Review: transient and steadystate response; DC gain and the FVT Today s topic: systemmodeling diagrams; prototype 2ndorder system


 Suzan Hardy
 2 years ago
 Views:
Transcription
1 Plan of the Lecture Review: transient and steadystate response; DC gain and the FVT Today s topic: systemmodeling diagrams; prototype 2ndorder system
2 Plan of the Lecture Review: transient and steadystate response; DC gain and the FVT Today s topic: systemmodeling diagrams; prototype 2ndorder system Goal: develop a methodology for representing and analyzing systems by means of block diagrams; start analyzing a prototype 2ndorder system.
3 Plan of the Lecture Review: transient and steadystate response; DC gain and the FVT Today s topic: systemmodeling diagrams; prototype 2ndorder system Goal: develop a methodology for representing and analyzing systems by means of block diagrams; start analyzing a prototype 2ndorder system. Reading: FPE, Sections ; lab manual
4 System Modeling Diagrams large system decompose smaller blocks (subsystems) compose
5 System Modeling Diagrams large system decompose smaller blocks (subsystems) compose this is the core of systems theory
6 System Modeling Diagrams large system decompose smaller blocks (subsystems) compose this is the core of systems theory We will take smaller blocks from some given library and play with them to create/build more complicated systems.
7 AllIntegrator Diagrams Our library will consist of three building blocks: ẏ 1/s y (or s ) (or ) u 1 + u 2 y = u 1 u 2 u a y = au integrator summing junction constant gain
8 AllIntegrator Diagrams Our library will consist of three building blocks: ẏ 1/s y (or s ) (or ) u 1 + u 2 y = u 1 u 2 u a y = au integrator summing junction constant gain Two warnings: We can (and will) work either with u, y (time domain) or with U, (sdomain) will often go back and forth When working with block diagrams, we typically ignore initial conditions.
9 AllIntegrator Diagrams Our library will consist of three building blocks: ẏ 1/s y (or s ) (or ) u 1 + u 2 y = u 1 u 2 u a y = au integrator summing junction constant gain Two warnings: We can (and will) work either with u, y (time domain) or with U, (sdomain) will often go back and forth When working with block diagrams, we typically ignore initial conditions. This is the lowest level we will go to in lectures; in the labs, you will implement these blocks using op amps.
10 Example 1 Build an allintegrator diagram for ÿ = u s 2 = U
11 Example 1 Build an allintegrator diagram for ÿ = u s 2 = U This is obvious: u ẏ 1/s 1/s y or U s 1/s 1/s
12 Example 2 (building on Example 1) ÿ + a 1 ẏ + a 0 y = u s 2 + a 1 s + a 0 = U U(s) or (s) = s 2 + a 1 s + a 0
13 Example 2 (building on Example 1) ÿ + a 1 ẏ + a 0 y = u s 2 + a 1 s + a 0 = U Always solve for the highest derivative: U(s) or (s) = s 2 + a 1 s + a 0 ÿ = a 1 ẏ a 0 y + u }{{} =v
14 Example 2 (building on Example 1) ÿ + a 1 ẏ + a 0 y = u s 2 + a 1 s + a 0 = U Always solve for the highest derivative: U(s) or (s) = s 2 + a 1 s + a 0 ÿ = a 1 ẏ a 0 y + u }{{} =v u + v ẏ 1/s 1/s y a 1 a 0
15 Example 2 (building on Example 1) ÿ + a 1 ẏ + a 0 y = u s 2 + a 1 s + a 0 = U Always solve for the highest derivative: U(s) or (s) = s 2 + a 1 s + a 0 ÿ = a 1 ẏ a 0 y + u }{{} =v U + V s 1/s 1/s a 1 a 0
16 Example 3 Build an allintegrator diagram for a system with transfer function H(s) = b 1s + b 0 s 2 + a 1 s + a 0
17 Example 3 Build an allintegrator diagram for a system with transfer function H(s) = b 1s + b 0 s 2 + a 1 s + a 0 1 Step 1: decompose H(s) = s 2 (b 1 s + b 0 ) + a 1 s + a 0
18 Example 3 Build an allintegrator diagram for a system with transfer function H(s) = b 1s + b 0 s 2 + a 1 s + a 0 1 Step 1: decompose H(s) = s 2 (b 1 s + b 0 ) + a 1 s + a 0 U 1 X b s 2 1 s + b 0 + a 1 s + a 0 here, X is an auxiliary (or intermediate) signal
19 Example 3 Build an allintegrator diagram for a system with transfer function H(s) = b 1s + b 0 s 2 + a 1 s + a 0 1 Step 1: decompose H(s) = s 2 (b 1 s + b 0 ) + a 1 s + a 0 U 1 X b s 2 1 s + b 0 + a 1 s + a 0 here, X is an auxiliary (or intermediate) signal Note: b 0 + b 1 s involves differentiation, which we cannot implement using an allintegrator diagram. But we will see that we don t need to do it directly.
20 Example 3, continued Step 1: decompose H(s) = 1 s 2 + a 1 s + a 0 (b 1 s + b 0 ) U 1 X b s 2 1 s + b 0 + a 1 s + a 0
21 Example 3, continued Step 1: decompose H(s) = 1 s 2 + a 1 s + a 0 (b 1 s + b 0 ) U 1 X b s 2 1 s + b 0 + a 1 s + a 0 Step 2: The transformation U X is from Example 2:
22 Example 3, continued Step 1: decompose H(s) = 1 s 2 + a 1 s + a 0 (b 1 s + b 0 ) U 1 X b s 2 1 s + b 0 + a 1 s + a 0 Step 2: The transformation U X is from Example 2: U + s 2 X sx 1/s 1/s X a 1 a 0
23 Example 3, continued Step 3: now we notice that (s) = b 1 sx(s) + b 0 X(s), and both X and sx are available signals in our diagram. So:
24 Example 3, continued Step 3: now we notice that (s) = b 1 sx(s) + b 0 X(s), and both X and sx are available signals in our diagram. So: b 1 U + s 2 X + sx X 1/s 1/s + b 0 a 1 a 0
25 Example 3, continued Allintegrator diagram for H(s) = b 1s + b 0 s 2 + a 1 s + a 0 b 1 U + s 2 X + sx X 1/s 1/s + b 0 a 1 a 0
26 Example 3, continued Allintegrator diagram for H(s) = b 1s + b 0 s 2 + a 1 s + a 0 b 1 U + s 2 X + sx X 1/s 1/s + b 0 a 1 a 0 Can we write down a statespace model corresponding to this diagram?
