ECE 211 · Lecture 20: Frequency Response & Bode Plots

§1 · The frequency response

The transfer function evaluated on the imaginary axis

The frequency response is the transfer function evaluated at $s = j\omega$ . A transfer function relates one signal to another in the $s$ domain, and the substitution $s = j\omega$ restricts attention to the sinusoidal steady state:

H(s) = \frac{Y(s)}{X(s)}, \qquad s = j\omega \;\;\Longrightarrow\;\; H(j\omega) = |H(j\omega)| \; \angle H(j\omega)

Two real functions of $\omega$ are therefore produced by one complex function, and two plots are required. By convention the magnitude is plotted in decibels and the angle is plotted in degrees, both against a logarithmic frequency axis.

Decibels

A decibel is a logarithmic measure of a power ratio. Because power is proportional to the square of a voltage or a current, the factor of 10 that applies to power becomes a factor of 20 when an amplitude ratio is measured:

10 \log_{10} \frac{P_2}{P_1} \;\;\Longrightarrow\;\; 20 \log_{10} A_v

Level (dB) 20

amplitude ratio 10.000 power ratio 100.000 every 20 dB is one factor of 10 in amplitude

Four values are worth committing to memory, because they recur in every sketch: $0\ \text{dB} \equiv 1$ , $\;\pm 3.01\ \text{dB} \equiv \sqrt{2}^{\,\pm 1}$ , $\;\pm 6.02\ \text{dB} \equiv 2^{\pm 1}$ , and $\;\pm 20\ \text{dB} \equiv 10^{\pm 1}$ .

Why the logarithm is used

A logarithm converts multiplication into addition, and that single property is the whole reason for the decibel axis. If a transfer function is written as a product and a quotient of simpler terms, then

20 \log \frac{A \cdot B}{C \cdot D} = 20 \log A + 20 \log B - 20 \log C - 20 \log D

\angle \frac{A \cdot B}{C \cdot D} = \angle A + \angle B - \angle C - \angle D

Both the magnitude in decibels and the angle in degrees are therefore additive over the factors. Individual pieces may be plotted separately and then added, provided that each kind of piece is understood. The remainder of this lecture identifies the four kinds and plots each one.

The function that motivates the method

The board notes open with a second-order transfer function and the question of how it is to be plotted:

H(s) = \frac{V_o}{V_s} = \frac{-9}{s^2 + \tfrac{17}{2} s + 18}

The denominator factors, because $s^2 + 8.5 s + 18 = (s + 4)(s + 4.5)$ , which may be verified from the sum $8.5$ and the product $18$ of the roots. The completed sketch is shown below as the destination; the sections that follow supply the method that produces it.

The finished Bode plot of $H(s) = -9/(s^2 + 8.5s + 18)$ . The straight-line sketch consists of segments whose slopes change only at the break frequencies; the exact response departs from it by a few decibels near those breaks and by nothing at all far from them.

straight-line approximation exact response break frequencies

§2 · Standard form

Four building blocks

Every rational transfer function with real coefficients factors into four kinds of term, and each kind has a plot that can be memorized. The four possible outcomes of factoring $H(s)$ are:

$K$ , a constant.
$s^N$ , poles and zeros at the origin.
$(s/\alpha + 1)^N$ , real poles and zeros.
$\left(s^2/\omega_0^2 + 2\zeta s/\omega_0 + 1\right)^N$ , complex conjugate pairs.

In each case a positive exponent $N$ denotes zeros and a negative exponent denotes poles, and $|N|$ counts how many coincide. The symbol $\zeta$ is the damping ratio.

The pole-zero plane of the example assembled in §7, $H(s) = 800\,s(s+5) / [(s+2)^2(s^2 + 10s + 100)]$ . A zero is drawn as a circle and a pole as a cross. Each cluster corresponds to one standard form.

Form 1 has no location in the plane at all. The constant $K$ is what remains after every pole and every zero has been extracted, so it appears nowhere as a cross or a circle. It nonetheless fixes the vertical position of the entire magnitude plot and, through its sign, contributes either $0°$ or $180°$ to the angle.

