Empirical Estimation of Transfer Functions for First Order Systems

We will now develop a formula for the motor DC gain constant $A$ in terms of a step input change $\Delta V$ and step output change $\Delta \omega$

problem a.

Using the final value theorem, find an expression for the steady state value of $\omega(t)$ when a step input of amplitude $V_x$ is applied.

note: $\tau > 0$ so the pole of $G_{\omega}(S)$ at $s=-\frac{1}{\tau}$ is in the open Left Half Plane (LHP) so the system is stable.

Given that the final value theorem, the steady state value of $f(t)$ is given by:

When a step input $V(s) = \frac{V_x}{s}$ is applied, the output in the Laplace domain is given by:

Since $\tau > 0$ when $s \to 0$:

The steady-state value of $\omega(t)$ under a step amplitude of $V_x$ is directly proportional to the product of DC gain and step input $V_x$

problem b.

Give the expression for $\omega (t)$ in response to a step input $V_x$ at time $t=0$. Assume a non-zero initial condition for $\omega (t)$, i.e., $\omega(t) = \omega_0$.

note: the response due to a non-zero initial condition $\omega_0$ can be modeled as the response due to input $v(t) = \omega_0 \delta(t)$ where $\delta(t)$ is the impulse function. Since $G_{\omega}(s)$ is a linear system, the response to step input $V_x$ with non-zero initial condition $\omega_0$ is just the sum of the responses due to the step and the initial condition.

The zero-state response to step input $V_x$ is given by the inverse Laplace transform of $G_{\omega}(s) \cdot V(s)$:

The zero-input response to initial condition $\omega_0$ can be modeled as the response to an impulse input $\omega_0 \delta(t)$. The Laplace transform of the impulse response is:

Which the zero-input response is given by: $\omega_{zi}(t) = \omega_0 \cdot e^{-\frac{t}{\tau}}$

The total response $\omega(t)$ is the sum of the zero-state and zero-input responses (due to linearity):

problem c.

For $\omega(t)$ in computed in part b, what is the $\lim_{t \to \infty} \omega(t)$? How does it compare to result in part a?

Since $e^{-\frac{t}{\tau}} \to 0$ as $t \to \infty$, the steady-state value of $\lim_{t \to \infty} \omega(t)$ is:

We see that the steady-state value of $\omega(t)$ is the same. The initial condition $\omega_0$ does not affect the steady-state value of $\omega(t)$, only influence transient response.

problem d.

Now assume that you run the motor with an initial step input of $V_{\text{min}}$ until time $t_0$. At time $t_0$, assume that the system has reached steady state and the step input is changed to $V_{\text{max}}$ at time $t_0$. In other words, the system input will take the form

$$v(t) = \begin{cases} v_{\text{min}} & \text{if } 0 \leq t < t_0 \\\ v_{\text{max}} & \text{if } t \geq t_0 \end{cases}$$

where $t_0 \gg \tau$ and $V_{\text{max}}$ and $V\_{\text{min}}$ may be non-zero.

Use the results above to show that:

$$A = \frac{\Delta \omega}{\Delta V}$$

where $\Delta V = V_{\text{max}} - V_{\text{min}}$ and $\Delta \omega = \omega_{ss} - \omega_0$ where $\omega_0$ is the steady-state response to a constant input $V_\text{min}$ and $\omega_{ss}$ is the steady-state response to the input $V_\text{max}$

For input $V_{\text{min}}$, the steady-state response is $\omega_0 = A \cdot V_{\text{min}}$

Similarly, for input $V_{\text{max}}$, the steady-state response is $\omega_{ss} = A \cdot V_{\text{max}}$

Thus, the change in steady-state response is

Thus $A = \frac{\Delta \omega}{\Delta V}$

Using the formula derived in Q1, use the following graphs to calculate A for this system.

Given $A = \frac{\Delta \omega}{\Delta V}$, from the graph $V_{\text{min}} = 1$ and $V_{\text{max}} = 5$ and $\omega_0 = 5$ and $\omega_{\text{ss}} = 25$, $A = 5$

For a first order system, the time it takes a step response to reach 63.2% of its steady state value ($t_1 − t_0$ in Fig. 1) is the response’s time constant $\tau$ . i.e., at time $t_1, \omega(t_1) = 0.632\Delta \omega + \omega_0$. Find the time constant $\tau$ for the above system.

From the graph, $\tau \approx 0.8 \sec$

Using $A$ and $\tau$ calculated in Q2 and Q3, find the transfer function in terms of $s$

The system quickly rises to a steady-state value without any oscillation, which suggests a first-order system. The transfer function: