计算机控制系统ch4-1.ppt

4-4 Transient and steady-state response analysis (3) Static Acceleration Error Constant * 4-4 Transient and steady-state response analysis * 4-4 Transient and steady-state response analysis * Note that Judge the stability of the system before computing the steady-state error of the system. For non-unit input signals, static error constants do not change, but steady-state error will be different according to the coefficients of the input signal. For multiple input signals, the steady-state error can be computed by superposing multiple steady-state errors. For example, if r(t) = a + bt, then The actuating error is the difference between the reference input and the feedback signal, not the difference between the reference input and the output. For a different closed-loop configuration, the static error constants can be determined by an analysis similar to the one just presented. If the closed-loop discrete-time control system does not have a closed-loop pulse transfer function, the static error constants cannot be defined, because the input signal cannot be separated from the system dynamics. 4-4 Transient and steady-state response analysis * 4-4 Transient and steady-state response analysis * * Example where a. T = 0.5, k = 6 and 10, compute the ess respectively; b. T = 0.5, determine the range of the k satisfying ess ? 0.05. 4-4 Transient and steady-state response analysis * Solution: Judge the stability of the system: 4-4 Transient and steady-state response analysis * a. when T = 0.5, k should satisfy 0 k 8.15, so we should only need to compute ess for k = 6. b. with the stability condition 0 k 8.15, we conclude that 4-4 Transient and steady-state response analysis 4-4 Transient and steady-state response analysis 3. Response to disturbances * 4-4 Transient and steady-state response analysis * 4-4 Transient and steady-state response analysis * Remark: 1. If a linear system is subjected to b