ME2142E Feedback Control Systems-Cheatsheet

ME2142E Feedback Control Systems Formula sheet ( )

( )

System Response – First-order systems Unit impulse

( )

Unit step

( )

(

Unit Ramp

( )

(

⁄

( ) ( )

) ⁄

)

Laplace Transform Time domain

S domain

( )

(

( ) (

)

( )

( ) ( )

)

( )

( ) ( )

∫

Time domain

( )

( ) ( )

S domain

( )⁄ ( ⁄

) ( ⁄ )

(

) (

.∫

( )

)

( )

Measures of transient performance Damped Nat freq.

√

Rise Time ( ⁄√

Max overshoot

⁄

⁄

Peak Time

√ (√

)

(

Settling Time

)

(

∫

)

∫(

(

)

)

⁄

(

)

)

Steady-State Errors Step input r=1 ⁄( )

Ramp input r=t

Type 0 system Type 1 system

0

⁄

Type 2 system

0

0

ess

( )

Accel input r=

⁄ ( )

( )

𝑒( )

𝑠

𝑠𝑅(𝑠) 𝐺(𝑠)

Block Diagram Algebra

Original Diagram

Equivalent Diagram

Original Diagram

Equivalent Diagram

𝑧

𝑅

𝑉𝑂 (𝑠) 𝑉𝑖 (𝑠) 𝑖𝑓 𝑠𝑒𝑟𝑖𝑒𝑠 𝑧

Routh’s Stability Criterion

(

)⁄

(

)⁄

)⁄

(

𝑅

𝐶𝑠

) ) 𝑅𝐶𝑠 𝐶𝑠

If all the coefficients, or the only one coefficient, in a derived row are zero, it means that there are roots of equal magnitude located symmetrically about the origin. For such cases, form an auxiliary polynomial with the coefficients of the row above the all-zero row and using the coefficients of the derivative of this polynomial to replace the all-zero row.

( ) ( )

(

𝑅 𝑋 𝑅𝐶𝑠 𝑅 (𝑅 𝐶 𝑠 𝑅 (𝑅 𝐶 𝑠

)⁄

Root Locus plot 1. 2.

Locate poles and zeros. Each branch starts from a pole and ends in a zero. If there are no zeros in the finite region, then the zeros are at infinity. No. of branches= No. of poles = order of characteristic equation. Loci exist on the real axis only to the left of an odd number of poles and/or zeros . Asymptotes angle:

∑

Asymptotes start from a point on real axis at ( ) ( )

At the break-in or breakaway points, ( )

4.

5.

Imaginary Axis Crossing, Two approaches: a. Use Routh Criteria to determine the value of K at which the system is critically stable. This is indicated by a value of zero in the first column but with no sign change in the first column of the Routh Array. b. Since the roots are on the imaginary axis, by letting in the characteristic equation and solve for and K. This is done by equating both the real and imaginary parts of the characteristic equation to zero. The angle of departure from a complex pole, is 180°+ (sum of angles between and all zeros) - (sum of angles between and all other poles).

6.

The angle of arrival at a complex pole,

characteristic equation:

, is 180°+ (sum of angles between

( ) ( )

( )

3.

A0066078X Lin Shaodun

( )

∑

and all other zeros) - (sum of angles between

( )

and all poles)

ME2142E Feedback Control Systems Formula sheet Bode plot ( )

Error=

| (

)|

√ (

(

( )

Gain

√

)|

Zero at Origin

Gain

function

)) | (

))

( (

Pole at Origin

( )

))

(

)

[

Zero

0dB at Slope = - 20 dB

( ( ( (

)) ]

))

Pole

Complex Pole

( )

( )

( )

0dB at Slope = 20 dB

Straight line

( (

0dB at Slope = 20 dB (

Phase

(

( )

0dB at Slope = - 40 dB

( ) ( ) ( )

) (

( )

0dB at Slope = - 20 dB

) )

Pure time delay

O dB line

( ) ( ) ( )

(

Polar plot Differentiator

function

Integrator

( )

(

Nyquist Gain Margin:

| (

First order Lead

2rd Order Lag

( )

( )

( )

(

Plot

First order lag

( )

)

)

) (

| (

)|

(

(

) (

) (

)|

(

(

)

) (

)

)

)

The gain crossover freq., wgc, is the freq. where the amplitude ratio is 1.The phase crossover freq., wpc , is the freq. where phase shift = -180o. ( )

( (

)( )(

) )

( )

(

)(

K pG

( )

)

(

)(

)

K pG

1 Y R N D 1  K pG 1  K pG 1  K pG

E  R Y 

A0066078X Lin Shaodun

K pG 1 1 R N D 1  K pG 1  K pG 1  K pG

( )

(

( )

)

K   U ( s)   K p  i  K d s  E ( s) s   t

u (t )  K p e(t )  K i  e(t )dt  K d 0

de(t ) dt

)