Feedback Control System

a) Open loop

b) Closed loop

𝑌(𝑠) 𝐺(𝑠)
=
𝑈(𝑠) 1 + 𝐻(𝑠)𝐺(𝑠)

𝐸(𝑠) 1
=
𝑈(𝑠) 1 + 𝐻𝐺
Our objective is to minimize the error E(s).
Now, by inspection one of the choices is to increase the value of HG !! (or H)
However, this cannot be done randomly; it must be done WRT some given specifications.

1) System Sensitivity (S)

𝛥𝑇/𝑇
𝑆𝐺𝑇 = 𝛥𝐺/𝐺
𝛥𝑇/𝑇
a) Open loop : 𝑆𝐺𝑇 = 𝛥𝐺 /𝐺 = 1 = 100%
𝑇(𝑠) = 𝐺(𝑠)
26
b) Closed loop

𝑌(𝑠) + 𝛥𝑌(𝑠) 𝐺(𝑠) + 𝛥𝐺(𝑠)

=
𝑈(𝑠) 1 + 𝐻(𝑠)(𝐺(𝑠) + 𝛥𝐺(𝑠))
𝑌(𝑠)
But we need or 𝛥𝑌(𝑠)
𝑈(𝑠)
𝑌(𝑠)
𝛥 𝐺(𝑠) + 𝛥𝐺(𝑠) 𝑌(𝑠)
= − ,
𝑈(𝑠) 1 + 𝐻(𝑠)(𝐺(𝑠) + 𝛥𝐺(𝑠)) 𝑈(𝑠)
𝑌(𝑠) 𝐺(𝑠)
𝐾𝑛𝑜𝑤𝑖𝑛𝑔 𝑡ℎ𝑎𝑡 ∶ 𝑇 = =
𝑈(𝑠) 1 + 𝐺(𝑠)𝐻(𝑠)

𝛥𝑌(𝑠) 𝛥𝐺(𝑠)
= , 𝛥𝐺𝐻 ≈ 0
𝑈(𝑠) (1 + 𝐺𝐻 + 𝛥𝐺𝐻)(1 + 𝐺𝐻)
𝛥𝐺(𝑠)
𝛥𝑌(𝑠) = 2 𝑈(𝑠)
(1+𝐺𝐻)

𝑌(𝑠)
Knowing that 𝑇 = 𝑈(𝑠)
⇒𝛥 𝑇 = 𝛥𝑌
𝑈
𝛥𝑇
𝛥𝐺 ,
⇒ 𝛥𝑇 =
𝐺(𝑠)
2 𝑆𝐺𝑇 = 𝑇
𝛥𝐺 , 𝑇 = 1+𝐺(𝑠)𝐻(𝑠)
(1+𝐺𝐻) 𝐺

𝛥𝐺
/𝑇 𝐺 (1+𝐻𝐺)
(1+𝐺𝐻)2
Hence, 𝑆𝐺𝑇 = = (1+𝐻𝐺)2
𝛥𝐺/𝐺 𝐺

1
𝑆𝐺𝑇 = 1+𝐻𝐺 -------- Approximation
Selecting HG to be a large value leads to lower sensitivity value.

27
2) Feedback Sensitivity

𝛥𝑇/𝑇 𝛥𝑇 𝐻
𝑆𝐻𝑇 = =
𝛥𝐻/𝐻 𝛥𝐻 𝑇
.
.
−𝐺𝐻
𝑆𝐻𝑇 =
1 + 𝐺𝐻
Notice that for a large value of GH, then |𝑆𝐻𝑇 | ≈ 1 which is similar to the
open loop.
Therefore, it is important to use feedback elements that do not change
easily.

