[4.9 BODE](#page-10-0) PLOTS

← Back to LINEAR SYSTEMS AND SIGNALS Overview

4.9 BODE PLOTS

Sketching frequency response plots (|H(jω)| and H(jω) versus ω) is considerably facilitated by the use of logarithmic scales. The amplitude and phase response plots as a function of ω on a logarithmic scale are known as Bode plots. By using the asymptotic behavior of the amplitude and the phase responses, we can sketch these plots with remarkable ease, even for higher-order transfer functions.

^† For simplicity, we have assumed nonrepeating characteristic roots. The procedure is readily modified for repeated roots, and the same conclusion results.

420 CHAPTER 4 CONTINUOUS-TIME SYSTEM ANALYSIS

Let us consider a system with the transfer function

$$H(s) = \frac{K(s+a_1)(s+a_2)}{s(s+b_1)(s^2+b_2s+b_3)}$$

\n(4.44)

where the second-order factor (s² + b2s + b3) is assumed to have complex conjugate roots.† We shall rearrange Eq. (4.44) in the form

$$H(s) = \frac{Ka_1a_2}{b_1b_3} \frac{\left(\frac{s}{a_1} + 1\right)\left(\frac{s}{a_2} + 1\right)}{s\left(\frac{s}{b_1} + 1\right)\left(\frac{s^2}{b_3} + \frac{b_2}{b_3} s + 1\right)}$$

and

$$H(j\omega) = \frac{Ka_1a_2}{b_1b_3} \frac{\left(1 + \frac{j\omega}{a_1}\right)\left(1 + \frac{j\omega}{a_2}\right)}{j\omega\left(1 + \frac{j\omega}{b_1}\right)\left[1 + j\frac{b_2\omega}{b_3} + \frac{(j\omega)^2}{b_3}\right]}$$

This equation shows that H(jω) is a complex function of ω. The amplitude response |H(jω)| and the phase response H(jω) are given by

$$|H(j\omega)| = \left| \frac{Ka_1a_2}{b_1b_3} \right| \frac{\left| 1 + \frac{j\omega}{a_1} \right| \left| 1 + \frac{j\omega}{a_2} \right|}{|j\omega| \left| 1 + \frac{j\omega}{b_1} \right| \left| 1 + j\frac{b_2\omega}{b_3} + \frac{(j\omega)^2}{b_3} \right|}$$

(4.45)

and

$$\angle H(j\omega) = \angle \left(\frac{Ka_1a_2}{b_1b_3}\right) + \angle \left(1 + \frac{j\omega}{a_1}\right) + \angle \left(1 + \frac{j\omega}{a_2}\right)$$ $$- \angle j\omega - \angle \left(1 + \frac{j\omega}{b_1}\right) - \angle \left[1 + \frac{jb_2\omega}{b_3} + \frac{(j\omega)^2}{b_3}\right]$$

(4.46)

From Eq. (4.46) we see that the phase function consists of the addition of terms of four kinds: (i) the phase of a constant, (ii) the phase of jω, which is 90◦ for all values of ω, (iii) the phase for the first-order term of the form 1+jω/a, and (iv) the phase of the second-order term

\[1+\frac{jb_2\omega}{b_3}+\frac{(j\omega)^2}{b_3}\]

We can plot these basic phase functions for ω in the range 0 to ∞ and then, using these plots, we can construct the phase function of any transfer function by properly adding these basic responses. Note that if a particular term is in the numerator, its phase is added, but if the term is in the

^† Coefficients a1, a² and b1, b2, b³ used in this section are not to be confused with those used in the representation of Nth-order LTIC system equations given earlier [Eqs. (2.1) or (4.26)].

denominator, its phase is subtracted. This makes it easy to plot the phase function H(jω) as a function of ω. Computation of |H(jω)|, unlike that of the phase function, however, involves the multiplication and division of various terms. This is a formidable task, especially when we have to plot this function for the entire range of ω (0 to ∞).

