Consider the 2-order system, without zero, whose transfer function is $$H (s) = \frac{51}{s^2 + 2s + 17}$$

An input signal $$u(t) = 4\varepsilon(t)$$ is applied to this system

Determine the maximum value reached by the system output signal

The solution says:

$$u = \exp\left(-\frac{\varsigma\pi}{\sqrt{1-\varsigma^2}}\right) = 0.456$$

$$y(ss)=\frac{4 \cdot 51}{17} = 12$$

$$y(max)=12 \cdot (1 + 0.456) = 17.47$$

I undestand that they calculate the superelevation using the formual of the 2-order system but i can't understand how they calculated \$y(ss)\$ and \$y(max)\$.

Could someone help me?

  • \$\begingroup\$ I expect that the step input they apply to the 2nd order filter is 4 volts i.e. 4 x 51/17 = 12. Double check your source. \$\endgroup\$
    – Andy aka
    Jan 8 at 10:37
  • \$\begingroup\$ You are using an input step u(t), but what is the quantity "u"? A kind of output? More than that, you have shown equation for y(ss) and y(max) with numbers inserted. How can we answer your question withozt knowing the general form for these equations and the name of these expressions? I have another equation for the maximum of the step response. \$\endgroup\$
    – LvW
    Jan 8 at 10:51
  • 1
    \$\begingroup\$ Action has been taken on deletions flagged by users. \$\endgroup\$
    – Russell McMahon
    Jan 15 at 9:34

The \$u\$ given in the solution seems to be the peak value of the swing. The maximum over-swing can be calculated using

$$\Delta h = K \exp\left(-\frac{\pi d}{\sqrt{1-d^2}}\right)$$

where \$K\$ is the DC gain of your 2nd order system and \$d\$ is the dampening of your system. This formula is K times your "u". This overswing apmplitude is defined for a step input from 0 to 1. You will supply a step from 0 to 4. Therefore, we will multiply this later by 4.

The dampening of a second order system can be read from its transfer function. Second order systems show the following function:

$$H(s) =\frac{K}{\frac{1}{\omega_0^2}s^2 + \frac{2d}{\omega_0}s+1}$$

Bringing your system into this form we get:

$$H(s) = \frac{\frac{51}{17}}{\frac{1}{17}s^2 + \frac{2 \cdot \frac{1}{\sqrt{17}}}{\sqrt{17}}s + 1}$$

Therefore the dampening \$d\$ is $$ d = \frac{1}{\sqrt{17}} $$ and the DC gain \$K\$ is $$ K = \frac{51}{17} = 3$$

Next we calculate the DC output of your system for \$t \to \infty\$. This can either be done using the proper end-value formula for the laplace transform, or in this case simpler by multiplying the amplitude of your step function (4) with the gain (K) of your system. This is your "\$y(ss)\$"

$$y(ss) = \lim_{t\to \infty} y(t) = K \cdot 4 = \frac{4 \cdot 51}{17} = 12$$

Because the dampening value \$d\$ is greater than 0, we know that the system will converge to an end value instead of just keep oscillating. Therefore, this approach is valid.

Your maximum value \$y_{max}\$ equals the maximum overswing amplitude plus \$y(ss)\$. (Whatever \$y(ss)\$ stands for. I'm just keeping your notation. I think it doesn't make any sense)

$$y_{max} = y(ss) + 4 \cdot \Delta h$$. The 4 comes again from your input signal which has a step of 4

Plugging in the numbers above leads to the specified results.

  • \$\begingroup\$ They're valid for all 2nd order systems that follow the pattern K/(as^2+bs+1). Formulas for the maximum overswing are of course only valid, if your system is stable. \$\endgroup\$
    – GNA
    Jan 10 at 18:33

From the differential equation of a 2nd-order system it is possible to derive the timely step response. From this equation it is possible to find the maximum Voutmax appearing at a certain time (t=tmax) of the step response.

Vout(t=tmax) = Voutmax = Vs*Ao[1+exp(tan(phi) * Pi)]

Vs: Step voltage; Ao=DC gain of the system; angle tan(phi)=Re(wp)/IM(wp)

The pole frequency wp is the magnitude of the pointer from the origin of the s-plane to the pole position and phi is the angle between this pointer and the imag. axis.

Comment: The angle phi can be expressed by the pole quality factor Qp resp. the dampimg factor d=1/2Qp. The time is tmax=Pi/Im(wp).


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