Review: transient and steady-state response; DC gain and the FVT Today s topic: system-modeling diagrams; prototype 2nd-order system

Plan of the Lecture Review: transient and steady-state response; DC gain and the FVT Today s topic: system-modeling diagrams; prototype 2nd-order system

Plan of the Lecture Review: transient and steady-state response; DC gain and the FVT Today s topic: system-modeling diagrams; prototype 2nd-order system Goal: develop a methodology for representing and analyzing systems by means of block diagrams; start analyzing a prototype 2nd-order system.

System Modeling Diagrams large system decompose smaller blocks (subsystems) compose

System Modeling Diagrams large system decompose smaller blocks (subsystems) compose this is the core of systems theory

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.

All-Integrator 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

All-Integrator 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, (s-domain) will often go back and forth When working with block diagrams, we typically ignore initial conditions.

Example 1 Build an all-integrator diagram for ÿ = u s 2 = U

Example 1 Build an all-integrator diagram for ÿ = u s 2 = U This is obvious: u ẏ 1/s 1/s y or U s 1/s 1/s

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

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

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

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

Example 3 Build an all-integrator diagram for a system with transfer function H(s) = b 1s + b 0 s 2 + a 1 s + a 0

Example 3 Build an all-integrator 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

Example 3 Build an all-integrator 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 all-integrator diagram. But we will see that we don t need to do it directly.

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

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:

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

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:

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

Example 3, continued All-integrator 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

Example 3, continued All-integrator 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 state-space model corresponding to this diagram?

Example 3, continued U + s 2 X + sx X 1/s 1/s + b 1 b 0 a 1 a 0 State-space model:

Example 3, continued U + s 2 X + sx X 1/s 1/s + b 1 b 0 a 1 a 0 State-space model: s 2 X = U a 1 sx a 0 X

Example 3, continued U + s 2 X + sx X 1/s 1/s + b 1 b 0 a 1 a 0 State-space model: s 2 X = U a 1 sx a 0 X ẍ = a 1 ẋ a 0 x + u

Example 3, continued U + s 2 X + sx X 1/s 1/s + b 1 b 0 a 1 a 0 State-space 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

Example 3, continued U + s 2 X + sx X 1/s 1/s + b 1 b 0 a 1 a 0 State-space 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

Example 3, continued State-space model: ẍ = a 1 ẋ a 0 x + u y = b 1 ẋ + b 0 x

Example 3, continued State-space model: ẍ = a 1 ẋ a 0 x + u y = b 1 ẋ + b 0 x x 1 = x, x 2 = ẋ

Example 3, continued State-space 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

Example 3, continued State-space 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

Example 3, continued State-space 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.

Example 3, continued State-space 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

Example 3, wrap-up All-integrator 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 State-space 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

Example 3, wrap-up All-integrator 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 State-space 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 state-space equations are unique (up to coordinate transformations).

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 all-integrator diagrams, etc.) Block diagrams are an abstraction (they hide unnecessary low-level detail...)

Basic System Interconnections: Series & Parallel

Basic System Interconnections: Series & Parallel Series connection

Basic System Interconnections: Series & Parallel Series connection U G 1 G 2 (G is common notation for t.f. s)

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)

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)

Basic System Interconnections: Negative Feedback R + U W G 1 Find the transfer function from R (reference) to G 2

Basic System Interconnections: Negative Feedback R + U W G 1 Find the transfer function from R (reference) to G 2 U = R W

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

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 )

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

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 1 1 + G 1 G 2 R G 1 1+G 1 G 2

Basic System Interconnections: Negative Feedback R + U W G 1 = = G 1 1 + G 1 G 2 R G 2

Basic System Interconnections: Negative Feedback R + U W G 1 = = G 1 1 + G 1 G 2 R G 2 The gain of a negative feedback loop: forward gain 1 + loop gain

Basic System Interconnections: Negative Feedback R + U W G 1 = = G 1 1 + 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.

Unity Feedback Other feedback configurations are also possible: + E R G 2 U G 1

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.

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)

Unity Feedback R + E G 2 U G 1

Unity Feedback R + E G 2 U G 1 Let s practice with deriving transfer functions: forward gain 1 + loop gain

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

Unity Feedback R + E G 2 U G 1 Let s practice with deriving transfer functions: Reference R to output : R = G 1G 2 1 + G 1 G 2 forward gain 1 + loop gain

Unity Feedback R + E G 2 U G 1 Let s practice with deriving transfer functions: Reference R to output : R = G 1G 2 1 + G 1 G 2 forward gain 1 + loop gain Reference R to control input U:

Unity Feedback R + E G 2 U G 1 Let s practice with deriving transfer functions: Reference R to output : R = G 1G 2 1 + G 1 G 2 forward gain 1 + loop gain Reference R to control input U: U R = G 2 1 + G 1 G 2

Unity Feedback R + E G 2 U G 1 Let s practice with deriving transfer functions: Reference R to output : R = G 1G 2 1 + G 1 G 2 forward gain 1 + loop gain Reference R to control input U: Error E to output : U R = G 2 1 + G 1 G 2

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

Prototype 2nd-Order 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

Prototype 2nd-Order 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 second-order systems, but this will be sufficient to develop some nontrivial intuition (dominant poles).

Prototype 2nd-Order System Consider the following transfer function: ω 2 n H(s) = s 2 + 2ζω n s + ωn 2

Prototype 2nd-Order System Consider the following transfer function: H(s) = ω 2 n s 2 + 2ζω n s + ω 2 n Comments:

Prototype 2nd-Order 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

Prototype 2nd-Order 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 2nd-degree monic polynomial, just written in a weird way

Prototype 2nd-Order System ω 2 n H(s) = s 2 + 2ζω n s + ωn 2

Prototype 2nd-Order 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

Prototype 2nd-Order 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:

Prototype 2nd-Order 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

Prototype 2nd-Order 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

Prototype 2nd-Order 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

Prototype 2nd-Order 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

Prototype 2nd-Order System H(s) = ω 2 n s 2 + 2ζω n s + ωn 2, ζ < 1

Prototype 2nd-Order 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

Prototype 2nd-Order 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

2nd-Order Response Let s compute the system s impulse and step response:

2nd-Order 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

2nd-Order 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 }

2nd-Order 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

2nd-Order 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]

2nd-Order 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

2nd-Order 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)

2nd-Order 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 1.5 1.0 The parameter ζ is called the damping ratio ζ > 1: system is overdamped 0.5 Ζ 0.1 Ζ 0.9 Ζ 1 2 4 6 8 10 12 14 t ζ < 1: system is underdamped ζ = 0: no damping (ω d = ω n )