We know that a log operation converts multiplication and division to addition and subtraction. So, instead of plotting |H(jω)|, why not plot log |H(jω)| to simplify our task? We can take advantage of the fact that logarithmic units are desirable in several applications, where the variables considered have a very large range of variation. This is particularly true in frequency response plots, where we may have to plot frequency response over a range from a very low frequency, near 0, to a very high frequency, in the range of 1010 or higher. A plot on a linear scale of frequencies for such a large range will bury much of the useful information at lower frequencies. Also, the amplitude response may have a very large dynamic range from a low of 10−⁶ to a high of 10⁶ . A linear plot would be unsuitable for such a situation. Therefore, logarithmic plots not only simplify our task of plotting, but, fortunately, they are also desirable in this situation.

There is another important reason for using logarithmic scale. The Weber–Fechner law (first observed by Weber in 1834) states that human senses (sight, touch, hearing, etc.) generally respond in a logarithmic way. For instance, when we hear sound at two different power levels, we judge one sound twice as loud when the ratio of the two sound powers is 10. Human senses respond to equal ratios of power, not equal increments in power [10]. This is clearly a logarithmic response.†

The logarithmic unit is the decibel and is equal to 20 times the logarithm of the quantity (log to the base 10). Therefore, 20log10 |H(jω)| is simply the log amplitude in decibels (dB).‡ Thus, instead of plotting |H(jω)|, we shall plot 20log10 |H(jω)| as a function of ω. These plots (log amplitude and phase) are called Bode plots. For the transfer function in Eq. (4.45), the log amplitude is

$$20\log|H(j\omega)| = 20\log\left|\frac{Ka_1a_2}{b_1b_3}\right| + 20\log\left|1 + \frac{j\omega}{a_1}\right| + 20\log\left|1 + \frac{j\omega}{a_2}\right| - 20\log|j\omega|$$ $$-20\log\left|1 + \frac{j\omega}{b_1}\right| - 20\log\left|1 + \frac{jb_2\omega}{b_3} + \frac{(j\omega)^2}{b_3}\right| \tag{4.47}$$

The term 20log(Ka1a2/b1b3) is a constant. We observe that the log amplitude is a sum of four basic terms corresponding to a constant, a pole or zero at the origin (20log|jω|), a first-order pole or zero (20log|1+jω/a|), and complex-conjugate poles or zeros (20log|1+jωb2/b³ +(jω)²/b3|).

^† Observe that the frequencies of musical notes are spaced logarithmically (not linearly). The octave is a ratio of 2. The frequencies of the same note in the successive octaves have a ratio of 2. On the Western musical scale, there are 12 distinct notes in each octave. The frequency of each note is about 6% higher than the frequency of the preceding note. Thus, the successive notes are separated not by some constant frequency, but by constant ratio of 1.06.

^‡ Originally, the unit bel (after the inventor of telephone, Alexander Graham Bell) was introduced to represent power ratio as log10 P2/P¹ bels. A tenth of this unit is a decibel, as in 10 log10 P2/P¹ decibels. Since the power ratio of two signals is proportional to the amplitude ratio squared, or |H(jω)| 2, we have 10 log10 P2/P¹ = 10 log10 |H(jω)| ² = 20 log10 |H(jω)| dB.

422 CHAPTER 4 CONTINUOUS-TIME SYSTEM ANALYSIS

We can sketch these four basic terms as functions of ω and use them to construct the log-amplitude plot of any desired transfer function. Let us discuss each of the terms.

4.9-1 Constant Ka1a2*/b1b***³**

The log amplitude of the constant Ka1a2/b1b² term is also a constant, 20log|Ka1a2/b1b3|. The phase contribution from this term is zero for positive value and π for negative value of the constant (complex constants can have different phases).

