Steady State Errors and Error Constants

Unit-3

Time Response Analysis

Lecture 3

Steady State Errors and Error Constants

• Steady state errors are an important parameter for evaluating the system performance.

• Steady state error is a measure of system accuracy.

• These errors arise form the nature of the inputs, type of system and from non-linearities of system componets such as static friction, backlash etc.

• The magnitudes of the steady state error due to the different individual inputs indicates the accuracy of the system.

• Consider a unity feedback system shown in Fig. 1.

• The closed-loop transfer function is 𝐶(𝑠)

𝑅(𝑠) = 𝐺(𝑠) 1 + 𝐺(𝑠)

Fig. 1 Unity feedback system

• The error signal be written as 𝐸 𝑠 = 𝐶 𝑠

𝐺 𝑠 = 𝑅 𝑠 1 + 𝐺 𝑠 𝐸 𝑠 = 1

1 + 𝐺 𝑠

The steady state error (e_ss) can be evaluated using the final value-theorem.

𝑒_𝑠𝑠 = lim

𝑡⟶∞ 𝑒 𝑡 = lim

𝑠⟶0 𝑠𝐸 𝑠 = lim

𝑠⟶0

𝑠𝑅 𝑠 1 + 𝐺 𝑠

Steady State error for Unit-step input For unit-step input, 𝑅 𝑠 = 1/𝑠

𝑒_𝑠𝑠 = lim

𝑠⟶0 𝑠𝐸 𝑠 = lim

𝑠⟶0 𝑠 𝑅 𝑠

1 + 𝐺 𝑠 = lim

𝑠⟶0

𝑠. (1 1 + 𝐺 𝑠𝑠)

= 1

1 + lim

𝑠⟶0𝐺 𝑠 = 1

1 + 𝐺 0 = 1 1 + 𝐾_𝑝 𝐾_𝑝 = lim

𝑠⟶0𝐺 𝑠 = 𝐺(0) K_p is called the 𝐩𝐨𝐬𝐢𝐭𝐢𝐨𝐧 𝐞𝐫𝐫𝐨𝐫 𝐜𝐨𝐧𝐬𝐭𝐚𝐧𝐭.

Steady State error for Unit-ramp input For unit-ramp input, 𝑅 𝑠 = 1/𝑠²

𝑒_𝑠𝑠 = lim

𝑠⟶0 𝑠𝐸 𝑠 = lim

𝑠⟶0 𝑠 𝑅 𝑠

1 + 𝐺 𝑠 = lim

𝑠⟶0

𝑠. ( 1 𝑠²) 1 + 𝐺 𝑠

= 1

𝑠⟶0lim 𝑠𝐺 𝑠 = 1 𝐾_𝑣 𝐾_𝑣 = lim

𝑠⟶0𝑠𝐺 𝑠 K_v is called the 𝐯𝐞𝐥𝐨𝐜𝐢𝐭𝐲 𝐞𝐫𝐫𝐨𝐫 𝐜𝐨𝐧𝐬𝐭𝐚𝐧𝐭.

Steady State error for Unit-parabolic input For unit-ramp input, 𝑅 𝑠 = 1/𝑠³

𝑒_𝑠𝑠 = lim

𝑠⟶0 𝑠𝐸 𝑠 = lim

𝑠⟶0 𝑠 𝑅 𝑠

1 + 𝐺 𝑠 = lim

𝑠⟶0

𝑠. ( 1 𝑠³) 1 + 𝐺 𝑠

= 1

𝑠⟶0lim 𝑠²𝐺 𝑠 = 1 𝐾_𝑎 𝐾_𝑎 = lim

𝑠⟶0𝑠²𝐺 𝑠

K_a is called the 𝐚𝐜𝐜𝐞𝐥𝐞𝐫𝐚𝐭𝐢𝐨𝐧 𝐞𝐫𝐫𝐨𝐫 𝐜𝐨𝐧𝐬𝐭𝐚𝐧𝐭.

‘Types’ of Control Systems

Consider a unity-feedback control system with the following open-loop transfer function

𝐺 𝑠 = 𝐾 𝑇_𝑎𝑠 + 1 𝑇_𝑏𝑠 + 1 … (𝑇_𝑚𝑠 + 1) 𝑠^𝑁 𝑇₁𝑠 + 1 𝑇₂𝑠 + 1 … (𝑇_𝑝𝑠 + 1)

It involves a term 𝑠^𝑁 in the denominator which represents the number of poles at the origin.

If N = 0, it is called a type 0 system. Similarly N=1, 2…will represent a type 1,2.. System respectively.

Therefore, a type 0 system will have no poles at the origin, type-1 system will have 1 pole and type 2 will have 2 poles at the origin.

