14.4 Bode Plots

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14.4 Bode Plots

Obtaining the frequenc y response from the transfer function as we did in Section 14.2 is an uphill task. The frequency range required in fre quency response is often so wide that it is incon venient to use a linear scale for the frequenc y axis. Also, there is a more systematic w ay of locating the important features of the magnitude and phase plots of the transfer function. For these reasons, it has become standard practice to plot the transfer function on a pair of semilogarithmic plots: The magnitude in decibels is plotted against the logarithm of the frequency; on a separate plot, the phase in degrees is plotted against the logarithm of the frequency. Such semilogarithmic plots of the transfer function—kno wn as Bode plots—have become the industry standard.

Bode plots are semilog plots of the magnitude (in decibels) and phase (in degrees) of a transfer function versus frequency.

Historical note: Named after Hendrik W. Bode (1905–1982), an engineer with the Bell Telephone Laboratories, for his pioneering work in the 1930s and 1940s.

Bode plots contain the same information as the nonlogarithmic plots discussed in the previous section, but they are much easier to construct, as we shall see shortly.

The transfer function can be written as

$$\mathbf{H} = H/\phi = He^{j\phi} \tag{14.12}$$

Taking the natural logarithm of both sides,

$$\ln H = \ln H + \ln e^{j\phi} = \ln H + j\phi \tag{14.13}$$

Thus, the real part of lnH is a function of the magnitude while the imaginary part is the phase. In a Bode magnitude plot, the gain

$$H_{\rm dB} = 20 \log_{10} H \tag{14.14}$$

is plotted in decibels (dB) v ersus frequency. Table 14.2 provides a few values of H with the corresponding v alues in decibels. In a Bode phase plot, ϕ is plotted in degrees versus frequency. Both magnitude and phase plots are made on semilog graph paper.

A transfer function in the form of Eq. (14.3) may be written in terms of factors that have real and imaginary parts. One such representation might be

A transfer function in the form of Eq. (14.3) may be written in terms of
\nors that have real and imaginary parts. One such representation might be
\n

$$\mathbf{H}(\omega) = \frac{K(j\omega)^{\pm 1}(1 + j\omega/z_1)[1 + j2\zeta_1\omega/\omega_k + (j\omega/\omega_k)^2] \cdots}{(1 + j\omega/p_1)[1 + j2\zeta_2\omega/\omega_n + (j\omega/\omega_n)^2] \cdots}$$

\n(14.15)

which is obtained by di viding out the poles and zeros in H(ω). The representation of H(ω) as in Eq. (14.15) is called the standard form. H(ω) may include up to seven types of different factors that can appear in various combinations in a transfer function. These are:

• 1. A gain K
• 2. A pole (jω) −1 or zero (jω) at the origin
• 3. A simple pole 1∕(1 + jω∕p1) or zero (1 + jω∕z1)
• 4. A quadratic pole 1 ∕[1 + j2ζ2ω∕ωn + ( jω∕ωn) 2 ] or zero [1 + j2ζ1ω∕ωk + ( jω∕ωk) 2 ]

In constructing a Bode plot, we plot each factor separately and then add them graphically. The factors can be considered one at a time and then combined additively because of the logarithms involved. It is this mathematical convenience of the logarithm that makes Bode plots a powerful engineering tool. 0 ϕ

We will now make straight-line plots of the factors listed above. We shall find that these straight-line plots known as Bode plots approximate the actual plots to a reasonable degree of accuracy. 0.1 1 10 100 ω

Constant term: For the g ain K, the magnitude is 20 log ¹⁰ K and the phase is 0°; both are constant with frequenc y. Thus, the magnitude and phase plots of the gain are shown in Fig. 14.9. If K is negative, the magnitude remains 20 log10 ∣K∣ but the phase is ±180°.

Pole/zero at the origin: For the zero (jω) at the origin, the magnitude is 20 log10 ω and the phase is 90°. These are plotted in Fig. 14.10, where we notice that the slope of the magnitude plot is 20 dB/decade, while the phase is constant with frequency.