4.9-2 Pole (or Zero) at the Origin

LOG MAGNITUDE

A pole at the origin gives rise to the term −20log|jω|, which can be expressed as

$$-20\log|j\omega| = -20\log\omega$$

This function can be plotted as a function of ω. However, we can effect further simplification by using the logarithmic scale for the variable ω itself. Let us define a new variable u such that

u = logω

Hence,

$$-20\log\omega = -20u$$

The log-amplitude function −20u is plotted as a function of u in Fig. 4.40a. This is a straight line with a slope of −20. It crosses the u axis at u = 0. The ω-scale (u = logω) also appears in Fig. 4.40a. Semilog graphs can be conveniently used for plotting, and we can directly plot ω on semilog paper. A ratio of 10 is a decade, and a ratio of 2 is known as an octave. Furthermore, a decade along the ω scale is equivalent to 1 unit along the u scale. We can also show that a ratio of 2 (an octave) along the ω scale equals to 0.3010 (which is log10 2) along the u scale.†

$$u_2 - u_1 = \log_{10} \omega_2 - \log_{10} \omega_1 = \log_{10} (\omega_2/\omega_1)$$

Thus, if

(ω2/ω1) = 10 (which is a decade)

then

u² −u¹ = log10 10 = 1

and if

(ω2/ω1) = 2 (which is an octave)

then

$$u_2 - u_1 = \log_{10} 2 = 0.3010$$

^† This point can be shown as follows. Let ω¹ and ω² along the ω scale correspond to u¹ and u² along the u scale so that logω¹ = u¹ and logω² = u2. Then

Figure 4.40 (a) Amplitude and (b) phase responses of a pole or a zero at the origin.

Note that equal increments in u are equivalent to equal ratios on the ω scale. Thus, 1 unit along the u scale is the same as one decade along the ω scale. This means that the amplitude plot has a slope of −20 dB/decade or −20(0.3010) = −6.02 dB/octave (commonly stated as −6 dB/octave). Moreover, the amplitude plot crosses the ω axis at ω = 1, since u = log10ω = 0 when ω = 1.

For the case of a zero at the origin, the log-amplitude term is 20 log ω. This is a straight line passing through ω = 1 and having a slope of 20 dB/decade (or 6 dB/octave). This plot is a mirror image about the ω axis of the plot for a pole at the origin and is shown dashed in Fig. 4.40a.

PHASE

The phase function corresponding to the pole at the origin is − jω [see Eq. (4.46)]. Thus,

$$\angle H(j\omega) = -\angle j\omega = -90^{\circ}$$

The phase is constant (−90◦) for all values of ω, as depicted in Fig. 4.40b. For a zero at the origin, the phase is jω = 90◦. This is a mirror image of the phase plot for a pole at the origin and is shown dashed in Fig. 4.40b.

4.9-3 First-Order Pole (or Zero)

THE LOG MAGNITUDE

The log amplitude of a first-order pole at −a is −20log|1+jω/a|. Let us investigate the asymptotic behavior of this function for extreme values of ω (ω a and ω a).

(a) For ω a,

$$-20\log\left|1+\frac{j\omega}{a}\right|\approx-20\log 1=0$$

Hence, the log-amplitude function → 0 asymptotically for ω a (Fig. 4.41a).

(a) For the other extreme case, where ω a,

$$-20\log\left|1+\frac{j\omega}{a}\right| \approx -20\log\left(\frac{\omega}{a}\right) = -20\log\omega + 20\log a = -20u + 20\log a$$

This represents a straight line (when plotted as a function of u, the log of ω) with a slope of −20 dB/decade (or −6 dB/octave). When ω = a, the log amplitude is zero. Hence, this line crosses the ω axis at ω = a, as illustrated in Fig. 4.41a. Note that the asymptotes in (a) and (b) meet at ω = a.

The exact log amplitude for this pole is

$$-20\log\left|1+\frac{j\omega}{a}\right| = -20\log\left(1+\frac{\omega^2}{a^2}\right)^{1/2} = -10\log\left(1+\frac{\omega^2}{a^2}\right)$$

This exact log magnitude function also appears in Fig. 4.41a. Observe that the actual and the asymptotic plots are very close. A maximum error of 3 dB occurs at ω = a. This frequency is known as the corner frequency or break frequency. The error everywhere else is less than 3 dB. A plot of the error as a function of ω is shown in Fig. 4.42a. This figure shows that the error at 1 octave above or below the corner frequency is 1 dB and the error at 2 octaves above or below the corner frequency is 0.3 dB. The actual plot can be obtained by adding the error to the asymptotic plot.