Form 2 collects the poles and zeros that sit exactly at the origin. The zero at $s = 0$ contributes the factor $s^1$ . A pole at the origin would contribute $s^{-1}$ ; such a pole is generally avoided in a stable design, because the response would grow without bound as the frequency approaches zero.

Form 3 collects the real poles and zeros, one factor for each. The zero at $s = -5$ contributes $(s/5 + 1)^{1}$ , and the double pole at $s = -2$ contributes $(s/2 + 1)^{-2}$ . The number $\alpha$ in the factor is the distance of the root from the origin, and it is called the break frequency.

Form 4 collects each complex conjugate pair into one real quadratic. The pair at $s = -5 \pm j8.66$ contributes $\left(s^2/10^2 + s/10 + 1\right)^{-1}$ , from which $\omega_0 = 10$ rad/s and $\zeta = 0.5$ are read directly. A conjugate pair is never split, because splitting it would introduce complex coefficients.

Putting a transfer function into standard form

Standard form requires the constant term of every factor to equal 1. The normalization is the step at which mistakes are made, so it is worked out here on the function introduced in §1.

The denominator is factored first:

H(s) = \frac{-9}{s^2 + 8.5 s + 18} = \frac{-9}{(s + 4)(s + 4.5)}

The roots are $-4$ and $-4.5$ , since $(-8.5 \pm \sqrt{8.5^2 - 72})/2 = (-8.5 \pm 0.5)/2$ . Both roots are real and negative, so only forms 1 and 3 will appear.

Each factor is then scaled so that its constant term becomes 1, and the scale factors are collected into $K$ :

H(s) = \frac{-9}{4\left(\dfrac{s}{4} + 1\right)\cdot 4.5\left(\dfrac{s}{4.5} + 1\right)} = \frac{-9}{18} \cdot \left(\frac{s}{4} + 1\right)^{-1}\left(\frac{s}{4.5} + 1\right)^{-1}

\boxed{\;H(s) = -0.5 \cdot \left(\frac{s}{4} + 1\right)^{-1} \left(\frac{s}{4.5} + 1\right)^{-1}\;}

Three numbers are now available by inspection: $K = -0.5$ , which places the low-frequency magnitude at $20\log_{10} 0.5 = -6.02$ dB and the low-frequency angle at $180°$ because $K$ is negative; and two real poles, which break at $\alpha = 4$ and $\alpha = 4.5$ rad/s.

A check is available at no cost: $H(0) = -9/18 = -0.5$ , which is $K$ itself, because every other factor equals 1 at $s = 0$ .

Why the constant term is normalized to 1

Each normalized factor equals 1, that is 0 dB, at low frequency. The entire low-frequency level of the magnitude plot is therefore carried by $K$ alone (together with any $s^N$ factor), and every other factor merely bends the curve at its own break frequency. Skipping the normalization moves the whole plot vertically by an unknown amount.

§3 · Form 1: the constant

A flat line in both plots, with no approximation involved

A constant is unchanged by the substitution $s = j\omega$ , so its magnitude and angle do not depend on frequency at all:

\tm{kc.lv}{|K| \Longrightarrow 20 \log_{10} |K| \ \text{dB}} \qquad\qquad \tm{kc.an}{\angle K \Longrightarrow 0° \ \text{or} \ 180°}

The angle is $0°$ when $K$ is positive and $180°$ when $K$ is negative, since a negative real number lies on the negative real axis. Both plots are therefore horizontal lines, and both are exact: no straight-line approximation is being made here.

The two plots of a constant gain. The sign is selected with the buttons and the size with the slider.

A positive constant contributes $0°$ to the angle at every frequency. Only the vertical placement of the magnitude plot is affected.