The Bode plots for the pole (jω) −1 are similar except that the slope of the magnitude plot is −20 dB/decade while the phase is −90°. In general,

TABLE 14.2

Specific gain and their decibel values.*

Magnitude H	20 log10 H (dB)
0.001	−60
0.01	−40
0.1	−20
0.5 __	−6
1∕ √ 2	−3
1	0
__ √ 2	3
2	6
10	20
20	26
100	40
1000	60

* Some of these values are approximate.

The origin is where ω = 1 or log ω = 0 and the gain is zero.

Figure 14.9

Bode plots for gain K: (a) magnitude plot, (b) phase plot.

for (jω) ^N, where N is an integer, the magnitude plot will have a slope of 20N dB/decade, while the phase is 90N degrees.

Simple pole/zero: For the simple zero (1 + jω∕z1), the magnitude is 20 log10 ∣1 + jω∕z1∣ and the phase is tan⁻¹ ω∕z1. We notice that

$$H_{\text{dB}} = 20 \log_{10} \left| 1 + \frac{j\omega}{z_1} \right| \qquad \Rightarrow \qquad 20 \log_{10} 1 = 0 \qquad \textbf{(14.16)}$$

\n

$$\text{as} \quad \omega \to 0$$

\n

$$H_{\text{dB}} = 20 \log_{10} \left| 1 + \frac{j\omega}{z_1} \right| \qquad \Rightarrow \qquad 20 \log_{10} \frac{\omega}{z_1} \qquad \textbf{(14.17)}$$

\n

$$\text{as} \quad \omega \to \infty$$

showing that we can approximate the magnitude as zero (a straight line with zero slope) for small v alues of ω and by a straight line with slope 20 dB/decade for large values of ω. The frequency ω = z1 where the two asymptotic lines meet is called the corner frequency or break frequency. Thus, the approximate magnitude plot is sho wn in Fig. 14.11(a), where the actual plot is also shown. Notice that the approximate plot is close to the actual plot except at the break frequency, where ω = z1 and the deviation is 20 log10 ∣(1 + j1)∣ = 20 log10 √ __ 2 ≃ 3 dB.

The phase tan⁻¹ (ω∕z1) can be expressed as

$$\phi = \tan^{-1}\left(\frac{\omega}{z_1}\right) = \begin{cases} 0, & \omega = 0 \\ 45^\circ, & \omega = z_1 \\ 90^\circ, & \omega \to \infty \end{cases}$$

(14.18)

As a straight-line approximation, we let ϕ≃ 0 for ω≤ z1∕10, ϕ ≃ 45° for ω = z1, and ϕ≃ 90° for ω ≥ 10z1. As shown in Fig. 14.11(b) along with the actual plot, the straight-line plot has a slope of 45° per decade.

The Bode plots for the pole 1 ∕(1 + jω∕p1) are similar to those in Fig. 14.11 except that the corner frequency is at ω = p1, the magnitude has a slope of −20 dB/decade, and the phase has a slope of −45° per decade.

Quadratic pole/zer o: The magnitude of the quadratic pole 1 ∕[1 + j2ζ2ω∕ωn + (jω∕ωn) 2 ] is −20 log10∣1 + j2ζ2ω∕ωn + ( jω∕ωn) 2 ∣ and the phase is −tan⁻¹ (2ζ2ω∕ωn)∕(1 − ω² ∕ωⁿ 2 ). But

A decade is an interval between two frequencies with a ratio of 10; e.g., between ω0 and 10ω0, or between 10 and 100 Hz. Thus, 20 dB/decade means that the magnitude changes 20 dB whenever the frequency changes tenfold or one decade.

The special case of dc (ω = 0) does not appear on Bode plots because log 0 = −∞, implying that zero frequency is infinitely far to the left of the origin of Bode plots.

Figure 14.10

Bode plot for a zero ( jω) at the origin: (a) magnitude plot, (b) phase plot.