EX:

Find 𝑆𝜏𝑇 .
𝑇
𝛥𝑇 𝜏
𝑆𝜏 =
𝛥𝜏 𝑇
100/(𝜏𝑆 + 1) 𝜏𝑆 + 1 100
𝑇= =
1 + 100/(𝜏𝑆 + 1) 𝜏𝑆 + 1 𝜏𝑆 + 101
𝛥𝑇 (𝜏𝑆 + 101)(0) − 100(𝑆) −100𝑆
= =
𝛥𝜏 (𝜏𝑆 + 101)2 (𝜏𝑆 + 101)2
−100𝑆 𝜏 −𝜏𝑆
𝑆𝜏𝑇 = =
(𝜏𝑆 + 101)2 100/(𝜏𝑆 + 101) 𝜏𝑆 + 101

28
DC Motor:

𝜔(𝑠) 𝐾
1
Let 𝐺(𝑠) = 𝑉 =
𝑎 (𝑠) 𝜏𝑆+1
𝐾𝑚 𝑅𝑎 𝐽
Where 𝐾1 = 𝑅 , 𝜏=𝑅
𝑎 𝑏+𝐾𝑏 𝐾𝑚 𝑎 𝑏+𝐾𝑏 𝐾𝑚
That is:

Control of transient response:

(A) Open Loop System

𝜔(𝑠) 𝐾1
𝐺(𝑠) = =
𝑉𝑎 (𝑠) 𝜏𝑆 + 1
𝐾𝑚
𝐾1 =
𝑅𝑎 𝑏 + 𝐾𝑏 𝐾𝑚
𝑅𝑎 𝐽
𝜏=
𝑅𝑎 𝑏 + 𝐾𝑚 𝐾2
𝐾2 𝐸
Let 𝑉𝑎 (𝑠) = 𝑆
𝐾1 𝐾2
⇒ 𝜔(𝑠) = ( 𝐸)
𝜏𝑆 + 1 𝑆

29
 −1
𝐾1 𝐾2
𝐿 {𝜔(𝑠) = 𝐸}
𝜏𝑆 + 1 𝑆
𝐾1 𝐾2 𝐸 𝐴1 𝐴2
𝜔(𝑠) = = +
(𝜏𝑆 + 1)𝑆 𝜏𝑆 + 1 𝑆
𝜔(𝑡) = (𝐴2 + 𝐴1 𝑒 −𝑡/𝜏 )𝑈(𝑡)
𝜔(𝑡) = 𝐾1 𝐾2 𝐸(1 − 𝑒 −𝑡/𝜏 )
As an open loop system, 𝐾1 𝐾2 𝐸 not much you can control.

(B) Closed Loop System

𝐾𝑡 : is determined by the physical construction of the tachometer (which

generates voltage that is proportional to the system’s speed).

𝜔(𝑠) 𝐾𝑎 𝐺(𝑠)
𝑇(𝑠) = =
𝑉𝑎 (𝑠) 1 + 𝐾𝑡 𝐾𝑎 𝐺(𝑠)
1 𝐾
Recall: 𝐺(𝑠) = 𝜏𝑆+1
𝜔(𝑠) 𝐾𝑎 𝐾1 /𝜏
=
𝑉𝑎 (𝑠) 𝑆 + [1 + 𝐾𝑡 𝐾𝑎 𝐾1 ]
If 𝑉𝑎 (𝑠) is a step function,
𝐾2
𝑉𝑎 (𝑠) = 𝐸
𝑆

30
 −1
𝐾𝑎 𝐾1 /𝜏 𝐾2 𝐸
𝐿 {𝜔(𝑠) = }
𝑆 + [1 + 𝐾𝑡 𝐾𝑎 𝐾1 ] 𝑆
−[1+𝐾𝑡 𝐾𝑎 𝐾1 ]
𝑡
𝜔(𝑡) = 𝐴1 + 𝐴2 𝑒 𝜏

Here, using the controller (K a ) you can control the system performance.

