EE2010_Gp2_Chap5_1011s2 - Sine i/p Freq Response Factors...

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Unformatted text preview: Sine i/p Freq Response Factors Summary Polar Intuition LT Analysis Response due to sinusoidal input What is the output signal of a stable system when the input signal is a sinusoidal function ? G(s) U sin( t) ϖ ??? I Intutively, the steady-state output will also be a sinusoidal signal as the solution to the system differential equations can be interpreted using two properties of sinusoidal signal : I Sinusoids can be differentiated indefinitely. Each differentiation results in another sinusoidal signal. sin ω t d dt-→ ω cos ω t d dt-→ - ω 2 sin ω t d dt-→ . .. I The result of adding two sinusoids of the same frequency together is another sinusoid with different amplitude and phase. However, the frequency of the signals is the same. Conclusion is consistent with the observation that the steady-state response is related to the input pole. EE2010 Systems & Control Notes  WW Tan Chapter 5 Page 1 Sine i/p Freq Response Factors Summary Polar Intuition LT Analysis Derivation of response due to sinusoidal input using Laplace Transform Assume that the transfer function of a stable system is G ( s ) = Y ( s ) U ( s ) , where the input u ( t ) is a sine wave with amplitude U i.e. u ( t ) = U sin ω t = ⇒ U ( s ) = U ω s 2 + ω 2 I The system output is Y ( s ) = G ( s ) U ( s ) = G ( s ) U ω s 2 + ω 2 I If G ( s ) has stable poles at s =- p 1 ,- p 2 ,... , then Y ( s ) = A 1 s + p 1 + A 2 s + p 2 + ... | {z } system poles + Cs + D s 2 + ω 2 | {z } input poles (1) EE2010 Systems & Control Notes  WW Tan Chapter 5 Page 2 Sine i/p Freq Response Factors Summary Polar Intuition LT Analysis I The time response is y ( t ) = A 1 e- p 1 t + A 2 e- p 2 t + ... | {z } transient + C cos ω t + D ω sin ω t | {z } steady-state I When t → ∞ , A 1 e- p 1 t + A 2 e- p 2 t + ... → i.e. the transient response decays to zero. I Hence, the output of the plant at steady state is y ss ( t ) = C cos ω t + D ω sin ω t = ˆ y sin( ω t + φ ) where ˆ y = q C 2 + ( D ω ) 2 φ = tan- 1 ω C D EE2010 Systems & Control Notes  WW Tan Chapter 5 Page 3 Sine i/p Freq Response Factors Summary Polar Intuition LT Analysis Example Calculate the output signal given that G ( s ) = 20 s + 1 and u ( t ) = sin(10 t ) Y ( s ) = G ( s ) U ( s ) = 200 ( s + 1)( s 2 + 100) y ( t ) = 200 101 e- t + 20 √ 101 sin(10 t- 1 . 47) 1 2 3 4 5 6 7 8-2-1 1 2 3 4 Time (seconds) Input, u(t), and Output, y(t) | → Steady-state Output, y(t) Input, u(t) Can the Final Value Theorem be used to predict the steady-state output if the input is a sinusoid ? EE2010 Systems & Control Notes  WW Tan Chapter 5 Page 4 Sine i/p Freq Response Factors Summary Polar Def & Motivation Theorem Example Bode Polar Frequency Response : Definition and Motivation Frequency response Steady state output of a system that is perturbed by a sinusoidal input of frequency ω (0 < ω < ∞ )....
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This note was uploaded on 07/11/2011 for the course ECE 2010 taught by Professor Tankaychen during the Spring '11 term at National University of Singapore.

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EE2010_Gp2_Chap5_1011s2 - Sine i/p Freq Response Factors...

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