EE203-SUNYBuffalo-23-Chapter13-04

EE203-SUNYBuffalo-23-Chapter13-04 - 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 Use of Superposition (1) EE 203 Circuit Analysis 2 Lecture 23 Chapter 13.3 Applications (continued) 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 23 | Chapter 13 | 4/8 | 1/12 We We assume that at the instant when the two sources are applied to the circuit, the inductor is carrying an initial current of ρ amperes and that the capacitor is carrying an initial voltage of γ volts. The desired response of the circuit is the voltage across the resistor resistor R2, labeled v2. We opted for the parallel equivalents for L and C because we anticipated for solving for V2 using the node-voltage method. EE 203 Circuit Analysis 2 | Spring 2008 | Prof. Kwang W. Oh | EE@SUNY-Buffalo Lecture 23 | Chapter 13 | 4/8 | 2/12 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 Use of Superposition (2) Use of Superposition (3) KCL1: I R1 = I L + I C KCL2: I C = I R 2 ⇒ ⇒ Vg − V1' R1 = V1' V1' − V2' + sL 1 sC V − V2' V2' = 1 sC C R2 ' 1 Y11 Vg R1 − Y12 Vg R1 Y 0 = V2' = 12 2 Y11 Y12 Y11Y22 − Y12 Y12 Y22 EE 203 Circuit Analysis 2 | Spring 2008 | Prof. Kwang W. Oh | EE@SUNY-Buffalo Lecture 23 | Chapter 13 | 4/8 | 3/12 EE 203 Circuit Analysis 2 | Spring 2008 | Prof. Kwang W. Oh | EE@SUNY-Buffalo Lecture 23 | Chapter 13 | 4/8 | 4/12 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 Use of Superposition (4) KCL1: I R1 = I L + I C ⇒ Use of Superposition (5) − V1'' V1'' V1'' − V2'' = + R1 sL 1 sC KCL1: I R1 = I L + I C + I ρ ⇒ s V '' − V2'' V2'' KCL2: I C = I R2 + (− I g ) ⇒ 1 = − Ig 1 sC R2 1 1 1 1 V1'' ( + + ) + V2'' (− )=0 1 sC R1 sL 1 sC V1'' (− V1''' ( 1 1 1 ) + V2'' ( + ) = Ig 1 sC R1 1 sC Y11 V2'' = KCL2: Ig ⇒ V1''' − V2''' V2''' = 1 sC R2 g 1 1 1 1 ρ ++ ) + V2''' (− )=− 1 sC R1 sL 1 sC s V1''' (− 1 1 1 ) + V2''' ( + )=0 1 sC R1 1 sC 0 Y12 I C = I R2 − V1''' V1''' V1''' − V2''' ρ = + + R1 sL 1 sC s Y11 Y12 Y12 Y22 Y11 = − ρ ρ s Y12 Y12 0 s V= = 2 Y11 Y12 Y11Y22 − Y12 Y12 Y22 Y11 I g ' '' 2 2 Y11Y22 − Y12 EE 203 Circuit Analysis 2 | Spring 2008 | Prof. Kwang W. Oh | EE@SUNY-Buffalo Lecture 23 | Chapter 13 | 4/8 | 5/12 EE 203 Circuit Analysis 2 | Spring 2008 | Prof. Kwang W. Oh | EE@SUNY-Buffalo Lecture 23 | Chapter 13 | 4/8 | 6/12 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 Use of Superposition (6) KCL1: I R1 + I γC = I L + I C ⇒ − V1'''' V '''' V '''' − V2'''' +γC = 1 + 1 R1 sL 1 sC KCL2: I C = I R + I γC 2 Use of Superposition (7) V1'''' − V2'''' V2'''' = +γC 1 sC R2 g V1''' ( ⇒ 1 1 1 1 ++ ) + V2''' (− ) =γC 1 sC R1 sL 1 sC V1''' (− 1 1 1 ) + V2''' ( + ) = −γ C 1 sC R1 1 sC Y11 Y V2''' = 12 Y11 Y12 γC −γ C Y12 Y22 = Y12 (−γ C ) − γ CY12 2 Y11Y22 − Y12 EE 203 Circuit Analysis 2 | Spring 2008 | Prof. Kwang W. Oh | EE@SUNY-Buffalo Lecture 23 | Chapter 13 | 4/8 | 7/12 EE 203 Circuit Analysis 2 | Spring 2008 | Prof. Kwang W. Oh | EE@SUNY-Buffalo Lecture 23 | Chapter 13 | 4/8 | 8/12 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 Use of Superposition (8) KCL1: I R1 + I γC = I L + I C + I ρ ⇒ Vg − V1 R1 s KCL2: I C = I R + I γC + (− I g ) 2 Y11 V2 = Y12 Vg R1 +γC − Ig −γ C Y11 Y12 Y12 Y22 ⇒ +γC = EE 203 Circuit Analysis 2 Lecture 23 Chapter 13.4 Transfer Function V1 V1 − V2 ρ + + sL 1 sC s V1 − V2 V2 = + γ C − Ig 1 sC R2 g 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 ρ s Y11 ( I g − γ C ) − Y12 ( = Vg R1 2 Y11Y22 − Y12 +γC − ρ s ) EE 203 Circuit Analysis 2 | Spring 2008 | Prof. Kwang W. Oh | EE@SUNY-Buffalo Lecture 23 | Chapter 13 | 4/8 | 9/12 EE 203 Circuit Analysis 2 | Spring 2008 | Prof. Kwang W. Oh | EE@SUNY-Buffalo Lecture 23 | Chapter 13 | 4/8 | 10/12 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 13.1 Transfer Function The The transfer function (H(s)) is defined as the s-domain ratio of the Laplace transform of the output (response, Y(s)) to the Laplace transform of the input (source, X(s)). In computing the transfer function, we restrict our attention to circuits where all initial conditions are zero. If a circuit has multiple independent sources, we can find the transfer function for each source and use superposition to find the response to all sources. Note that the transfer function depends on what is defined as the output signal. If the current I is defined as the response signal of the circuit If the voltage V is defined as the output signal of the circuit EE 203 Circuit Analysis 2 | Spring 2008 | Prof. Kwang W. Oh | EE@SUNY-Buffalo Lecture 23 | Chapter 13 | 4/8 | 11/12 Vg − Vo 1000 = Vo V +o 250 + 0.05s 106 s Vg 1 1 s + + 6)= 1000 1000 250 + 0.05s 10 1 V 1000(250 + 0.05s )(106 ) 1000 H (s) = o = × s 1000(250 + 0.05s )(106 ) 1 1 Vg + + 1000 250 + 0.05s 106 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 23 | Chapter 13 | 4/8 | 12/12 ...
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