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 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 ...) Block diagrams describe the flow of information
Basic System Interconnections: Series & Parallel
Basic System Interconnections: Series & Parallel Series connection
Basic System Interconnections: Series & Parallel Series connection ( G is common G 1 G 2 U Y notation for t.f.’s)
Basic System Interconnections: Series & Parallel Series connection ( G is common G 1 G 2 U Y notation for t.f.’s) G 1 G 2 U Y Y 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 ( G is common G 1 G 2 U Y notation for t.f.’s) G 1 G 2 U Y Y U = G 1 G 2 (for SISO systems, the order of G 1 and G 2 does not matter) Parallel connection
Basic System Interconnections: Series & Parallel Series connection ( G is common G 1 G 2 U Y notation for t.f.’s) G 1 G 2 U Y Y U = G 1 G 2 (for SISO systems, the order of G 1 and G 2 does not matter) Parallel connection G 1 + U Y + G 2
Basic System Interconnections: Series & Parallel Series connection ( G is common G 1 G 2 U Y notation for t.f.’s) G 1 G 2 U Y Y U = G 1 G 2 (for SISO systems, the order of G 1 and G 2 does not matter) Parallel connection G 1 + U Y + G 2 Y U G 1 + G 2 Y U = G 1 + G 2
Basic System Interconnections: Negative Feedback + U G 1 R Y − Find the transfer function from R W (reference) to Y G 2
Basic System Interconnections: Negative Feedback + U G 1 R Y − Find the transfer function from R W (reference) to Y G 2 U = R − W
Basic System Interconnections: Negative Feedback + U G 1 R Y − Find the transfer function from R W (reference) to Y G 2 U = R − W Y = G 1 U
Basic System Interconnections: Negative Feedback + U G 1 R Y − Find the transfer function from R W (reference) to Y G 2 U = R − W Y = G 1 U = G 1 ( R − W )
Basic System Interconnections: Negative Feedback + U G 1 R Y − Find the transfer function from R W (reference) to Y G 2 U = R − W Y = G 1 U = G 1 ( R − W ) = G 1 R − G 1 G 2 Y
Basic System Interconnections: Negative Feedback + U G 1 R Y − Find the transfer function from R W (reference) to Y G 2 G 1 = ⇒ Y = R 1 + G 1 G 2 U = R − W Y = G 1 U = G 1 ( R − W ) G 1 U Y = G 1 R − G 1 G 2 Y 1 + G 1 G 2
Basic System Interconnections: Negative Feedback + U G 1 R Y G 1 − = ⇒ Y = R W 1 + G 1 G 2 G 2
Basic System Interconnections: Negative Feedback + U G 1 R Y G 1 − = ⇒ Y = R W 1 + G 1 G 2 G 2 The gain of a negative feedback loop: forward gain 1 + loop gain
Basic System Interconnections: Negative Feedback + U G 1 R Y G 1 − = ⇒ Y = R W 1 + G 1 G 2 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 U R G 2 G 1 Y −
Unity Feedback Other feedback configurations are also possible: + E U R G 2 G 1 Y − This is called unity feedback — no component on the feedback path.
Unity Feedback Other feedback configurations are also possible: + E U R G 2 G 1 Y − This is called unity feedback — no component on the feedback path. Common structure (saw this in Lecture 1): ◮ R = reference ◮ U = control input ◮ Y = output ◮ E = error ◮ G 1 = plant (also denoted by P ) ◮ G 2 = controller or compensator (also denoted by C or K )
Unity Feedback + E U G 2 G 1 R Y −
Unity Feedback + E U G 2 G 1 R Y − Let’s practice with deriving transfer functions: forward gain 1 + loop gain
Unity Feedback + E U G 2 G 1 R Y − Let’s practice with deriving transfer functions: forward gain 1 + loop gain ◮ Reference R to output Y :
Unity Feedback + E U G 2 G 1 R Y − Let’s practice with deriving transfer functions: forward gain 1 + loop gain ◮ Reference R to output Y : Y G 1 G 2 R = 1 + G 1 G 2
Unity Feedback + E U G 2 G 1 R Y − Let’s practice with deriving transfer functions: forward gain 1 + loop gain ◮ Reference R to output Y : Y G 1 G 2 R = 1 + G 1 G 2 ◮ Reference R to control input U :
Unity Feedback + E U G 2 G 1 R Y − Let’s practice with deriving transfer functions: forward gain 1 + loop gain ◮ Reference R to output Y : Y G 1 G 2 R = 1 + G 1 G 2 ◮ Reference R to control input U : U G 2 R = 1 + G 1 G 2
Unity Feedback + E U G 2 G 1 R Y − Let’s practice with deriving transfer functions: forward gain 1 + loop gain ◮ Reference R to output Y : Y G 1 G 2 R = 1 + G 1 G 2 ◮ Reference R to control input U : U G 2 R = 1 + G 1 G 2 ◮ Error E to output Y :
Unity Feedback + E U G 2 G 1 R Y − Let’s practice with deriving transfer functions: forward gain 1 + loop gain ◮ Reference R to output Y : Y G 1 G 2 R = 1 + G 1 G 2 ◮ Reference R to control input U : U G 2 R = 1 + G 1 G 2 ◮ Error E to output Y : Y E = G 1 G 2 (no feedback path)
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.
