AME-558

Syllabus | Technical Information | Class | Homework and Exams | References and Links | FAQ

Topic 1: Review of Classical Control Theory

• 1.1 Open and closed loop poles
• 1.2 Stability analysis
• 1.3 Steady state errors
• 1.4 Transient response
• 1.5 Frequency response

Transient response

Consider a system described by a transfer function

$\frac{Y\left(s\right)}{U\left(s\right)}=H\left(s\right)$

If this system is forced by an input $u\left(t\right)$, then the resultant output $y\left(t\right)$ can be decomposed into a residual response $y_{\text{r}}\left(t\right)$ and a decaying response $\rho \left(t\right)$

$y\left(t\right)=y_{\text{r}}\left(t\right)+\rho \left(t\right)$

where $\underset{t\to \infty }{lim} \rho \left(t\right)=0$. The early response where $\rho \left(t\right)$ cannot be neglected is known as the transient response. In general the transient response depends on the initial conditions and the particular input $u\left(t\right)$. However, for linear systems we can learn a lot about the transient response for arbitrary inputs by looking at the transient response for standard inputs, especially step inputs.

We shall look primarily at systems having only poles and no zeros.

First-order systems

Consider the first-order system with transfer function

$H\left(s\right)=K\mathrm{ }\frac{1}{\tau s+1}$

Assuming zero initial condition, the response of this system to a step input $u\left(t\right)=\bar{u}$ is

$y\left(t\right)=y_{\text{ss}}(1-e^{-t/\tau })$

where the steady-state output is $y_{\text{ss}}=K\bar{u}$. A normalized plot of this response is shown below

Note that the pole of this system is $p=-1/\tau$. As pole $p$ becomes more negative, the time constant $\tau$ becomes smaller, so that the response time becomes faster.

The output is within 37% of steady state after one time constant, 14% after two time constants, 5% after three time constants, and 2% after four time constants.

The transient response of more complex inputs is predicted to a large extent by the transient response to step inputs. Shown below is the response of the same first order system to a step-plus-sinusoidal input $u\left(t\right)=1+0.2sin\left(4t\right)$.

Second-order systems

Consider next the second-order system with transfer function

$H\left(s\right)=K\mathrm{ }\frac{{\omega }_{\text{n}}^2}{s^2+2\zeta {\omega }_{\text{n}}s+{\omega }_{\text{n}}^2}$

Parameter ${\omega }_{\text{n}}$ is known as the natural frequency. Parameter $\zeta$ is known as the damping ratio. If these quantities are both positive then the system is stable.

If $\zeta >1$ the system is said to be overdamped. Assuming zero initial conditions, the response to a step input $u\left(t\right)=\bar{u}$ is

$y\left(t\right)=y_{\text{ss}}\frac{{\omega }_{\text{n}}^2}{p_2-p_1}[\frac{1}{p_1}(1-e^{p_{1}t})-\frac{1}{p_2}(1-e^{p_{2}t})]$

where the steady-state output is $y_{\text{ss}}=K\bar{u}$; and where the poles (eigenvalues) are $p_1=-\zeta {\omega }_{\text{n}}+{\omega }_{\text{n}}\sqrt{{\zeta }^2-1}$ and $p_2=-\zeta {\omega }_{\text{n}}-{\omega }_{\text{n}}\sqrt{{\zeta }^2-1}$.

If $\zeta =1$ the system is said to be critically damped. The two poles are equal, $p_1=p_2=-{\omega }_{\text{n}}$. Assuming zero initial conditions, the response to a step input $u\left(t\right)=\bar{u}$ is

$y\left(t\right)=y_{\text{ss}}[1-(1+{\omega }_{\text{n}}t)e^{-{\omega }_{\text{n}}t}]$

where the steady-state output is $y_{\text{ss}}=K\bar{u}$.

If $0\le \zeta \le 1$ the system is said to be underdamped, and the two poles are complex conjugate pairs. Assuming zero initial conditions, the response to a step input $u\left(t\right)=\bar{u}$ is

$y\left(t\right)=y_{\text{ss}}[1-\frac{1}{\sqrt{1-{\zeta }^2}}{e}^{-\zeta {\omega }_{\text{n}}t}sin({\omega }_{\text{d}}t+\alpha )]$

where the steady-state output is $y_{\text{ss}}=K\bar{u}$, the damped frequency is ${\omega }_{\text{d}}={\omega }_{\text{n}}\sqrt{1-{\zeta }^2}$, and the phase angle is $\alpha ={cos}^{-1}\zeta$.