A negative constant contributes $180°$ at every frequency, and it leaves the magnitude plot untouched, because $|K|$ discards the sign. An inverting amplifier is the ordinary source of such a term.

|K| 10

K = 10.0 level 20.00 dB angle 0°

A level of 0 dB is not a gain of zero

A magnitude of 0 dB means $|H| = 1$ , that is, the output equals the input. A gain of zero would be $-\infty$ dB and could not be drawn. Negative decibels denote attenuation, positive decibels denote amplification.

§4 · Form 2: poles and zeros at the origin

A straight line of slope 20N decibels per decade

The factor $s^N$ becomes $(j\omega)^N$ , whose magnitude is $\omega^N$ and whose angle is $N$ times $90°$ :

s^N \Longrightarrow (j\omega)^N \Longrightarrow \tm{sn.sl}{20 \log_{10} \omega^N = 20 N \log_{10} \omega} \qquad \tm{sn.an}{\angle (j\omega)^N = 90° \cdot N}

Because the horizontal axis is $\log_{10}\omega$ , the magnitude is a straight line of slope $20N$ decibels per decade. The line passes through 0 dB at $\omega = 1$ for every $N$ , since $\log_{10} 1 = 0$ . The angle is a constant, so the phase plot is again a horizontal line. Both plots are exact.

The factor $s^N$ . A positive $N$ counts zeros at the origin and tilts the line upward; a negative $N$ counts poles at the origin and tilts it downward.

N 1

one zero at the origin slope +20 dB/dec angle +90° at every frequency

Numerator up, denominator down

A factor in the numerator raises the angle by $90°$ per zero; a factor in the denominator lowers it by $90°$ per pole. The same rule of signs applies to the slope: $+20$ dB/dec per zero at the origin, $-20$ dB/dec per pole. Poles at the origin are generally avoided in practice, because the magnitude then grows without bound as $\omega \to 0$ .

The decade, and why

\log_{10}

is convenient

A decade is a factor of 10 in frequency. On a logarithmic axis a decade occupies a fixed distance, which is what makes a constant slope meaningful: $\log_{10}(1) = 0$ , $\log_{10}(10) = 1$ , $\log_{10}(100) = 2$ , and so on. A slope quoted in decibels per decade is therefore a genuine straight line on these axes.

§5 · Form 3: real poles and zeros

The break frequency, and two segments joined there

The factor $(s/\alpha + 1)^N$ becomes $(j\omega/\alpha + 1)^N$ , whose magnitude follows from the Pythagorean sum of the real and imaginary parts:

\left|\left(\frac{j\omega}{\alpha} + 1\right)^{\!N}\right| = \left(1 + \left(\frac{\omega}{\alpha}\right)^{2}\right)^{\!N/2} \;\Longrightarrow\; \tm{rl.as}{20 \log_{10}\left(1 + \left(\frac{\omega}{\alpha}\right)^{2}\right)^{\!N/2}}

The approximation consists of replacing this expression by its two limits, one on each side of $\alpha$ :

\omega \ll \alpha: \quad 20 \log_{10} 1^{N/2} = 0 \ \text{dB} \qquad\qquad \omega \gg \alpha: \quad 20 \log_{10}\left(\frac{\omega}{\alpha}\right)^{\!N} = 20 N \log_{10}\frac{\omega}{\alpha}

The magnitude is therefore flat at 0 dB below $\alpha$ and a straight line of slope $20N$ dB/dec above it. The frequency $\alpha$ at which the two segments meet is called the break frequency, and it equals the distance of the root from the origin.

The phase, and the two-decade ramp

The angle of the same factor is $N \tan^{-1}(\omega/\alpha)$ , which is $0°$ well below the break and $90N°$ well above it. The standard approximation joins those two levels by a straight line drawn between one decade below and one decade above the break:

\omega \ll \alpha \Rightarrow 0° \qquad \omega = \alpha \Rightarrow N \tan^{-1}(1) = 45N° \qquad \omega \gg \alpha \Rightarrow 90N°

A real pole or zero. The break frequency $\alpha$ and the exponent $N$ are set by the sliders. The marker on the magnitude plot sits at the exact value at the break; the marker on the phase plot sits at the midpoint of the ramp, where the approximation happens to be exact.

