Equation 6\u20111<\/p>\r\n<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n
[latex]G(s)[\/latex] has only one pole, and no zeros. Its unit step response can be derived as shown in Equation 6\u20112:<\/p>\r\n\r\n
| \r\n [latex]Y(s) = \\frac{1}{s}G(s) = \\frac{K}{s(s+a)} = \\frac{K_{1}}{s} + \\frac{K_{2}}{(s+a)}[\/latex]<\/p>\r\n[latex]K_1=\\left .\\begin{matrix} \\frac{K}{s} \\end{matrix}\\right|_{s=0}=\\frac{K}{a}[\/latex]\r\n\r\n[latex]K_2=\\left .\\begin{matrix} \\frac{K}{s+a} \\end{matrix}\\right|_{s=-a}=\\frac{-K}{a}[\/latex]\r\n [latex]Y(s) = \\frac{K}{a}. \\frac{1}{s} - \\frac{K}{a}. \\frac{1}{(s+a)}[\/latex]<\/p>\r\n [latex]y(t) = \\frac{K}{a}. (1-e^{-at}. 1(t) = K_{dc}\\left ( 1-e^{1\\frac{t}{\\tau}} \\right ). 1(t)[\/latex]<\/p>\r\n<\/td>\r\n | \r\n Equation 6\u20112<\/p>\r\n<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n Note that:<\/span><\/p>\r\n\r\n [latex]\\frac{dy}{dt} = \\frac{K_{dc}}{\\tau}e^{-\\frac{t}{\\tau}}. 1(t)[\/latex]<\/p>\r\n [latex]\\frac{dy}{dt} (0)= \\frac{K_{dc}}{\\tau} \\neq 0[\/latex]<\/p>\r\n<\/td>\r\n Equation 6\u20113<\/p>\r\n<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n If the unit step input is used, the process DC gain and time constant can be evaluated directly from the graph, as shown in the following example.<\/p>\r\n\r\n Consider a plot of the response of a certain unknown process, shown in Figure 6\u20111. We would like to derive a model for this unknown system, i.e. a transfer function that would give a response closest to that of our system, let's call it [latex]G_{m}(s)[\/latex]. The response looks like an exponential rise with a non-zero slope at t=0 and is therefore identified as the response of a first order process (system). As such, the response can be described by the following equation:<\/p>\r\n [latex]y(t) = K_{dc}(1-e^{-\\frac{t}{\\tau}}). 1(t)[\/latex]<\/p>\r\n\r\n\r\n[caption id=\"attachment_750\" align=\"aligncenter\" width=\"491\"] A first order system is described by the transfer function in Equation 6\u20111:<\/p>\n Equation 6\u20111<\/p>\n<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n [latex]G(s)[\/latex] has only one pole, and no zeros. Its unit step response can be derived as shown in Equation 6\u20112:<\/p>\n [latex]Y(s) = \\frac{1}{s}G(s) = \\frac{K}{s(s+a)} = \\frac{K_{1}}{s} + \\frac{K_{2}}{(s+a)}[\/latex]<\/p>\n [latex]K_1=\\left .\\begin{matrix} \\frac{K}{s} \\end{matrix}\\right|_{s=0}=\\frac{K}{a}[\/latex]<\/p>\n [latex]K_2=\\left .\\begin{matrix} \\frac{K}{s+a} \\end{matrix}\\right|_{s=-a}=\\frac{-K}{a}[\/latex]<\/p>\n [latex]Y(s) = \\frac{K}{a}. \\frac{1}{s} - \\frac{K}{a}. \\frac{1}{(s+a)}[\/latex]<\/p>\n [latex]y(t) = \\frac{K}{a}. (1-e^{-at}. 1(t) = K_{dc}\\left ( 1-e^{1\\frac{t}{\\tau}} \\right ). 1(t)[\/latex]<\/p>\n<\/td>\n Equation 6\u20112<\/p>\n<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n Note that:<\/span><\/p>\n [latex]\\frac{dy}{dt} = \\frac{K_{dc}}{\\tau}e^{-\\frac{t}{\\tau}}. 1(t)[\/latex]<\/p>\n [latex]\\frac{dy}{dt} (0)= \\frac{K_{dc}}{\\tau} \\neq 0[\/latex]<\/p>\n<\/td>\n Equation 6\u20113<\/p>\n<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n If the unit step input is used, the process DC gain and time constant can be evaluated directly from the graph, as shown in the following example.<\/p>\n Consider a plot of the response of a certain unknown process, shown in Figure 6\u20111. We would like to derive a model for this unknown system, i.e. a transfer function that would give a response closest to that of our system, let’s call it [latex]G_{m}(s)[\/latex]. The response looks like an exponential rise with a non-zero slope at t=0 and is therefore identified as the response of a first order process (system). As such, the response can be described by the following equation:<\/p>\n [latex]y(t) = K_{dc}(1-e^{-\\frac{t}{\\tau}}). 1(t)[\/latex]<\/p>\n |
![\"Figure](\"http:\/\/pressbooks.library.ryerson.ca\/controlsystems\/wp-content\/uploads\/sites\/75\/2019\/07\/fig6_1-300x243.png\")
Figure 6-1: First Order System response[\/caption]","rendered":"