How to use time-domain for calculation of Transient response specifications

Hello everyone, I hope that all of you are doing good in your life. The topic for today’s discussion is time domain specifications of 2nd order under damp systems. However, the topic will involve the definition of factors like peak overshoot, Peak Time, Settling time, Delay Time, and rise time. Besides we are also going to study the numerals on time-domain specifications. The output response of the system is;Output response of a second order system in time domain

However all of the factors of the time domain specifications are mention in the above graph.

Response of Second Order System

However, before starting with the time domain specifications let’s first have a short overview of the response of a second-order system. Consider an open-loop transfer function ωn2/s(s+2ξωn) in connection with a unity feedback. This results in the formation of a close-loop Thus its transfer function is given as;

    \[\frac {C(s)} {R(s)} = \frac {G(s)} {1 + G(s) } \end.\]

    \[However~By ~ Substituting ~ G(s)\end.\]

    \[= \frac {\omega _n^2} {s (s +2 \xi \omega_n)}\end.\]

    \[\frac {C(s)} {R(s)} = \frac {\frac {\omega _n^2} {s (s +2 \xi \omega_n)}} {1 + \frac {\omega _n^2} {s(s +2 \xi \omega_n)} } \end.\]

    \[\therefore \frac {C(s)} {R(s)} = \frac {\omega_n^2} {s^2 + 2 \xi \omega_n s + \omega_n^2) } \end.\]

Since the highest power of the variable s is two the system becomes second order and thus it is referred to as a second-order system.

Now lets discuss its characteristic equation of the system i.e

    \[s^2 + 2 \xi \omega_n s + \omega^2 = 0\end.\]

Thus the roots of the characteristic equation is;

    \[s = \frac {-2 \omega \xi_n \pm \sqrt{(2 \xi \omega_n)^2 - 4 \omega_n^2}} {2}\end.\]

    \[= \frac {-2 (\xi \omega_n \pm \omega \sqrt{\xi^2 - 1})} {2}\end.\]

    \[s = -\xi \omega_n \pm \sqrt{\xi^2 - 1}\end.\]

There mainly four possibilities of the roots that can be obtain;

  1. Roots are imaginary when ξ = 0 which means that the system is completely oscillatory in nature.
  2. The two roots are real and equal when ξ = 1 which results in the formation of a highly damp system in nature.
  3. The two roots are real but not equal when ξ >1, which results in the formation of an over the damp system in nature.
  4. And lastly when two roots are complex conjugate when 0 < ξ < 1, this results in an under the damp system.

From the transfer function we can get the value of C(s) i.e. Laplace transform of output signal as;

    \[ C(s) = \frac {\omega_n^2} {s^2 + 2 \xi \omega_n s + \omega_n^2 } * R(s) \end.\]

Where R(s) is the Laplace transform of input signal, ξ is the damping ratio and lastly ωn as the natural frequency

Time-domain specifications

Delay Time (Td)

It is the time required by the output response to reach half of the maximum final value from the zeroth instance. However, it is express in seconds.

    \[T_d = \frac {1 + 0.7\xi} {\omega _n } \end.\]

Thus from the above equation, we can conclude that the Delay time is inversely proportional to the natural frequency (ωn) and directly proportional to the damping ratio (ξ).

Rise Time (Tr)

It is the time taken by the output response to reach or say rise to 90% of the maximum final value with respect to the zeroth level. However, for an over the damp system, it ranges from 0 to 90%. It is express in seconds

    \[T_r = \frac {\pi - \theta} {\omega _d } \end.\]

Where θ must be in radians. Besides in the above equation, we can see that the rise time (Tr) is inversely proportional to the damped frequency (ωd)

Peak Time(Tp)

The time require to reach the first peak value or the first overshoot of the time response.

    \[T_p = \frac {\pi} {\omega _d } = \frac {\pi} {\omega _n \sqrt{1- \xi^2} }\end.\]

However, from the above equation, we can see that Tp is inversely proportional to both damp frequency and the natural frequency i.e. ωd and ωn.

Peak Overshoot (Mp)

It is nothing but the maximum value of possible error between the reference input and the output. Most certainly it is express in percentage.

    \[M_p = \varrho^{\frac {-\pi \xi} {\sqrt{1-\xi^2}}}\end.\]

Thus from peak overshoot we can easily calculate the formula for percentage of peak overshoot.

    \[\Percentage ~ M_p = \varrho^{\frac {-\pi \xi} {\sqrt {1-\xi^2}}} * 100\end.\]

However, from the above equation, we can conclude that the % of peak overshoot (Mp) is inversely proportional to the damping ratio (ξ)

Settling Time (Ts)

It can be specified as the time require to reach a steady state value and stay within a specific range of tolerance through out around an output.

    \[For~Tollerance~of~2%\ end}.\]

    \[T_s = \frac {4} {\xi \omega_n}\end.\]

    \[For~Tollerance~of~5%\ end}.\]

    \[T_s = \frac {3} {\xi \omega_n}\end.\]

Thus from the above equation we get to know that Settling time (Ts) and natural frequency (ωn) are inversely proportional to each other.

Problems on time-domain Specifications

Q.1 A second-order system is given by T.F = 25/(s2 + 6s +25) Find; its time-domain specifications if it is influenced by a unit step response. Besides doing also calculate o/p response for a step input.