The amplitude response for a zero at −a (shown dotted in Fig. 4.41a) is identical to that of the pole at −a with a sign change and therefore is the mirror image (about the 0 dB line) of the amplitude plot for a pole at −a.

PHASE

The phase for the first-order pole at −a is

$$\angle H(j\omega) = -\angle \left(1 + \frac{j\omega}{a}\right) = -\tan^{-1}\left(\frac{\omega}{a}\right)$$

Let us investigate the asymptotic behavior of this function. For ω a,

$$-\tan^{-1}\left(\frac{\omega}{a}\right) \approx 0$$

and, for ω a,

$$-\tan^{-1}\left(\frac{\omega}{a}\right) \approx -90^{\circ}$$

Figure 4.41 (a) Amplitude and (b) phase responses of a first-order pole or zero at s = −a.

The actual plot along with the asymptotes is depicted in Fig. 4.41b. In this case, we use a three-line segment asymptotic plot for greater accuracy. The asymptotes are a phase angle of 0◦ for ω ≤ a/10, a phase angle of −90◦ for ω ≥ 10a, and a straight line with a slope −45◦/decade connecting these two asymptotes (from ω = a/10 to 10a) crossing the ω axis at ω = a/10. It can be seen from Fig. 4.41b that the asymptotes are very close to the curve and the maximum error is 5.7◦. Figure 4.42b plots the error as a function of ω; the actual plot can be obtained by adding the error to the asymptotic plot.

Figure 4.42 Errors in asymptotic approximation of a first-order pole at s = −a.

The phase for a zero at −a (shown dotted in Fig. 4.41b) is identical to that of the pole at −a with a sign change, and therefore is the mirror image (about the 0◦ line) of the phase plot for a pole at −a.

4.9-4 Second-Order Pole (or Zero)

Let us consider the second-order pole in Eq. (4.44). The denominator term is s² + b2s + b3. We shall introduce the often-used standard form s² + 2ζωns + ω^{2 n} instead of s² + b2s + b3. With this form, the log amplitude function for the second-order term in Eq. (4.47) becomes

$$-20\log\left|1+2j\zeta\frac{\omega}{\omega_n}+\left(\frac{j\omega}{\omega_n}\right)^2\right|$$

and the phase function is

$$-\angle \left[1+2j\zeta \frac{\omega}{\omega_n}+\left(\frac{j\omega}{\omega_n}\right)^2\right]$$

\n(4.48)

THE LOG MAGNITUDE

The log amplitude is given by

log amplitude =

$$-20 \log \left| 1 + 2j\zeta \left( \frac{\omega}{\omega_n} \right) + \left( \frac{j\omega}{\omega_n} \right)^2 \right|$$

(4.49)

For ω ωn, the log amplitude becomes

$$log amplitude \approx -20 log 1 = 0$$

For ω ωn, the log amplitude is

$$\log amplitude \approx -20 \log \left| \left( -\frac{\omega}{\omega_n} \right)^2 \right| = -40 \log \left( \frac{\omega}{\omega_n} \right)$$

= -40 log $\omega$ - 40 log $\omega_n$ = -40u - 40 log $\omega_n$ (4.50)

The two asymptotes are zero for ω<ω*ⁿ* and −40u−40logω*ⁿ* for ω>ωn. The second asymptote is a straight line with a slope of −40 dB/decade (or −12 dB/octave) when plotted against the log ω scale. It begins at ω = ω*ⁿ* [see Eq. (4.50)]. The asymptotes are depicted in Fig. 4.43a. The exact log amplitude is given by [see Eq. (4.49)]

log amplitude =

$$-20 \log \left\{ \left[ 1 - \left( \frac{\omega}{\omega_n} \right)^2 \right]^2 + 4 \zeta^2 \left( \frac{\omega}{\omega_n} \right)^2 \right\}^{1/2}$$

(4.51)

The log amplitude in this case involves a parameter ζ , resulting in a different plot for each value of ζ . For complex-conjugate poles,† ζ < 1. Hence, we must sketch a family of curves for a number of values of ζ in the range 0 to 1. This is illustrated in Fig. 4.43a. The error between the actual plot and the asymptotes is shown in Fig. 4.44. The actual plot can be obtained by adding the error to the asymptotic plot.

