Question 10.16: Bode Plot Sketching Plot the Bode magnitude and phase for th...
Bode Plot Sketching
Plot the Bode magnitude and phase for the system with the transfer function
K G(s)=K \frac{s+2}{s\left(s^{2}+8 s+400\right)},
where K=1.
Step 1: Convert the transfer function to the frequency response function
K G(j \omega)=\frac{j \omega+2}{j \omega\left[(j \omega)^{2}+8 j \omega+400\right]}=\frac{0.005((j \omega / 2)+1)}{j \omega\left[(j \omega / 20)^{2}+2(0.2)(j \omega / 20)+1\right]} .
Note that first-order and second-order terms are expressed in their corresponding basic forms, j \omega \tau+1 and \left(j \omega / \omega_{n}\right)^{2}+2 \zeta j \omega / \omega_{n}+1.
Step 2: Identify the basic terms and the corner frequencies associated with first-order and second-order terms.
The basic terms in this example are listed as follows:
1. One constant term 0.005
2. One integral term 1 / j \omega
3. One first-order term j \omega / 2+1 in the numerator with 1 / \tau=2 \mathrm{rad} / \mathrm{s}
4. One second-order term in the denominator with \omega_{n}=20 \mathrm{rad} / \mathrm{s} and \zeta=0.2
Step 3: Draw the asymptotes for the magnitude curve.
We start with indicating the corner frequencies on the frequency axis. Then, the asymptote for the derivative term is drawn through the point (1 \mathrm{rad} / \mathrm{s}, 0 \mathrm{~dB}) with a slope of -20 \mathrm{~dB} / \mathrm{decade}. The asymptote is extended until the first corner frequency 2 \mathrm{rad} / \mathrm{s} is met, which is associated with the first-order term in the numerator. At the first corner frequency, the slope increases by 20 \mathrm{~dB} / decade and changes to 0 \mathrm{~dB} / decade. We then continue extending the asymptote until the second corner frequency 20 \mathrm{rad} / \mathrm{s} is met, which is associated with the second-order term in the denominator. At the second corner frequency, the slope decreases by -40 \mathrm{~dB} / \mathrm{decade} and changes from 0 \mathrm{~dB} / decade to -40 \mathrm{~dB} / decade. Finally, we consider the effect of the constant term 0.005 by sliding the asymptotes downward by 46 \mathrm{~dB} (i.e., 20 \log _{10} 0.005=-46 \mathrm{~dB} ). This completes the composite magnitude asymptotes.
The approximate Bode magnitude plot can be obtained by drawing a smooth curve following the asymptotes. The magnitude is 3 \mathrm{~dB} above the asymptote at the first-order numerator corner frequency. A resonant peak is sketched at the second-order denominator corner frequency. Note that the associated damping ratio \zeta is 0.2 . Thus, the magnitude above the asymptote is approximately 1 / 0.4=2.5 or 7.9588 \mathrm{~dB}. The accurate value of the peak above the asymptote is 1 /(20 \left.\log _{10}(2 \zeta)\right)=8 \mathrm{~dB}. Figure 10.70a shows the magnitude plot and the asymptotes.
Step 4: Draw the asymptotes for the phase curve.
Following the same procedure as in Step 3, we can sketch the asymptotes for the composite phase plot. We start with sketching the asymptote for the derivative term with a horizontal line at -90^{\circ}. The phase changes by 90^{\circ} at the first-order numerator corner frequency 2 \mathrm{rad} / \mathrm{s}, and -180^{\circ} at the second-order denominator corner frequency 20 \mathrm{rad} / \mathrm{s}. Because the phase of a constant is 0^{\circ} for any frequency, the constant is not considered when sketching the phase plot. Figure 10.70b shows the phase plot and the asymptotes.
