EE203-SUNYBuffalo-24-Chapter13-05

EE203-SUNYBuffalo-24-Chapter13-05 - 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 24 Chapter 13.5 Transfer Function in Partial Fraction Expansions 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 24 | Chapter 13 | 5/8 | 1/11 Transfer Function in Partial Fraction Expansions Expanding Expanding the right-hand side of Eq. 13.96 into a sum of partial fractions produces a term for each pole of H(s) and X(s). Poles are the roots of the denominator polynomial Zeros are the roots of the numerator pol nomial. ly The terms generated by the poles of H(s) give rise to the transient component of the total response. The terms generated by the poles of X(s) give rise to the steady-state component of the response. By steady-state response, we mean the response that exists after the transient components have become negligible. EE 203 Circuit Analysis 2 | Spring 2008 | Prof. Kwang W. Oh | EE@SUNY-Buffalo Lecture 24 | Chapter 13 | 5/8 | 2/11 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 Example 13.2 (1) Example 13.2 (2) 1000(( −3000 + j 4000) + 5000) 50 (( −3000 + j 4000) + 3000 + j 4000) ( −3000 + j 4000) 2 50000(2000 + j 4000) 100 + j 200 = = j8000( −7000000 − j 24000000) 8000 × 24 − j8000 × 7 11 = ( 2 + j11) × 10−4 = ( 4 + 121) × 10−4 ∠ tan −1 ( ) = 5 5 × 10−4 ∠79.70° 2 v0 = [10 5 × 10−4 e −3000t cos(4000t + 79.70°) + 10t − 4 × 10−4 ]u (t ) K1 = H ( s) = Vo 1000( s + 5000) 50 , v g = 50tu (t ) ⇒ Vg = 2 = Vg s 2 + 6000 s + 25 × 106 s Vo = H ( s )Vg = = 1000( s + 5000) 50 1000( s + 5000) 50 = s 2 + 6000 s + 25 × 106 s 2 ( s + 3000 − j 4000)( s + 3000 + j 4000) s 2 10 5 × 10 −4 e −3000 t cos( 4000t + 79.70°) This term is generated by the poles (-3000+j4000) and (-3000-j4000) of the transfer function H(s). 10t − 4 × 10−4 These two terms are generated by the second-order pole (K/s2) of the driving voltage. K K1 K1* K + + 22 + 3 s + 3000 − j 4000 s + 3000 + j 4000 s s K1 = 5 5 × 10 − 4 ∠79.70°, K1* = 5 5 × 10 − 4 ∠ − 79.70° K 2 = 10, K 3 = −4 × 10 − 4 , v0 = [10 5 ×10 − 4 e −3000t cos(4000t + 79.70°) + 10t − 4 × 10 − 4 ]u (t ) EE 203 Circuit Analysis 2 | Spring 2008 | Prof. Kwang W. Oh | EE@SUNY-Buffalo Lecture 24 | Chapter 13 | 5/8 | 3/11 EE 203 Circuit Analysis 2 | Spring 2008 | Prof. Kwang W. Oh | EE@SUNY-Buffalo Lecture 24 | Chapter 13 | 5/8 | 4/11 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 Observations on the Use of H(s) in Circuit Analysis (1) If If the input is delayed by a seconds If input = a unit impulse source The response becomes If Therefore, delaying the input by a seconds simply delays the response function by a seconds. A circuit that exhibits this characteristic is said to be time time invariant. system, circuit, black box… input Observations on the Use of H(s) in Circuit Analysis (2) If input = a unit impulse source x(t) = δ(t) the output = the unit impulse response y(t) = h(t) = the inverse transform of the transfer function. input output, response x(t-a) y(t-a) h(t) δ(t) impulse source EE 203 Circuit Analysis 2 | Spring 2008 | Prof. Kwang W. Oh | EE@SUNY-Buffalo output, response Lecture 24 | Chapter 13 | 5/8 | 5/11 system, circuit, black box… impulse response EE 203 Circuit Analysis 2 | Spring 2008 | Prof. Kwang W. Oh | EE@SUNY-Buffalo Lecture 24 | Chapter 13 | 5/8 | 6/11 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 24 Chapter 13.6 Transfer Function & Convolution Integral 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 24 | Chapter 13 | 5/8 | 7/11 Let’s Think of a Black Box Triggered by an Impulse Input Impulse sources x1(t) = δ(t) Impulse responses y1(t) = h(t) h(t) = e-t u(t) Black Box y2(t) = h(t-2) x2(t) = δ(t-2) t x3(t) = δ(t-5) EE 203 Circuit Analysis 2 | Spring 2008 | Prof. Kwang W. Oh | EE@SUNY-Buffalo y3(t) = h(t-5) Lecture 24 | Chapter 13 | 5/8 | 8/11 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 Example: LR Circuit Impulse Response (2) Example: LR Circuit Impulse Response (1) Many of following examples use the impulse response of a simple LR (or RC) voltage divider time-domain s-domain s 1 V 1 Vi ⇒ H ( s ) = 0 = s +1 Vi s + 1 vi = x (t ) = δ (t ) ⇒ y (t ) = h(t ) Vo = vo = y (t ) = h(t ) = e −t u(t ) h(t) = e-t u(t) input output, response h(t) δ(t) impulse source system, circuit, black box… EE 203 Circuit Analysis 2 | Spring 2008 | Prof. Kwang W. Oh | EE@SUNY-Buffalo impulse response Lecture 24 | Chapter 13 | 5/8 | 9/11 SMALL for Big Things University at Buffalo nanobioSensors & MicroActuators Learning Lab The State University of New York Example: LR Circuit Impulse Response (3) EE 203 Circuit Analysis 2 | Spring 2008 | Prof. Kwang W. Oh | EE@SUNY-Buffalo Black Box Lecture 24 | Chapter 13 | 5/8 | 11/11 t EE 203 Circuit Analysis 2 | Spring 2008 | Prof. Kwang W. Oh | EE@SUNY-Buffalo Lecture 24 | Chapter 13 | 5/8 | 10/11 ...
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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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