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Lecture Control system design: Frequency response methods - Nguyễn Công Phương
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Lecture Control system design: Frequency response methods presents the following content: Frequency response plots, performance specifications in the frequency domain, frequency response methods using control system software.
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Nội dung Text: Lecture Control system design: Frequency response methods - Nguyễn Công Phương
- Nguyễn Công Phương CONTROL SYSTEM DESIGN Frequency Response Methods
- Contents I. Introduction II. Mathematical Models of Systems III. State Variable Models IV. Feedback Control System Characteristics V. The Performance of Feedback Control Systems VI. The Stability of Linear Feedback Systems VII. The Root Locus Method VIII.Frequency Response Methods IX. Stability in the Frequency Domain X. The Design of Feedback Control Systems XI. The Design of State Variable Feedback Systems XII. Robust Control Systems XIII.Digital Control Systems s i tes.google.com/site/ncpdhbkhn 2
- Frequency Response Methods 1. Introduction 2. Frequency Response Plots 3. Performance Specifications in the Frequency Domain 4. Frequency Response Methods Using Control System Software s i tes.google.com/site/ncpdhbkhn 3
- Introduction (1) • The frequency response of a system: the steady – state response of the system to a sinusoidal input signal. • The sinusoid is a unique input signal, and the resulting output signal for a linear system, as well as signals throughout the system, is sinusoidal in the steady state. • It differs from the input waveform only in amplitude and phase angle. s i tes.google.com/site/ncpdhbkhn 4
- Introduction (2) m (s ) m (s ) T ( s) = = n ∏ (s + p ) q( s ) i i =1 Aω r(t ) = Asin ωt → R(s ) = s2 + ω 2 k1 k αs + β → Y (s) = R (s )T ( s) = + ... + n + 2 s + p1 s + pn s + ω 2 αs + β → y (t ) = k1e− p1t + ... + kne − pnt + L−1 2 2 s +ω αs + β ysteady − state (t ) = lim y(t ) = lim L−1 2 2 t →∞ t →∞ s +ω = A T ( jω ) sin(ω t + φ ), φ = ∠T ( jω ) s i tes.google.com/site/ncpdhbkhn 5
- Frequency Response Methods 1. Introduction 2. Frequency Response Plots 3. Performance Specifications in the Frequency Domain 4. Frequency Response Methods Using Control System Software s i tes.google.com/site/ncpdhbkhn 6
- Frequency Response Plots (1) G ( jω ) = G ( s ) s = jω = Re[G ( jω )] + Im[G ( jω )] = R(ω ) + jX (ω ) = G ( jω ) e jφ (ω ) = G ( jω ) ∠[φ (ω )] = R 2 (ω ) + X 2 (ω )∠{tan −1 [ X (ω ) / R (ω )]} s i tes.google.com/site/ncpdhbkhn 7
- Frequency Response Plots (2) Ex. 1 1 1 V1( s) + + V2 (s ) = Vcapacitor (s ) = I (s) = . R Cs Cs 1 + R V1 ( s ) V2 ( s ) Cs C V2 (s ) 1 → G( s) = = − − V1 (s ) RCs + 1 1 → G( jω ) = G ( s ) s = jω = 0.5 Positive ω jω ( RC ) + 1 0.4 Negative ω 1 1 0.3 = , ω1 = j ( ω / ω1 ) + 1 0.2 RC 0.1 X(ω) 0 1 ω / ω1 = − -0.1 j 1 + (ω / ω1) 2 1 + (ω / ω1 ) 2 -0.2 -0.3 -0.4 1 = ∠ tan (−ω / ω1 ) −1 -0.5 1 + (ω / ω1 ) 2 0 0.2 0.4 0.6 0.8 1 R(ω) s i tes.google.com/site/ncpdhbkhn 8