α (rad/s) 10

N −1

one real pole slope above α −20 dB/dec error at α −3.01 dB total phase swing −90° phase ramp 1.0 to 100 rad/s phase at α −45.0°

straight-line approximation exact response

Where the approximation costs the most

The largest magnitude error occurs exactly at the break, where the true value is $10N\log_{10} 2 = 3.01N$ dB rather than 0 dB. One decade above the break the error has fallen to $0.04N$ dB. The phase approximation errs by $5.71N°$ at the lower end of the ramp and by $-5.71N°$ at the upper end, and by nothing at the break itself, where both the true angle and the straight line pass through $45N°$ .

§6 · Form 4: complex conjugate pairs

One quadratic, two roots, and twice the slope

The quadratic factor is treated as a unit. The substitution $s = j\omega$ gives

\left(\frac{s^2}{\omega_0^2} + 2\zeta\frac{s}{\omega_0} + 1\right)^{\!N} \Longrightarrow \left(1 - \frac{\omega^2}{\omega_0^2} + j\,2\zeta\frac{\omega}{\omega_0}\right)^{\!N}

\tm{qd.as}{20 \log_{10} \left[\left(1 - \frac{\omega^2}{\omega_0^2}\right)^{\!2} + \left(2\zeta\frac{\omega}{\omega_0}\right)^{\!2}\right]^{N/2}}

The same two limits are taken. Below $\omega_0$ the bracket tends to 1 and the magnitude to 0 dB. Above $\omega_0$ the term $\omega^4/\omega_0^4$ dominates, so

\omega \gg \omega_0: \quad 20 \log_{10}\left[\left(\frac{\omega}{\omega_0}\right)^{\!4}\right]^{N/2} = 40 N \log_{10}\frac{\omega}{\omega_0}

The high-frequency slope is $40N$ dB/dec, that is, twice the slope of a real factor, because the conjugate pair contains two roots. The break frequency is $\omega_0$ , and the phase swings by $180N°$ rather than $90N°$ for the same reason.

The damping ratio decides what actually happens at the break

The straight-line sketch does not depend on $\zeta$ at all. The exact curve does, and near $\omega_0$ the difference between them can be large:

\text{at } \omega = \omega_0: \quad \tm{qd.atw}{20 N \log_{10}(2\zeta) \ \text{dB}} \qquad\qquad \angle H = 90N°

For $\zeta = 0.5$ the factor $2\zeta$ equals 1 and the exact curve passes through 0 dB, so the sketch is momentarily perfect. For smaller $\zeta$ a resonant peak appears, and for $\zeta$ approaching zero the peak is unbounded.

A complex conjugate pair with $\omega_0 = 10$ rad/s. Reducing $\zeta$ raises the resonant peak without moving a single asymptote.

ζ 0.5

N −1

slope above ω₀ −40 dB/dec total phase swing −180° exact value at ω₀ 0.00 dB extremum 1.25 dB at ω = 7.07 rad/s

straight-line approximation exact response

The angle needs both arguments, not the ratio alone

The board notes give the angle as $N \tan^{-1}\!\left[\,2\zeta\tfrac{\omega}{\omega_0} \big/ \left(1 - \tfrac{\omega^2}{\omega_0^2}\right)\right]$ . Above $\omega_0$ the denominator turns negative while the numerator stays positive, so the ratio alone is ambiguous and the single-argument arc tangent returns an angle in the wrong quadrant. The two-argument form must be used, which is what keeps the curve on this page continuous through $90N°$ at $\omega_0$ instead of folding back.

Where the resonant peak sits

Minimizing $(1 - u^2)^2 + (2\zeta u)^2$ with $u = \omega/\omega_0$ gives $u^2 = 1 - 2\zeta^2$ , so an extremum exists only when $\zeta < 1/\sqrt{2} \approx 0.707$ . Its value is $20N\log_{10}\left(2\zeta\sqrt{1-\zeta^2}\right)$ dB, which is a peak when $N < 0$ and a notch when $N > 0$ . The readout above reports both numbers as $\zeta$ is varied.