Note: because of the load inertia is assumed to be very large, we take the
response by increasing [K a ] and approximate it by
1 −[𝐾𝑡 𝐾𝑎 𝐾1 ]
𝑡
𝜔(𝑡) ≃ 𝐾2 𝐸[1 + 𝑒 𝜏 ]
𝐾𝑡
Disturbance Signals For Feedback Control:

Our objective here is to see the effect of the disturbance (𝑇𝑑 ) on the error (𝐸 ). Hence, we need
𝐸(𝑠)
to find the transfer function ( ).
𝑇𝑑 (𝑠)
(A) Open Loop:

𝑉𝑎 (𝑠) = 𝑅(𝑠) ∶ 𝐼𝑛𝑝𝑢𝑡 and 𝑌 (𝑠) = 𝜔(𝑠): 𝑂𝑢𝑡𝑝𝑢𝑡

𝐸𝑎𝑐𝑡𝑢𝑎𝑙 (𝑠) = 𝑅(𝑠) − 𝑌(𝑠) = 𝑉𝑎 (𝑠) − 𝜔(𝑠) -------- (1)

Let R(s) be zero
⇒ 𝐸(𝑠) = −𝜔(𝑠) -------- (2)
1 𝐾
Also, 𝐸 = −𝜔 = −[𝐽𝑆+𝑏 ( 𝑅𝑚 𝐸𝑎 − 𝑇𝑑 )] -------- (3)
𝑎
𝐸𝑎 = −𝐾𝑏 𝜔 -------- (4)
𝜔 = −𝐸 -------- (5)

31
 𝐾𝑚 𝐾𝑏 𝐸 𝑑 𝑇
𝐸 = −𝑅 + -------- (6)
𝑎 (𝐽𝑆+𝑏) 𝐽𝑆+𝑏
𝐾𝑚 𝐾𝑏 𝑇𝑑
[1 + ]𝐸 =
𝑅𝑎 (𝐽𝑆 + 𝑏) 𝐽𝑆 + 𝑏
𝐸(𝑠) 1
= 𝐾 𝐾𝑚 -------- (7)
𝑇𝑑 (𝑠) 𝐽𝑆+𝑏+( 𝑏 )
𝑅𝑎

⇒ 𝐸(𝑠) = 1 1𝐷
𝐾𝑏 𝐾𝑚 𝑆
𝐽𝑆 + 𝑏 + ( )
𝑅𝑎
1 1
𝑒𝑠𝑠 = 𝑙𝑖𝑚 𝐸(𝑡) = 𝑙𝑖𝑚𝑆𝐸(𝑠) = 𝑙𝑖𝑚𝑆 𝐾 𝐾 𝐷
𝑡→∞ 𝑆→0 𝑆→0 𝐽𝑆+𝑏+( 𝑏 𝑚 ) 𝑆
𝑅𝑎
𝐷
𝑒𝑠𝑠 =
𝐾 𝐾
𝑏 + 𝑏𝑅 𝑚
𝑎

(B) Closed Loop:

1
Let 𝐺1 = 𝐾𝑎 𝑅 𝐾𝑚
𝑎
1
𝐺2 = 𝐽𝑆+𝑏
𝐾
𝐻 = 𝐾𝑡 + 𝐾𝑏
𝑎

32
𝐸 (𝑠) = 𝑅(𝑠) − 𝜔(𝑠) = −𝜔(𝑠), assuming 𝑅(𝑠) = 0
𝐸(𝑠) = −[𝐺2 (𝑠){𝐺1 (𝑠)𝐸𝑎 (𝑠) − 𝑇𝑑 (𝑠)}]
𝐸(𝑠) = −[𝐺2 (𝑠){𝐺1 (𝑠)𝐻(𝑠)𝐸(𝑠) − 𝑇𝑑 (𝑠)}]
𝐺2
𝐸(𝑠) = 𝑇
1 + 𝐺1 𝐺2 𝐻 𝑑
𝐸 𝐺2
=
𝑇𝑑 1 + 𝐺1 𝐺2 𝐻

Our objective is (𝑒 → 0)
1
Note: If 𝐺1 𝐺2 𝐻 ≫ 1 then 𝐸 = 𝐺 𝑇𝑑
1𝐻
Hence, the error can be minimized if we can obtain a large 𝐺1 𝐻 which can
𝐾𝑎 𝐾𝑚
be achieved based on the selection of 𝐾𝑎 (𝐺1 = ).
𝑅𝑎

