite frequency the roles are exchanged. The resistor holds neither extreme, and its voltage therefore peaks somewhere in between, which is the band-pass case.

Where the taps come from, in detail

One current flows in the series loop, $I_s = V_s / Z$ , and each element voltage is that current times the element impedance. Dividing through by $sL$ puts each result in the standard form:

\frac{V_C}{V_s} = \frac{1/sC}{R + sL + 1/sC} = \frac{1/LC}{s^2 + \tfrac{R}{L}s + \tfrac{1}{LC}} \qquad \frac{V_R}{V_s} = \frac{(R/L)\,s}{s^2 + \tfrac{R}{L}s + \tfrac{1}{LC}} \qquad \frac{V_L}{V_s} = \frac{s^2}{s^2 + \tfrac{R}{L}s + \tfrac{1}{LC}}

Matching the denominator against the standard form identifies both parameters, and the three numerators identify the three constants:

\omega_0^2 = \frac{1}{LC}, \qquad 2\zeta\omega_0 = \frac{R}{L} \qquad\Longrightarrow\qquad \zeta = \frac{R}{2}\sqrt{\frac{C}{L}}

The capacitor tap has $K = \omega_0^2$ , so its DC gain is $1$ ; the resistor tap has $K = 2\zeta\omega_0$ , so its peak value is $1$ ; the inductor tap has $K = 1$ , so its infinite-frequency gain is $1$ . A passive ladder again only attenuates, except for the resonant rise discussed below.

The three numerators plotted on one axis at $\omega_0 = 100$ rad/s, each scaled to a passband gain of 1: LPF in blue, BPF in green, HPF in orange. Lowering the damping ratio narrows the band-pass and lifts a resonant peak on the low-pass and the high-pass.

damping ratio ζ 0.5

2ζω₀ = 100.0 rad/s low-pass maximum = 1.155 (the low-pass overshoots its DC gain only while ζ < 0.707)

§6 · The band-pass peak

Where the maximum sits, and what it is worth

The band-pass case, $N = 1$ , is the one that requires work, because its maximum is not at either extreme of the frequency axis. The location of the maximum is found by rewriting the response so that the numerator is constant.

H(j\omega) = \frac{jK\omega}{-\omega^2 + j2\zeta\omega_0\omega + \omega_0^2} \cdot \frac{1/\omega}{1/\omega} = \frac{jK}{-\omega + j2\zeta\omega_0 + \dfrac{\omega_0^2}{\omega}}

Taking the magnitude separates the real and the imaginary parts of the denominator:

|H(j\omega)| = \frac{K} {\left[\left(\tm{bw.w0}{\dfrac{\omega_0^2}{\omega} - \omega}\right)^2 + \left(\tm{bw.peak}{2\zeta\omega_0}\right)^2\right]^{1/2}}

The numerator is now a constant, so the magnitude is largest exactly where the denominator is smallest. The second term of the denominator does not depend on $\omega$ , and the first term is a square, so the smallest denominator is obtained by driving that square to zero:

\frac{\omega_0^2}{\omega} - \omega = 0 \;\Longrightarrow\; \omega^2 = \omega_0^2 \;\Longrightarrow\; \boxed{\omega = \omega_0}

\boxed{\;|H(j\omega)|_{\max} = \frac{K}{2\zeta\omega_0}\;}

The maximum therefore sits at the undamped natural frequency, and its value is the numerator constant divided by the middle coefficient of the denominator. Setting the derivative to zero produces the same answer, but the argument above avoids the differentiation entirely.

The band-pass response normalized to its peak. The maximum sits at $\omega_0$ , and the half-power level is crossed twice, at $\omega_{c1}$ and at $\omega_{c2}$ . The shaded strip is the passband between the two cutoffs.

damping ratio ζ 0.3

ω_c1 = 74.4 ω_c2 = 134.4 rad/s ω_c2 − ω_c1 = 60.0 = 2ζω₀ = 60.0 ω_c1 · ω_c2 = 10000 = ω₀²

Two relations worth keeping

Applying the half-power condition to the expression above gives $\omega_0^2/\omega - \omega = \pm 2\zeta\omega_0$ , whose positive roots are

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

Two consequences follow immediately, and both are visible in the readout above: the two cutoffs are separated by $\omega_{c2} - \omega_{c1} = 2\zeta\omega_0$ , and their product is $\omega_{c1}\omega_{c2} = \omega_0^2$ . The second relation says that $\omega_0$ is the geometric mean of the two cutoffs, not the arithmetic mean, so the peak does not sit at the middle of the shaded strip on a linear axis. Both relations serve as checks on any band-pass calculation.

§7 · Worked example B: a second-order band-pass

$H(s) = \dfrac{500s}{s^2 + 200s + 10^4}$

Read the standard form

The transfer function is already in the standard form, so the three parameters are read off by matching coefficients:

\omega_0^2 = 10\,000 \;\Longrightarrow\; \omega_0 = 100\ \text{rad/s}, \qquad 2\zeta\omega_0 = 200 \;\Longrightarrow\; \zeta = 1, \qquad K = 500

The numerator carries $s^1$ , so $N = 1$ and this is a second-order band-pass filter. Its order is two because the denominator is quadratic, which corresponds to two energy storage devices.