27 Example 3, continued U + s 2 X + sx X 1/s 1/s + b 1 b 0 a 1 a 0 Statespace model:
28 Example 3, continued U + s 2 X + sx X 1/s 1/s + b 1 b 0 a 1 a 0 Statespace model: s 2 X = U a 1 sx a 0 X
29 Example 3, continued U + s 2 X + sx X 1/s 1/s + b 1 b 0 a 1 a 0 Statespace model: s 2 X = U a 1 sx a 0 X ẍ = a 1 ẋ a 0 x + u
30 Example 3, continued U + s 2 X + sx X 1/s 1/s + b 1 b 0 a 1 a 0 Statespace model: s 2 X = U a 1 sx a 0 X ẍ = a 1 ẋ a 0 x + u = b 1 sx + b 0 X
31 Example 3, continued U + s 2 X + sx X 1/s 1/s + b 1 b 0 a 1 a 0 Statespace model: s 2 X = U a 1 sx a 0 X ẍ = a 1 ẋ a 0 x + u = b 1 sx + b 0 X y = b 1 ẋ + b 0 x
32 Example 3, continued Statespace model: ẍ = a 1 ẋ a 0 x + u y = b 1 ẋ + b 0 x
33 Example 3, continued Statespace model: ẍ = a 1 ẋ a 0 x + u y = b 1 ẋ + b 0 x x 1 = x, x 2 = ẋ
34 Example 3, continued Statespace model: ẍ = a 1 ẋ a 0 x + u y = b 1 ẋ + b 0 x x 1 = x, x 2 = ẋ ) ( ) ( ) (ẋ1 0 1 x1 = + ẋ 2 a 0 a 1 x 2 ( ) 0 u 1
35 Example 3, continued Statespace model: ẍ = a 1 ẋ a 0 x + u y = b 1 ẋ + b 0 x x 1 = x, x 2 = ẋ ) ( ) ( ) (ẋ1 0 1 x1 = + ẋ 2 a 0 a 1 x 2 ( ) 0 u y = ( ) ( ) x b 1 0 b 1 1 x 2
36 Example 3, continued Statespace model: ẍ = a 1 ẋ a 0 x + u y = b 1 ẋ + b 0 x x 1 = x, x 2 = ẋ ) ( ) ( ) (ẋ1 0 1 x1 = + ẋ 2 a 0 a 1 x 2 ( ) 0 u y = ( ) ( ) x b 1 0 b 1 1 x 2 This is called controller canonical form.
37 Example 3, continued Statespace model: ẍ = a 1 ẋ a 0 x + u y = b 1 ẋ + b 0 x x 1 = x, x 2 = ẋ ) ( ) ( ) (ẋ1 0 1 x1 = + ẋ 2 a 0 a 1 x 2 ( ) 0 u y = ( ) ( ) x b 1 0 b 1 1 x 2 This is called controller canonical form. Easily generalizes to dimension > 1 The reason behind the name will be made clear later in the semester
38 Example 3, wrapup Allintegrator diagram for H(s) = b 1s + b 0 s 2 + a 1 s + a 0 b 1 U + s 2 X + sx X 1/s 1/s + b 0 a 1 a 0 Statespace model: ) ( ) ( ) (ẋ1 0 1 x1 = + ẋ 2 a 0 a 1 x 2 ( ) 0 u y = ( ) ( ) x b 1 0 b 1 1 x 2
39 Example 3, wrapup Allintegrator diagram for H(s) = b 1s + b 0 s 2 + a 1 s + a 0 b 1 U + s 2 X + sx X 1/s 1/s + b 0 a 1 a 0 Statespace model: ) ( ) ( ) (ẋ1 0 1 x1 = + ẋ 2 a 0 a 1 x 2 ( ) 0 u y = ( ) ( ) x b 1 0 b 1 1 x 2 Important: for a given H(s), the diagram is not unique. But, once we build a diagram, the statespace equations are unique (up to coordinate transformations).
40 Basic System Interconnections Now we will take this a level higher we will talk about building complex systems from smaller blocks, without worrying about how those blocks look on the inside (they could themselves be allintegrator diagrams, etc.)
41 Basic System Interconnections Now we will take this a level higher we will talk about building complex systems from smaller blocks, without worrying about how those blocks look on the inside (they could themselves be allintegrator diagrams, etc.) Block diagrams are an abstraction (they hide unnecessary lowlevel detail...)
42 Basic System Interconnections Now we will take this a level higher we will talk about building complex systems from smaller blocks, without worrying about how those blocks look on the inside (they could themselves be allintegrator diagrams, etc.) Block diagrams are an abstraction (they hide unnecessary lowlevel detail...) Block diagrams describe the flow of information
43 Basic System Interconnections: Series & Parallel
44 Basic System Interconnections: Series & Parallel Series connection
45 Basic System Interconnections: Series & Parallel Series connection U G 1 G 2 (G is common notation for t.f. s)
46 Basic System Interconnections: Series & Parallel Series connection U G 1 G 2 (G is common notation for t.f. s) U = G 1G 2 U G 1 G 2 (for SISO systems, the order of G 1 and G 2 does not matter)
47 Basic System Interconnections: Series & Parallel Series connection U G 1 G 2 (G is common notation for t.f. s) U = G 1G 2 Parallel connection U G 1 G 2 (for SISO systems, the order of G 1 and G 2 does not matter)
48 Basic System Interconnections: Series & Parallel Series connection U G 1 G 2 (G is common notation for t.f. s) U = G 1G 2 Parallel connection U G 1 G 2 (for SISO systems, the order of G 1 and G 2 does not matter) U G G 2
49 Basic System Interconnections: Series & Parallel Series connection U G 1 G 2 (G is common notation for t.f. s) U = G 1G 2 Parallel connection U G 1 G 2 (for SISO systems, the order of G 1 and G 2 does not matter) U G G 2 U = G U G 1 + G G 2
50 Basic System Interconnections: Negative Feedback R + U W G 1 Find the transfer function from R (reference) to G 2
51 Basic System Interconnections: Negative Feedback R + U W G 1 Find the transfer function from R (reference) to G 2 U = R W
52 Basic System Interconnections: Negative Feedback R + U W G 1 Find the transfer function from R (reference) to G 2 U = R W = G 1 U
53 Basic System Interconnections: Negative Feedback R + U W G 1 Find the transfer function from R (reference) to G 2 U = R W = G 1 U = G 1 (R W )
54 Basic System Interconnections: Negative Feedback R + U W G 1 Find the transfer function from R (reference) to G 2 U = R W = G 1 U = G 1 (R W ) = G 1 R G 1 G 2
55 Basic System Interconnections: Negative Feedback R + U W G 1 Find the transfer function from R (reference) to G 2 U = R W = G 1 U = G 1 (R W ) = G 1 R G 1 G 2 U = = G G 1 G 2 R G 1 1+G 1 G 2
56 Basic System Interconnections: Negative Feedback R + U W G 1 = = G G 1 G 2 R G 2
57 Basic System Interconnections: Negative Feedback R + U W G 1 = = G G 1 G 2 R G 2 The gain of a negative feedback loop: forward gain 1 + loop gain
58 Basic System Interconnections: Negative Feedback R + U W G 1 = = G G 1 G 2 R G 2 The gain of a negative feedback loop: forward gain 1 + loop gain This is an important relationship, easy to derive no need to memorize it.
59 Unity Feedback Other feedback configurations are also possible: + E R G 2 U G 1
60 Unity Feedback Other feedback configurations are also possible: + E R G 2 U G 1 This is called unity feedback no component on the feedback path.
