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For node of V, V V. (1e jex R For node V2 + jaC(V, -V,)+ j@C(V, –V.)=0 26 + jet jaCV, + jeCV, (17 For node of V. V. -v.)+ joc(V, -V,)-0 (18 Simplifying the above equations and using the substitution X = cRC (19 We get V, = (20 41+ jx }(V, +V_) jX V, (21 2(1+ jX )(V, +V.) (22 V.-E xV + jXv,) Substituting equations (20) & (21) into Eq (22) and solving for VV| we get the transfer function V X-1 (23 V X-4 jX -1 It is clear that the value of |VVị becomes zero at X=1 which gives RC =1 from which the center frequency fe is f. = 2RC (24 The magnitude frequency response is v. V (x -1} +16X² x-1 (25 From which we can find the cutoff frequencies 27 5-2 fi = V5+2 2RC f = (2e 2RC

For node of V, V V. (1e jex R For node V2 + jaC(V, -V,)+ j@C(V, –V.)=0 26 + jet jaCV, + jeCV, (17 For node of V. V. -v.)+ joc(V, -V,)-0 (18 Simplifying the above equations and using the substitution X = cRC (19 We get V, = (20 41+ jx }(V, +V_) jX V, (21 2(1+ jX )(V, +V.) (22 V.-E xV + jXv,) Substituting equations (20) & (21) into Eq (22) and solving for VV| we get the transfer function V X-1 (23 V X-4 jX -1 It is clear that the value of |VVị becomes zero at X=1 which gives RC =1 from which the center frequency fe is f. = 2RC (24 The magnitude frequency response is v. V (x -1} +16X² x-1 (25 From which we can find the cutoff frequencies 27 5-2 fi = V5+2 2RC f = (2e 2RC

Introductory Circuit Analysis (13th Edition)
Introductory Circuit Analysis (13th Edition)
13th Edition
ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
Chapter1: Introduction
Section: Chapter Questions
Problem 1P: Visit your local library (at school or home) and describe the extent to which it provides literature...
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For node of V 2 jøCV, +(V, -V.)+(V, -v.)=0 V, V. + joc (16) For node V2 + joC(v, -V,)+ joC(V, -v.)= 0 26 joc jaCV, + j@CV, (17) For node of V. V. -v.)+ jac(V, -V,)=0 + jo + jaCV, (18) Simplifying the above equations and using the substitution X = RC (19) We get (20) 2(1+ jX) jX V, = (21) 2(1+ jx}W, +V.) (V, + jXV;) (1+, (22) Substituting equations (20) & (21) into Eq (22) and solving for V V we get the transfer function V. X-1 (23) V, X-4 jX -1 It is clear that the value of |V/V becomes zero at X=1 which gives RC =1 from which the center frequency fe is f. 2RC (24) The magnitude frequency response is |v. V (x-1) +16Xx² x'-1 (25) From which we can find the cutoff frequencies 27 V5-2 V5+2 fi = 2RC (26) 2.RC
Transcribed Image Text:For node of V 2 jøCV, +(V, -V.)+(V, -v.)=0 V, V. + joc (16) For node V2 + joC(v, -V,)+ joC(V, -v.)= 0 26 joc jaCV, + j@CV, (17) For node of V. V. -v.)+ jac(V, -V,)=0 + jo + jaCV, (18) Simplifying the above equations and using the substitution X = RC (19) We get (20) 2(1+ jX) jX V, = (21) 2(1+ jx}W, +V.) (V, + jXV;) (1+, (22) Substituting equations (20) & (21) into Eq (22) and solving for V V we get the transfer function V. X-1 (23) V, X-4 jX -1 It is clear that the value of |V/V becomes zero at X=1 which gives RC =1 from which the center frequency fe is f. 2RC (24) The magnitude frequency response is |v. V (x-1) +16Xx² x'-1 (25) From which we can find the cutoff frequencies 27 V5-2 V5+2 fi = 2RC (26) 2.RC
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