(14.19)

Bode plots of zero (1 + jω∕z1): (a) magnitude plot, (b) phase plot.

and

$$H_{\text{dB}} = -20 \log_{10} \left| 1 + \frac{j2\zeta_2 \omega}{\omega_n} + \left( \frac{j\omega}{\omega_n} \right)^2 \right| \qquad \Rightarrow \qquad -40 \log_{10} \frac{\omega}{\omega_n}$$

as $\omega \to \infty$ (14.20)

Thus, the amplitude plot consists of tw o straight asymptotic lines: one with zero slope for ω< ωn and the other with slope −40 dB/decade for ω > ωn, with ωn as the corner frequenc y. Figure 14.12(a) sho ws the approximate and actual amplitude plots. Note that the actual plot depends on the damping factor ζ2 as well as the corner frequency ωn. The significant peaking in the neighborhood of the corner frequency should be added to the straight-line approximation if a high le vel of accurac y is desired. However, we will use the straight-line approximation for the sake of simplicity.

Figure 14.12 Bode plots of quadratic pole [1 + j2ζω∕ωn − ω² ∕ ωⁿ 2 ] −1 : (a) magnitude plot, (b) phase plot.

There is another procedure for obtaining Bode plots that is faster and perhaps more efficient than the one we have just discussed. It consists in realizing that zeros cause an increase in slope, while poles cause a decrease. By starting with the low-frequency asymptote of the Bode plot, moving along the frequency axis, and increasing or decreasing the slope at each corner frequency, one can sketch the Bode plot immediately from the transfer function without the effort of making individual plots and adding them. This procedure can be used once you become proficient in the one discussed here.

Digital computers have rendered the procedure discussed here almost obsolete. Several software packages such as PSpice, MATLAB, Mathcad, and Micro-Cap can be used to generate frequency response plots. We will discuss PSpice later in the chapter.

The phase can be expressed as

$$\phi = -\tan^{-1} \frac{2\zeta_2 \omega / \omega_n}{1 - \omega^2 / \omega_n^2} = \begin{cases} 0, & \omega = 0 \\ -90^\circ, & \omega = \omega_n \\ -180^\circ, & \omega \to \infty \end{cases}$$

(14.21)

The phase plot is a straight line with a slope of −90° per decade starting at ωn∕10 and ending at 10ωn, as shown in Fig. 14.12(b). We see again that the difference between the actual plot and the straight-line plot is due to the damping f actor. Notice that the straight-line approximations for both magnitude and phase plots for the quadratic pole are the same as those for a double pole, that is, (1 + jω∕ωn) −2 . We should e xpect this because the double pole (1 + jω∕ωn) −2 equals the quadratic pole 1∕[1 + j2ζ2ω∕ωn + (jω∕ωn) 2 ] when ζ2 = 1. Thus, the quadratic pole can be treated as a double pole as far as straight-line approximation is concerned.

For the quadratic zero [1 + j2ζ1ω∕ωk + (jω∕ωk) 2 ], the plots in Fig. 14.12 are in verted because the magnitude plot has a slope of 40 dB/decade while the phase plot has a slope of 90° per decade.

Table 14.3 presents a summary of Bode plots for the se ven factors. Of course, not every transfer function has all seven factors. To sketch the Bode plots for a function H(ω) in the form of Eq. (14.15), for e xample, we first record the corner frequencies on the semilog graph paper, sketch the factors one at a time as discussed above, and then combine additively

the graphs of the f actors. The combined graph is often dra wn from left to right, changing slopes appropriately each time a corner frequenc y is encountered. The following examples illustrate this procedure.

Example 14.3 Construct the Bode plots for the transfer function

ots for the transfer function
\n

$$\mathbf{H}(\omega) = \frac{200j\omega}{(j\omega + 2)(j\omega + 10)}$$

Solution:

We first put H(ω) in the standard form by dividing out the poles and zeros. Thus,

We first put

$$\mathbf{H}(\omega)$$

in the standard form by dividing out the poles and
zeros. Thus,

$$\mathbf{H}(\omega) = \frac{10j\omega}{(1 + j\omega/2)(1 + j\omega/10)}$$ $$= \frac{10 |j\omega|}{|1 + j\omega/2||1 + j\omega/10|} \frac{(90^\circ - \tan^{-1} \omega/2 - \tan^{-1} \omega/10)}{1 - j\omega/10}$$

