EE203-SUNYBuffalo-26-Chapter13-07

EE203-SUNYBuffalo-26-Chapter13-07 - SMALL for Big Things...

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Unformatted text preview: SMALL for Big Things University at Buffalo SMALL for Big Things University at Buffalo nanobioSensors & MicroActuators Learning Lab The State University of New York nanobioSensors & MicroActuators Learning Lab The State University of New York EE 203 Circuit Analysis 2 Lecture 26 Chapter 13.7 Transfer Transfer Function and SteadyState Sinusoidal Response Transfer Function and Steady-State Sinusoidal Response (1) Once Once we have computed a circuit's transfer function H(s), we no longer need to perform a separate phasor analysis of the circuit to determine its steady-state response. Instead, we use the transfer function to relate the steady-state response response to the excitation source. Kwang W. Oh, Ph.D., Assistant Professor SMALL (nanobioSensors and MicroActuators Learning Lab) Department of Electrical Engineering University at Buffalo, The State University of New York 215E Bonner Hall, SUNY-Buffalo, Buffalo, NY 14260-1920 Tel: (716) 645-3115 Ext. 1149, Fax: (716) 645-3656 E-mail: kwangoh@buffalo.edu, http://www.SMALL.Buffalo.edu EE 203 Circuit Analysis 2 | Spring 2008 | Prof. Kwang W. Oh | EE@SUNY-Buffalo Lecture 26 | Chapter 13 | 7/8 | 1/6 EE 203 Circuit Analysis 2 | Spring 2008 | Prof. Kwang W. Oh | EE@SUNY-Buffalo Lecture 26 | Chapter 13 | 7/8 | 2/6 SMALL for Big Things University at Buffalo SMALL for Big Things University at Buffalo nanobioSensors & MicroActuators Learning Lab The State University of New York nanobioSensors & MicroActuators Learning Lab The State University of New York Transfer Function and Steady-State Sinusoidal Response (2) α = 0, β = ω , θ = θ (ω ) + φ EE 203 Circuit Analysis 2 | Spring 2008 | Prof. Kwang W. Oh | EE@SUNY-Buffalo Lecture 26 | Chapter 13 | 7/8 | 3/6 Transfer Function and Steady-State Sinusoidal Response (3) which indicates how to use the transfer function H(s) to find the steadystate sinusoidal response yss(t) of a circuit. The amplitude of the response equals the amplitude of the source, A, times th the magnitude of the transfer function, |H(jω)|. The phase angle of the response, φ + θ(ω), equals the phase angle of the source, φ, plus the phase angle of the transfer function, θ(ω). EE 203 Circuit Analysis 2 | Spring 2008 | Prof. Kwang W. Oh | EE@SUNY-Buffalo Lecture 26 | Chapter 13 | 7/8 | 4/6 SMALL for Big Things University at Buffalo SMALL for Big Things University at Buffalo nanobioSensors & MicroActuators Learning Lab The State University of New York nanobioSensors & MicroActuators Learning Lab The State University of New York Recall: Example 13.1 Vg − Vo 1000 = Example 13.4 Vo V +o 250 + 0.05s 106 s v0 = y ss (t ) = A | H ( jω ) | cos[ωt + φ + θ (ω )] = 120 Vg 1 1 s )= + + 1000 1000 250 + 0.05s 106 1 Vo 1000(250 + 0.05s )(106 ) 1000 H (s) = = × 6 s 1 1 Vg + + 6 1000(250 + 0.05s )(10 ) 1000 250 + 0.05s 10 1 (250 + 0.05s )(106 ) 1000( s + 5000) = × 50 = 2 6 6 s + 6000 s + 25 × 106 (250 + 0.05s )(10 ) + 1000(10 ) + 1000(250 + 0.05s ) s 1 50 1000( s + 5000) 1000( s + 5000) 1000( s + 5000) =2 = = s + 2 × 3000 s + 9000000 + 16000000 ( s + 3000) 2 + (4000) 2 ( s + 3000 − j 4000)( s + 3000 + j 4000) Vo ( EE 203 Circuit Analysis 2 | Spring 2008 | Prof. Kwang W. Oh | EE@SUNY-Buffalo Lecture 26 | Chapter 13 | 7/8 | 5/6 A = 120 V, | H ( jω ) |= 2 cos(5000t + 30° − 45°) = 20 2 cos(5000t − 15°) 6 2 , ω = 5000 rad/s, φ = 30°, θ (ω ) = −45° 6 v g = x(t ) = A cos(ωt + φ ) = 120 cos(5000t + 30°) 1000( s + 5000) (from Example 13.1) s 2 + 6000 s + 25 × 10 6 1000( j 5000 + 5000) 1000( j 5000 + 5000) H ( jω ) = = ( j 5000) 2 + 6000( j 5000) + 25 × 10 6 − 25 × 10 6 + 6000( j 5000) + 25 × 10 6 H (s) = = 1 + j (1 + j )(− j ) 1 − j 2 = = = ∠ − 45° j6 6 6 6 EE 203 Circuit Analysis 2 | Spring 2008 | Prof. Kwang W. Oh | EE@SUNY-Buffalo Lecture 26 | Chapter 13 | 7/8 | 6/6 ...
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This note was uploaded on 10/19/2011 for the course EE 203 taught by Professor Staff during the Spring '08 term at SUNY Buffalo.

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