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1. AE23001
Anubhab Pal
North Eastern Regional Institute of Science and
Technology
Nirjuli – Arunachal Pradesh
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2. Lecture 10
• Lecture topics
1. Numerical problem solution
2. Dynamic error in first order system
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3. Numerical problem related to
second order system
1. Step Response
A force sensor has a mass of 0.5 kg, stiffness of 2 × 102 Nm−1 and a damping
constant of 6.0 N s m−1.
A. Calculate the steady-state sensitivity, natural frequency and damping ratio
for the sensor.
B. Calculate the displacement of the sensor for a steady input force of 2 N.
C. If the input force is suddenly increased from 2 to 3 N, derive an expression
for the resulting displacement of the sensor
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4. • Solution
A.
i. Steady state sensitivity of a second order system can be given by,
ii. Natural frequency of a second order system is,
iii. And, damping ratio is
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𝑲 =
𝟏
𝒌
…(7.20)
𝝎𝒏 = 𝒌
𝒎 …(7.8)
𝝃 =
𝝀
𝟐 𝒌 ∙ 𝒎
…(7.9)

5. • Solution
Given data,
• Spring stiffness, k = 2 × 102 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
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𝑲 =
𝟏
𝟐 × 𝟏𝟎𝟐
= 𝟓 × 𝟏𝟎−𝟑 mN−1
𝝎𝒏 = 𝟐 × 𝟏𝟎𝟐
𝟎. 𝟓 = 𝟐𝟎 rad/s
𝝃 =
𝟔. 𝟎
𝟐 𝟐 × 𝟏𝟎𝟐 ∙ 𝟎. 𝟓
= 𝟎. 𝟑

6. • Solution
B.
i. 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, Is = 2 N
Therefore, the sensor output will be,
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𝑶𝒔 = 𝑲 ∙ 𝑰𝒔
𝑶𝒔 = 𝟓 × 𝟏𝟎−𝟑
∙ 𝟐 = 𝟎. 𝟎𝟏 m

7. • Solution
C.
i. 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, ωn= 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
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Or, 𝒇𝟎 𝒕 = 𝟏 − 𝒆−𝝃𝝎𝒏𝒕 cos 𝝎𝒅𝑡 +
𝝃
𝟏 − 𝝃𝟐
sin 𝝎𝒅𝑡
…(10.25)

8. • Solution
C.
• Therefore, the expression for sensor output will be,
• Putting the values,
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𝑶 𝒕 = 𝑶 𝒕 − + 𝑲 ∙ 𝑯 ∙ 𝟏 − 𝒆−𝝃𝝎𝒏𝒕
cos 𝝎𝒏 𝟏 − 𝝃𝟐𝑡 +
𝝃
𝟏 − 𝝃𝟐
sin 𝝎𝒏 𝟏 − 𝝃𝟐𝒕
𝑶 𝒕 = 𝟏 × 𝟏𝟎−𝟐
+ 𝟓 × 𝟏𝟎−𝟑
∙ 𝟏 ∙ 𝟏 − 𝒆−𝟎.𝟑∙𝟐𝟎𝒕
cos 𝟐𝟎 𝟏 − 𝟎. 𝟑𝟐𝑡 +
𝟎. 𝟑
𝟏 − 𝟎. 𝟑𝟐
sin 𝟐𝟎 𝟏 − 𝟎. 𝟑𝟐𝒕
𝑶 𝒕 = 𝟏 × 𝟏𝟎−𝟐 + 𝟓 × 𝟏𝟎−𝟑 𝟏 − 𝒆−𝟔𝒕 cos 𝟏𝟗. 𝟎8𝑡 + 𝟎. 𝟑𝟑 sin 𝟏𝟗. 𝟎𝟖𝒕