Hence, the magnitude and phase are

$$H_{\text{dB}} = 20 \log_{10} 10 + 20 \log_{10} |j\omega| - 20 \log_{10} \left| 1 + \frac{j\omega}{2} \right|$$ $$- 20 \log_{10} \left| 1 + \frac{j\omega}{10} \right|$$ $$\phi = 90^{\circ} - \tan^{-1} \frac{\omega}{2} - \tan^{-1} \frac{\omega}{10}$$

We notice that there are two corner frequencies at ω = 2,10. For both the magnitude and phase plots, we sketch each term as shown by the dotted lines in Fig. 14.13. We add them up graphically to obtain the overall plots shown by the solid curves.

Figure 14.13 For Example 14.3: (a) magnitude plot, (b) phase plot.

$$\mathbf{H}(\omega) = \frac{5(j\omega + 2)}{j\omega(j\omega + 10)}$$

Answer: See Fig. 14.14.

$$\mathbf{H}(\omega) = \frac{j\omega + 10}{j\omega(j\omega + 5)^2}$$

Solution:

Putting H(ω) in the standard form, we get

$$\text{H}(\omega) = \frac{0.4(1 + j\omega/10)}{j\omega(1 + j\omega/5)^2}$$

From this, we obtain the magnitude and phase as

$$H_{\text{dB}} = 20 \log_{10} 0.4 + 20 \log_{10} \left| 1 + \frac{j\omega}{10} \right| - 20 \log_{10} |j\omega|$$ $$- 40 \log_{10} \left| 1 + \frac{j\omega}{5} \right|$$ $$\phi = 0^{\circ} + \tan^{-1} \frac{\omega}{10} - 90^{\circ} - 2 \tan^{-1} \frac{\omega}{5}$$

There are two corner frequencies at ω = 5, 10 rad/s. For the pole with corner frequency at ω = 5, the slope of the magnitude plot is −40 dB/decade and that of the phase plot is −90° per decade due to the power of 2. The

Obtain the Bode plots for Example 14.4

magnitude and the phase plots for the individual terms (in dotted lines) and the entire H( jω) (in solid lines) are in Fig. 14.15.

Figure 14.15

Bode plots for Example 14.4: (a) magnitude plot, (b) phase plot.

Figure 14.16

For Practice Prob. 14.4: (a) magnitude plot, (b) phase plot.

Example 14.5 Draw the Bode plots for

$$H(s) = \frac{s+1}{s^2 + 12s + 100}$$

Solution:

• 1. Define. The problem is clearly stated and we follo w the technique outlined in the chapter.
• 2. Present. We are to develop the approximate Bode plot for the given function, H(s).
• 3. Alternative. The two most effective choices would be the approximation technique outlined in the chapter , which we will use here, and MATLAB, which can actually give us the exact Bode plots.

4. Attempt. We express H(s) as

express H(s) as

$$\mathbf{H}(\omega) = \frac{1/100(1 + j\omega)}{1 + j\omega 1.2/10 + (j\omega/10)^2}$$

For the quadratic pole, ωn = 10 rad/s, which serves as the corner frequency. The magnitude and phase are

$$H_{\text{dB}} = -20 \log_{10} 100 + 20 \log_{10} |1 + j\omega|$$ $$- 20 \log_{10} \left| 1 + \frac{j\omega 1.2}{10} - \frac{\omega^2}{100} \right|$$ $$\phi = 0^\circ + \tan^{-1} \omega - \tan^{-1} \left[ \frac{\omega 1.2/10}{1 - \omega^2/100} \right]$$

Figure 14.17 shows the Bode plots. Notice that the quadratic pole is treated as a repeated pole at ωk, that is, (1 + jω∕ωk) 2 , which is an approximation.

Figure 14.17 Bode plots for Example 14.5: (a) magnitude plot, (b) phase plot.