Steady State Error

(A) Open Loop

𝐸(𝑠) = 𝑅(𝑠) − 𝑌(𝑠) , 𝑌(𝑠) = 𝐺(𝑠)𝑅(𝑠)

𝐸(𝑠) = (1 − 𝐺(𝑠))𝑅(𝑠)
Note: If 𝐸(𝑠) = 0, then 𝐺(𝑠) = 1

(B) Closed Loop

33
𝐸(𝑠) = 𝑅(𝑠) − 𝑌(𝑠) → 𝑌(𝑠) = 𝑅(𝑠) − 𝐸(𝑠)
𝑌(𝑠) = 𝐺(𝑠)[𝑅(𝑠) − 𝐻(𝑠)𝑌(𝑠)]
𝐸(𝑠) 1 − 𝐺(𝑠) + 𝐺(𝑠)𝐻(𝑠)
=
𝑅(𝑠) 1 + 𝐺(𝑠)𝐻(𝑠)

Note: If 𝐻(𝑠) = 1, then

𝐸(𝑠) 1
= 1+𝐺(𝑠) in this case 𝐸 = 𝐸𝑎
𝑅(𝑠)

𝐸(𝑠) 1+𝐺(𝑠)𝐻(𝑠) 𝐺(𝑠)

However, = −
𝑅(𝑠) 1+𝐺(𝑠)𝐻(𝑠) 1+𝐺(𝑠)𝐻(𝑠)
𝐺(𝑠)
𝐸(𝑠) = [1 − ]𝑅(𝑠)
1+𝐺(𝑠)𝐻(𝑠)
𝐸(𝑠) = [1 − 𝑇(𝑠)]𝑅(𝑠)

Hence,

(1) Open Loop

1
𝑒𝑠𝑠 = 𝑙𝑖𝑚𝑆[1 − 𝐺(𝑠)] = 1 − 𝐺(0) assuming R(s) a unit step.
𝑆→0 𝑆

(2) Closed Loop

𝐺(𝑠) 1
𝑒𝑠𝑠 = 𝑙𝑖𝑚𝑆(1 − 1+𝐻(𝑠)𝐺(𝑠)) 𝑆
𝑆→0
𝐺(0)
𝑒𝑠𝑠 = 1 − 1+𝐻(0)𝐺(0)

34
Note: 𝐺(0) = 𝐺(𝑠) @ 𝑆 = 0 is called dc gain.

Hence, for a closed loop with unity feedback

1 1 1
𝑒𝑠𝑠 = 𝑙𝑖𝑚𝑆( ) =
𝑆→0 1 + 𝐺(𝑠) 𝑆 1 + 𝐺(0)
Ex:
A- Open Loop:
𝑒𝑠𝑠 = 1 − 𝐺(0) = 1 − 𝐾

B- Closed Looop:

1 1
𝑒𝑠𝑠 = =
1 + 𝐺(0) 1 + 𝐾

* For the open loop for K=1, we have 𝑒𝑠𝑠 = 0

1
* For the closed loop for K=100, we have 𝑒𝑠𝑠 = 1+100 = 0.009
Now, assume there is change of 10% of the assumed K parameter
Then, for the open loop, 𝑒𝑠𝑠 = 1 − 0.9 = 0.1
1
& for the closed loop,𝑒𝑠𝑠 = 1+90 = 0.0109

35
Note: Cost of the feedback may be listed as:
(1) Cost of the feedback sensors.
(2) Noise added to the system caused by new elements.
(3) Loss of gain.