A root locus of poles for ${\omega }_{\text{n}}=1$ and for five different values of $\zeta$ is shown below.

Three cases are underdamped, one is critically damped, and one is overdamped. Step responses corresponding to the same cases are shown below.

Here are some characteristics of second-order system step responses:

The return time is a characteristic time constant

$\tau =\frac{1}{|Re(p\left)\right|}=\frac{1}{\zeta {\omega }_{\text{n}}}$

For an underdamped system, the rise time is the time it takes the output to first pass through the steady state value.

$t_{\text{r}}=\frac{\pi -\alpha }{{\omega }_{\text{n}}\sqrt{1-{\zeta }^2}}$

Critically damped and overdamped systems have infinite rise times.

The 2% settling time is the time it takes the output to settle permanently within 2% of its steady state value. For an underdamped system $0<\zeta <1$ this value is approximately

$t_{\text{s}}=-\frac{ln(0.02\sqrt{1-{\zeta }^2})}{\zeta {\omega }_{\text{n}}}$

The peak time is the time it takes the output to reach its peak value (maximum overshoot). For an underdamped system the peak time is

$t_{\text{p}}=\frac{\pi }{{\omega }_{\text{n}}\sqrt{1-{\zeta }^2}}$

The corresponding maximum overshoot expressed as a fraction of the steady state value is

$\text{OS}=\frac{y_{\text{max}}-y_{\text{ss}}}{y_{\text{ss}}}=exp\left(\frac{-\zeta \pi }{\sqrt{1-{\zeta }^2}}\right)$

Dominant poles

The right-most poles of a system transfer function are said to dominate the other poles. For stable systems these are the poles nearest the imaginary axis. For zero initial conditions, the dominant poles determine to a large extent the transient response.

Consider for example the system with transfer function

$\frac{Y\left(s\right)}{U\left(s\right)}=H\left(s\right)=\frac{1}{(0.5s+1)(s^2+0.8s+1)}$

The poles of this system are $p_1=-0.4+0.9165i$, $p_2=-0.4-0.9165i$, and $p_3=-2$, as shown below.

The dominant poles are the complex conjugate pair $p_1$ and $p_2$, with associated transfer function

$\mathrm{H\text{'}}\left(s\right)=\frac{1}{(s^2+0.8s+1)}$

A comparison of the step responses of $H\left(s\right)$ and $\mathrm{H\text{'}}\left(s\right)$ is shown below.

Next consider the system with transfer function

$\frac{Y\left(s\right)}{U\left(s\right)}=H\left(s\right)=\frac{1}{(10s+1)(s^2+0.8s+1)}$

The poles of this system are $p_1=-0.4+0.9165i$, $p_2=-0.4-0.9165i$, and $p_3=-0.1$, as shown below.

The dominant pole is the real pole $p_3$, with associated transfer function

$\mathrm{H\text{'}}\left(s\right)=\frac{1}{(10s+1)}$

A comparison of the step responses of $H\left(s\right)$ and $\mathrm{H\text{'}}\left(s\right)$ is shown below.

Zeros

The presence of zeros, especially non-minimum-phase zeros (zeros in the right-half plane), can change the step response considerably. We shall not attempt to characterize all such systems, but do consider the case of near pole-zero cancellation by example.

Consider the system with transfer function

$\frac{Y\left(s\right)}{U\left(s\right)}=H\left(s\right)=\frac{(8s+1)}{(10s+1)(s^2+0.8s+1)}$

The poles of this system are $p_1=-0.4+0.9165i$, $p_2=-0.4-0.9165i$, and $p_3=-0.1$. The zero is $z=-0.125$. The poles and zero are shown below.

The right-most pole is $p_3=-0.1$. This pole is nearly cancelled by the zero $z=-0.125$. Thus the system response is dominated by the complex conjugate pair $p_1$ and $p_2$, with associated transfer function

$\mathrm{H\text{'}}\left(s\right)=\frac{1}{(s^2+0.8s+1)}$

A comparison of the step responses of $H\left(s\right)$ and $\mathrm{H\text{'}}\left(s\right)$ is shown below.

Note that the effect of the slow pole $p_3=-0.1$ is not entirely cancelled, and that the settling time of $H\left(s\right)$ is considerably longer than that of $\mathrm{H\text{'}}\left(s\right)$.

What's next?

Go on to page 5, where we'll look at frequency response.