5. Evaluate. Although we could use MATLAB to validate the solution, we will use a more straightforward approach. First, we must realize that the denominator assumes that ζ = 0 for the approximation, so we will use the following equation to check our answer:

$$\mathbf{H}(s) \simeq \frac{s+1}{s^2 + 10^2}$$

We also note that we need to actually solve for HdB and the corresponding phase angle ϕ. First, let ω = 0.

$$H_{\text{dB}} = 20 \log_{10}(1/100) = -40$$

and $\phi = 0^{\circ}$

Now try ω = 1.

$$H_{\rm dB} = 20 \log_{10}(1.4142/99) = -36.9 \text{ dB}$$

which is the expected 3 dB up from the corner frequency.

$$\phi = 45^{\circ}$$

from $\mathbf{H}(j) = \frac{j+1}{-1+100}$

Now try ω = 100.

HdB = 20 log10 (100) − 20 log10 (9900) = 39.91 dB

ϕ is 90° from the numerator minus 180°, which gives −90°. We now have checked three different points and got close agreement, and, because this is an approximation, we can feel confident that we have worked the problem successfully.

You can reasonably ask why did we not check at ω = 10? If we just use the approximate value we used above, we end up with an infinite value, which is to be expected from ζ = 0 (see Fig. 14.12a). If we used the actual value of H( j10) we will still end up being far from the approximate values, since ζ = 0.6 and Fig. 14.12a shows a significant deviation from the approximation. We could have reworked the problem with ζ = 0.707, which would have gotten us closer to the approximation. However, we really have enough points without doing this.

6. Satisfactory? We are satisfied the problem has been worked successfully and we can present the results as a solution to the problem.

Practice Problem 14.5 Construct the Bode plots for

Answer: See Fig. 14.18.

$$H(s) = \frac{10}{s(s^2 + 80s + 400)}$$

For Practice Prob. 14.5: (a) magnitude plot, (b) phase plot.

Example 14.6 Given the Bode plot in Fig. 14.19, obtain the transfer function H(ω).

Solution:

To obtain H(ω) from the Bode plot, we keep in mind that a zero al ways causes an upward turn at a corner frequency, while a pole causes

40 dB

H

0

Figure 14.19 For Example 14.6.

a downward turn. We notice from Fig. 14.19 that there is a zero jω at the origin, which should have intersected the frequency axis at ω = 1. This is indicated by the straight line with slope +20 dB/decade. The fact that this straight line is shifted by 40 dB indicates that there is a 40-dB gain; that is,

$$40 = 20 \log_{10} K \qquad \Rightarrow \qquad \log_{10} K = 2$$

or

$$K = 10^2 = 100$$

In addition to the zero jω at the origin, we notice that there are three factors with corner frequencies at ω = 1, 5, and 20 rad/s. Thus, we have:

• 1. A pole at p = 1 with slope −20 dB/decade to cause a downward turn and counteract the zero at the origin. The pole at p = 1 is determined as 1∕(1 + jω∕1).
• 2. Another pole at p = 5 with slope −20 dB/decade causing a do wnward turn. The pole is 1∕(1 + jω∕5).
• 3. A third pole at p = 20 with slope −20 dB/decade causing a further downward turn. The pole is 1∕(1 + jω∕20).

Putting all these together gives the corresponding transfer function as

H turn. The pole is

$$1/(1 + j\omega/20)$$

.
\nthese together gives the corresponding transfer in
\n

$$\mathbf{H}(\omega) = \frac{100 j\omega}{(1 + j\omega/1)(1 + j\omega/5)(1 + j\omega/20)}$$

\n

$$= \frac{j\omega 10^4}{(j\omega + 1)(j\omega + 5)(j\omega + 20)}$$

or

$$(j\omega + 1)(j\omega + 3)(j\omega + 20)$$ $$\mathbf{H}(s) = \frac{10^4 s}{(s+1)(s+5)(s+20)}, \qquad s = j\omega$$

Obtain the transfer function H( ^ω) corresponding to the Bode plot in Practice Problem 14.6 Fig. 14.20.

Answer: ^H(ω) = 2,000,000(s + 5) _______________ (s + 10)(s + 100)2 .

To see how to use MATLAB to produce Bode plots, refer to Section 14.11.

0.1 1 5 10 20 100

‒40 dB/decade

+20 dB/decade