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


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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 )