Frequency Response and Bode Plots in Control Systems, Summaries of Finance

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ESE 499 – Feedback Control Systems

SECTION 4: FIRST- AND

SECOND-ORDER SYSTEMS

2

First- and Second-Order Systems

 All transfer functions can be decomposed into 1 st- and 2 nd-order terms by factoring Δ 𝑠𝑠

𝐺𝐺 𝑠𝑠 =

𝑁𝑁𝑁𝑁𝑁𝑁 𝑠𝑠 𝑠𝑠 − 𝑝𝑝 1 ⋯ 𝑠𝑠 − 𝑝𝑝𝑛𝑛 𝑠𝑠 2 + 𝑎𝑎 11 𝑠𝑠 + 𝑎𝑎 10 ⋯ 𝑠𝑠 2 + 𝑎𝑎 (^) 𝑚𝑚1 𝑠𝑠 + 𝑎𝑎 (^) 𝑚𝑚

 Real poles – 1 st^ -order terms  Complex-conjugate poles – 2 nd^ -order terms

 These terms and, therefore, the poles determine the nature of the time- domain response  Real poles – decaying exponentials  Complex-conjugate poles - decaying sinusoids

 All time-domain responses will be a superposition of decaying exponentials and decaying sinusoids  These are the natural modes or eigenmodes of the system

(^4) Response of First-Order Systems

5

First-Order System – Impulse Response

 First-order transfer function:

𝐺𝐺 𝑠𝑠 =

𝐴𝐴 𝑠𝑠+𝜎𝜎

 Single real pole at

𝑠𝑠 = −𝜎𝜎 = −

1

𝜏𝜏

where 𝜏𝜏 is the system time constant

 Impulse response:

𝑔𝑔 𝑡𝑡 = ℒ −1^ 𝐺𝐺 𝑠𝑠 = 𝐴𝐴𝑒𝑒 −𝜎𝜎𝜎𝜎^ = 𝐴𝐴𝑒𝑒

−

𝑡𝑡 𝜏𝜏

𝑔𝑔 𝑡𝑡 = 𝐴𝐴𝑒𝑒

−

𝑡𝑡 𝜏𝜏

7

Impulse Response vs. Pole Location

 Increasing 𝜎𝜎 corresponds to decreasing 𝜏𝜏 and a faster response

8

First-Order System – Step Response

 Step response in the Laplace domain

𝑌𝑌 𝑠𝑠 =

1 𝑠𝑠 � 𝐺𝐺 𝑠𝑠^ =^

𝐴𝐴 𝑠𝑠 𝑠𝑠+𝜎𝜎

 Inverse transform back to time domain via partial fraction expansion

𝑌𝑌 𝑠𝑠 =

𝐴𝐴 𝑠𝑠 𝑠𝑠+𝜎𝜎 =^

𝑟𝑟 1 𝑠𝑠 +^

𝑟𝑟 2 𝑠𝑠+𝜎𝜎

𝐴𝐴 = 𝑟𝑟 1 + 𝑟𝑟 2 𝑠𝑠 + 𝜎𝜎𝑟𝑟 1

𝑠𝑠 0 : 𝜎𝜎𝑟𝑟 1 = 𝐴𝐴 → 𝑟𝑟 1 =

𝐴𝐴 𝜎𝜎

𝑠𝑠 1 : 𝑟𝑟 1 + 𝑟𝑟 2 = 0 → 𝑟𝑟 2 = −

𝐴𝐴 𝜎𝜎

𝑌𝑌 𝑠𝑠 = 𝐴𝐴/𝑠𝑠 𝜎𝜎− 𝐴𝐴𝑠𝑠+𝜎𝜎/𝜎𝜎

 Time-domain step response

𝑒𝑒 −𝜎𝜎𝜎𝜎^ = 𝐵𝐵 − 𝐵𝐵𝑒𝑒

− 𝜏𝜏𝜎𝜎

10

Step Response vs. Pole Location

 Increasing 𝜎𝜎 corresponds to decreasing 𝜏𝜏 and a faster response

11

Pole Location and Stability

 First-order transfer function

where 𝑝𝑝 is the system pole  Impulse response is

𝑔𝑔 𝑡𝑡 = 𝐴𝐴𝑒𝑒 𝑝𝑝𝜎𝜎

 If 𝑝𝑝 < 0, 𝑔𝑔 𝑡𝑡 decays to zero  Pole in the left half-plane  System is stable

 If 𝑝𝑝 > 0, 𝑔𝑔 𝑡𝑡 grows without bound  Pole in the right half-plane  System is unstable

