Part 4 - Time Respone | rezarezvan.com

Introduction

Once we have obtained a mathematical representation of a system, our next step is to analyze its transient and steady-state response.

Transient and Steady-State Response

The purpose of a closed loop control is to make a system’s output follow the input as cloesly as possible.

Ideally the output would at all times correspond exactly to the input, but this is not possible in real systems which (due to the effects or inertia, inductance, heat transfer, etc.) do not respond instantaneously to changes in the input.

In practice, a system is judged by a number of criteria, the three most important of which are,

Is the System Absolutely Stable?
- After an input disturbance, the output should settle down to a steady value.
How Accurate is the System in Steady State?
- The steady-state error should be small.
How Quickly Does the System Reach Steady State?
- Steady-state should be reached quickly without exessive overshoot or oscillation.

Steady-State Response to Step/Ramp Input

The steady-state response of a system to a step input is very often of interest.

For this, the final value theorem is applied.

\lim_{t \to \infty} f(t) = \lim_{s \to 0} sF(s)

If the transfer function of the system is denoted $G_{\text{sys}}(s)$ , then the final value theorem tells us that for a step of size $\bar{U}$ ,

u(t) = \bar{U} \quad \forall t > 0 \newline U(s) = \frac{\bar{U}}{s} \newline \lim_{t \to \infty} y(t) = \lim_{s \to 0} sU(s)G_{\text{sys}}(s) = \lim_{s \to 0} s \cdot \frac{\bar{U}}{s}G_{\text{sys}}(s) = \bar{U} \lim_{s \to 0} G_{\text{sys}}(s)

For a ramp input of slope $\bar{U}$ ,

u(t) = \bar{U}t \quad \forall t > 0 \newline U(s) = \frac{\bar{U}}{s^2} \newline \lim_{t \to \infty} y(t) = \lim_{s \to 0} sU(s)G_{\text{sys}}(s) = \lim_{s \to 0} s \cdot \frac{\bar{U}}{s^2}G_{\text{sys}}(s) = \bar{U} \lim_{s \to 0} \frac{G_{\text{sys}}(s)}{s}

Poles and Zeroes

The poles and zeroes of a system are important in determining the system’s response.

Let’s see how we analyze a systems poles and zeroes to determine a system’s response.

Consider,

G(s) = \frac{b_{n - 1} s^{n - 1} + b_{n - 2} s^{n - 2} + \ldots + + b_0}{s^n + a_{n - 1} s^{n - 1} + \ldots + + a_0} = \frac{N(s)}{D(s)}

Poles of $G(s)$ are the roots of $D(s) = 0$ and zeroes of $G(s)$ are the roots of $N(s) = 0$ .

Generally, at poles $G(s) = \infty$ unless the pole is cancelled by a matching zero.

At zeroes, $G(s) = 0$ unless the zero is cancelled by a matching pole.

Poles and Zeroes of a First Order System

A system’s output response contains two parts,

Forced or steady-state response, this is caused by the poles of the input function, $R(s)$ .
Natural or homogenous response, this is caused by the poles of the transfer function, $G(s)$ .

Example

We have our transfer function $G(s) = \frac{s + 2}{s + 5}$ , and our input is $R(s)$ .

Let the input be a step response, $R(s) = \frac{1}{s}$ . Thus, our output is,

C(s) = R(s) G(s) = \frac{1}{s} \cdot \frac{s + 2}{s + 5} = \frac{s + 2}{s(s + 5)}

Using partial fraction decomposition, we get,

C(s) = \frac{2/5}{s} + \frac{3/5}{s + 5}

In the time domain,

c(t) = \frac{2}{5} + \frac{3}{5}e^{-5t}

Conclusion from the Example

Pole of input function generated forced response.
Pole of transfer function generated natural response, $e^{-5t}$ . This is not affected at all by the zeroes.
Pole on real axis, say at $-\alpha$ , generates an exponential response, $e^{-\alpha t}$ . Farther to the left on negative axis, the faster the response decays.
Both the poles and zeroes contribute to the amplitude of the response. For us, this will not be important.

First Order Systems

We use systems of the form $G(s) = \frac{a}{s + a}$ as our base form for our definitions.

If our input is the step function, $R(s) = \frac{1}{s}$ , we get,

c(t) = c_f(t) + c_n(t) = 1 - e^{-at}

Time Constant

Our first specification is the system’s time constant, $\tau = \frac{1}{a}$ .

The time constant is the time required for the step response to rise to 63% percent of its final value,

c(\tau) = 1 - e^{-a\tau} = 1 - e^{-1} = 1 - 0.37 = 0.63

As $\frac{d c(t)}{dt} = ae^{-at}$ , we thus have $a$ equal to the slope at $t = 0$ .

We call $a$ the exponential frequency.

Rise and Settling Time

Rise time, $T_r$ , is the time for the output to go from 10% to 90% of its final value,

T_r = \frac{2.2}{a}.

Settling time, $T_s$ , is the time for the output to reach 98% of its final value.

Setting $c(T_s) = 0.98$ , we find that,

T_s = \frac{4}{a}

First Order Transfer Function via Testing

It is quite often not possible or practical to obtain a system’s transfer function by analytical means. Perhaps the system is closed, and the component parts are not easily identifiable.

The transfer function is a representation of the system from input to output, the system’s step response can lead to a representation even though the inner construction is not known.

With a step input, we can measure the time constant and the steady state value, from which the transfer function can be calculated

Second Order Systems

For first order systems, varying the systems parameters only changed the speed of the response.

Form of a second order system we will analyze is,

G(s) = \frac{b}{s^2 + as + b}

Changes in these parameters can actually change the form of the system’s response.

