Discuss the time specifications of the underdamped systems.

1. Control Systems
Lecture - 8
E. Ramkumar
Assistant Professor / EEE
12-04-2022 UEEC004 - Control Systems 1

2. Time response of Second Order System
• 𝐶 𝑆 =
ω𝑛 2
𝑆2
+2δω𝑛𝑆+ω𝑛 2
•
𝐶(𝑆)
𝑅(𝑆)
=
100
𝑆2
+0𝑆+100
ω𝑛 2 = 100
ω𝑛 = 10
2δω𝑛 = 0
2δ = 0
δ = 0
12-04-2022 UEEC004 - Control Systems 2

3. Time response of Second Order System
• 𝐶 𝑆 =
ω𝑛 2
𝑆2
+2δω𝑛𝑆+ω𝑛 2
•
𝐶(𝑆)
𝑅(𝑆)
=
100
𝑆2
+10𝑆+100
ω𝑛 2 = 100
ω𝑛 = 10
2δω𝑛 = 10
2δ = 1
δ = 0.5
12-04-2022 UEEC004 - Control Systems 3

4. Time response of Second Order System
• 𝐶 𝑆 =
ω𝑛 2
𝑆2
+2δω𝑛𝑆+ω𝑛 2
•
𝐶(𝑆)
𝑅(𝑆)
=
100
𝑆2
+20𝑆+100
ω𝑛 2 = 100
ω𝑛 = 10
2δω𝑛 = 20
2δ = 2
δ = 1
12-04-2022 UEEC004 - Control Systems 4

5. Time response of Second Order System
• 𝐶 𝑆 =
ω𝑛 2
𝑆2
+2δω𝑛𝑆+ω𝑛 2
•
𝐶(𝑆)
𝑅(𝑆)
=
100
𝑆2
+40𝑆+100
ω𝑛 2 = 100
ω𝑛 = 10
2δω𝑛 = 40
2δ = 4
δ = 2
12-04-2022 UEEC004 - Control Systems 5

6. Time Response Specifications
12-04-2022 UEEC004 - Control Systems 6
Delay time, td:
Time required for the
response to reach 50%
of the final value at first
instance.
td =
1 + 0.7𝛿
𝜔𝑛
Rise time, tr:
Time required for the
response to rise from
10% to 90% of the final
value for ODS and 0 to
100% of the final value
for UDS, at first
instance. .
Peak time, tp:
Time required for the
response to reach peak
value of time response.
Peak Overshoot, Mp:
It is the normalized
difference between the
peak value of time
response and the steady
state value.
Mp =
𝑦 𝑡𝑝 − 𝑦(∞)
𝑦(∞)
Settling time, ts:
Time required for the
response to reach and
stay within a specified
tolerance band of its
final or steady state
value of time response.
Usually, tolerance band
is 2 or 5%.
td tr tp
Mp
ts
Allowable
tolerance

7. Rise Time, tr
• 𝑦 𝑡𝑟 = 1 = 1 −
𝑒−𝛿𝜔𝑛
𝑡𝑟
√(1−𝛿2
)
sin(𝜔𝑑𝑡𝑟 + 𝜃)
•
𝑒−𝛿𝜔𝑛
𝑡𝑟
√(1−𝛿2
)
sin(𝜔𝑑𝑡𝑟 + 𝜃) = 0
• For above eqn to be ZERO,
• sin 𝜔𝑑𝑡𝑟 + 𝜃 = 0
• 𝜔𝑑𝑡𝑟 + 𝜃 = 𝜋
• 𝑡𝑟 =
𝜋−𝜃
𝜔𝑑
• 𝑡𝑟 =
𝜋−cos−1 𝛿
𝜔𝑛
√(1−𝛿2
)
12-04-2022 UEEC004 - Control Systems 7
tr
𝑡𝑟 =
𝜋 − cos−1
𝛿
𝜔𝑛√(1 − 𝛿2)

8. Peak Time, tp
12-04-2022 UEEC004 - Control Systems 8
tp
• 𝑦 𝑡𝑝 = 1 −
𝑒−𝛿𝜔𝑛
𝑡
√(1−𝛿2
)
sin(𝜔𝑑𝑡 + 𝜃)
• To find the first peak
•
𝑑𝑦 𝑡𝑝
𝑑𝑡
= 0
•
𝛿𝜔𝑛
𝑒−𝛿𝜔𝑛
𝑡𝑝
√(1−𝛿2
)
sin(𝜔𝑑𝑡𝑝 + 𝜃) −
𝑒−𝛿𝜔𝑛
𝑡𝑝
1−𝛿2 𝜔𝑑cos 𝜔𝑑𝑡𝑝 + 𝜃
• 𝛿 sin 𝜔𝑑𝑡𝑝 + 𝜃 − √(1 − 𝛿2) cos(𝜔𝑑𝑡𝑝 + 𝜃) = 0
• sin 𝜔𝑑𝑡𝑝 + 𝜃 cos 𝜃 − cos 𝜔𝑑𝑡𝑝 + 𝜃 sin 𝜃 = 0
• sin 𝜔𝑑𝑡𝑝 = 0
• 𝜔𝑑𝑡𝑝 = 0, 𝜋, 2𝜋 …
• 𝑡𝑝 =
𝜋
𝜔𝑑
• 𝑡𝑝 =
𝜋
𝜔𝑛
√(1−𝛿2
)
𝑡𝑝 =
𝜋
𝜔𝑛√(1 − 𝛿2)

9. Peak Overshoot, Mp
12-04-2022 UEEC004 - Control Systems 9
Mp
• 𝑀𝑝 =
𝑦 𝑡𝑝
−1
1
• 𝑦 𝑡𝑝 − 1 = 1 −
𝑒−𝛿𝜔𝑛
𝑡𝑝
1−𝛿2 sin 𝜔𝑑𝑡𝑝 + 𝜃 − 1
• 𝑦 𝑡𝑝 = −
𝑒−𝛿𝜔𝑛
𝑡𝑝
√(1−𝛿2
)
sin(𝜔𝑑𝑡𝑝 + 𝜃)
• 𝑡𝑝 =
𝜋
𝜔𝑑
• 𝑀𝑝 = −
𝑒
−𝛿𝜔𝑛
𝜋
𝜔𝑑
√(1−𝛿2
)
sin(𝜔𝑑
𝜋
𝜔𝑑
+ 𝜃)
• 𝑀𝑝 =
𝑒
−
𝛿𝜋
√(1−𝛿
2
)
√(1−𝛿2
)
sin( 𝜃)
• 𝑀𝑝 = 𝑒
−
𝛿𝜋
√(1−𝛿2
)
𝑀𝑝 = 100𝑒
𝛿𝜋
√(1−𝛿2
)%