Steady-state error for ‘Type-0’ system For a type-0 system,

𝐺 𝑠 = 𝐾 1 + 𝑇_𝑧1𝑠 1 + 𝑇_𝑧2𝑠 … 1 + 𝑇_𝑝1𝑠 1 + 𝑇_𝑝2𝑠 … 𝐾_𝑝 = lim

𝑠⟶0

𝐾 1 + 𝑇_𝑧1𝑠 1 + 𝑇_𝑧2𝑠 …

1 + 𝑇_𝑝1𝑠 1 + 𝑇_𝑝2𝑠 … = 𝐾 𝑒_𝑠𝑠 position = 1

1 + 𝐾_𝑝 = 1

1 + 𝐾 = finite value

𝐾_𝑣 = lim

𝑠⟶0𝑠𝐺 𝑠 = lim

𝑠⟶0 𝑠 𝐾 1 + 𝑇_𝑧1𝑠 1 + 𝑇_𝑧2𝑠 …

1 + 𝑇_𝑝1𝑠 1 + 𝑇_𝑝2𝑠 … = 0

⇒ 𝑒_𝑠𝑠 velocity = 1

𝐾_𝑣 = 1

0 = ∞ 𝐾_𝑎 = lim

𝑠⟶0𝑠²𝐺 𝑠 = lim

𝑠⟶0 𝑠² 𝐾 1 + 𝑇_𝑧1𝑠 1 + 𝑇_𝑧2𝑠 …

1 + 𝑇_𝑝1𝑠 1 + 𝑇_𝑝2𝑠 … = 0

⇒ 𝑒_𝑠𝑠 acceleration = 1

𝐾_𝑎 = 1

0 = ∞

Steady-state error for ‘Type-1’ system For a type-1 system,

𝐺 𝑠 = 𝐾 1 + 𝑇_𝑧1𝑠 1 + 𝑇_𝑧2𝑠 … 𝑠 1 + 𝑇_𝑝1𝑠 1 + 𝑇_𝑝2𝑠 … 𝐾_𝑝 = lim

𝑠⟶0

𝐾 1 + 𝑇_𝑧1𝑠 1 + 𝑇_𝑧2𝑠 …

𝑠 1 + 𝑇_𝑝1𝑠 1 + 𝑇_𝑝2𝑠 … = ∞ 𝑒_𝑠𝑠 position = 1

1 + 𝐾_𝑝 = 1

1 + ∞ = 0

𝐾_𝑣 = lim

𝑠⟶0𝑠𝐺 𝑠 = lim

𝑠⟶0 𝑠 𝐾 1 + 𝑇_𝑧1𝑠 1 + 𝑇_𝑧2𝑠 …

𝑠 1 + 𝑇_𝑝1𝑠 1 + 𝑇_𝑝2𝑠 … = 𝐾

⇒ 𝑒_𝑠𝑠 velocity = 1

𝐾_𝑣 = 1

𝐾 = finite value 𝐾_𝑎 = lim

𝑠⟶0𝑠²𝐺 𝑠 = lim

𝑠⟶0 𝑠² 𝐾 1 + 𝑇_𝑧1𝑠 1 + 𝑇_𝑧2𝑠 …

𝑠 1 + 𝑇_𝑝1𝑠 1 + 𝑇_𝑝2𝑠 … = 0

⇒ 𝑒_𝑠𝑠 acceleration = 1

𝐾_𝑎 = 1

0 = ∞

Steady-state error for ‘Type-2’ system For a type-2 system,

𝐺 𝑠 = 𝐾 1 + 𝑇_𝑧1𝑠 1 + 𝑇_𝑧2𝑠 … 𝑠² 1 + 𝑇_𝑝1𝑠 1 + 𝑇_𝑝2𝑠 … 𝐾_𝑝 = lim

𝑠⟶0

𝐾 1 + 𝑇_𝑧1𝑠 1 + 𝑇_𝑧2𝑠 …

𝑠 1 + 𝑇_𝑝1𝑠 1 + 𝑇_𝑝2𝑠 … = ∞ 𝑒_𝑠𝑠 position = 1

1 + 𝐾_𝑝 = 1

1 + ∞ = 0

𝐾_𝑣 = lim

𝑠⟶0𝑠𝐺 𝑠 = lim

𝑠⟶0 𝑠 𝐾 1 + 𝑇_𝑧1𝑠 1 + 𝑇_𝑧2𝑠 …

𝑠² 1 + 𝑇_𝑝1𝑠 1 + 𝑇_𝑝2𝑠 … = ∞

⇒ 𝑒_𝑠𝑠 velocity = 1

𝐾_𝑣 = 1

∞ = 0 𝐾_𝑎 = lim

𝑠⟶0𝑠²𝐺 𝑠 = lim

𝑠⟶0 𝑠² 𝐾 1 + 𝑇_𝑧1𝑠 1 + 𝑇_𝑧2𝑠 …

𝑠² 1 + 𝑇_𝑝1𝑠 1 + 𝑇_𝑝2𝑠 … = 𝐾

⇒ 𝑒_𝑠𝑠 acceleration = 1

𝐾_𝑎 = 1

𝐾 = finite value

• Type-0 system has a constant position error, infinite velocity and infinite acceleration errors.

