Final answer:
The steady state error in an open loop DC motor due to a step input can be determined using the final value theorem. The error is given by e(t) = y(t) - x(t), where y(t) is the output of the motor and x(t) is the input. The steady state error can be calculated by evaluating the difference between the desired output and the actual output.
Step-by-step explanation:
To find the steady state error, we need to determine the difference between the desired output and the actual output of the motor. In this case, the input is a step function given by x(t) = Au(t), where u(t) is the unit step function and A is the amplitude of the step. The transfer function of the open loop motor is G(s) = 3/s(s+3). We can use the final value theorem to find the steady state value of the output.
The final value theorem states that the steady state value of a time domain function can be found by evaluating the Laplace transform of the function at s=0. In this case, we take the Laplace transform of the input function x(t) to get X(s) and multiply it by the transfer function G(s). The output Y(s) is then obtained by multiplying X(s)G(s) and taking the inverse Laplace transform.
Let's calculate the steady state error for the given transfer function and input:
X(s) = A/s
Y(s) = X(s)G(s) = A/s * 3/s(s+3)
By performing partial fraction decomposition, we can write Y(s) as:
Y(s) = -A/3 * (1/s - 1/(s+3))
Taking the inverse Laplace transform, we obtain the output y(t) as:
y(t) = -A/3 * (1 - e^(-3t))
Finally, the steady state error is given by:
e(t) = y(t) - x(t)
= -A/3 * (1 - e^(-3t)) - A
where A is the amplitude of the step input and t is the time.