61 Unity Feedback Other feedback configurations are also possible: + E R G 2 U G 1 This is called unity feedback no component on the feedback path. Common structure (saw this in Lecture 1): R = reference U = control input = output E = error G 1 = plant (also denoted by P ) G 2 = controller or compensator (also denoted by C or K)
62 Unity Feedback R + E G 2 U G 1
63 Unity Feedback R + E G 2 U G 1 Let s practice with deriving transfer functions: forward gain 1 + loop gain
64 Unity Feedback R + E G 2 U G 1 Let s practice with deriving transfer functions: Reference R to output : forward gain 1 + loop gain
65 Unity Feedback R + E G 2 U G 1 Let s practice with deriving transfer functions: Reference R to output : R = G 1G G 1 G 2 forward gain 1 + loop gain
66 Unity Feedback R + E G 2 U G 1 Let s practice with deriving transfer functions: Reference R to output : R = G 1G G 1 G 2 forward gain 1 + loop gain Reference R to control input U:
67 Unity Feedback R + E G 2 U G 1 Let s practice with deriving transfer functions: Reference R to output : R = G 1G G 1 G 2 forward gain 1 + loop gain Reference R to control input U: U R = G G 1 G 2
68 Unity Feedback R + E G 2 U G 1 Let s practice with deriving transfer functions: Reference R to output : R = G 1G G 1 G 2 forward gain 1 + loop gain Reference R to control input U: Error E to output : U R = G G 1 G 2
69 Unity Feedback R + E G 2 U G 1 Let s practice with deriving transfer functions: Reference R to output : R = G 1G G 1 G 2 forward gain 1 + loop gain Reference R to control input U: Error E to output : U R = G G 1 G 2 E = G 1G 2 (no feedback path)
70 Block Diagram Reduction Given a complicated diagram involving series, parallel, and feedback interconnections, we often want to write down an overall transfer function from one of the variables to another.
71 Block Diagram Reduction Given a complicated diagram involving series, parallel, and feedback interconnections, we often want to write down an overall transfer function from one of the variables to another. This requires lots of practice: read FPE, Section 3.2 for examples.
72 Block Diagram Reduction Given a complicated diagram involving series, parallel, and feedback interconnections, we often want to write down an overall transfer function from one of the variables to another. This requires lots of practice: read FPE, Section 3.2 for examples. General strategy: Name all the variables in the diagram Write down as many relationships between these variables as you can Learn to recognize series, parallel, and feedback interconnections Replace them by their equivalents Repeat
73 Prototype 2ndOrder System So far, we have only seen transfer functions that have either real poles or purely imaginary poles: 1 s + a, 1 (s + a)(s + b), 1 s 2 + ω 2
74 Prototype 2ndOrder System So far, we have only seen transfer functions that have either real poles or purely imaginary poles: 1 s + a, 1 (s + a)(s + b), 1 s 2 + ω 2 We also need to consider the case of complex poles, i.e., ones that have Re(s) 0 and Im(s) 0.
75 Prototype 2ndOrder System So far, we have only seen transfer functions that have either real poles or purely imaginary poles: 1 s + a, 1 (s + a)(s + b), 1 s 2 + ω 2 We also need to consider the case of complex poles, i.e., ones that have Re(s) 0 and Im(s) 0. For now, we will only look at secondorder systems, but this will be sufficient to develop some nontrivial intuition (dominant poles).
76 Prototype 2ndOrder System So far, we have only seen transfer functions that have either real poles or purely imaginary poles: 1 s + a, 1 (s + a)(s + b), 1 s 2 + ω 2 We also need to consider the case of complex poles, i.e., ones that have Re(s) 0 and Im(s) 0. For now, we will only look at secondorder systems, but this will be sufficient to develop some nontrivial intuition (dominant poles). Plus, you will need this for Lab 1.
77 Prototype 2ndOrder System Consider the following transfer function: ω 2 n H(s) = s 2 + 2ζω n s + ωn 2
78 Prototype 2ndOrder System Consider the following transfer function: H(s) = ω 2 n s 2 + 2ζω n s + ω 2 n Comments:
79 Prototype 2ndOrder System Consider the following transfer function: H(s) = ω 2 n s 2 + 2ζω n s + ω 2 n Comments: ζ > 0, ω n > 0 are arbitrary parameters
80 Prototype 2ndOrder System Consider the following transfer function: H(s) = ω 2 n s 2 + 2ζω n s + ω 2 n Comments: ζ > 0, ω n > 0 are arbitrary parameters the denominator is a general 2nddegree monic polynomial, just written in a weird way
81 Prototype 2ndOrder System Consider the following transfer function: H(s) = ω 2 n s 2 + 2ζω n s + ω 2 n Comments: ζ > 0, ω n > 0 are arbitrary parameters the denominator is a general 2nddegree monic polynomial, just written in a weird way H(s) is normalized to have DC gain = 1 (provided DC gain exists)
82 Prototype 2ndOrder System ω 2 n H(s) = s 2 + 2ζω n s + ωn 2
83 Prototype 2ndOrder System ω 2 n H(s) = s 2 + 2ζω n s + ωn 2 By the quadratic formula, the poles are: s = ζω n ± ω n ζ 2 1 = ω n (ζ ± ) ζ 2 1
84 Prototype 2ndOrder System ω 2 n H(s) = s 2 + 2ζω n s + ωn 2 By the quadratic formula, the poles are: s = ζω n ± ω n ζ 2 1 = ω n (ζ ± ) ζ 2 1 The nature of the poles changes depending on ζ:
85 Prototype 2ndOrder System ω 2 n H(s) = s 2 + 2ζω n s + ωn 2 By the quadratic formula, the poles are: s = ζω n ± ω n ζ 2 1 = ω n (ζ ± ) ζ 2 1 The nature of the poles changes depending on ζ: ζ > 1