Performance Of Feedback Control

Input Testing Signals

(1) Step input signal

𝑟(𝑡) = 𝐴 , 𝑡 > 0
𝐴
𝑅(𝑠) = 𝑆

(2) Ramp input

𝑟(𝑡) = 𝐴𝑡 , 𝑡>0
𝐴
𝑅(𝑠) = 2
𝑆
(3) Parabolic input

𝑟(𝑡) = 𝐴𝑡 2 , 𝑡 > 0
2𝐴
𝑅(𝑠) = 3
𝑆
(4) Unit impulse

36
 ∞
∫−∞ 𝛿(𝑡) = 1
𝛿(𝑠) = 1
Feedback second order system

𝑌(𝑠) 𝐾 𝐾
= 𝑆(𝑆+𝑃)+𝐾 = 𝑆 2 +𝑆𝑃+𝐾 This is the standard second order system
𝑅(𝑠)

The general form for the second order system:

𝑌(𝑠) 𝜔𝑛2
=
𝑅(𝑠) 𝑆 2 + 2𝜁𝜔𝑛 𝑆 + 𝜔𝑛2
Where 𝜔𝑛 : natural frequency
𝜁 : damping factor

𝑆1,2 = −𝛼 ± √𝛼 2 − 𝜔𝑜2
𝑆 = −𝛼 ± 𝑗𝜔𝑑 Under damped

𝑆 2 + 2𝜁𝜔𝑛 𝑆 + 𝜔𝑛2 ≡ Characteristic equation

𝑆1,2 = −𝜁𝜔𝑛 ± 𝑗𝜔𝑛 √1 − 𝜁 2

⇒ 𝛼 = 𝜁𝜔𝑛
⇒ 𝜔𝑑 = 𝜔𝑛 √1 − 𝜁 2

37
To a unit step input:
1 −𝜁𝜔 𝑡
𝑦(𝑡) = 1 − 𝑒 𝑛 𝑠𝑖𝑛(𝜔 𝛽𝑡 + 𝜃 )
𝑛
𝛽

Where 𝛽 = √1 − 𝜁 2 , 𝜃 = 𝑐𝑜𝑠 −1 (𝜁)

Notes:
(1) If 𝜁 > 1,then we have an over damped system.
(2) If 𝜁 = 1, then we have a critically damped system.
(3) If (0 < 𝜁 < 1), then we have an under damped system.
(4) If 𝜁 = 0, then we have a pure imaginary system.
(5) If 𝜁 < 0, then we have unstable root.

Final value = average value = 1

To impulse input:
38
 𝜔𝑛 −𝜁𝜔 𝑡
𝑦(𝑡) = 𝑒 𝑛 𝑠𝑖𝑛(𝜔 𝛽𝑡)
𝑛
𝛽
If (𝜁
= 0), then
𝜔
𝑦(𝑡) = 𝛽𝑛 𝑠𝑖𝑛(𝜔𝑛 𝛽𝑡), where the average is 0

𝑀𝑃𝑡 − 𝑓𝑉
𝑃. 𝑂. = × 100%
𝑓𝑉
𝑃. 𝑂.: Percent overshoot.
𝑀𝑃𝑡 : Peak value of the time response.
𝑓𝑉 : Final value.

𝑇𝑆 : Settling time: the time when the response fluctuates with 2% of the
final value.

1 −𝜁𝜔 𝑡
𝑦(𝑡) = 1 − 𝑒 𝑛 𝑠𝑖𝑛(𝜔 𝛽𝑡 + 𝜃)
𝑛
𝛽
= 1 ± 0.02

39
𝑒 −𝜁𝜔𝑛 𝑡𝑠 = 0.02
⇒ 𝜁𝜔𝑛 𝑡𝑠 = 3.912 ≃ 4
𝟏
⇒ 𝒕𝒔 = 𝟒 = 𝟒𝝉
𝜻𝝎𝒏
Now, to find the maximum overshoot
𝑑𝑦(𝑡) 𝑑𝑦(𝑡)
𝐿{ } = 0 → 𝐿{ } = 𝑆𝑌(𝑠)
𝑑𝑡 𝑑𝑡
𝜔𝑛2 1
𝑆𝑌(𝑠) = 𝑆 2
𝑆 + 2𝜁𝜔𝑛 𝑆 + 𝜔𝑛2 𝑆
2
𝜔𝑛 𝜔𝑛 −𝜁𝜔
𝐿−1 { 2 2 } = 𝑒 𝑛 𝑠𝑖𝑛(𝜔 𝛽𝑡 ) = 0
𝑛 𝑝
𝑆 + 2𝜁𝜔𝑛 𝑆 + 𝜔𝑛 𝛽

