In this case, 'GH' value is negative because the gain of feedback path is negative.If the value of (1+GH) is greater than 1, then sensitivity decreases.

So, the control system becomes unstable.Therefore, we have to properly choose the feedback in order to make the control system stable.To know the effect of feedback on noise, let us compare the transfer function relations with and without feedback due to noise signal alone.It is obtained by making the other input $R(s)$ equal to zero.It is obtained by making the other input $R(s)$ equal to zero.In the closed loop control system, the gain due to noise signal is decreased by a factor of $(1+G_a G_b H)$ provided that the term $(1+G_a G_b H)$ is greater than one. Rather, we shall investigate a type of steady-state error that is caused by the incapability of a system to follow particular types of inputs.Steady-state error is the difference between the input and the output for a prescribed test input as time tends to infinity. A system is called type 0, type 1, type 2,…if =0, 1, 2…respectively. System Metrics . So, the feedback will increase the overall gain of the system in one frequency range and decrease in the other frequency range.$S_{G}^{T} = \frac{\frac{\partial T}{T}}{\frac{\partial G}{G}}=\frac{Percentage\: change \: in \:T}{Percentage\: change \: in \:G}$ $S_{G}^{T}=\frac{\partial T}{\partial G}\frac{G}{T}$ Do partial differentiation with respect to G on both sides of Equation 2.$\frac{\partial T}{\partial G}=\frac{\partial}{\partial G}\left (\frac{G}{1+GH} \right )=\frac{(1+GH).1-G(H)}{(1+GH)^2}=\frac{1}{(1+GH)^2}$ Substitute Equation 5 and Equation 6 in Equation 4.$$S_{G}^{T}=\frac{1}{(1+GH)^2}(1+GH)=\frac{1}{1+GH}$$If the value of (1+GH) is less than 1, then sensitivity increases.

represent constant-velocity inputs to a position control system by their linearly increasing amplitude. In this case, 'GH' value is positive because the gain of feedback path is positive.In general, 'G' and 'H' are functions of frequency. As the type increases, accuracy is improved. In order to have zero steady-state error,And for the limit to be infinite the denominator must be equal to zero a If there are no integrations, the n=0, and it yields a finite error. If there one or zero integrators in the forward path then . Changes in the reference input will cause unavoidable errors during transient periods and may also cause steady-state errors. Consider the unity-feedback control system with the following open-loop transfer function in the denominator, representing a pole of multiplicity at the origin. When a system is being designed and analyzed, it doesn't make any sense to test the system with all manner of strange input functions, or to measure all sorts of arbitrary performance metrics. Let's say that we have a system with a disturbance that enters in the manner shown below. A system may have no steady-state error to a step input, but the same system may exhibit nonzero steady-state error to a ramp input. Stability: It is an important characteristic of the control system. The table shows the static error constants and the steady-state error as a functions of Control systems often do not have unity feedback because of the compensation used to practical way to analyze the steady-state error is to take the system and form a unity feedback system by adding and subtracting unity feedback paths as shown in Figure 7.15:Donde G(s)=G1(s)G2(s) y H(s)=H1(s)/G1(s). In other words, there must be at least three integrations in the forward path. When H (s) is not equal to 1, u (t) may or may not be the error, depending on the form and purpose of H (s). If is written so that each term in the numerator and denominator, except the term .

If either the output or some part of the output is returned to the input side and utilized as part of the system input, then it is known as feedback.Feedback plays an important role in order to improve the performance of the control systems. This is the case shown in Figure 7-2a, output 2.In summary, for a step input to a unity feedback system, the steady-state error will be zero if there is at least one pure integration in the forward path. If there is only one integrator in the forward path then lim , and the steady-state error will be infinite and lead to diverging ramp, as shown in Figure 7.2b, output 3.In order to have zero steady-state error for a parabolic input, we must have:. The step response of a system in a given initial state consists of the time evolution of its outputs when its control inputs are Heaviside step functions.In electronic engineering and control theory, step response is the time behaviour of the outputs of a general system when its inputs change from zero to one in a very short time. Since:This last equation allows us to calculate the steady-state error and then draw conclusions about the relationship that exists between the open-loop system Is the gain of the forward transfer function. The feedback concept has been the foundation for control system analysis and design. Errors in a control system can be attributed to many factors.

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