86 Prototype 2ndOrder System ω 2 n H(s) = s 2 + 2ζω n s + ωn 2 By the quadratic formula, the poles are: s = ζω n ± ω n ζ 2 1 = ω n (ζ ± ) ζ 2 1 The nature of the poles changes depending on ζ: ζ > 1 both poles are real and negative
87 Prototype 2ndOrder System ω 2 n H(s) = s 2 + 2ζω n s + ωn 2 By the quadratic formula, the poles are: s = ζω n ± ω n ζ 2 1 = ω n (ζ ± ) ζ 2 1 The nature of the poles changes depending on ζ: ζ > 1 ζ = 1 both poles are real and negative
88 Prototype 2ndOrder System ω 2 n H(s) = s 2 + 2ζω n s + ωn 2 By the quadratic formula, the poles are: s = ζω n ± ω n ζ 2 1 = ω n (ζ ± ) ζ 2 1 The nature of the poles changes depending on ζ: ζ > 1 ζ = 1 both poles are real and negative one negative pole
89 Prototype 2ndOrder System ω 2 n H(s) = s 2 + 2ζω n s + ωn 2 By the quadratic formula, the poles are: s = ζω n ± ω n ζ 2 1 = ω n (ζ ± ) ζ 2 1 The nature of the poles changes depending on ζ: ζ > 1 ζ = 1 ζ < 1 both poles are real and negative one negative pole
90 Prototype 2ndOrder System ω 2 n H(s) = s 2 + 2ζω n s + ωn 2 By the quadratic formula, the poles are: s = ζω n ± ω n ζ 2 1 = ω n (ζ ± ) ζ 2 1 The nature of the poles changes depending on ζ: ζ > 1 ζ = 1 ζ < 1 both poles are real and negative one negative pole two complex poles with negative real parts
91 Prototype 2ndOrder System ω 2 n H(s) = s 2 + 2ζω n s + ωn 2 By the quadratic formula, the poles are: s = ζω n ± ω n ζ 2 1 = ω n (ζ ± ) ζ 2 1 The nature of the poles changes depending on ζ: ζ > 1 ζ = 1 ζ < 1 both poles are real and negative one negative pole two complex poles with negative real parts s = σ ± jω d where σ = ζω n, ω d = ω n 1 ζ 2
92 Prototype 2ndOrder System H(s) = ω 2 n s 2 + 2ζω n s + ωn 2, ζ < 1
93 Prototype 2ndOrder System The poles are H(s) = ω 2 n s 2 + 2ζω n s + ωn 2, ζ < 1 s = ζω n ± jω n 1 ζ 2 = σ ± jω d
94 Prototype 2ndOrder System The poles are H(s) = ω 2 n s 2 + 2ζω n s + ωn 2, ζ < 1 s = ζω n ± jω n 1 ζ 2 = σ ± jω d Im! d =! n p 1 2 Note that! n =! n ' 0 Re σ 2 + ω 2 d = ζ2 ω 2 n + ω 2 n ζ 2 ω 2 n = ω 2 n cos ϕ = ζω n ω n = ζ
95 2ndOrder Response Let s compute the system s impulse and step response:
96 2ndOrder Response Let s compute the system s impulse and step response: H(s) = ω 2 n s 2 + 2ζω n s + ω 2 n ω 2 n = (s + σ) 2 + ωd 2
97 2ndOrder Response Let s compute the system s impulse and step response: H(s) = ω 2 n s 2 + 2ζω n s + ω 2 n = ω 2 n (s + σ) 2 + ω 2 d Impulse response: { h(t) = L 1 {H(s)} = L 1 ω 2 n (s + σ) 2 + ω 2 d }
98 2ndOrder Response Let s compute the system s impulse and step response: H(s) = ω 2 n s 2 + 2ζω n s + ω 2 n = ω 2 n (s + σ) 2 + ω 2 d Impulse response: { } (ω h(t) = L 1 {H(s)} = L 1 2 n /ω d )ω d (s + σ) 2 + ωd 2
99 2ndOrder Response Let s compute the system s impulse and step response: H(s) = ω 2 n s 2 + 2ζω n s + ω 2 n = ω 2 n (s + σ) 2 + ω 2 d Impulse response: { } (ω h(t) = L 1 {H(s)} = L 1 2 n /ω d )ω d (s + σ) 2 + ωd 2 = ω2 n ω d e σt sin(ω d t)
100 2ndOrder Response Let s compute the system s impulse and step response: H(s) = ω 2 n s 2 + 2ζω n s + ω 2 n = ω 2 n (s + σ) 2 + ω 2 d Impulse response: { } (ω h(t) = L 1 {H(s)} = L 1 2 n /ω d )ω d (s + σ) 2 + ωd 2 = ω2 n ω d e σt sin(ω d t) (table, # 20)
101 2ndOrder Response Let s compute the system s impulse and step response: H(s) = ω 2 n s 2 + 2ζω n s + ω 2 n = ω 2 n (s + σ) 2 + ω 2 d Impulse response: { } (ω h(t) = L 1 {H(s)} = L 1 2 n /ω d )ω d (s + σ) 2 + ωd 2 Step response: = ω2 n ω d e σt sin(ω d t) (table, # 20) { } { H(s) L 1 = L 1 ω 2 } n s s[(s + σ) 2 + ωd 2]
102 2ndOrder Response Let s compute the system s impulse and step response: H(s) = ω 2 n s 2 + 2ζω n s + ω 2 n = ω 2 n (s + σ) 2 + ω 2 d Impulse response: { } (ω h(t) = L 1 {H(s)} = L 1 2 n /ω d )ω d (s + σ) 2 + ωd 2 Step response: = ω2 n ω d e σt sin(ω d t) (table, # 20) { } { H(s) L 1 = L 1 σ 2 + ω 2 } d s s[(s + σ) 2 + ωd 2]
103 2ndOrder Response Let s compute the system s impulse and step response: H(s) = ω 2 n s 2 + 2ζω n s + ω 2 n = ω 2 n (s + σ) 2 + ω 2 d Impulse response: { } (ω h(t) = L 1 {H(s)} = L 1 2 n /ω d )ω d (s + σ) 2 + ωd 2 = ω2 n ω d e σt sin(ω d t) (table, # 20) Step response: { } { H(s) L 1 = L 1 σ 2 + ω 2 } d s s[(s + σ) 2 + ωd 2] = 1 e (cos(ω σt d t) + σ ) sin(ω d t) ω d
104 2ndOrder Response Let s compute the system s impulse and step response: H(s) = ω 2 n s 2 + 2ζω n s + ω 2 n = ω 2 n (s + σ) 2 + ω 2 d Impulse response: { } (ω h(t) = L 1 {H(s)} = L 1 2 n /ω d )ω d (s + σ) 2 + ωd 2 = ω2 n ω d e σt sin(ω d t) (table, # 20) Step response: { } { H(s) L 1 = L 1 σ 2 + ω 2 } d s s[(s + σ) 2 + ωd 2] = 1 e (cos(ω σt d t) + σ ) sin(ω d t) (table, #21) ω d
105 2ndOrder Step Response ω 2 n H(s) = s 2 + 2ζω n s + ωn 2 = (s + σ) 2 + ωd 2 u(t) = 1(t) y(t) = 1 e (cos(ω σt d t) + σ ) sin(ω d t) ω d ω 2 n where σ = ζω n and ω d = ω n 1 ζ 2 (damped frequency)
106 2ndOrder Step Response ω 2 n H(s) = s 2 + 2ζω n s + ωn 2 = (s + σ) 2 + ωd 2 u(t) = 1(t) y(t) = 1 e (cos(ω σt d t) + σ ) sin(ω d t) ω d ω 2 n where σ = ζω n and ω d = ω n 1 ζ 2 (damped frequency) y t The parameter ζ is called the damping ratio ζ > 1: system is overdamped 0.5 Ζ 0.1 Ζ 0.9 Ζ t ζ < 1: system is underdamped ζ = 0: no damping (ω d = ω n )
Systems Analysis and Control
Systems Analysis and Control Matthew M. Peet Illinois Institute of Technology Lecture 8: Response Characteristics Overview In this Lecture, you will learn: Characteristics of the Response Stability Real
More informationPlan of the Lecture. Goal: wrap up lead and lag control; start looking at frequency response as an alternative methodology for control systems design.
