In the open-loop process shown in the figure, the input $U(s)$, the transfer function $G_p(s)$ and the output $Y(s)$ are given in the Laplace domain in terms of the Laplace variable $s$. For this process, which of the following is true? (where $M, \tau_p, K_p$, are the magnitude of the input, the characteristic time and the gain for the process, respectively) $
$y(t) = M K_p \left(1 - e^{-t/\tau_p}\right)$
To determine the correct expression for the output \( y(t) \) of the given open-loop process, we need to perform the following steps:
- Identify the transfer function \( G_p(s) \) and the input \( U(s) \) from the given diagram.
- Compute the output \( Y(s) \) in the Laplace domain using the relationship: \(Y(s) = G_p(s) \cdot U(s)\).
- Perform an inverse Laplace transform to find \( y(t) \).
Step 1: Identify Transfer Function and Input
From the diagram above:
- Input: \(U(s) = \frac{M}{s}\)
- Transfer Function: \(G_p(s) = \frac{K_p}{\tau_p s + 1}\)
Step 2: Compute the Output in the Laplace Domain
Using the formula for the output:
\(Y(s) = G_p(s) \cdot U(s) = \left(\frac{K_p}{\tau_p s + 1}\right) \cdot \left(\frac{M}{s}\right)\)
Simplifying, we get:
\(Y(s) = \frac{M K_p}{s (\tau_p s + 1)}\)
Step 3: Perform Inverse Laplace Transform
The inverse Laplace transform of \(\frac{M K_p}{s (\tau_p s + 1)}\) can be found using partial fraction decomposition:
\(\frac{M K_p}{s (\tau_p s + 1)} = \frac{A}{s} + \frac{B}{\tau_p s + 1}\)
Solving for \( A \) and \( B \), we find:
- \(A = M K_p\)
- \(B = -M K_p\)
Thus, the inverse Laplace transform yields:
\(y(t) = M K_p (1 - e^{-t/\tau_p})\)
Therefore, the correct expression for the output \( y(t) \) is:
Correct Answer: \(y(t) = M K_p \left(1 - e^{-t/\tau_p}\right)\)
Important Questions from First and Second Order Systems
- The transfer function of a process is $G(s) = \frac{K_p}{ \tau_p s+1}$, where $K_p$ is the gain and $ \tau_p$ is the time constant. This is a __________ process.
A thermometer measuring body temperature follows a first-order response with a time constant of 40 seconds. The instrument will reach 95% of its steady-state output at __________ seconds.
(Round off to the nearest integer)
The output $y(t)$ of a first-order process is governed by the following differential equation
$\tau_p\frac{dy}{ dt} + y = K_p f(t)$
where $T_p$ is a non-zero time constant, $K_p$ is the gain and $f(t)$ is the input with $f(0) = 0$.
Assume $y(0) = 0$. The transfer function for this process is (consider $s$ as the independent variable in the Laplace domain)
