Sketching Bode plots

In section 13.4, we learned the Bode plot shapes of the standard building blocks: constant gain, poles and zeros at the origin, real poles and zeros, and complex conjugate pole and zero pairs. In this section, we learn to combine them to sketch the Bode plot of any transfer function that can be factored into these building blocks.

The key property is that for $H(s) = \prod_i H_i(s)$, the magnitude and phase satisfy \[\begin{aligned} 20 \log_{10} \abs{H(j\omega)} = \sum_i 20 \log_{10} \abs{H_i(j\omega)} && \text{and} && \angle{H(j\omega)} = \sum_i \angle{H_i(j\omega)}. \end{aligned}\] That is, when magnitudes are expressed in decibels, both magnitudes and phases add. This is the entire basis for Bode plot sketching: factor the transfer function, sketch each factor’s Bode plot, and graphically sum them.

Procedure

The following procedure produces an asymptotic (straight-line approximation) Bode plot for a transfer function $H(s)$.

Factor $H(s)$ into standard-form building blocks. Write $H(s) = K \prod_i H_i(s)$, where each $H_i$ is one of:
- a real pole $1/(\tau s + 1)$ or real zero $(\tau s+1)$, where $\omega_b = 1/\tau$ is the break frequency,
- a complex conjugate pole pair in the standard form $\omega_n^2 / (s^2 + 2\zeta \omega_n s + \omega_n^2$ or a complex conjugate zero pair, its reciprocal, where $\omega_n$ is the break frequency,
- a pole $1/s$ or a zero $s$ at the origin,
and $K$ is the remaining constant gain. Each factor has unity DC gain (evaluates to $1$ at $s = 0$) except for $K$ and the origin factors.
Identify the DC gain and all break frequencies. The DC gain is $K$, contributing $20\log_{10}\abs{K}$ dB at all frequencies. List the break frequencies in ascending order.
Sketch the asymptotic magnitude plot. Sketch the magnitude of each factor by straight-line approximation of the graph of each, according to section 13.4, and graphically sum.
Sketch the asymptotic phase plot. Sketch the phase of each factor by straight-line approximation of the graph of each, according to section 13.4, and graphically sum.
Refine near break frequencies. The asymptotic sketch is exact far from break frequencies but deviates near them:
- At each real pole or zero break frequency, the true magnitude is $3$ dB from the asymptote (below for a pole, above for a zero).
- For an underdamped complex conjugate pole pair ($\zeta < 1/\sqrt{2}$), there is a resonance peak near $\omega_n$ with height approximately $1/(2\zeta)$ above the asymptote (in linear magnitude). The peak is more pronounced for smaller $\zeta$. A zero pair has the reciprocal effect.

Example 13.3

Sketch the Bode plot of $$\begin{aligned} H(s) = \frac{s + 1}{10 s^2 + 101 s + 10}. \end{aligned}$$

Step 1: factor into standard form

Factor the denominator. The roots of 10s² + 101s + 10 = 0 are $$\begin{aligned} s = \frac{-101 \pm \sqrt{101^2 - 400}}{20} = \frac{-101 \pm 99}{20}, \end{aligned}$$ giving s = − 0.1 and s = − 10. Therefore, $$\begin{aligned} 10 s^2 + 101 s + 10 = 10(s + 0.1)(s + 10). \end{aligned}$$

Now convert each factor to standard form (s/ω_b+1):

(s+0.1) = 0.1 (s/0.1+1), a pole with break frequency ω_b = 0.1 rad/s.
(s+10) = 10 (s/10+1), a pole with break frequency ω_b = 10 rad/s.
(s+1) = 1 ⋅ (s/1+1), a zero with break frequency ω_b = 1 rad/s.

Collecting the constants that were factored out of the denominator, $$\begin{aligned} H(s) = \frac{s + 1}{10 \cdot 0.1\,(s/0.1 + 1) \cdot 10\,(s/10 + 1)} = \underbrace{0.1}_{K} \cdot \frac{s/1 + 1}{(s/0.1 + 1)(s/10 + 1)}. \end{aligned}$$

Step 2: identify DC gain and break frequencies

DC gain: K = 0.1, so 20log₁₀(0.1) = − 20 dB.
Break frequencies (ascending): 0.1, 1, 10 rad/s.