Plan of the Lecture Review: design using Root Locus; dynamic compensation; PD and lead control Today s topic: PI and lag control; introduction to frequencyresponse design method Goal: wrap up lead and
More informationTime Response Analysis (Part II)
Time Response Analysis (Part II). A critically damped, continuoustime, second order system, when sampled, will have (in Z domain) (a) A simple pole (b) Double pole on real axis (c) Double pole on imaginary
More informationSystems Analysis and Control
Systems Analysis and Control Matthew M. Peet Arizona State University Lecture 21: Stability Margins and Closing the Loop Overview In this Lecture, you will learn: Closing the Loop Effect on Bode Plot Effect
More informationSystems Analysis and Control
Systems Analysis and Control Matthew M. Peet Arizona State University Lecture 8: Response Characteristics Overview In this Lecture, you will learn: Characteristics of the Response Stability Real Poles
More informationCourse roadmap. Step response for 2ndorder system. Step response for 2ndorder system
ME45: Control Systems Lecture Time response of ndorder systems Prof. Clar Radcliffe and Prof. Jongeun Choi Department of Mechanical Engineering Michigan State University Modeling Laplace transform Transfer
More informationAN INTRODUCTION TO THE CONTROL THEORY
OpenLoop controller An OpenLoop (OL) controller is characterized by no direct connection between the output of the system and its input; therefore external disturbance, nonlinear dynamics and parameter
More informationSTABILITY. Have looked at modeling dynamic systems using differential equations. and used the Laplace transform to help find step and impulse
SIGNALS AND SYSTEMS: PAPER 3C1 HANDOUT 4. Dr David Corrigan 1. Electronic and Electrical Engineering Dept. corrigad@tcd.ie www.sigmedia.tv STABILITY Have looked at modeling dynamic systems using differential
More informationOneSided Laplace Transform and Differential Equations
OneSided Laplace Transform and Differential Equations As in the dcretetime case, the onesided transform allows us to take initial conditions into account. Preliminaries The onesided Laplace transform
More informationTransient Response of a SecondOrder System
Transient Response of a SecondOrder System ECEN 830 Spring 01 1. Introduction In connection with this experiment, you are selecting the gains in your feedback loop to obtain a wellbehaved closedloop
More informationNotes for ECE320. Winter by R. Throne
Notes for ECE3 Winter 45 by R. Throne Contents Table of Laplace Transforms 5 Laplace Transform Review 6. Poles and Zeros.................................... 6. Proper and Strictly Proper Transfer Functions...................
More informationTopic # Feedback Control Systems
Topic #1 16.31 Feedback Control Systems Motivation Basic Linear System Response Fall 2007 16.31 1 1 16.31: Introduction r(t) e(t) d(t) y(t) G c (s) G(s) u(t) Goal: Design a controller G c (s) so that the
More informationHomework 7  Solutions
Homework 7  Solutions Note: This homework is worth a total of 48 points. 1. Compensators (9 points) For a unity feedback system given below, with G(s) = K s(s + 5)(s + 11) do the following: (c) Find the
More informationSystems Analysis and Control
Systems Analysis and Control Matthew M. Peet Illinois Institute of Technology Lecture 2: Drawing Bode Plots, Part 2 Overview In this Lecture, you will learn: Simple Plots Real Zeros Real Poles Complex
More informationINTRODUCTION TO DIGITAL CONTROL
ECE4540/5540: Digital Control Systems INTRODUCTION TO DIGITAL CONTROL.: Introduction In ECE450/ECE550 Feedback Control Systems, welearnedhow to make an analog controller D(s) to control a lineartimeinvariant
More informationTransient response via gain adjustment. Consider a unity feedback system, where G(s) = 2. The closed loop transfer function is. s 2 + 2ζωs + ω 2 n
Design via frequency response Transient response via gain adjustment Consider a unity feedback system, where G(s) = ωn 2. The closed loop transfer function is s(s+2ζω n ) T(s) = ω 2 n s 2 + 2ζωs + ω 2
More informationLecture 5: Linear Systems. Transfer functions. Frequency Domain Analysis. Basic Control Design.
ISS0031 Modeling and Identification Lecture 5: Linear Systems. Transfer functions. Frequency Domain Analysis. Basic Control Design. Aleksei Tepljakov, Ph.D. September 30, 2015 Linear Dynamic Systems Definition
More informationOutline. Classical Control. Lecture 1
Outline Outline Outline 1 Introduction 2 Prerequisites Block diagram for system modeling Modeling Mechanical Electrical Outline Introduction Background Basic Systems Models/Transfers functions 1 Introduction
More informationTopic # Feedback Control
Topic #5 6.3 Feedback Control StateSpace Systems Fullstate Feedback Control How do we change the poles of the statespace system? Or,evenifwecanchangethepolelocations. Where do we put the poles? Linear
More informationCYBER EXPLORATION LABORATORY EXPERIMENTS
CYBER EXPLORATION LABORATORY EXPERIMENTS 1 2 Cyber Exploration oratory Experiments Chapter 2 Experiment 1 Objectives To learn to use MATLAB to: (1) generate polynomial, (2) manipulate polynomials, (3)
More informationProblem Set 3: Solution Due on Mon. 7 th Oct. in class. Fall 2013
EE 56: Digital Control Systems Problem Set 3: Solution Due on Mon 7 th Oct in class Fall 23 Problem For the causal LTI system described by the difference equation y k + 2 y k = x k, () (a) By first finding
More informationChapter 12. Feedback Control Characteristics of Feedback Systems
Chapter 1 Feedbac Control Feedbac control allows a system dynamic response to be modified without changing any system components. Below, we show an openloop system (a system without feedbac) and a closedloop
More informationControl Systems I Lecture 10: System Specifications
Control Systems I Lecture 10: System Specifications Readings: Guzzella, Chapter 10 Emilio Frazzoli Institute for Dynamic Systems and Control DMAVT ETH Zürich November 24, 2017 E. Frazzoli (ETH) Lecture
More informationControls Problems for Qualifying Exam  Spring 2014
Controls Problems for Qualifying Exam  Spring 2014 Problem 1 Consider the system block diagram given in Figure 1. Find the overall transfer function T(s) = C(s)/R(s). Note that this transfer function
More informationChapter 5 HW Solution
Chapter 5 HW Solution Review Questions. 1, 6. As usual, I think these are just a matter of text lookup. 1. Name the four components of a block diagram for a linear, timeinvariant system. Let s see, I
More informationChapter 7: Time Domain Analysis
Chapter 7: Time Domain Analysis Samantha Ramirez Preview Questions How do the system parameters affect the response? How are the parameters linked to the system poles or eigenvalues? How can Laplace transforms
More informationControl Systems I. Lecture 5: Transfer Functions. Readings: Emilio Frazzoli. Institute for Dynamic Systems and Control DMAVT ETH Zürich
Control Systems I Lecture 5: Transfer Functions Readings: Emilio Frazzoli Institute for Dynamic Systems and Control DMAVT ETH Zürich October 20, 2017 E. Frazzoli (ETH) Lecture 5: Control Systems I 20/10/2017
More informationCHAPTER 5 : REDUCTION OF MULTIPLE SUBSYSTEMS
CHAPTER 5 : REDUCTION OF MULTIPLE SUBSYSTEMS Objectives Students should be able to: Reduce a block diagram of multiple subsystems to a single block representing the transfer function from input to output
More informationModule 3F2: Systems and Control EXAMPLES PAPER 2 ROOTLOCUS. Solutions
Cambridge University Engineering Dept. Third Year Module 3F: Systems and Control EXAMPLES PAPER ROOTLOCUS Solutions. (a) For the system L(s) = (s + a)(s + b) (a, b both real) show that the rootlocus
More informationTransform Solutions to LTI Systems Part 3
Transform Solutions to LTI Systems Part 3 Example of second order system solution: Same example with increased damping: k=5 N/m, b=6 Ns/m, F=2 N, m=1 Kg Given x(0) = 0, x (0) = 0, find x(t). The revised
More informationECE317 : Feedback and Control
ECE317 : Feedback and Control Lecture : Steadystate error Dr. Richard Tymerski Dept. of Electrical and Computer Engineering Portland State University 1 Course roadmap Modeling Analysis Design Laplace
More informationRaktim Bhattacharya. . AERO 422: Active Controls for Aerospace Vehicles. Dynamic Response
.. AERO 422: Active Controls for Aerospace Vehicles Dynamic Response Raktim Bhattacharya Laboratory For Uncertainty Quantification Aerospace Engineering, Texas A&M University. . Previous Class...........