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.
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 1 1 s + a, ( s + a )( s + b ) , 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 1 1 s + a, ( s + a )( s + b ) , 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.
Prototype 2nd-Order System So far, we have only seen transfer functions that have either real poles or purely imaginary poles: 1 1 1 s + a, ( s + a )( s + b ) , 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 So far, we have only seen transfer functions that have either real poles or purely imaginary poles: 1 1 1 s + a, ( s + a )( s + b ) , 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). Plus, you will need this for Lab 1.
Prototype 2nd-Order System Consider the following transfer function: ω 2 n H ( s ) = s 2 + 2 ζω n s + ω 2 n
Prototype 2nd-Order System Consider the following transfer function: ω 2 n H ( s ) = s 2 + 2 ζω n s + ω 2 n Comments:
Prototype 2nd-Order System Consider the following transfer function: ω 2 n H ( s ) = s 2 + 2 ζω n s + ω 2 n Comments: ◮ ζ > 0 , ω n > 0 are arbitrary parameters
Prototype 2nd-Order System Consider the following transfer function: ω 2 n H ( s ) = s 2 + 2 ζω 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 Consider the following transfer function: ω 2 n H ( s ) = s 2 + 2 ζω 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 ◮ H ( s ) is normalized to have DC gain = 1 (provided DC gain exists)
Prototype 2nd-Order System ω 2 n H ( s ) = s 2 + 2 ζω n s + ω 2 n
Prototype 2nd-Order System ω 2 n H ( s ) = s 2 + 2 ζω n s + ω 2 n By the quadratic formula, the poles are: � ζ 2 − 1 s = − ζω n ± ω n � � � ζ 2 − 1 = − ω n ζ ±
Prototype 2nd-Order System ω 2 n H ( s ) = s 2 + 2 ζω n s + ω 2 n By the quadratic formula, the poles are: � ζ 2 − 1 s = − ζω n ± ω n � � � ζ 2 − 1 = − ω n ζ ± The nature of the poles changes depending on ζ :
Prototype 2nd-Order System ω 2 n H ( s ) = s 2 + 2 ζω n s + ω 2 n By the quadratic formula, the poles are: � ζ 2 − 1 s = − ζω n ± ω n � � � ζ 2 − 1 = − ω n ζ ± The nature of the poles changes depending on ζ : ◮ ζ > 1
Prototype 2nd-Order System ω 2 n H ( s ) = s 2 + 2 ζω n s + ω 2 n By the quadratic formula, the poles are: � ζ 2 − 1 s = − ζω n ± ω n � � � ζ 2 − 1 = − ω n ζ ± 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 + 2 ζω n s + ω 2 n By the quadratic formula, the poles are: � ζ 2 − 1 s = − ζω n ± ω n � � � ζ 2 − 1 = − ω n ζ ± The nature of the poles changes depending on ζ : ◮ ζ > 1 both poles are real and negative ◮ ζ = 1
Prototype 2nd-Order System ω 2 n H ( s ) = s 2 + 2 ζω n s + ω 2 n By the quadratic formula, the poles are: � ζ 2 − 1 s = − ζω n ± ω n � � � ζ 2 − 1 = − ω n ζ ± The nature of the poles changes depending on ζ : ◮ ζ > 1 both poles are real and negative ◮ ζ = 1 one negative pole