For second-order zeros (complex-conjugate zeros), the plots are mirror images (about the 0 dB line) of the plots depicted in Fig. 4.43a. Note the resonance phenomenon of the complex-conjugate poles. This phenomenon is barely noticeable for ζ > 0.707 but becomes pronounced as ζ → 0.

PHASE

The phase function for second-order poles, as apparent in Eq. (4.48), is

$$\angle H(j\omega) = -\tan^{-1}\left[\frac{2\zeta\left(\frac{\omega}{\omega_n}\right)}{1 - \left(\frac{\omega}{\omega_n}\right)^2}\right]$$

(4.52)

For ω ωn,

$$\angle H(j\omega) \approx 0$$

^† For ^{ζ ≥} 1, the two poles in the second-order factor are no longer complex but real, and each of these two real poles can be dealt with as a separate first-order factor.

Figure 4.43 Amplitude and phase response of a second-order pole.

For ω ωn,

$$\angle H(j\omega) \simeq -180^\circ$$

Hence, the phase → −180◦ as ω → ∞. As in the case of amplitude, we also have a family of phase plots for various values of ζ , as illustrated in Fig. 4.43b. A convenient asymptote for the phase of complex-conjugate poles is a step function that is 0◦ for ω<ω*ⁿ* and −180◦ for ω>ωn.

Figure 4.44 Errors in the asymptotic approximation of a second-order pole.

Error plots for such an asymptote are shown in Fig. 4.44 for various values of ζ . The exact phase is the asymptotic value plus the error.

For complex-conjugate zeros, the amplitude and phase plots are mirror images of those for complex conjugate-poles.

We shall demonstrate the application of these techniques with two examples.

EXAMPLE 4.29 Bode Plots for Second-Order Transfer Function with Real Roots

Sketch Bode plots for the transfer function

$$H(s) = \frac{20s(s+100)}{(s+2)(s+10)}$$

MAGNITUDE PLOT

First, we write the transfer function in normalized form

$$H(s) = \frac{20 \times 100}{2 \times 10} \frac{s \left(1 + \frac{s}{100}\right)}{\left(1 + \frac{s}{2}\right)\left(1 + \frac{s}{10}\right)} = 100 \frac{s \left(1 + \frac{s}{100}\right)}{\left(1 + \frac{s}{2}\right)\left(1 + \frac{s}{10}\right)}$$

Here, the constant term is 100; that is, 40 dB (20 log 100 = 40). This term can be added to the plot by simply relabeling the horizontal axis (from which the asymptotes begin) as the 40 dB line (see Fig. 4.45a). Such a step implies shifting the horizontal axis upward by 40 dB. This is precisely what is desired.

In addition, we have two first-order poles at −2 and −10, one zero at the origin, and one zero at −100.

Step 1. For each of these terms, we draw an asymptotic plot as follows (shown in Fig. 4.45a by dashed lines):

• (a) For the zero at the origin, draw a straight line with a slope of 20 dB/decade passing through ω = 1.
• (b) For the pole at −2, draw a straight line with a slope of −20 dB/decade (for ω > 2) beginning at the corner frequency ω = 2.
• (c) For the pole at −10, draw a straight line with a slope of −20 dB/decade beginning at the corner frequency ω = 10.
• (d) For the zero at −100, draw a straight line with a slope of 20 dB/decade beginning at the corner frequency ω = 100.
• Step 2. Add all the asymptotes, as depicted in Fig. 4.45a by solid line segments.
• Step 3. Apply the following corrections (see Fig. 4.42a):
  • (a) The correction at ω = 1 because of the corner frequency at ω = 2 is −1 dB. The correction at ω = 1 because of the corner frequencies at ω = 10 and ω = 100 is quite small (see Fig. 4.42a) and may be ignored. Hence, the net correction at ω = 1 is −1 dB.

Figure 4.45 (a) Amplitude and (b) phase responses of the second-order system.

• (b) The correction at ω = 2 because of the corner frequency at ω = 2 is −3 dB, and the correction because of the corner frequency at ω = 10 is −0.17 dB. The correction because of the corner frequency ω = 100 can be safely ignored. Hence the net correction at ω = 2 is −3.17 dB.