More informationEE C128 / ME C134 Fall 2014 HW 6.2 Solutions. HW 6.2 Solutions
EE C28 / ME C34 Fall 24 HW 6.2 Solutions. PI Controller For the system G = K (s+)(s+3)(s+8) HW 6.2 Solutions in negative feedback operating at a damping ratio of., we are going to design a PI controller
More informationIntroduction to Feedback Control
Introduction to Feedback Control Control System Design Why Control? OpenLoop vs ClosedLoop (Feedback) Why Use Feedback Control? ClosedLoop Control System Structure Elements of a Feedback Control System
More informationGEORGIA INSTITUTE OF TECHNOLOGY SCHOOL of ELECTRICAL & COMPUTER ENGINEERING FINAL EXAM. COURSE: ECE 3084A (Prof. Michaels)
GEORGIA INSTITUTE OF TECHNOLOGY SCHOOL of ELECTRICAL & COMPUTER ENGINEERING FINAL EXAM DATE: 09Dec13 COURSE: ECE 3084A (Prof. Michaels) NAME: STUDENT #: LAST, FIRST Write your name on the front page
More informationControl of Manufacturing Processes
Control of Manufacturing Processes Subject 2.830 Spring 2004 Lecture #18 Basic Control Loop Analysis" April 15, 2004 Revisit Temperature Control Problem τ dy dt + y = u τ = time constant = gain y ss =
More informationAPPLICATIONS FOR ROBOTICS
Version: 1 CONTROL APPLICATIONS FOR ROBOTICS TEX d: Feb. 17, 214 PREVIEW We show that the transfer function and conditions of stability for linear systems can be studied using Laplace transforms. Table
More informationDynamic Response. Assoc. Prof. Enver Tatlicioglu. Department of Electrical & Electronics Engineering Izmir Institute of Technology.
Dynamic Response Assoc. Prof. Enver Tatlicioglu Department of Electrical & Electronics Engineering Izmir Institute of Technology Chapter 3 Assoc. Prof. Enver Tatlicioglu (EEE@IYTE) EE362 Feedback Control
More informationDr Ian R. Manchester Dr Ian R. Manchester AMME 3500 : Review
Week Date Content Notes 1 6 Mar Introduction 2 13 Mar Frequency Domain Modelling 3 20 Mar Transient Performance and the splane 4 27 Mar Block Diagrams Assign 1 Due 5 3 Apr Feedback System Characteristics
More informationTime Response of Systems
Chapter 0 Time Response of Systems 0. Some Standard Time Responses Let us try to get some impulse time responses just by inspection: Poles F (s) f(t) splane Time response p =0 s p =0,p 2 =0 s 2 t p =
More informationControl of Manufacturing Processes
Control of Manufacturing Processes Subject 2.830 Spring 2004 Lecture #19 Position Control and Root Locus Analysis" April 22, 2004 The Position Servo Problem, reference position NC Control Robots Injection
More informationProportional plus Integral (PI) Controller
Proportional plus Integral (PI) Controller 1. A pole is placed at the origin 2. This causes the system type to increase by 1 and as a result the error is reduced to zero. 3. Originally a point A is on
More informationControl Systems I. Lecture 6: Poles and Zeros. Readings: Emilio Frazzoli. Institute for Dynamic Systems and Control DMAVT ETH Zürich
Control Systems I Lecture 6: Poles and Zeros Readings: Emilio Frazzoli Institute for Dynamic Systems and Control DMAVT ETH Zürich October 27, 2017 E. Frazzoli (ETH) Lecture 6: Control Systems I 27/10/2017
More informationLaplace Transform Analysis of Signals and Systems
Laplace Transform Analysis of Signals and Systems Transfer Functions Transfer functions of CT systems can be found from analysis of Differential Equations Block Diagrams Circuit Diagrams 5/10/04 M. J.
More informationAMME3500: System Dynamics & Control
Stefan B. Williams May, 211 AMME35: System Dynamics & Control Assignment 4 Note: This assignment contributes 15% towards your final mark. This assignment is due at 4pm on Monday, May 3 th during Week 13
More informationEssence of the Root Locus Technique
Essence of the Root Locus Technique In this chapter we study a method for finding locations of system poles. The method is presented for a very general setup, namely for the case when the closedloop
More informationIntroduction & Laplace Transforms Lectures 1 & 2
Introduction & Lectures 1 & 2, Professor Department of Electrical and Computer Engineering Colorado State University Fall 2016 Control System Definition of a Control System Group of components that collectively
More informationDiscrete Systems. Step response and pole locations. Mark Cannon. Hilary Term Lecture
Discrete Systems Mark Cannon Hilary Term 22  Lecture 4 Step response and pole locations 4  Review Definition of transform: U() = Z{u k } = u k k k= Discrete transfer function: Y () U() = G() = Z{g k},
More informationIntroduction to Modern Control MT 2016
CDT Autonomous and Intelligent Machines & Systems Introduction to Modern Control MT 2016 Alessandro Abate Lecture 2 Firstorder ordinary differential equations (ODE) Solution of a linear ODE Hints to nonlinear
More informationClassify a transfer function to see which order or ramp it can follow and with which expected error.
Dr. J. Tani, Prof. Dr. E. Frazzoli 505900 Control Systems I (Autumn 208) Exercise Set 0 Topic: Specifications for Feedback Systems Discussion: 30.. 208 Learning objectives: The student can grizzi@ethz.ch,
More informationDigital Control: Part 2. ENGI 7825: Control Systems II Andrew Vardy
Digital Control: Part 2 ENGI 7825: Control Systems II Andrew Vardy Mapping the splane onto the zplane We re almost ready to design a controller for a DT system, however we will have to consider where
More information16.31 Homework 2 Solution
16.31 Homework Solution Prof. S. R. Hall Issued: September, 6 Due: September 9, 6 Problem 1. (Dominant Pole Locations) [FPE 3.36 (a),(c),(d), page 161]. Consider the second order system ωn H(s) = (s/p
More informationEE C128 / ME C134 Fall 2014 HW 9 Solutions. HW 9 Solutions. 10(s + 3) s(s + 2)(s + 5) G(s) =
1. Pole Placement Given the following openloop plant, HW 9 Solutions G(s) = 1(s + 3) s(s + 2)(s + 5) design the statevariable feedback controller u = Kx + r, where K = [k 1 k 2 k 3 ] is the feedback
More informationCourse Summary. The course cannot be summarized in one lecture.