13

Second-Order Systems

 Second-order transfer function

𝑁𝑁𝑁𝑁𝑚𝑚 𝑠𝑠 𝑠𝑠 2 +𝑎𝑎 1 𝑠𝑠+𝑎𝑎 0

𝑁𝑁𝑁𝑁𝑚𝑚 𝑠𝑠 𝑠𝑠+𝜎𝜎 2 +𝜔𝜔𝑑𝑑^2

where 𝜔𝜔𝑑𝑑 is the damped natural frequency

 Can also express the 2 nd-order transfer function as

𝑁𝑁𝑁𝑁𝑚𝑚 𝑠𝑠 𝑠𝑠 2 +2𝜁𝜁𝜔𝜔𝑛𝑛 𝑠𝑠+𝜔𝜔 (^) 𝑛𝑛^2

where 𝜔𝜔𝑛𝑛 is the un-damped natural frequency , and 𝜁𝜁 is the damping ratio

𝜎𝜎 𝜔𝜔 (^) 𝑛𝑛  Two poles at

𝑠𝑠 1 , 2 = −𝜎𝜎 ± 𝜎𝜎 2 − 𝜔𝜔𝑛𝑛^2 = −𝜁𝜁𝜔𝜔𝑛𝑛 ± 𝜔𝜔𝑛𝑛 𝜁𝜁 2 − 1

14

Categories of Second-Order Systems

 The 2nd^ -order system poles are

𝑠𝑠 1 , 2 = −𝜁𝜁𝜔𝜔𝑛𝑛 ± 𝜔𝜔𝑛𝑛 𝜁𝜁 2 − 1

 Value of 𝜁𝜁 determines the nature of the poles and, therefore, the response

 𝜻𝜻 > 𝟏𝟏: Over-damped  Two distinct, real poles – sum of decaying exponentials – treat as two first-order terms  𝑠𝑠 1 = −𝜎𝜎 1 , 𝑠𝑠 2 = −𝜎𝜎 2

 𝜻𝜻 = 𝟏𝟏: Critically-damped  Two identical, real poles – time-scaled decaying exponentials  𝑠𝑠 1 , 2 = −𝜎𝜎 = −𝜁𝜁𝜔𝜔𝑛𝑛 = −𝜔𝜔𝑛𝑛

 𝟎𝟎 < 𝜻𝜻 < 𝟏𝟏: Under-damped  Complex-conjugate pair of poles – sum of decaying sinusoids  𝑠𝑠 1 , 2 = −𝜎𝜎 ± 𝑗𝑗𝜔𝜔 (^) 𝑑𝑑 = −𝜁𝜁𝜔𝜔𝑛𝑛 ± 𝑗𝑗𝜔𝜔𝑛𝑛 1 − 𝜁𝜁 2

 𝜻𝜻 = 𝟎𝟎: Un-damped  Purely-imaginary, conjugate pair of poles – sum of non-decaying sinusoids  𝑠𝑠 1 , 2 = ±𝑗𝑗𝜔𝜔𝑛𝑛

16

Second-Order Poles - 0 ≤ 𝜁𝜁 ≤ 1

 Can relate 𝜎𝜎, 𝜔𝜔 (^) 𝑑𝑑, 𝜔𝜔𝑛𝑛,

and 𝜁𝜁 to pole location

geometry

 𝜎𝜎 is the real part

 𝜔𝜔𝑑𝑑 is the imaginary part

 𝜔𝜔𝑛𝑛 is the pole magnitude

 𝜁𝜁 is a measure of system

damping

𝜁𝜁 =

= sin 𝜃𝜃

17

Impulse Response – Critically-Damped

 For 𝜁𝜁 = 1, the transfer function reduces to

𝑠𝑠 2 + 2𝜔𝜔𝑛𝑛 𝑠𝑠 + 𝜔𝜔𝑛𝑛^2

 Impulse response

𝑔𝑔 𝑡𝑡 = ℒ −1^ 𝐺𝐺 𝑠𝑠

𝑔𝑔 𝑡𝑡 = 𝐴𝐴𝑡𝑡𝑒𝑒 −𝜎𝜎𝜎𝜎

19

Impulse Response – Under-Damped

 For 0 < 𝜁𝜁 < 1, the transfer function is

𝐺𝐺 𝑠𝑠 =

𝐴𝐴 𝑠𝑠 2 + 2𝜁𝜁𝜔𝜔𝑛𝑛 𝑠𝑠 + 𝜔𝜔𝑛𝑛^2

 Complete the square on the denominator

𝐺𝐺 𝑠𝑠 =

𝐴𝐴

𝑠𝑠 + 𝜁𝜁𝜔𝜔𝑛𝑛 2 + 𝜔𝜔𝑛𝑛 1 − 𝜁𝜁 2

2 =^

𝐴𝐴 𝑠𝑠 + 𝜁𝜁𝜔𝜔𝑛𝑛 2 + 𝜔𝜔 (^) 𝑑𝑑^2

 Rewrite in the form of a damped sinusoid

𝐺𝐺 𝑠𝑠 =

𝐴𝐴 𝜔𝜔 (^) 𝑑𝑑

𝜔𝜔 (^) 𝑑𝑑 𝑠𝑠 + 𝜁𝜁𝜔𝜔𝑛𝑛 2 + 𝜔𝜔 (^) 𝑑𝑑^2

=

𝐴𝐴 𝜔𝜔 (^) 𝑑𝑑

𝜔𝜔 (^) 𝑑𝑑 𝑠𝑠 + 𝜎𝜎 2 + 𝜔𝜔 (^) 𝑑𝑑^2

 Inverse Laplace transform for the time-domain impulse response

𝑔𝑔 𝑡𝑡 =

𝐴𝐴

𝜔𝜔𝑑𝑑

𝑒𝑒 −𝜎𝜎𝜎𝜎sin(𝜔𝜔𝑑𝑑 𝑡𝑡)

20

Under-Damped Impulse Response vs. 𝜔𝜔𝑛𝑛

𝑒𝑒 −𝜎𝜎𝜎𝜎^ sin 𝜔𝜔𝑑𝑑 𝑡𝑡 = 𝐵𝐵𝑒𝑒 −𝜁𝜁𝜔𝜔𝑛𝑛𝜎𝜎^ sin 𝜔𝜔𝑛𝑛 1 − 𝜁𝜁 2 𝑡𝑡