𝑠𝑖𝑛(𝜔𝑛 𝛽𝑡𝑝 ) = 0
Taking
⇒ 𝜔𝑛 𝛽𝑡𝑝 = 𝜋
𝝅
⇒ 𝒕𝒑 =
𝝎𝒏 √𝟏 − 𝜻𝟐
At𝑡 = 𝑡𝑝 ,
𝑦(𝑡) = 𝑀𝑡𝑝
𝜋
1 −𝜁𝜔𝑛 (𝜔 √1−𝜁 2 ) 𝜋
=− 𝑒 𝑛 𝑠𝑖𝑛(𝜔𝑛 √1 − 𝜁 2 ( + 𝜃)
𝛽 𝜔𝑛 √1 − 𝜁 2
−𝜻𝝅
𝟐
⇒ 𝑴𝒑𝒕 = 𝟏 + 𝒆√𝟏−𝜻
−𝜁𝜋
√1−𝜁2
⇒ 𝑃. 𝑂 = 100𝑒 %
EX :

40
Determine the value of K and P so that:
1) P.O < 5%
2) ts with 2%
3) ts < 4 seconds

𝑌(𝑠) 𝐾 2
𝜔𝑛
Sol: = 𝑆 2 +𝑃𝑆+𝐾 = 𝑆 2 +2𝜁𝜔 2
𝑅(𝑠) 𝑛 𝑆+𝜔𝑛
−𝜁𝜋
√1−𝜁2
⇒𝑒 < 0.05
−0.707𝜋
1
Let 𝜁= = 0.707 , 𝑒 √1−0.7072 = 0.043 < 0.05
√2
−𝜁𝜔𝑛 𝑡𝑠
⇒𝑒 < 0.02
⇒ 𝜁𝜔𝑛 𝑡𝑠 < 3.9 ≃ 4
4 4
𝑡𝑠 < 𝜁𝜔 = 𝜏 , 𝑖𝑓 𝜏 = 1 𝑡ℎ𝑒𝑛 𝑡𝑠 < 4
𝑛
1 1 1
𝜏 = 𝜁𝜔 ⇒ 1 = 0.707𝜔 ⇒ 𝜔𝑛 = 0.707 = √2
𝑛 𝑛
⇒𝐾= 𝜔𝑛2 =2
1
⇒ 𝑃 = 2𝜁𝜔𝑛 = 2 √2 = 2
√2

In General:

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Now, in general
𝐾 ∏𝑀
𝑖=1 (𝑆+𝑍𝑖 ) 𝑁(𝑠)
𝐺(𝑠) = 𝑄 = 𝑆 𝑁 𝐷(𝑠)
𝑆 𝑁 ∏𝐾=1 (𝑆+𝑃𝐾 )

Note: consider the following system

For H(s) =1 (unity feedback)

1
𝐸(𝑠) = 𝑅(𝑠)
1 + 𝐺(𝑠)
𝑒𝑆𝑆 = 𝑙𝑖𝑚 𝑆𝐸(𝑠)
𝑆→0
1
𝑒𝑆𝑆 = 𝑙𝑖𝑚 𝑆[1+𝐺(𝑠) 𝑅(𝑠)]
𝑆→0

𝐴
(1) For a step input: (𝑅(𝑠) = 𝑆 )