Course Summary Unit 1: Introduction Unit 2: Modeling in the Frequency Domain Unit 3: Time Response Unit 4: Block Diagram Reduction Unit 5: Stability Unit 6: SteadyState Error Unit 7: Root Locus Techniques
More informationGoals for today 2.004
Goals for today Block diagrams revisited Block diagram components Block diagram cascade Summing and pickoff junctions Feedback topology Negative vs positive feedback Example of a system with feedback Derivation
More informationLast week: analysis of pinionrack w velocity feedback
Last week: analysis of pinionrack w velocity feedback Calculation of the steady state error Transfer function: V (s) V ref (s) = 0.362K s +2+0.362K Step input: V ref (s) = s Output: V (s) = s 0.362K s
More informationEE3CL4: Introduction to Linear Control Systems
1 / 17 EE3CL4: Introduction to Linear Control Systems Section 7: McMaster University Winter 2018 2 / 17 Outline 1 4 / 17 Cascade compensation Throughout this lecture we consider the case of H(s) = 1. We
More informationROOT LOCUS. Consider the system. Root locus presents the poles of the closedloop system when the gain K changes from 0 to. H(s) H ( s) = ( s)
C1 ROOT LOCUS Consider the system R(s) E(s) C(s) + K G(s)  H(s) C(s) R(s) = K G(s) 1 + K G(s) H(s) Root locus presents the poles of the closedloop system when the gain K changes from 0 to 1+ K G ( s)
More informationSoftware Engineering 3DX3. Slides 8: Root Locus Techniques
Software Engineering 3DX3 Slides 8: Root Locus Techniques Dr. Ryan Leduc Department of Computing and Software McMaster University Material based on Control Systems Engineering by N. Nise. c 2006, 2007
More informationResponse to a pure sinusoid
Harvard University Division of Engineering and Applied Sciences ES 145/215  INTRODUCTION TO SYSTEMS ANALYSIS WITH PHYSIOLOGICAL APPLICATIONS Fall Lecture 14: The Bode Plot Response to a pure sinusoid
More informationPoles, Zeros and System Response
Time Response After the engineer obtains a mathematical representation of a subsystem, the subsystem is analyzed for its transient and steady state responses to see if these characteristics yield the desired
More informationLec 6: State Feedback, Controllability, Integral Action
Lec 6: State Feedback, Controllability, Integral Action November 22, 2017 Lund University, Department of Automatic Control Controllability and Observability Example of Kalman decomposition 1 s 1 x 10 x
More informationSECTION 2: BLOCK DIAGRAMS & SIGNAL FLOW GRAPHS
SECTION 2: BLOCK DIAGRAMS & SIGNAL FLOW GRAPHS MAE 4421 Control of Aerospace & Mechanical Systems 2 Block Diagram Manipulation Block Diagrams 3 In the introductory section we saw examples of block diagrams
More informationReview: stability; Routh Hurwitz criterion Today s topic: basic properties and benefits of feedback control
Plan of the Lecture Review: stability; Routh Hurwitz criterion Today s topic: basic properties and benefits of feedback control Goal: understand the difference between openloop and closedloop (feedback)
More informationTransient Stability Analysis with PowerWorld Simulator
Transient Stability Analysis with PowerWorld Simulator T11: Single Machine Infinite Bus Modeling (SMIB) 21 South First Street Champaign, Illinois 6182 +1 (217) 384.633 support@powerworld.com http://www.powerworld.com
More information12.7 Steady State Error
Lecture Notes on Control Systems/D. Ghose/01 106 1.7 Steady State Error For first order systems we have noticed an overall improvement in performance in terms of rise time and settling time. But there
More informationControl Systems I. Lecture 7: Feedback and the Root Locus method. Readings: Jacopo Tani. Institute for Dynamic Systems and Control DMAVT ETH Zürich
Control Systems I Lecture 7: Feedback and the Root Locus method Readings: Jacopo Tani Institute for Dynamic Systems and Control DMAVT ETH Zürich November 2, 2018 J. Tani, E. Frazzoli (ETH) Lecture 7:
More informationModeling and Experimentation: MassSpringDamper System Dynamics
Modeling and Experimentation: MassSpringDamper System Dynamics Prof. R.G. Longoria Department of Mechanical Engineering The University of Texas at Austin July 20, 2014 Overview 1 This lab is meant to
More informationLecture 9 Timedomain properties of convolution systems
EE 12 spring 2122 Handout #18 Lecture 9 Timedomain properties of convolution systems impulse response step response fading memory DC gain peak gain stability 9 1 Impulse response if u = δ we have y(t)
More informationControl Systems. University Questions
University Questions UNIT1 1. Distinguish between open loop and closed loop control system. Describe two examples for each. (10 Marks), Jan 2009, June 12, Dec 11,July 08, July 2009, Dec 2010 2. Write
More informationRaktim Bhattacharya. . AERO 632: Design of Advance Flight Control System. Preliminaries
. AERO 632: of Advance Flight Control System. Preliminaries Raktim Bhattacharya Laboratory For Uncertainty Quantification Aerospace Engineering, Texas A&M University. Preliminaries Signals & Systems Laplace
More informationControl System Design
ELEC ENG 4CL4: Control System Design Notes for Lecture #4 Monday, January 13, 2003 Dr. Ian C. Bruce Room: CRL229 Phone ext.: 26984 Email: ibruce@mail.ece.mcmaster.ca Impulse and Step Responses of ContinuousTime
More informationEE480.3 Digital Control Systems. Part 7. Controller Design I.  Pole Assignment Method  State Estimation
EE480.3 Digital Control Systems Part 7. Controller Design I.  Pole Assignment Method  State Estimation Kunio Takaya Electrical and Computer Engineering University of Saskatchewan February 10, 2010 **
More informationDefinition of the Laplace transform. 0 x(t)e st dt
Definition of the Laplace transform Bilateral Laplace Transform: X(s) = x(t)e st dt Unilateral (or onesided) Laplace Transform: X(s) = 0 x(t)e st dt ECE352 1 Definition of the Laplace transform (cont.)
