# read how to control the Response of a system

Response of a system a major aspect of system modeling and control system in appliances and industries. In a modeled system that has it’s variables controlled, then the manipulated variables are inputted at different conditions to check the exactness of the system output. The response of a system is a function of time domain. Theses response are us all represented in a Laplace transform method for proper evaluation and modeling. This a very crucial aspects that engineers pay close attention to.

The transfer or response function of a modeled system in  Laplace transform domain is given by the formula below

T.S =       { Laplace of output}/{Laplace transform of input}

For a response of a system to be evaluated properly, there are some forcing functions that works on system as method in which their responses were modeled to be controlled. These forcing functions includes: unit step function which can be represented as 1/S in Laplace domain, impulse function which can be represented as 1/S*2 in Laplace domain, pulse function, sinusoidal functions. These functions can be attached  to a response of a system based on the condition subjected to the system. If the system above contains a flowing fluid and it’s modeled ( having it’s formula of flow) , the response  will be the respect of the following, mass flow rate, volume flow rate as measured. Some industries may be measuring it mechanically( that is open the tap and allow the fluid to flow for some measured time and the tap is being turned off and the flow rates are calculated) while some can use modeled the system with a programming software like MATLAB, has a definite relationship between the  constant variable and the manipulated variables, therefore, all the will do is to get some volume of the fluid and input their data in the formula and get the response, this will help to detect when the system is faulty  when the results is compared with the mechanically calculated response.

Example: if the system above has a transfer function  1/V – 4EXP(-2t)  where T and V are time and volume of the fluid , what will be the response when t2 seconds and V10m^3

Solution

T.S    1/V – 4EXP(-2t)

Substituting the variables

T.S 1/10  4EXP(-2*2)

T.S 0.10.0734

T.S 0.0267

If  when this is evaluated and the mechanically calculated does not correspond then check out the system there is problem.

Also a problem can be represented in a differential form,  depending on the degree, all you do is to solve the equation and place it back to the time domain by finding the Laplace transform of the solution and evaluate.

Example

Find the response of a system with the differential equation   =0 at t=0 and 2.

Solution:

First is to factorize the denominator and resolve into partial fraction

Resolving the right hand side into partial fraction

10= A(S+4) +B(S-1)

By comparing coefficients

A+B=0……………………….(I)

4A-B=10……………………(II)

From (I); A= -B

Substitute into (II)

4A- (-A) =10

A=2, b= -2

Therefore this becomes

Taking the Laplace transform of the solution

L{S-1} =EXP(t) and L{S+4} =EXP(-4t)

Therefore T.S = 2(EXP(t) – EXP(-4t))

Evaluating when t =o and 2

T.S =0 and 14.777