Prototype 2nd-Order System ω 2 n H ( s ) = s 2 + 2 ζω n s + ω 2 n By the quadratic formula, the poles are: � ζ 2 − 1 s = − ζω n ± ω n � � � ζ 2 − 1 = − ω n ζ ± The nature of the poles changes depending on ζ : ◮ ζ > 1 both poles are real and negative ◮ ζ = 1 one negative pole ◮ ζ < 1
Prototype 2nd-Order System ω 2 n H ( s ) = s 2 + 2 ζω n s + ω 2 n By the quadratic formula, the poles are: � ζ 2 − 1 s = − ζω n ± ω n � � � ζ 2 − 1 = − ω n ζ ± The nature of the poles changes depending on ζ : ◮ ζ > 1 both poles are real and negative ◮ ζ = 1 one negative pole ◮ ζ < 1 two complex poles with negative real parts
Prototype 2nd-Order System ω 2 n H ( s ) = s 2 + 2 ζω n s + ω 2 n By the quadratic formula, the poles are: � ζ 2 − 1 s = − ζω n ± ω n � � � ζ 2 − 1 = − ω n ζ ± The nature of the poles changes depending on ζ : ◮ ζ > 1 both poles are real and negative ◮ ζ = 1 one negative pole ◮ ζ < 1 two complex poles with negative real parts s = − σ ± jω d � 1 − ζ 2 where σ = ζω n , ω d = ω n
Prototype 2nd-Order System ω 2 n H ( s ) = , ζ < 1 s 2 + 2 ζω n s + ω 2 n
Prototype 2nd-Order System ω 2 n H ( s ) = , ζ < 1 s 2 + 2 ζω n s + ω 2 n The poles are � 1 − ζ 2 = − σ ± jω d s = − ζω n ± jω n
Prototype 2nd-Order System ω 2 n H ( s ) = , ζ < 1 s 2 + 2 ζω n s + ω 2 n The poles are � 1 − ζ 2 = − σ ± jω d s = − ζω n ± jω n Im p Note that ω d = ω n 1 − ζ 2 ω n σ 2 + ω 2 d = ζ 2 ω 2 n + ω 2 n − ζ 2 ω 2 n ϕ Re σ = ζ! n 0 = ω 2 n cos ϕ = ζω n = ζ ω n
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: ω 2 ω 2 n n H ( s ) = = s 2 + 2 ζω n s + ω 2 ( s + σ ) 2 + ω 2 n d
2nd-Order Response Let’s compute the system’s impulse and step response: ω 2 ω 2 n n H ( s ) = = s 2 + 2 ζω n s + ω 2 ( s + σ ) 2 + ω 2 n d ◮ Impulse response: � � ω 2 h ( t ) = L − 1 { H ( s ) } = L − 1 n ( s + σ ) 2 + ω 2 d
2nd-Order Response Let’s compute the system’s impulse and step response: ω 2 ω 2 n n H ( s ) = = s 2 + 2 ζω n s + ω 2 ( s + σ ) 2 + ω 2 n d ◮ Impulse response: � ( ω 2 � n /ω d ) ω d h ( t ) = L − 1 { H ( s ) } = L − 1 ( s + σ ) 2 + ω 2 d
2nd-Order Response Let’s compute the system’s impulse and step response: ω 2 ω 2 n n H ( s ) = = s 2 + 2 ζω n s + ω 2 ( s + σ ) 2 + ω 2 n d ◮ Impulse response: � ( ω 2 � n /ω d ) ω d h ( t ) = L − 1 { H ( s ) } = L − 1 ( s + σ ) 2 + ω 2 d = ω 2 e − σt sin( ω d t ) n ω d
2nd-Order Response Let’s compute the system’s impulse and step response: ω 2 ω 2 n n H ( s ) = = s 2 + 2 ζω n s + ω 2 ( s + σ ) 2 + ω 2 n d ◮ Impulse response: � ( ω 2 � n /ω d ) ω d h ( t ) = L − 1 { H ( s ) } = L − 1 ( s + σ ) 2 + ω 2 d = ω 2 e − σt sin( ω d t ) n (table, # 20) ω d
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