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 Ki and transfer function of Gi(s), then overall transfer function of
system becomes,
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∆𝑶(𝒔)
∆𝑰(𝒔)
= 𝑮 𝒔 = 𝑮𝟏(𝒔) ∙ 𝑮𝟐(𝒔) ∙ ⋯ ∙ 𝑮𝒊(𝒔) ∙ 𝑮𝒏(𝒔) …(12.1)
Or, ∆𝑶(𝒔) = ∆𝑰(𝒔) ∙ 𝑮 𝒔
Or, ∆𝑶(𝒕) = ℒ−𝟏
∆𝑰(𝒔) ∙ 𝑮 𝒔 …(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. 𝐼 𝑠𝑖𝑛 𝜔𝑡 , The output signal will be,
The error will be
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𝑬 𝒕 = ∆𝑶 𝒕 − ∆𝑰 𝒕
𝑬 𝒕 = ℒ−𝟏 ∆𝑰(𝒔) ∙ 𝑮 𝒔 − ∆𝑰 𝒕 …(12.3)
∆𝑶(𝒕) = 𝑮 𝒋𝜔 ∙ 𝐼 sin 𝜔𝑡 + 𝝋 …(12.4)
𝑬 𝒕 = 𝑮 𝒋𝜔 ∙ 𝐼 sin 𝜔𝑡 + 𝝋 − 𝐼 sin 𝜔𝑡
Or, 𝑬 𝒕 = 𝑰 ∙ 𝑮 𝒋𝜔 sin 𝜔𝑡 + 𝝋 − sin 𝜔𝑡 …(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
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∆𝑶(𝒕) =
𝒏=𝟏
∞
𝑰𝒏 𝑮 𝒋𝒏𝜔𝟏 sin 𝒏𝜔𝟏𝑡 + 𝝋𝒏 …(12.7)
𝑬 𝒕 =
𝒏=𝟏
∞
𝑰𝒏 𝑮 𝒋𝒏𝜔𝟏 sin 𝒏𝜔𝟏𝑡 + 𝝋𝒏 − sin 𝒏𝜔𝟏𝑡 …(12.8)
∆𝑰(𝒕) =
𝒏=𝟏
∞
𝑰𝒏 sin 𝒏𝜔𝟏𝑡 …(12.6)

12. Numerical problem related to first
order system dynamic error
1. 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 ω0 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.
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𝑻 𝒕 = 𝟏𝟎 sin 𝜔0𝑡 +
1
2
sin 2𝜔0𝑡 +
𝟏
𝟑
sin 3𝜔0𝑡 +
𝟏
𝟒
sin 4𝜔0𝑡

13. 13

14. • 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, ω0 = 2π/T = 0.1 rad/sec
Therefore, sensor transfer function will become,
And phase angle will be,
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∆𝑶(𝒕) =
𝒏=𝟏
∞
𝑰𝒏 𝑮 𝒋𝒏𝜔𝟎 sin 𝒏𝜔𝟎𝑡 + 𝝋𝒏 …(12.7)
𝑮 jnω0 =
1
1 + 𝜏𝟐(𝒏ω0)𝟐
…(10.5)
𝝋 = tan−𝟏
−nω0𝜏 …(10.6)

15. • Solution
When 𝒏 = 𝟏;
Similarly, When 𝒏 = 𝟐; 𝑮 jnω0 = 𝟎. 𝟕𝟎𝟕; φ = -45.00°
And, When 𝒏 = 𝟑; 𝑮 jnω0 = 𝟎. 𝟓𝟓𝟒; φ = -56.31°
And, When 𝒏 = 𝟒; 𝑮 jnω0 = 𝟎. 𝟒𝟒𝟕; φ = -63.43°
Given data,
I0= 10; I1 = 5; I2 = 10/3; I3 = 2.5
Now calculating individual terms of Eq. 12.7
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𝑮 jnω0 =
1
1 + 𝟓𝟐 ∙ 𝟏 ∙ 𝟎. 𝟏 𝟐
= 𝟎. 𝟖𝟗𝟒
𝝋 = tan−𝟏
−𝟏 ∙ 𝟎. 1 ∙ 𝟓 = −𝟐𝟔. 𝟓𝟕°
𝑰𝟎 𝑮 𝒋𝟎. 𝟏 sin 𝟎. 𝟏𝑡 + 𝝋𝟎 = 𝟏𝟎 ∙ 𝟎. 𝟖𝟗𝟒 ∙ sin 0.1𝑡 − 26.57° …(a)
𝑰𝟏 𝑮 𝒋𝟎. 𝟐 sin 𝟎. 𝟐𝑡 + 𝝋𝟏 = 𝟓 ∙ 𝟎. 𝟕𝟎𝟕 ∙ sin 0.2𝑡 − 45° …(b)
𝑰𝟐 𝑮 𝒋𝟎. 𝟑 sin 𝟎. 𝟑𝑡 + 𝝋𝟐 =
𝟏𝟎
𝟑
∙ 𝟎. 𝟓𝟓𝟒 ∙ sin 0.3𝑡 − 56.31° …(c)
𝑰𝟑 𝑮 𝒋𝟎. 𝟒 sin 𝟎. 𝟒𝑡 + 𝝋𝟑 = 𝟐. 𝟓 ∙ 𝟎. 𝟒𝟒𝟕 ∙ sin 0.4𝑡 − 63.43° …(d)

