As the normalized time variable and select a value for ζ. Using the omputer, we solve for the val...
as the normalized time variable and select a value for ζ. Using the omputer, we solve for the values of»,I that yield c(1) = 0.9 and c() 0.1. Subtracting the two values of aht yields the normalized rise time, aT, for that value of 5. Continuing in like fashion with other values of ζ, we obtain the results ploded in Figure 4.16.5 Let us look at an example. Damping Normalized rise time 1.104 1.203 1.321 1 .463 1.638 1·854 2.126 2.467 2.883 3.0 0.1 03 0.5 2.8 2.2 0.7 08 0.9 1.8 1.6 1.2 1.0 0.1 02 0.7 Damping ratio FIGURE 4.16 Nomalized rise timevrsas damping ratio for aseaond orderunderdamped response Example 4.5 Finding T, %OST, and T, from a Transfer Function Virtul Experiment 42 System Resporse e efcthatazu taquency PROBLEM: Civen the transfer function 100 (4.43) find TP: %OS, T1, and Tr. of the Quaiser lsear Sevoa pplie o SOLUTION: aJn and ζ are calculated as 10 and 0.75, respectively. Now substitute ζ and um ink) EQ& (4.34), (4.38), and (4.42) and find, respectively, that TP-0.475 second, %OS-2.838, and T,-0.533 second. Using the table in Figure 4.16, de normalized rise time is approximaely 2.3 seconds. Dividing by an yields T 023 second. This problem demonstrates that we can find TP, %OS, T. and T, without the tedious task of taking an inverse Laplace transform, plotting the output response, and taking measurements from the plot.
as the normalized time variable and select a value for ζ. Using the omputer, we solve for the values of»,I that yield c(1) = 0.9 and c() 0.1. Subtracting the two values of aht yields the normalized rise time, aT, for that value of 5. Continuing in like fashion with other values of ζ, we obtain the results ploded in Figure 4.16.5 Let us look at an example. Damping Normalized rise time 1.104 1.203 1.321 1 .463 1.638 1·854 2.126 2.467 2.883 3.0 0.1 03 0.5 2.8 2.2 0.7 08 0.9 1.8 1.6 1.2 1.0 0.1 02 0.7 Damping ratio FIGURE 4.16 Nomalized rise timevrsas damping ratio for aseaond orderunderdamped response Example 4.5 Finding T, %OST, and T, from a Transfer Function Virtul Experiment 42 System Resporse e efcthatazu taquency PROBLEM: Civen the transfer function 100 (4.43) find TP: %OS, T1, and Tr. of the Quaiser lsear Sevoa pplie o SOLUTION: aJn and ζ are calculated as 10 and 0.75, respectively. Now substitute ζ and um ink) EQ& (4.34), (4.38), and (4.42) and find, respectively, that TP-0.475 second, %OS-2.838, and T,-0.533 second. Using the table in Figure 4.16, de normalized rise time is approximaely 2.3 seconds. Dividing by an yields T 023 second. This problem demonstrates that we can find TP, %OS, T. and T, without the tedious task of taking an inverse Laplace transform, plotting the output response, and taking measurements from the plot.