The peak, from Section 6

|H|_{\max} = \frac{K}{2\zeta\omega_0} = \frac{500}{200} \;\Longrightarrow\; \boxed{|H|_{\max} = 2.5} \qquad\text{at}\qquad \omega = \omega_0 = 100\ \text{rad/s}

The passband gain is therefore $2.5$ , and the peak is marked on the curve at $\omega_0$ .

The half-power level

|H(j\omega_c)| = \frac{2.5}{\sqrt{2}} \approx 1.77

The dashed level cuts the curve twice, so two cutoff frequencies are expected, as the band-pass shape requires.

The half-power condition becomes a quadratic in $\omega_c^2$

\left|\frac{j500\omega_c}{-\omega_c^2 + j200\omega_c + 10^4}\right|^2 = \left(\frac{2.5}{\sqrt{2}}\right)^2 = \frac{25}{8}

\tm{e2.mag}{\frac{500^2\,\omega_c^2} {\left(10^4 - \omega_c^2\right)^2 + 200^2\omega_c^2}} = \frac{25}{8}

Multiplying out gives $8 \cdot 500^2 \omega_c^2 = 25\left[10^8 - 2 \cdot 10^4 \omega_c^2 + \omega_c^4 + 4 \cdot 10^4 \omega_c^2\right]$ , and dividing by 25 leaves a quadratic in $\omega_c^2$ :

\boxed{\;\omega_c^4 - 60\,000\,\omega_c^2 + 10^8 = 0\;}

The two cutoff frequencies

\omega_c^2 = \frac{60\,000 \pm \sqrt{60\,000^2 - 4 \cdot 10^8}}{2} = \frac{60\,000 \pm 56\,568.5}{2}

\Longrightarrow\; \boxed{\omega_{c1} = 41.4\ \text{rad/s}} \qquad \boxed{\omega_{c2} = 241.4\ \text{rad/s}}

Only the positive roots are kept, because a negative frequency has no meaning here. The shaded strip between $\omega_{c1}$ and $\omega_{c2}$ is the passband.

Type: band-pass Order: 2 Passband gain: 2.5 ω₀ = 100 rad/s ω_c1, ω_c2 = 41.4, 241.4 rad/s

Check: the two relations of Section 6

Both relations of Section 6 are satisfied by the two roots, which confirms them independently of the quartic:

\omega_{c2} - \omega_{c1} = 241.4 - 41.4 = 200 = 2\zeta\omega_0\ \checkmark \qquad \omega_{c1}\,\omega_{c2} = 41.42 \cdot 241.42 = 10\,000 = \omega_0^2\ \checkmark

The closed form of Section 6 reproduces the same pair exactly. With $\zeta = 1$ and $\omega_0 = 100$ ,

\omega_{c1,c2} = 100\left(\sqrt{2} \mp 1\right) = 41.42\ \text{and}\ 241.42\ \text{rad/s}\ \checkmark

The rounding to one decimal place in the board result is the only difference.

A damping ratio of one does not make the filter useless

Here $\zeta = 1$ , so the denominator is critically damped and factors as $(s + 100)^2$ . The response is still a perfectly good band-pass; the two real poles simply place the cutoffs far apart, giving a wide passband. A narrow band-pass requires a small $\zeta$ , as the slider in Section 6 shows.

§8 · Your turn

Practice

Both problems should be attempted on paper before the hints are opened, in order. Each asks for all four descriptors: type, order, passband gain, and cutoff frequency.

Problem 1: a first-order transfer function

H(s) = \frac{2s + 60}{s + 10}

Hint 1: Find the two extreme values first

The extreme values are read by setting $\omega = 0$ and by letting $\omega \to \infty$ in $H(j\omega) = (2j\omega + 60)/(j\omega + 10)$ . One of the two is the maximum, and the half-power level is computed from that one.

Hint 2: The shape is the mirror image of Example A

In Example A the response rose from a nonzero floor to a larger ceiling. Here it does the opposite. The magnitude squared is $|H|^2 = (4\omega^2 + 3600)/(\omega^2 + 100)$ ; its behaviour with $\omega$ can be settled by comparing the coefficients, without differentiating.

Solution

The two extreme values are

H(j0) = \frac{60}{10} = 6, \qquad H(j\infty) = \frac{2j\omega}{j\omega} = 2

The magnitude falls monotonically from $6$ to $2$ , so the filter is low-pass in character, of order 1, with a passband gain of 6. The half-power condition is applied to the maximum, $6$ :

\frac{4\omega_c^2 + 3600}{\omega_c^2 + 100} = \left(\frac{6}{\sqrt{2}}\right)^2 = 18

4\omega_c^2 + 3600 = 18\omega_c^2 + 1800 \;\Longrightarrow\; 14\omega_c^2 = 1800 \;\Longrightarrow\; \boxed{\omega_c = \sqrt{\tfrac{900}{7}} = \tfrac{30}{\sqrt{7}} \approx 11.34\ \text{rad/s}}

The level $6/\sqrt{2} = 4.24$ lies between the floor $2$ and the ceiling $6$ , so the crossing exists and is unique.

Check. Substituting $\omega_c^2 = 900/7$ returns $(4 \cdot \tfrac{900}{7} + 3600)/(\tfrac{900}{7} + 100) = (\tfrac{3600 + 25200}{7})/(\tfrac{900 + 700}{7}) = 28800/1600 = 18\ \checkmark$

Problem 2: a series RLC with numbers

A series RLC circuit driven by $V_s$ , with $R = 20\ \Omega$ , $L = 0.1$ H, and $C = 10\ \mu\text{F}$ . The output is taken across the resistor. Find the type, the order, the passband gain, and the cutoff frequencies.

Hint 1: Identify the tap before computing anything

The output is taken across the resistor, which is row $N = 1$ of the table in Section 5. The type follows from the table alone, and the passband gain of that row was worked out in the collapsible derivation there.

Hint 2: The two parameters come from the component values

$\omega_0 = 1/\sqrt{LC}$ and $2\zeta\omega_0 = R/L$ . Both cutoff frequencies then follow from the closed form of Section 6, and the two relations of Section 6 provide the check.

Solution

The output across the resistor is the $N = 1$ row, so the filter is band-pass, of order 2 (one inductor and one capacitor). The parameters are

\omega_0 = \frac{1}{\sqrt{LC}} = \frac{1}{\sqrt{0.1 \cdot 10 \times 10^{-6}}} = \frac{1}{\sqrt{10^{-6}}} = 1000\ \text{rad/s}

2\zeta\omega_0 = \frac{R}{L} = \frac{20}{0.1} = 200 \;\Longrightarrow\; \zeta = \frac{200}{2 \cdot 1000} = 0.1

The transfer function is therefore $H(s) = 200s / (s^2 + 200s + 10^6)$ , and the passband gain is

|H|_{\max} = \frac{K}{2\zeta\omega_0} = \frac{200}{200} = 1 \qquad\text{at}\qquad \omega = \omega_0 = 1000\ \text{rad/s}

The cutoff frequencies come from the closed form of Section 6:

\omega_{c1,c2} = \omega_0\left(\sqrt{\zeta^2 + 1} \mp \zeta\right) = 1000\left(\sqrt{1.01} \mp 0.1\right)

\boxed{\omega_{c1} = 905.0\ \text{rad/s}} \qquad \boxed{\omega_{c2} = 1105.0\ \text{rad/s}}

Check. The separation is $1105.0 - 905.0 = 200 = 2\zeta\omega_0 = R/L$ , and the product is $905.0 \cdot 1105.0 = 1.000 \times 10^6 = \omega_0^2$ . Both relations of Section 6 hold. The narrow passband, $200$ rad/s around a centre of $1000$ rad/s, is the direct consequence of the small damping ratio $\zeta = 0.1$ .

The same answer from the quartic, as a second check

Applying the half-power condition directly, as in Example B:

\frac{200^2\,\omega_c^2}{\left(10^6 - \omega_c^2\right)^2 + 200^2\omega_c^2} = \frac{1}{2} \;\Longrightarrow\; \omega_c^4 - 2.04 \times 10^6\,\omega_c^2 + 10^{12} = 0

\omega_c^2 = \frac{2.04 \times 10^6 \pm 402\,007}{2} \;\Longrightarrow\; \omega_c = 905.0\ \text{and}\ 1105.0\ \text{rad/s}\ \checkmark

The quartic route and the closed form agree, as they must.

A filter is a transfer function chosen for its magnitude response

The four descriptors

The ideal and the realizable

One pole, two options

The cutoff frequency is the pole

H(s)=4s+1s+2H(s) = \dfrac{4s + 1}{s + 2}H(s)=s+24s+1​

The problem, and the order

The DC gain

The infinite-frequency gain, which is the maximum

The half-power level

Solving the half-power condition

Two resistors and one capacitor, read at the two extremes

The two ladders

At DC the capacitor is an open circuit

At infinite frequency the capacitor is a short circuit

The two transfer functions

The series RLC and the standard form

The standard form

Where the maximum sits, and what it is worth

H(s)=500ss2+200s+104H(s) = \dfrac{500s}{s^2 + 200s + 10^4}H(s)=s2+200s+104500s​

Read the standard form

The peak, from Section 6

The half-power level

The half-power condition becomes a quadratic in ωc2\omega_c^2ωc2​

The two cutoff frequencies

Practice

Problem 1: a first-order transfer function

Problem 2: a series RLC with numbers

$H(s) = \dfrac{4s + 1}{s + 2}$

$H(s) = \dfrac{500s}{s^2 + 200s + 10^4}$

The half-power condition becomes a quadratic in $\omega_c^2$