§7 · Assembling a complete Bode plot

Every piece, added break by break

The example from the board notes exercises all four forms at once:

H(s) = \frac{800 \, s \, (s + 5)}{(s + 2)^2 \, (s^2 + 10 s + 100)}

Two answers are available before any work is done, simply by counting. There are 2 zeros and 4 poles, so at high frequency $H$ behaves as $s^2/s^4 = s^{-2}$ : the final slope is $-40$ dB/dec and the final angle is $-180°$ .

2 zeros 4 poles final slope −40 dB/dec final angle −180°

Standard form

Each factor is normalized and the scale factors are gathered into $K$ :

H(s) = 800 \cdot s \cdot \underbrace{5\left(\frac{s}{5}+1\right)}_{s+5} \cdot \underbrace{\frac{1}{4}\left(\frac{s}{2}+1\right)^{-2}}_{(s+2)^{-2}} \cdot \underbrace{\frac{1}{100}\left(\frac{s^2}{10^2} + \frac{s}{10} + 1\right)^{-1}}_{(s^2+10s+100)^{-1}}

\boxed{\;H(s) = 10 \cdot s \cdot \left(\frac{s}{5}+1\right) \left(\frac{s}{2}+1\right)^{-2} \left(\frac{s^2}{10^2} + \frac{s}{10} + 1\right)^{-1}\;}

since $K = 800 \cdot 5 / (4 \cdot 100) = 10$ . The quadratic gives $\omega_0 = 10$ rad/s and $2\zeta/\omega_0 = 1/10$ , hence $\zeta = 0.5$ . Each factor may now be read off:

Factor	Form	Break	Slope	Phase
10	1 · constant	—	—	0°
s	2 · zero at origin	none	+20	+90°
(s/2 + 1)⁻²	3 · double real pole	2	−40	−180°
(s/5 + 1)	3 · real zero	5	+20	+90°
(s²/10² + s/10 + 1)⁻¹	4 · pair, ζ = 0.5	10	−40	−180°

Slopes are quoted in decibels per decade. The first two rows carry no break frequency: their contributions apply at every frequency, which is why the low-frequency asymptote already has a slope of $+20$ dB/dec and an angle of $+90°$ before any break is reached.

The magnitude, segment by segment

Start below every break

Below the lowest break only $K$ and $s$ are active, so the asymptote is $20\log_{10} 10 + 20 \log_{10}\omega = 20 + 20\log_{10}\omega$ : a line of slope $+20$ dB/dec passing through 20 dB at $\omega = 1$ rad/s.

Draw the low-frequency segment out to the first break

20 + 20\log_{10}\omega \quad \text{for} \quad \omega \le 2 \;\;\Longrightarrow\;\; \text{at } \omega = 2: \; 20 + 6.02 = 26.02 \ \text{dB}

The segment is drawn from the left edge to $\omega = 2$ rad/s, where the first break occurs. Its endpoint value is the starting point of the next segment.

At ω = 2 the double pole bends the curve by −40

+20 - 40 = \tm{bm.g2}{-20 \ \text{dB/dec}}

A double pole changes the slope by $-20$ dB/dec twice over. From 2 to 5 rad/s the asymptote therefore falls at 20 dB/dec, reaching $26.02 - 20\log_{10}(5/2) = 18.06$ dB at $\omega = 5$ .

At ω = 5 the real zero bends it back by +20

-20 + 20 = \tm{bm.g3}{0 \ \text{dB/dec}}

The asymptote is flat at 18.06 dB from 5 to 10 rad/s. A zero always bends the curve upward, a pole always downward, and only the slope changes at a break; the value itself is continuous.

At ω = 10 the conjugate pair bends it by −40

0 - 40 = \tm{bm.g4}{-40 \ \text{dB/dec}}

The quadratic contributes twice the slope of a real factor, because it carries two roots. No break remains beyond this one, so $-40$ dB/dec is the final slope, which agrees with the count of 2 zeros against 4 poles made at the outset.