• (c) The correction at ω = 10 because of the corner frequency at ω = 10 is −3 dB, and the correction because of the corner frequency at ω = 2 is −0.17 dB. The correction because of ω = 100 can be ignored. Hence the net correction at ω = 10 is −3.17 dB.

• (d) The correction at ω = 100 because of the corner frequency at ω = 100 is 3 dB, and the corrections because of the other corner frequencies may be ignored.

• (e) In addition to the corrections at corner frequencies, we may consider corrections at intermediate points for more accurate plots. For instance, the corrections at ω = 4 because of corner frequencies at ω = 2 and 10 are −1 and about −0.65, totaling −1.65 dB. In the same way, the corrections at ω = 5 because of corner frequencies at ω = 2 and 10 are −0.65 and −1, totaling −1.65 dB.

With these corrections, the resulting amplitude plot is illustrated in Fig. 4.45a.

PHASE PLOT

We draw the asymptotes corresponding to each of the four factors:

• (a) The zero at the origin causes a 90◦ phase shift.
• (b) The pole at s = −2 has an asymptote with a zero value for −∞ <ω< 0.2 and a slope of −45◦/decade beginning at ω = 0.2 and going up to ω = 20. The asymptotic value for ω > 20 is −90◦.
• (c) The pole at s = −10 has an asymptote with a zero value for −∞ <ω< 1 and a slope of −45◦/decade beginning at ω = 1 and going up to ω = 100. The asymptotic value for ω > 100 is −90◦.
• (d) The zero at s = −100 has an asymptote with a zero value for −∞ <ω< 10 and a slope of 45◦/decade beginning at ω = 10 and going up to ω = 1000. The asymptotic value for ω > 1000 is 90◦. All the asymptotes are added, as shown in Fig. 4.45b. The appropriate corrections are applied from Fig. 4.42b, and the exact phase plot is depicted in Fig. 4.45b.

EXAMPLE 4.30 Bode Plots for Second-Order Transfer Function with Complex Poles

Sketch the amplitude and phase response (Bode plots) for the transfer function

$$H(s) = \frac{10(s+100)}{s^2 + 2s + 100} = 10 \frac{1 + \frac{s}{100}}{1 + \frac{s}{50} + \frac{s^2}{100}}$$

MAGNITUDE PLOT

Here, the constant term is 10: that is, 20 dB(20 log 10 = 20). To add this term, we simply label the horizontal axis (from which the asymptotes begin) as the 20 dB line, as before (see Fig. 4.46a).

Figure 4.46 (a) Amplitude and (b) phase responses of the second-order system.

In addition, we have a real zero at s = −100 and a pair of complex conjugate poles. When we express the second-order factor in standard form,

$$s^2 + 2s + 100 = s^2 + 2\zeta \omega_n s + \omega_n^2$$

434 CHAPTER 4 CONTINUOUS-TIME SYSTEM ANALYSIS

we have

$$\omega_n = 10 \quad \text{and} \quad \zeta = 0.1$$

Step 1. Draw an asymptote of −40 dB/decade (−12 dB/octave) starting at ω = 10 for the complex conjugate poles, and draw another asymptote of 20 dB/decade starting at ω = 100 for the (real) zero.

Step 2. Add both asymptotes.

Step 3. Apply the correction at ω = 100, where the correction because of the corner frequency ω = 100 is 3 dB. The correction because of the corner frequency ω = 10, as seen from Fig. 4.44a for ζ = 0.1, can be safely ignored. Next, the correction at ω = 10 because of the corner frequency ω = 10 is 13.90 dB (see Fig. 4.44a for ζ = 0.1). The correction because of the real zero at −100 can be safely ignored at ω = 10. We may find corrections at a few more points. The resulting plot is illustrated in Fig. 4.46a.