May see respones similar to first-order system, damped oscillations, or undamped oscillations.

For second order systems, we will encounter four cases, let’s review them.

Overdamped Response

For overdamped response, we have a system with two non equal real poles.

For example,

\begin{aligned} C(s) & = \frac{9}{s(s^2 + 9s + 9)} \newline & = \frac{9}{s(s + 7.854)(s + 1.146)} \end{aligned}

From inspection of poles, we know the form of the system’s response will be,

c(t) = K_1 + K_2 e^{-\sigma_1 t} + K_3 e^{-\sigma_2 t}

where $-\sigma_1 = -7.854$ and $-\sigma_2 = -1.146$ , are our two real poles.

Critically Damped Respose

For critically damped response, we have a system with two equal real poles.

For example,

\begin{aligned} C(s) & = \frac{9}{s(s^2 + 6s + 9)} \newline & = \frac{9}{s(s + 3)^2} \newline & = \frac{K_1}{s} + \frac{K_2}{(s + 3)} + \frac{K_3}{(s + 3)^2} \end{aligned}

From inspection of poles, we know that the form of the system’s response will be,

c(t) = K_1 + K_2 e^{-\sigma_1 t} + K_3 \cdot te^{-\sigma_1 t}

where $-\sigma_1 = -3$ is our real pole.

Underdamped Response

For underdamped response, we have a system with two complex conjugate poles (non zero real and imaginary parts).

For example,

\begin{aligned} C(s) & = \frac{9}{s(s^2 + 2s + 9)} \newline & = \frac{9}{s(s + 1 + j\sqrt{8})(s + 1 - j \sqrt{8}} \newline & = \frac{K_1}{s} + \frac{\alpha + j\beta}{s + 1 + j\sqrt{8}} + \frac{a - j\beta}{s + 1 - j\sqrt{8}} \end{aligned}

Thus, the form of the system’s response will be,

c(t) = K_1 + e^{-\sigma_d t}[2\alpha \cos(\omega_d t) + 2\beta \sin(\omega_d t)]

where $-\sigma_d \pm j\omega_d = -1 \pm j\sqrt{8}$ .

For systems with poles at $s = -\sigma_d \pm j\omega_d$ , the real part $\sigma_d$ determines the exponential frequency (decay rate) for the exponential envelope.

The imaginary part $\omega_d$ determines the oscillation frequency of the sinusoids, and is called the damped frequency of oscillation.

Note, we can also write,

e^{-\sigma_d t}[2\alpha \cos(\omega_d t) + 2\beta \sin(\omega_d t)] = K_4 e^{-\sigma_d t} \cos(\omega_d t - \phi)

where $\phi = \tan^{-1}\left(\frac{\beta}{\alpha}\right)$ and $K_4 = \sqrt{(2\alpha)^2 + (2\beta)^2}$

Undamped Response

For undamped response, we have a system with two imaginary poles (zero real part).

For example,

\begin{aligned} C(s) & = \frac{9}{s(s^2 + 9)} \newline & = \frac{9}{s(s + j3)(s - j3)} \newline & = \frac{K_1}{s} + \frac{\alpha + j\beta}{s + j3} + \frac{\alpha - j\beta}{s - j3} \end{aligned}

Thus, the form of the system’s response will be,

c(t) = K_1 + e^{-(0)t}[2\alpha \cos(\omega_d t) + 2\beta \sin(\omega_d t)] = K_1 + 2\alpha \cos(\omega_d t) + 2\beta \sin(\omega_d t) = K_1 + K_4 \cos(\omega_d t - \phi)

where $\pm j\omega_d = \pm j3$ .

General Second Order System

Let’s now generalize our discussion of second order systems and develop specifications to describe the response of the system.

The natural frequency $\omega_n$ of a second order system is the frequency of oscillation of the system with damping removed.
The damping ratio $\zeta$ of a second order system is a way to describe a system’s damped oscillation, independent of time scale.

We want to rewrite the second order system in terms of $\omega_n$ and $\zeta$ .

I’ll skip the steps but,

G(s) = \frac{\omega_n^2}{s^2 + 2\zeta\omega_n s + \omega_n^2}

Using this, we can find that the poles are,

s_{1, 2} = -\zeta\omega_n \pm \omega_n \sqrt{\zeta^2 - 1}

Underdamped Response Specifications

Let $c_{\text{final}} = \lim_{t \to \infty} c(t)$ .

Rise time $T_r$ is the time for the output to go from 10% $(0.1 c_{\text{final}})$ to 90% $(0.9 c_{\text{final}})$ of its final value.
Peak time $T_p$ is the time required to reach the first and largest peak $c_{\text{max}}$ .
Percent overshoot, %OS it the percentage that the output overshoots the final value at $t = T_p$ .

\text{OS} = \frac{c_{\text{max}} - c_{\text{final}}}{c_{\text{final}}} \cdot 100\%.

Settling time $T_s$ is time required for the output to reach and stay within $\pm 2\%$ of $c_{\text{final}}$ .

Transient Response of Higher Order System

If we’re dealing with higher order systems, we can just write this as a sum of first and second order,

X_0(s) = \frac{1}{s} + \sum_{r = 1}^N \frac{A_r}{(s - \sigma_r)} + \sum_{c = 1}^N \frac{A_c}{(s - \sigma_c)^2 + \omega_c^2}

so that the time domain response is of the form,

x_0(t) = 1 + \sum_{r = 1}^N B_r e^{\sigma_r t} + \sum_{c = 1}^N B_c e^{\sigma_c t} \sin(\omega_c t)