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1. AE23001 Anubhab Pal North Eastern Regional Institute of Science and Technology Nirjuli – Arunachal Pradesh 1

2. Lecture 12 Lecture topics Numerical problem solution Dynamic error in first order system 2

3. Numerical problem related to second order system Step Response A force sensor has a mass of 0.5 kg, stiffness of 2 × 10 2 Nm −1 and a damping constant of 6.0 N s m −1 . Calculate the steady-state sensitivity, natural frequency and damping ratio for the sensor. Calculate the displacement of the sensor for a steady input force of 2 N. If the input force is suddenly increased from 2 to 3 N, derive an expression for the resulting displacement of the sensor 3

4. Solution Steady state sensitivity of a second order system can be given by, Natural frequency of a second order system is, And, damping ratio is 4 …(7.20) …(7.20) …(7.8) …(7.8) …(7.9) …(7.9)

5. Solution Given data, Spring stiffness, k = 2 × 10 2 Nm −1 Sensor mass, m = 0.5 kg Damping constant, λ = 6.0 N s m −1 Putting the values in respective equations Steady state sensitivity Natural frequency Damping ratio 5 mN −1 mN −1 rad/s rad/s

6. Solution Steady state response of second order system can be given by, Previously calculated, Steady state sensitivity, K = 5 ×10 -3 mN -1 Given data, Steady state input, I s = 2 N Therefore, the sensor output will be, 6 m m

7. Solution The system is underdamped as damping ratio = 0.3. Therefore Step response of the system can be given by, Previously calculated, Damping ratio, ξ = 0.3 Natural frequency, ω n = 20 rad s -1 Given data Initial output, O(t-) = 0.01 m (This is steady state output) Initial input, I(t-) = 2 N Final input, I(t) = 3 N Therefore, step height, H = I(0) - I(0-) = 3 – 2 = 1 N 7 Or, …(10.25) Or, …(10.25)

8. Solution Therefore, the expression for sensor output will be, Putting the values, 8

9. Dynamic errors in first order system If there are n number of elements in the system, each of which is having steady state sensitivity of K i and transfer function of G i (s) , then overall transfer function of system becomes, 9 …(12.1) …(12.1) Or, Or, Or, …(12.2) Or, …(12.2)

10. Dynamic errors in first order system Now error can be given by, Using Eq. 12.2 Now if the input signal is sinusoidal i.e. , The output signal will be, The error will be   10 …(12.3) …(12.3) …(12.4) …(12.4) Or, …(12.5) Or, …(12.5)

11. Dynamic errors in first order system Now if the input signal is periodic (combination of more than one sinusoid), The output signal will become 11 …(12.7) …(12.7) …(12.8) …(12.8) …(12.6) …(12.6)

12. Numerical problem related to first order system dynamic error Sinusoidal Response A temperature measurement system for a gas reactor consists of linear elements and has an overall steady-state sensitivity of unity. The temperature sensor has a time constant of 5.0 s; an ideal low-pass filter with a cut-off frequency of 0.05 Hz is also present. The input temperature signal is periodic with period 63 s and can be approximated by the Fourier series: Where ω is the angular frequency of the fundamental component. Calculate expressions for the time response of: (i) The system output signal (ii) The system dynamic error. 12

13. Solution The expression for output signal for a first order system with periodic input may be given by, Given data, Time constant, τ = 5 sec Angular frequency, ω = 2 π /T = 0.1 rad/sec Therefore, sensor transfer function will become, And phase angle will be, 13 …(12.7) …(12.7) …(10.5) …(10.5) …(10.6) …(10.6)

14. Solution When ; Similarly, When ; ; φ = -45.00° And, When ; ; φ = -56.31° And, When ; ; φ = -63.43° Given data, I = 10; I 1 = 5; I 2 = 10/3; I 3 = 2.5 Now calculating individual terms of Eq. 12.7   14 …(a) …(a) …(b) …(b) …(c) …(c) …(d) …(d)

15. Solution Now summing up Eq. a to d we get, Next, error can be calculated by, Putting the values, 15 …(12.8) …(12.8)

16. Solution Now if a low pass filter is introduced Given data, Cut off frequency = 0.05 Hz = 0.314 rad/s Therefore, output signal will be Due to the presence of filter, the harmonic is chopped off. Now, the error will be,   16

17. Thank you End of Lecture 12 17