The results indicated to show that system (a) had a significantly larger absolute steady state error as compared to the system (b). It is equally clear that system (B) displayed a greater accuracy with less absolute steady state error as compared to that of the system (a). This leads to a conclusion that the positive error value of system (b) established the system output’s stability. Prior to the introduction and calculation of the error formulas, it is possible to determine the system’s performance by the use of its constant (K) value. A perfect system with no steady state error, ideally, Kp would be equal to zero, Kv would be equal to one and Ka would be equal to one. It is however extremely challenging to attain this kind of a system.
If Kv or Ka would be equal to zero, then the steady state error would be infinite as illustrated in the results in the table. If Kpos would be equal to negative one, then the steady state error would become infinite. The results from the table show that the Kpos is very close to negative one. This is basically the reason as to why it has a large value of the steady state error. The sign recorded by the steady state error gives an indication of whether the output is greater than the input or vice versa. For instance, the table shows that system (a) displays a negative error value making the output of the system to be greater than the input. This designates some degree of amplification. On the other hand, the system (b) displays a positive error making the output smaller than the input. ...Show more