More informationEL2520 Control Theory and Practice
So far EL2520 Control Theory and Practice r Fr wu u G w z n Lecture 5: Multivariable systems Fy Mikael Johansson School of Electrical Engineering KTH, Stockholm, Sweden SISO control revisited: Signal
More informationME 375 EXAM #1 Friday, March 13, 2015 SOLUTION
ME 375 EXAM #1 Friday, March 13, 2015 SOLUTION PROBLEM 1 A system is made up of a homogeneous disk (of mass m and outer radius R), particle A (of mass m) and particle B (of mass m). The disk is pinned
More informationControl Systems Design
ELEC4410 Control Systems Design Lecture 18: State Feedback Tracking and State Estimation Julio H. Braslavsky julio@ee.newcastle.edu.au School of Electrical Engineering and Computer Science Lecture 18:
More informationPlan of the Lecture. Review: stability; Routh Hurwitz criterion Today s topic: basic properties and benefits of feedback control
Plan of the Lecture Review: stability; Routh Hurwitz criterion Today s topic: basic properties and benefits of feedback control Plan of the Lecture Review: stability; Routh Hurwitz criterion Today s topic:
More informationECE 486 Control Systems
ECE 486 Control Systems Spring 208 Midterm #2 Information Issued: April 5, 208 Updated: April 8, 208 ˆ This document is an info sheet about the second exam of ECE 486, Spring 208. ˆ Please read the following
More informationLecture 5: Frequency domain analysis: Nyquist, Bode Diagrams, second order systems, system types
Lecture 5: Frequency domain analysis: Nyquist, Bode Diagrams, second order systems, system types Venkata Sonti Department of Mechanical Engineering Indian Institute of Science Bangalore, India, 562 This
More informationECE Circuit Theory. Final Examination. December 5, 2008
ECE 212 H1F Pg 1 of 12 ECE 212  Circuit Theory Final Examination December 5, 2008 1. Policy: closed book, calculators allowed. Show all work. 2. Work in the provided space. 3. The exam has 3 problems
More information2.161 Signal Processing: Continuous and Discrete Fall 2008
MIT OpenCourseWare http://ocw.mit.edu 2.6 Signal Processing: Continuous and Discrete Fall 2008 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms. MASSACHUSETTS
More informationVideo 5.1 Vijay Kumar and Ani Hsieh
Video 5.1 Vijay Kumar and Ani Hsieh Robo3x1.1 1 The Purpose of Control Input/Stimulus/ Disturbance System or Plant Output/ Response Understand the Black Box Evaluate the Performance Change the Behavior
More informationEE Homework 12  Solutions. 1. The transfer function of the system is given to be H(s) = s j j
EE3054  Homework 2  Solutions. The transfer function of the system is given to be H(s) = s 2 +3s+3. Decomposing into partial fractions, H(s) = 0.5774j s +.5 0.866j + 0.5774j s +.5 + 0.866j. () (a) The
More informationME 304 CONTROL SYSTEMS Spring 2016 MIDTERM EXAMINATION II
ME 30 CONTROL SYSTEMS Spring 06 Course Instructors Dr. Tuna Balkan, Dr. Kıvanç Azgın, Dr. Ali Emre Turgut, Dr. Yiğit Yazıcıoğlu MIDTERM EXAMINATION II May, 06 Time Allowed: 00 minutes Closed Notes and
More informationSingleInputSingleOutput Systems
TF 502 SingleInputSingleOutput Systems SIST, ShanghaiTech Introduction OpenLoop ControlResponse Proportional Control General PID Control Boris Houska 11 Contents Introduction OpenLoop ControlResponse
More informationChapter 3. 1 st Order Sine Function Input. General Solution. Ce t. Measurement System Behavior Part 2
Chapter 3 Measurement System Behavior Part 2 1 st Order Sine Function Input Examples of Periodic: vibrating structure, vehicle suspension, reciprocating pumps, environmental conditions The frequency of
More informationEE480.3 Digital Control Systems. Part 7. Controller Design I.  Pole Assignment Method
EE480.3 Digital Control Systems Part 7. Controller Design I.  Pole Assignment Method Kunio Takaya Electrical and Computer Engineering University of Saskatchewan March 3, 2008 ** Go to fullscreen mode
More informationChapter 7. Digital Control Systems
Chapter 7 Digital Control Systems 1 1 Introduction In this chapter, we introduce analysis and design of stability, steadystate error, and transient response for computercontrolled systems. Transfer functions,
More informationECE 3793 Matlab Project 3
ECE 3793 Matlab Project 3 Spring 2017 Dr. Havlicek DUE: 04/25/2017, 11:59 PM What to Turn In: Make one file that contains your solution for this assignment. It can be an MS WORD file or a PDF file. Make
More informationTest 2 SOLUTIONS. ENGI 5821: Control Systems I. March 15, 2010
Test 2 SOLUTIONS ENGI 5821: Control Systems I March 15, 2010 Total marks: 20 Name: Student #: Answer each question in the space provided or on the back of a page with an indication of where to find the
More informationControl Systems I. Lecture 4: Diagonalization, Modal Analysis, Intro to Feedback. Readings: Emilio Frazzoli
Control Systems I Lecture 4: Diagonalization, Modal Analysis, Intro to Feedback Readings: Emilio Frazzoli Institute for Dynamic Systems and Control DMAVT ETH Zürich October 13, 2017 E. Frazzoli (ETH)
More informationso mathematically we can say that x d [n] is a discretetime signal. The output of the DT system is also discrete, denoted by y d [n].
ELEC 36 LECURE NOES WEEK 9: Chapters 7&9 Chapter 7 (cont d) Discreteime Processing of Continuousime Signals It is often advantageous to convert a continuoustime signal into a discretetime signal so
More informationSECTION 4: STEADY STATE ERROR
SECTION 4: STEADY STATE ERROR MAE 4421 Control of Aerospace & Mechanical Systems 2 Introduction Steady State Error Introduction 3 Consider a simple unity feedback system The error is the difference between
More informationBangladesh University of Engineering and Technology. EEE 402: Control System I Laboratory
Bangladesh University of Engineering and Technology Electrical and Electronic Engineering Department EEE 402: Control System I Laboratory Experiment No. 4 a) Effect of input waveform, loop gain, and system
More informationParameter Derivation of Type2 DiscreteTime PhaseLocked Loops Containing Feedback Delays
Parameter Derivation of Type DiscreteTime PhaseLocked Loops Containing Feedback Delays Joey Wilson, Andrew Nelson, and Behrouz FarhangBoroujeny joey.wilson@utah.edu, nelson@math.utah.edu, farhang@ece.utah.edu
More information(b) A unity feedback system is characterized by the transfer function. Design a suitable compensator to meet the following specifications:
1. (a) The open loop transfer function of a unity feedback control system is given by G(S) = K/S(1+0.1S)(1+S) (i) Determine the value of K so that the resonance peak M r of the system is equal to 1.4.
More informationPD, PI, PID Compensation. M. Sami Fadali Professor of Electrical Engineering University of Nevada
PD, PI, PID Compensation M. Sami Fadali Professor of Electrical Engineering University of Nevada 1 Outline PD compensation. PI compensation. PID compensation. 2 PD Control L= loop gain s cl = desired closedloop
More informationDue Wednesday, February 6th EE/MFS 599 HW #5
Due Wednesday, February 6th EE/MFS 599 HW #5 You may use Matlab/Simulink wherever applicable. Consider the standard, unityfeedback closed loop control system shown below where G(s) = /[s q (s+)(s+9)]
More informationAnalysis of Stability &
INC 34 Feedback Control Sytem Analyi of Stability & SteadyState Error S Wonga arawan.won@kmutt.ac.th Summary from previou cla Firtorder & econd order ytem repone τ ωn ζω ω n n.8.6.4. ζ ζ. ζ.5 ζ ζ.5 ct.8.6.4...4.6.8..4.6.8
More information