• The position error constant is equal to the open loop gain of the transfer function in the time-constant form.

• Type-1 system has a zero position error, constant velocity and infinite acceleration errors.

• The velocity error constant is equal to the open loop gain of the transfer function in the time-constant form.

• Type-2 system has a zero position error, zero velocity and constant acceleration errors. The accelaration error constant is equal to the open loop gain of the transfer function in the time- constant form.

• The error constants indicates the steady-state performance. As the type of system increases, more steady-state errors are eliminated.

• Systems with type greater than two are generally not employed because they are more difficult to stabilize.

Type of Input Steady-State Error

Type-0 System Type-1 System Type-2 System

Unit-Step 1/(1+K_p) 0 0

Unit-Ramp ∞ 1/K_v 0

Unit-Parabolic ∞ ∞ 1/K_a

𝐾_𝑝 = lim

𝑠⟶0𝐺 𝑠 𝐾_𝑣 = lim

𝑠⟶0𝑠𝐺 𝑠 𝐾_𝑎 = lim

𝑠⟶0𝑠²𝐺 𝑠 Steady-state errors for various inputs and systems

• Example – Determine the type and order of the systems with following transfer functions:

a G s H s = K(s + 3)

s²(s + 2)(s + 5) b G s H s = (s + 2)

(s − 3)(s + 0.1) c G s H s = 20

s³(s⁴ + 8s² + 16)

• Solution -

(a) Since the number of open-loop at the origin = 2, therefore the type of system is 2. The highest power of s present in the denominator = 4, therefore the order of the system is 4.

(fourth-order system).

(b) It is type-0 system because the number of open-loop at the origin is 0. The order of the system is 2 (second-order system) since the highest power of s in the denominator is 2.

(c) It is a type-3 system (number of open-loop at the origin is 3).

The order of the system is 7 since the highest power of s in the denominator is 7.

• Example – A unity-feedback system is characterized by the following open-loop transfer function:

G s = 1

s(0.5s + 1)(0.2s + 1)

Determine the steady state errors for unit-step, unit-ramp and unit-acceleration input.

Solution – The position error constant, 𝐾_𝑝 = lim

𝑠⟶0𝐺 𝑠 = lim

𝑠⟶0

1

s(0.5s + 1)(0.2s + 1) = ∞

⇒ steady state error 𝑒_𝑠𝑠= 1

1 + 𝐾_𝑝 = 1

1 + ∞ = 0 The velocity error constant,

𝐾_𝑣 = lim

𝑠⟶0𝑠𝐺 𝑠 = lim

𝑠⟶0 𝑠. 1

s(0.5s + 1)(0.2s + 1) = 1

⇒ steady state error 𝑒_𝑠𝑠= 1

𝐾_𝑣 = 1

1 = 1 The acceleration error constant,

𝐾_𝑎 = lim

𝑠⟶0𝑠²𝐺 𝑠 = lim

𝑠⟶0 𝑠². 1

s(0.5s + 1)(0.2s + 1) = 0

⇒ steady state error 𝑒_𝑠𝑠= 1

𝐾 = 1

0 = ∞

• Example – The closed-loop transfer function of a unity- feedback system is given as:

C s

R s = Ks + b

(𝑠²+as + b)

Determine the open-loop transfer function G(s). Show that the steady-state error with unit-ramp input is given by ^a−K

b .

Solution – For the given unity feedback system, the closed-loop transfer function is

C s

R s = 𝐺 𝑠

1 + 𝐺 𝑠 = Ks + b (𝑠²+as + b)

Cross-multiplying the equations

(𝑠²+as + b)G s = Ks + b + Ks + b G s

⇒ G s (𝑠²+as + b − (Ks + b)] = (Ks + b)

⇒ G s = Ks + b s s + a a − K 𝐾_𝑣 = lim

𝑠⟶0𝑠𝐺 𝑠 = lim

𝑠⟶0𝑠. Ks + b

s s + a a − K = 𝑏 𝑎 − 𝐾

⇒ steady state error 𝑒_𝑠𝑠= 1

𝐾 = 𝑎 − 𝐾 𝑏

Solve the following problems:

1. For a unity feedback system having an open-loop transfer function:

G s = K s + 2 𝑠 + 3 𝑠²(𝑠²+8s + 15)

Determine (a) type of system (b) error constants 𝐾_𝑝, 𝐾_𝑣 and 𝐾_𝑎 and (c) steady-state error for unit-step, unit-ramp and unit-parabolic inputs.

2. A unity-feedback system is characterized by the following open-loop transfer function:

G s = 1

s(0.4s + 1)(0.15s + 1)

Determine the steady state errors for unit-step, unit-ramp and unit-acceleration input.

References

1. Control Systems by A. Anand Kumar.

2. Modern Control Engineering by Katsuhiko Ogata.

3. Control Systems by M. Gopal

4. NPTEL Video Lectures on Control Engineering (https://nptel.ac.in/courses/108/106/108106098/)

5. Control Systems Engineering by Norman S. Nise.