As a sanity check, H(0) = 1/(10⋅0.1⋅10) = 0.1. ✓

Step 3: sketch asymptotic magnitude

The resulting asymptotic magnitude sketch is shown in figure 13.5.

Step 4: sketch asymptotic phase

The resulting asymptotic phase sketch is shown in figure 13.6.

Step 5: refine

At each break frequency, the true magnitude is 3 dB from the asymptote. In this example, the break frequencies are well separated (each a decade apart), so the corrections do not overlap significantly.

figure 13.7 compares the asymptotic sketch to the actual Bode plot. The sketch captures the essential shape; the main deviations are the 3 dB rounded corners at each break frequency.

Figure 13.7: Asymptotic sketch (dashed) vs. actual Bode plot (solid).

Example 13.4

Sketch the Bode plot of $$\begin{aligned} H(s) = \frac{200000\,(s+1)}{s^3+110\,s^2+11000\,s+100000}. \end{aligned}$$

Step 1: factor into standard form

The denominator is a cubic. We can find one real root by inspection or the rational root theorem: s = − 10 is a root, since − 1000 + 11000 − 110000 + 100000 = 0. Dividing out (s+10), $$\begin{aligned} s^3+110\,s^2+11000\,s+100000 = (s+10)(s^2+100\,s+10000). \end{aligned}$$ The quadratic s² + 100 s + 10000 has discriminant 100² − 4(10000) = − 30000 < 0, so its roots are a complex conjugate pair: $s = -50 \pm j\,50\sqrt{3}$.

Convert to standard form. The real pole: $$\begin{aligned} (s + 10) = 10\,(s/10 + 1), \quad \omega_b = 10. \end{aligned}$$ The complex conjugate pair has ω_n² = 10000 and 2ζω_n = 100, giving ω_n = 100 and ζ = 0.5: $$\begin{aligned} s^2 + 100\,s + 10000 = 100^2\!\left(\frac{s^2}{100^2} + \frac{2(0.5)\,s}{100} + 1\right). \end{aligned}$$ The numerator zero: (s+1) = 1 ⋅ (s/1+1), ω_b = 1.

Collecting constants, $$\begin{aligned} H(s) &= \frac{200000\,(s/1 + 1)}{10\,(s/10 + 1) \cdot 100^2\!\left(\frac{s^2}{100^2} + \frac{s}{100} + 1\right)} \\ &= \underbrace{2}_{K} \cdot (s/1 + 1) \cdot \frac{1}{s/10 + 1} \cdot \frac{1}{\dfrac{s^2}{100^2} + \dfrac{s}{100} + 1}. \end{aligned}$$

Step 2: identify DC gain and break frequencies

DC gain: K = 2, so 20log₁₀(2) ≈ 6 dB.
Break frequencies: ω_b = 1 (zero), ω_b = 10 (real pole), ω_n = 100 (complex pole pair, ζ = 0.5).

Sanity check: H(0) = 200000/100000 = 2. ✓

Step 3: sketch asymptotic magnitude

The asymptotic magnitude sketch is shown in figure 13.8.

Step 4: sketch asymptotic phase

The asymptotic phase sketch is shown in figure 13.9.

Step 5: refine

The complex pole pair has ζ = 0.5. Since $\zeta < 1/\sqrt{2}$, there is a resonance peak near ω_n = 100 rad/s. The peak magnitude is approximately $$\begin{aligned} \frac{1}{2\zeta\sqrt{1 - \zeta^2}} = \frac{1}{2(0.5)\sqrt{0.75}} = \frac{1}{\sqrt{0.75}} \approx 1.15, \end{aligned}$$ which is about 1.2 dB above the asymptote — a mild bump. For smaller ζ, this peak would be much more pronounced (see figure 13.4).

The real pole at ω = 10 and the zero at ω = 1 each introduce a 3 dB correction at their respective break frequencies.

figure 13.10 compares the asymptotic sketch to the actual Bode plot. The sketch is a good approximation; the resonance peak near ω_n = 100 is visible but mild at ζ = 0.5.

Online Resources for Section 13.5

No online resources.