Compare with the exact response

The exact curve is now drawn over the sketch. It agrees with the asymptotes far from every break and departs from them by a few decibels nearby, the largest departures occurring where two breaks lie close together. The straight-line sketch is what is expected on paper; the exact curve is what a numerical tool produces.

The phase, segment by segment

Each factor contributes a ramp of $90N°$ (or $180N°$ for a quadratic) spread over the two decades that straddle its break. The corner frequencies are therefore 0.2 and 20 for the double pole, 0.5 and 50 for the zero, and 1 and 100 for the conjugate pair.

Collect the corner frequencies

Six corners are in play: 0.2 and 20 from the double pole, 0.5 and 50 from the zero, and 1 and 100 from the conjugate pair. Between consecutive corners the total phase is a straight line, so only the slope in degrees per decade has to be tracked.

Below 0.2 rad/s only the zero at the origin contributes

The constant $K = 10$ is positive and adds nothing. The factor $s$ adds a constant $+90°$ . No ramp has begun, so the phase is flat at $+90°$ .

At 0.2 the double-pole ramp begins: −90°/dec

A ramp of $-180°$ spread over two decades has slope $-90°$ per decade. At $\omega = 0.5$ the phase has fallen to $90 - 90\log_{10}(0.5/0.2) = 54.2°$ .

At 0.5 the zero's ramp begins: +45°/dec is added

The zero contributes $+90°$ over two decades, that is $+45°$ per decade, so the net slope becomes $-90 + 45 = -45°$ per decade. At $\omega = 1$ the phase is $40.6°$ .

At 1 the conjugate pair's ramp begins: −90°/dec is added

The pair contributes $-180°$ over two decades. The net slope is now $-90 + 45 - 90 = -135°$ per decade, which is maintained until the first ramp ends. At $\omega = 20$ the phase reaches $40.6 - 135\log_{10}(20) = -135.0°$ .

The ramps end one by one

Above 20 the double-pole ramp is finished and the slope relaxes to $-45°$ /dec; above 50 the zero's ramp is finished and the slope steepens to $-90°$ /dec; above 100 every ramp is complete and the phase is flat. The final value is $-152.9 - 90\log_{10}(2) = -180°$ , exactly as predicted by the pole and zero count.

Compare with the exact response

The exact phase follows the straight-line sketch closely, with the largest disagreement where several corners crowd together. At $\omega = 10$ rad/s the two curves very nearly coincide, at $-93.9°$ and $-94.4°$ respectively, because the conjugate pair contributes precisely $-90°$ there under both descriptions.

Reading one frequency off both plots

The assembled sketch and the exact response together. The slider moves the cursor across both plots and reports what each curve gives at that frequency.

ω (rad/s) 1.00

exact 18.28 dB sketch 20.00 dB error −1.72 dB

exact 42.4° sketch 40.6° error 1.8°

The five-step recipe, in summary

Factor the numerator and the denominator.
Normalize every factor so that its constant term is 1, and collect the scale factors into $K$ .
List the breaks in increasing order, with the slope change and the phase change that each one carries.
Start below the lowest break, where only $K$ and $s^N$ act, and change the slope at each break in turn.
Check the ends: the low-frequency slope is $20N$ dB/dec from the $s^N$ factor alone, and the high-frequency slope is 20 dB/dec times (zeros minus poles).

§8 · Your turn

Practice: sketch the Bode plot of $H(s)$

H(s) = \frac{4000\,(s + 10)}{s\,(s + 40)(s + 100)}

The problem should be attempted on paper before the hints are opened, in order. Both the magnitude and the phase are required, over 0.1 to 1000 rad/s.

Hint 1: Count before you factor

There is 1 zero and there are 3 poles, so the high-frequency slope is $20 \times (1 - 3) = -40$ dB/dec and the final angle is $-180°$ . One pole sits at the origin, so form 2 is present with $N = -1$ : the low-frequency slope is $-20$ dB/dec and the low-frequency angle is $-90°$ .

