Question 10.2: Draw the Bode plots for the system shown in Figure 10.10, wh...

We will make a Bode plot for the open-loop function G(s) = K (s + 3)/[s (s + 1) (s + 2)]. The Bode plot is the sum of the Bode plots for each first-order term. Thus, it is convenient to use the normalized plot for each of these terms so that the low-frequency asymptote of each term, except the pole at the origin, is at 0 dB, making it easier to add the components of the Bode plot. We rewrite G(s) showing each term normalized to a low-frequency gain of unity. Hence,

G (s) = \frac{\frac{3}{2} K  (\frac{s}{3}  +  1 )}{s  (s  +  1) (\frac{s}{2}  +  1 )}       (10.25)

Now determine that the break frequencies are at 1, 2, and 3. The magnitude plot should begin a decade below the lowest break frequency and extend a decade above the highest break frequency. Hence, we choose 0.1 radian to 100 radians, or three decades, as the extent of our plot.

At ω = 0.1, the low-frequency value of the function is found from Eq. (10.25) using the low-frequency values for all of the [(s/a) + 1] terms (i.e., s = 0) and the actual value for the s term in the denominator. Thus, G ( j0.1) ≈ \frac{3}{2} K/0.1 = 15 K. The effect of K is to move the magnitude curve up (increasing K) or down (decreasing K) by the amount of 20 log K. K has no effect upon the phase curve. If we choose K = 1, the magnitude plot can be denormalized later for any value of K that is calculated or known.

Figure 10.11(a) shows each component of the Bode log-magnitude frequency response. Summing the components yields the composite plot shown in Figure 10.11(b). The results are summarized in Table 10.2, which can be used to obtain the slopes. Each pole and zero is itemized in the first column. Reading across the table shows its contribution at each frequency. The last row is the sum of the slopes and correlates with Figure 10.11(b). The Bode magnitude plot for K = 1 starts at ω = 0.1 with a value of 20 log 15 = 23.52 dB, and decreases immediately at a rate of −20 dB/decade, due to the s term in the denominator. At ω = 1, the (s + 1) term in the denominator begins its 20 dB/decade downward slope and causes an additional 20 dB/decade negative slope, or a total of −40 dB/decade. At ω = 2, the term [(s/2) + 1] begins its −20 dB/decade slope, adding yet another −20 dB/decade to the resultant plot, or a total of −60 dB/decade slope that continues until ω = 3. At this frequency, the [(s/3) + 1] term in the numerator begins its positive 20 dB/decade slope. The resultant magnitude plot, therefore, changes from a slope of −60 dB/decade to −40 dB/decade at ω = 3, and continues at that slope, since there are no other break frequencies.

TABLE 10.2 Bode magnitude plot: slope contribution from each pole and zero in Example 10.2

The slopes are easily drawn by sketching straight-line segments decreasing by 20 dB over a decade. For example, the initial −20 dB/decade slope is drawn from 23.52 dB at ω = 0.1, to 3.52 dB (a 20 dB decrease) at ω = 1. The −40 dB/decade slope starting at ω = 1 is drawn by sketching a line segment from 3.52 dB at ω = 1, to −36.48 dB (a 40-dB decrease) at ω = 10, and using only the portion from ω = 1 to ω = 2. The next slope of −60 dB/decade is drawn by first sketching a line segment from ω = 2 to ω = 20 (1 decade) that drops down by 60 dB, and using only that portion of the line from ω = 2 to ω = 3. The final slope is drawn by sketching a line segment from ω = 3 to ω = 30 (1 decade) that drops by 40 dB . This slope continues to the end of the plot.

Phase is handled similarly. However, the existence of breaks, a decade below and a decade above the break frequency, requires a little more bookkeeping. Table 10.3 shows the starting and stopping frequencies of the 45°/decade slope for each of the poles and zeros. For example, reading across for the pole at −2, we see that the −45° slope starts at a frequency of 0.2 and ends at 20. Filling in the rows for each pole and then summing the columns yields the slope portrait of the resulting phase plot. Looking at the row marked Total slope, we see that the phase plot will have a slope of −45°/decade from a frequency of 0.1 to 0.2. The slope will then increase to −90°/decade from 0.2 to 0.3. The slope will return to −45°/decade from 0.3 to 10 rad/s. A slope of 0 ensues from 10 to 20 rad/s, followed by a slope of +45°/decade from 20 to 30 rad/s. Finally, from 30 rad/s to infinity, the slope is 0°/decade.

TABLE 10.3 Bode phase plot: slope contribution from each pole and zero in Example 10.2

The resulting component and composite phase plots are shown in Figure 10.12. Since the pole at the origin yields a constant −90° phase shift, the plot begins at −90° and follows the slope portrait just described.