Solution :- However by comparing with standard transfer function of second order system i.e.

    \[T.F = \frac {\omega_n^2} {s^2 + 2 \xi \omega_n s + \omega_n^2}\end.\]

we get;

    \[\omega_n^2 = 25 ~ \therefore \omega_n = 5\end.\]

    \[And~2 \xi \omega_n = 6 ~ \therefore \xi = 0.6\end.\]

However before starting with the time domain calculation we must calculate the value of θ

    \[\theta = \tan^{-1}(\frac {\sqrt {1- \xi^2}} {\xi})\end.\]

    \[=~\tan^{-1}(\frac {\sqrt {1- 0.6^2}} {0.6})~=~0.9272 \end.\]

certainly θ is express in radians most of the times.

So now lets start the calculations;

    \[\omega_d = \omega_n \sqrt {1- \xi^2} \end.\]

    \[= 5\sqrt {1- 0.6^2} = 4 rad/sec\end.\]

    \[T_r = \frac {\pi - \theta} {\omega_d}\end.\]

    \[\thereore T_r = \frac {\pi - 0.9272} {4} =0.5535 sec\end.\]

Thus the value of Rise time Tr is 0.5535 secs.

    \[T_d = \frac {1 + 0.7\xi} {\omega_n}\end.\]

    \[\thereore T_d = \frac {1 + 0.7*0.6} {5} = 0.284 sec\end.\]

Thus the value of Delay time Td is 0.28 secs.

    \[T_p = \frac {\pi} {\omega_d}\end.\]

    \[\thereore T_p = \frac{\pi} {4} = 0.785 sec\end.\]

Thus the value of Peak Time Tp is 0.785 secs.

    \[T_s = \frac {4} {\xi \omega_n}\end.\]

    \[\thereore T_s = \frac {4} {0.6*4} = 1.33 sec\end.\]

Thus the value value of Settling time Ts is 1.333 secs.

    \[%M_p = \varrho^{\frac {-\pi \xi} {\sqrt{1-\xi^2}}}*100\end.\]

    \[\thereore %M_p = \varrho^{\frac {-\pi * 0.6} {\sqrt{1-0.6^2}}}*100 = 9.48%\end.\]

Lastly the value of Peak overshoot Mp is 9.48%

However for a second order under damp system for step input and output response is;

    \[C(t) = 1-  \frac {\varrho^-\omega_n \xi t} {\sqrt{1-\xi^2}} * sin(\omega_dt + \theta)\end.\]

    \[C(t) = 1-  \frac {\varrho^-0.6*5 t} {\sqrt{1-0.6^2}} * sin(\4t + 0.9272)\end.\]

    \[C(t) = 1-  1.5625 \varrho^-3t * Sin (4t + 0.92)\end.\]

Q.2 For a Control System below find the value K1 & K2 so that Mp = 25% and Tp = 4sec Assume unit step input.Block diagram of a system for the calculation of time domain specification

Solution:- The transfer function of the system above is;

    \[\frac {c(s)} {R(s)} = \frac {\frac{K_1}{s^2}} {1- \frac {K_1} {s^2}(1 + K_2s)}\end.\]

    \[~ = \frac {K_1} {s^2 + K_1 + K_1K_2s}\end.\]

However by comparing with standard equation of transfer function we get;

    \[\omega_n^2 = K_1 ~ \therefore \omega_n = \sqrt{K_1}\end.\]

    \[And~2 \xi \omega_n = K_1K_2 ~ \therefore \xi = \frac {\sqrt {K_1} K2} {2}\end.\]

As the value of Peak overshoot is mention in the question thus we are going to use it to find the value of ξ,

    \[M_p = \varrho^{\frac {-\pi \xi} {\sqrt{1-\xi^2}}}*100\end.\]

    \[\therefore 25 = \varrho^{\frac {-\pi \xi} {\sqrt{1-\xi^2}}}*100\end.\]

    \[\therefore log(25) = \frac {-\pi \xi} {\sqrt{1-\xi^2}}\end.\]

    \[\therefore -1.3862\sqrt{1-\xi^2} = -\pi \xi \end.\]

    \[\therefore 1.9215(1-\xi^2) = -\pi^2 \xi^2 \end.\]

    \[\therefore 1.9215(1-\xi^2) = -\pi^2 \xi^2 \end.\]

    \[\therefore 1.9215 =  \xi^2 11.7905 \end.\]

    \[\therefore\xi = 0.4037 \end.\]

Thus the value of ξ is 0.4037.

    \[T_p = \frac {\pi} {\omega_d}\end.\]

    \[T_p = \frac {\pi} {\omega_n \sqrt {1 - \xi^2} }\end.\]

    \[\thereore 4 = \frac{\pi} {\omega_n \sqrt {1 - \xi^2} }\end.\]

    \[\thereore \frac {4} {\pi} \sqrt{1 - \xi^2} = \frac{1}{\omega_n } \end.\]

    \[\thereore \omega_n = 0.8584 \end.\]

Thus the value of natural frequency is 0.8584

Since now all of the require values are present with us we can use it to find K1 and K2

    \[\since \omega_n = \sqrt {K_1}\end.\]

    \[We ~Get~K_1~as\end.\]

    \[K_1 = \omega_n^2 = 0.7369 \end.\]

    \[\since \xi = \frac {1} {2} \sqrt {K_1}K_2 = \frac {1} {2} \sqrt {0.7369}K_2 \end.\]

    \[\therefore K_2 = 0.9405 \end.\]


However here we are at the end of the blog. I hope that you liked the blog and all your doubts are clear. If you did like the blog then please do make sure to share it with others and mention the part which you like the most. Above we have only mention a few of the sample problems on the time-domain specification but we have covered every aspect of it in those numerical. Besides if do want more references then please do let us know. We will be please to help you with that.

Have a nice day 🙂


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