- Frequency Response Plots (3) Semilog plots of the magnitude (in decibels) and phase (in degrees) of a transfer function versus frequency TdB = 20log10 T T =T φ → φ ( )( T = T1T2 T3 ... = T1 φ1 T2 φ2 T3 φ3 ... )( ) = ( T1T2T3... ) φ1 + φ2 + φ3 + ... 20 log10 T = 20log10 T1 + 20log10 T2 + 20log10 T3 + ... → φ = φ1 + φ2 + φ3 + ... s i tes.google.com/site/ncpdhbkhn 9
- Frequency Response Plots (4) jω j 2ζ ω jω 2 K ( jω) ±1 1 + 1 + 1 + ... z1 ωk ωk T(ω) = jω j2ζ 2ω jω 2 1 + 1 + + ... p1 ωn ωn K : gain 1 1 1 : pole at the origin : simple pole : quadratic pole jω 2 jω j 2ζ 2ω jω 1+ 1+ + p1 ωn ω n jω 2 jω : zero at the origin 1+ : simple zero 1 + j2ζ 1ω + jω : quadratic zero z1 ωk ωk s i tes.google.com/site/ncpdhbkhn 10
- Frequency Response Plots (5) TdB = 20 log10 K T(ω) = K → φ = 0 TdB φ 20 log10 K 0 0.1 1 10 100 ω 0.1 1 10 100 ω s i tes.google.com/site/ncpdhbkhn 11
- Frequency Response Plots (6) 1 TdB = −20 log10 ω T(ω) = → jω φ = −90o TdB 20 φ 0 0o 0.1 1 10 ω 0.1 1 10 ω −20 − 90o s i tes.google.com/site/ncpdhbkhn 12
- Frequency Response Plots (7) TdB = 20 log10 ω T(ω) = jω → φ = 90 o φ TdB 90o 20 0 0o 0.1 1 10 ω 0.1 1 10 ω −20 s i tes.google.com/site/ncpdhbkhn 13
- Frequency Response Plots (8) jω TdB = −20log10 1 + 1 p1 T(ω ) = → jω 1+ −1 ω p1 φ = tan − TdB p1 5 0 -5 -10 -15 -20 -25 ω 0.1p1 p1 10 p1 s i tes.google.com/site/ncpdhbkhn 14
- Frequency Response Plots (9) jω TdB = −20log10 1 + 1 p1 T(ω ) = → jω 1+ −1 ω p1 φ = tan − p1 φ 0.1p1 p1 10 p1 100 p1 0 ω -10 -20 -30 -40 -50 -60 -70 -80 -90 -100 s i tes.google.com/site/ncpdhbkhn 15
- Frequency Response Plots (10) jω TdB = 20log10 1 + jω z1 T(ω ) = 1 + → z1 ω φ = tan −1 z1 TdB 25 20 15 10 5 0 -5 ω 0.1z1 z1 10z1 s i tes.google.com/site/ncpdhbkhn 16
- Frequency Response Plots (11) jω TdB = 20log10 1 + jω z1 T(ω ) = 1 + → z1 ω φ = tan −1 φ z1 100 90 80 70 60 50 40 30 20 10 ω 0 0.1z1 z1 10z1 100z1 s i tes.google.com/site/ncpdhbkhn 17
- Frequency Response Plots (12) 2 TdB = −20log10 1 + j 2ζ ω 2 + jω 1 ωn ωn T(ω ) = 2 → j2ζ 2ω jω −1 j 2ζ 2ω / ωn 1+ + φ = − 2 tan ωn ωn − ω 2 ω 1 / n 20 TdB ζ 2 = 0.05 10 ζ 2 = 0.2 0 ζ 2 = 0.4 -10 ζ 2 = 0.707 -20 ζ 2 = 1.5 -30 ω -40 0.1ωn ωn 10ωn 100ωn s i tes.google.com/site/ncpdhbkhn 18
- Frequency Response Plots (13) 2 TdB = −20log10 1 + j 2ζ ω 2 + jω 1 ωn ωn T(ω ) = 2 → j2ζ 2ω jω −1 j 2ζ 2ω / ωn 1+ + φ = − 2 tan ωn ωn − ω 2 ω 1 / n φ 0.1ωn ωn 10ωn 100ωn 0 -20 ω -40 ζ 2 = 1.5 -60 -80 ζ 2 = 0.707 -100 ζ 2 = 0.4 -120 -140 ζ 2 = 0.2 -160 ζ 2 = 0.05 -180 s i tes.google.com/site/ncpdhbkhn 19
- Frequency Response Plots (14) 2 TdB = 20log10 1 + j 2ζ ω 2 + jω j 2ζ 2ω jω 2 ωn ωn T(ω ) = 1 + + → ωn ω n −1 j 2ζ 2ω / ωn φ = tan 2 − ω 2 ω 1 / n 40 TdB 30 ζ 2 = 1.5 20 ζ 2 = 0.707 10 ζ 2 = 0.4 0 -10 ζ 2 = 0.05 ζ 2 = 0.2 ω -20 0.1ωn ωn 10ωn 100ωn s i tes.google.com/site/ncpdhbkhn 20
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