PHASE PLOT

The asymptote for the complex conjugate poles is a step function with a jump of −180◦ at ω = 10. The asymptote for the zero at s = −100 is zero for ω ≤ 10 and is a straight line with a slope of 45◦/decade, starting at ω = 10 and going to ω = 1000. For ω ≥ 1000, the asymptote is 90◦. The two asymptotes add to give the sawtooth shown in Fig. 4.46b. We now apply the corrections from Figs. 4.42b and 4.44b to obtain the exact plot.

Figure 4.47 MATLAB-generated Bode plots for Ex. 4.30.

BODE PLOTS WITH MATLAB

Bode plots make it relatively simple to hand-draw straight-line approximations to a system’s magnitude and frequency responses. To produce exact Bode plots, we turn to MATLAB and its bode command.

bode(tf([10 1000],[1 2 100]),‘k-’);

The resulting MATLAB plots, shown in Fig. 4.47, match the plots shown in Fig. 4.46.

Comment. These two examples demonstrate that actual frequency response plots are very close to asymptotic plots, which are so easy to construct. Thus, by mere inspection of H(s) and its poles and zeros, one can rapidly construct a mental image of the frequency response of a system. This is the principal virtue of Bode plots.

POLES AND ZEROS IN THE RIGHT HALF-PLANE

In our discussion so far, we have assumed the poles and zeros of the transfer function to be in the left half-plane. What if some of the poles and/or zeros of H(s) lie in the RHP? If there is a pole in the RHP, the system is unstable. Such systems are useless for any signal-processing application. For this reason, we shall consider only the case of the RHP zero. The term corresponding to RHP zero at s = a is (s/a) −1, and the corresponding frequency response is (jω/a) −1. The amplitude response is

$$\left|\frac{j\omega}{a} - 1\right| = \left(\frac{\omega^2}{a^2} + 1\right)^{1/2}$$

This shows that the amplitude response of an RHP zero at s = a is identical to that of an LHP zero or s = −a. Therefore, the log amplitude plots remain unchanged whether the zeros are in the LHP or the RHP. However, the phase corresponding to the RHP zero at s = a is

$$\angle \left(\frac{j\omega}{a} - 1\right) = \angle -\left(1 - \frac{j\omega}{a}\right) = \pi + \tan^{-1}\left(\frac{-\omega}{a}\right) = \pi - \tan^{-1}\left(\frac{\omega}{a}\right)$$

whereas the phase corresponding to the LHP zero at s = −a is tan−¹(ω/a).

The complex-conjugate zeros in the RHP give rise to a term s²−2ζωns+ω^{2 n}, which is identical to the term s² +2ζωns+ω^{2 n} with a sign change in ζ . Hence, from Eqs. (4.51) and (4.52), it follows that the amplitudes are identical, but the phases are of opposite signs for the two terms.

Systems whose poles and zeros are restricted to the LHP are classified as minimum phase systems. Minimum phase systems are particularly desirable because the system and its inverse are both stable.

4.9-5 The Transfer Function from the Frequency Response

In the preceding section we were given the transfer function of a system. From a knowledge of the transfer function, we developed techniques for determining the system response to sinusoidal inputs. We can also reverse the procedure to determine the transfer function of a minimum phase

436 CHAPTER 4 CONTINUOUS-TIME SYSTEM ANALYSIS

system from the system’s response to sinusoids. This application has significant practical utility. If we are given a system in a black box with only the input and output terminals available, the transfer function has to be determined by experimental measurements at the input and output terminals. The frequency response to sinusoidal inputs is one of the possibilities that is very attractive because the measurements involved are so simple. One needs only to apply a sinusoidal signal at the input and observe the output. We find the amplitude gain |H(jω)| and the output phase shift H(jω) (with respect to the input sinusoid) for various values of ω over the entire range from 0 to ∞. This information yields the frequency response plots (Bode plots) when plotted against log ω. From these plots we determine the appropriate asymptotes by taking advantage of the fact that the slopes of all asymptotes must be multiples of ±20 dB/decade if the transfer function is a rational function (function that is a ratio of two polynomials in s). From the asymptotes, the corner frequencies are obtained. Corner frequencies determine the poles and zeros of the transfer function. Because of the ambiguity about the location of zeros since LHP and RHP zeros (zeros at s = ±a) have identical magnitudes, this procedure works only for minimum phase systems.