Hint 2: Normalize

H(s) = \frac{4000 \cdot 10\left(\frac{s}{10}+1\right)} {s \cdot 40\left(\frac{s}{40}+1\right)\cdot 100\left(\frac{s}{100}+1\right)}

The scale factors give $K = 40000/4000 = 10$ , that is $+20$ dB.

Hint 3: The standard form and the break list

H(s) = 10 \cdot s^{-1}\left(\frac{s}{10}+1\right) \left(\frac{s}{40}+1\right)^{-1}\left(\frac{s}{100}+1\right)^{-1}

Breaks at 10 (a zero, $+20$ ), 40 (a pole, $-20$ ), and 100 (a pole, $-20$ ) rad/s. The low-frequency asymptote is $20 - 20\log_{10}\omega$ , which is 40 dB at $\omega = 0.1$ and 0 dB at $\omega = 10$ .

Solution: the magnitude

The asymptote starts at $-20$ dB/dec and passes through 0 dB at $\omega = 10$ rad/s, where the zero flattens it. It stays at 0 dB until $\omega = 40$ , then falls at $-20$ dB/dec to $ $0 - 20\log_{10}\frac{100}{40} = -7.96 \ \text{dB at } \omega = 100,$ $after which the second pole steepens it to$ -40 $dB/dec, reaching$ -47.96 $dB at$ \omega = 1000 $rad/s. The final slope of$ -40$ dB/dec matches the count made in hint 1.

Check at $\omega = 20$ rad/s, midway along the flat segment: $ $|H(j20)| = \frac{4000\,|10 + j20|}{|j20|\,|40+j20|\,|100+j20|} = \frac{89442.7}{91214.4} = 0.981 \;\Longrightarrow\; -0.17 \ \text{dB}$ $ The sketch predicts 0 dB there, so the error is well under a quarter of a decibel. ✓

Solution: the phase

The phase begins at $-90°$ , held by the pole at the origin. Ramps of $+45°$ /dec (the zero, from 1 to 100), $-45°$ /dec (the pole at 40, from 4 to 400), and $-45°$ /dec (the pole at 100, from 10 to 1000) are then superposed, and the phase ends at $-180°$ .

Check at $\omega = 20$ rad/s: $ $\angle H(j20) = \tan^{-1}\frac{20}{10} - 90° - \tan^{-1}\frac{20}{40} - \tan^{-1}\frac{20}{100}$ $= 63.43° - 90° - 26.57° - 11.31° = -64.44°$ $The straight-line sketch gives$ -76.4° $at that frequency, an error of$ 12°$. The phase approximation is the coarser of the two, and it is coarsest wherever the two-decade ramps of neighboring breaks overlap, as all three do here.

The completed sketch

The finished plots, for comparison with the sketch made on paper. The break frequencies are 10, 40, and 100 rad/s.

straight-line approximation exact response

The transfer function evaluated on the imaginary axis

Decibels

Why the logarithm is used

The function that motivates the method

Four building blocks

Putting a transfer function into standard form

A flat line in both plots, with no approximation involved

A straight line of slope 20N decibels per decade

The break frequency, and two segments joined there

The phase, and the two-decade ramp

One quadratic, two roots, and twice the slope

The damping ratio decides what actually happens at the break

Every piece, added break by break

Standard form

The magnitude, segment by segment

Start below every break

Draw the low-frequency segment out to the first break

At ω = 2 the double pole bends the curve by −40

At ω = 5 the real zero bends it back by +20

At ω = 10 the conjugate pair bends it by −40

Compare with the exact response

The phase, segment by segment

Collect the corner frequencies

Below 0.2 rad/s only the zero at the origin contributes

At 0.2 the double-pole ramp begins: −90°/dec

At 0.5 the zero's ramp begins: +45°/dec is added

At 1 the conjugate pair's ramp begins: −90°/dec is added

The ramps end one by one

Compare with the exact response

Reading one frequency off both plots

Practice: sketch the Bode plot of H(s)H(s)H(s)

The completed sketch

Practice: sketch the Bode plot of $H(s)$