16. • Solution
Now summing up Eq. a to d we get,
Next, error can be calculated by,
Putting the values,
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∆𝑶 𝒕 =
𝟏𝟎 ∙ 𝟎. 𝟖𝟗𝟒 ∙ sin 0.1𝑡 − 26.57° + 𝟓 ∙ 𝟎. 𝟕𝟎𝟕 ∙ sin 0.2𝑡 − 45° +
𝟏𝟎
𝟑
∙ 𝟎. 𝟓𝟓𝟒
∙ sin 0.3𝑡 − 56.31° + 𝟐. 𝟓 ∙ 𝟎. 𝟒𝟒𝟕 ∙ sin 0.4𝑡 − 63.43°
∆𝑶 𝒕 =
𝟏𝟎 𝟎. 𝟖𝟗𝟒 ∙ sin 0.1𝑡 − 26.57° +
𝟏
𝟐
∙ 𝟎. 𝟕𝟎𝟕 ∙ sin 0.2𝑡 − 45° +
𝟏
𝟑
∙ 𝟎. 𝟓𝟓𝟒
𝑬 𝒕 =
𝒏=𝟏
∞
𝑰𝒏 𝑮 𝒋𝒏𝜔𝟏 sin 𝒏𝜔𝟏𝑡 + 𝝋𝒏 − sin 𝒏𝜔𝟏𝑡 …(12.8)
𝑬 𝒕 =
𝟏𝟎 𝟎. 𝟖𝟗𝟒 ∙ sin 0.1𝑡 − 26.57° +
𝟏
𝟐
∙ 𝟎. 𝟕𝟎𝟕 ∙ sin 0.2𝑡 − 45° +
𝟏
𝟑
∙ 𝟎. 𝟓𝟓𝟒

17. • 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
𝟏
𝟒
∙ 𝟎. 𝟒𝟒𝟕 ∙ sin 0.4𝑡 − 63.43° is
chopped off.
Now, the error will be,
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∆𝑶 𝒕 =
𝟏𝟎 ∙ 𝟎. 𝟖𝟗𝟒 ∙ sin 0.1𝑡 − 26.57° + 𝟓 ∙ 𝟎. 𝟕𝟎𝟕 ∙ sin 0.2𝑡 − 45° +
𝟏𝟎
𝟑
∙ 𝟎. 𝟓𝟓𝟒
∙ sin 0.3𝑡 − 56.31°
∆𝑶 𝒕 =
𝟏𝟎 𝟎. 𝟖𝟗𝟒 ∙ sin 0.1𝑡 − 26.57° +
𝟏
𝟐
∙ 𝟎. 𝟕𝟎𝟕 ∙ sin 0.2𝑡 − 45° +
𝟏
𝟑
∙ 𝟎. 𝟓𝟓𝟒
𝑬 𝒕 =
𝟏𝟎 𝟎. 𝟖𝟗𝟒 ∙ sin 0.1𝑡 − 26.57° +
𝟏
𝟐
∙ 𝟎. 𝟕𝟎𝟕 ∙ sin 0.2𝑡 − 45° +
𝟏
𝟑
∙ 𝟎. 𝟓𝟓𝟒

18. Thank you
End of Lecture 12
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