42
 1 𝐴 𝐴
⇒ 𝑒𝑆𝑆 = 𝑙𝑖𝑚 𝑆 1+𝐺(𝑠) 𝑆 = 𝑁(𝑠)
𝑆→0 1+𝑙𝑖𝑚 𝑁
𝑆→0𝑆 𝐷(𝑠)
(a) N=0 (zero type)
𝐴 𝐴
𝑒𝑆𝑆 = 𝑁(𝑠) = 𝐴 =𝐵≠0 Not good
1+𝑙𝑖𝑚 1+ 1
𝑆→0𝐷(𝑠) 𝐴2

(b) N=1 (type one) or N>1

𝐴 𝐴
𝑒𝑆𝑆 = 𝑁(𝑠) = 1+∞ = 0 Good
1+𝑙𝑖𝑚
𝑆→0𝑆𝐷(𝑠)
1 1
⇒ 𝑒𝑆𝑆 = 1+𝑙𝑖𝑚𝐺(𝑠) = 1+𝐾
𝑃
𝑆→0
Where Kp is position-error-constant
𝐾𝑃 = 𝑙𝑖𝑚 𝐺(𝑠)
𝑆→0
𝐴
(2) For a ramp input: (𝑅(𝑠) = 𝑆 2 )
1 𝐴 𝐴
⇒ 𝑒𝑆𝑆 = 𝑙𝑖𝑚𝑆[1+𝐺(𝑠) 𝑆 2 ] = 𝑙𝑖𝑚 𝑁(𝑠)
𝑆→0 𝑆→0 𝑆+𝑆 𝑁
𝑆 𝐷(𝑠)
(a) N=0
𝑒𝑆𝑆 = ∞ Not good

(b) N≥2
𝑒𝑆𝑆 = 0
𝐴
⇒ 𝑒𝑆𝑆 = 𝐾
𝑉
Kv : velocity error constant = 𝑙𝑖𝑚 𝑆𝐺(𝑠)
𝑆→0
𝐴
(3) For a parabolic input: (𝑅(𝑠) = 𝑆 3)
For N≥3
𝑒𝑆𝑆 = 0
𝐴
⇒ 𝑒𝑆𝑆 = 𝐾
𝑎

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 Where 𝐾𝑎 = 𝑙𝑖𝑚 𝑆 2 𝐺(𝑠) and known as acceleration error
𝑆→0
constant.

EX:

For a unit step input, design the controller (K) such that 𝑒𝑆𝑆 =0
Solution:
1
Since the feedback is not unity, we can’t apply 𝐸(𝑠) = 1+𝐺(𝑠) 𝑅(𝑠)
⇒Convert to open loop system
𝑌(𝑠) 𝐾(𝑆+4)
𝐺(𝑠) = 𝑅(𝑠) = (𝑆+4)(𝑆+2)+2𝐾
𝐸(𝑠)𝑜𝑝𝑒𝑛 𝑙𝑜𝑜𝑝 = [1 − 𝑇(𝑠)]𝑅(𝑠)
𝐾(𝑆+4) 1
𝑒𝑆𝑆 = lim 𝑆[1 − (𝑆+4)(𝑆+2)+2𝐾] 𝑆 = 0
𝑆→0
4𝐾
1 − 4(2)+2𝐾 = 0
⇒𝐾=4
Model Order Reduction (MOR) :

𝐾
Given 𝑇(𝑠) = 𝑆(𝑆+2)(𝑆+30)

𝑦(𝑡) = 𝐴 + 𝐴1 𝑒 −2𝑡 + 𝐴2 𝑒 −30𝑡

𝐾
Hence, final value of 𝑇(𝑠) = 60

𝐾 𝐾
𝑇(𝑠)𝑟𝑒𝑑𝑢𝑐𝑒𝑑 = ⇒ 𝑓𝑖𝑛𝑎𝑙 𝑣𝑎𝑙𝑢𝑒 = !!
𝑆(𝑆 + 2) 2

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 𝐾 𝐾
𝑇(𝑠)𝑟𝑒𝑑𝑢𝑐𝑒𝑑 = ⇒ 𝑓𝑖𝑛𝑎𝑙 𝑣𝑎𝑙𝑢𝑒 =
𝑆(𝑆 + 2)(30) 60

45