EE 422G Notes: Chapter 6 Instructor: Cheung Page 6-1Applications of the Laplace Transform Application in Circuit Analysis 1. Review of Resistive Network 1) Elements 2) Superposition 1 PDF Created with deskPDF PDF Writer - Trial :: http://www.docudesk.comEE 422G Notes: Chapter 6 Instructor: Cheung Page 6-23) KVL and KCL – Select a node for ground. Watch out for signs! 4) Equivalent Circuits Open Circuit Voltages OCV V= = Rs = Equivalent Resistance Short Circuit Currents SCI I= = Thevenin Equivalent Circuit Norton Equivalent Circuit Rs = Same as before PDF Created with deskPDF PDF Writer - Trial :: http://www.docudesk.comEE 422G Notes: Chapter 6 Instructor: Cheung Page 6-35) Nodal Analysis and Mesh Analysis Mesh analysis (use KVL) ++=−+−+=224111213212111)()(SSSVIRIRVIIRIIRIRV Solve for I1 and I2. 2. Characteristics of Dynamic Networks 1) Inductor 2) Capacitor ∫∞−==tCCCCdiCtvtvdtdCtiττ)(1)(or )()( ∫∞−==tLLLLdvLtitidtdLtvττ)(1)(or )()( (Use KCL) PDF Created with deskPDF PDF Writer - Trial :: http://www.docudesk.comEE 422G Notes: Chapter 6 Instructor: Cheung Page 6-43) Operation Amplifier A general op-amp model is described above. In practice, the input resistance, Rin, is very large (> 1012 Ω) and the gain, A, is very large (>105). Thus, we will use the ideal model in the analysis: 1. Input current Ii(t) = 0 (due to the large input impedence) 2. Input voltage difference vi(t) = 0 and output voltage vo(t) is dictated by external circuit (due to the large gain) Based on the ideal op-amp model, v2(t) = v1(t) (1) Also, as the op-amp does not have any input current, applying KCL at the inverting port, we have v2(t)/Ra= (vo(t)-v2(t))/Rb vo(t)/v2(t) = 1+Rb/Ra Plug in (1), we have vo(t)/v1(t) = 1+Rb/Ra This circuit is called Non-Inverting Amplifier. - + ≡ + - + - vo(t) vi(t) + - vi(t) - + - Avi(t) Rin Ii(t) Inverting input Non-Inverting input - + + - v1(t) Ra Rb vo(t) io(t) v2(t) + - + - vo(t) Example: PDF Created with deskPDF PDF Writer - Trial :: http://www.docudesk.comEE 422G Notes: Chapter 6 Instructor: Cheung Page 6-54) Mutual Inductor – used in transformer. Two separate circuits with coupling currents. To link the two circuits together, introduce a combined current term (i1+i2): dtdiMLiidtdMdtdiMdtdiLdtdiMdtdiMtviidtdMdtdiMLdtdiMdtdiMdtdiMdtdiLtv22212222122111211111)()()()()()(−++=−++=++−=++−= Equivalent circuit: Make sure both i1 and i2 point either away or toward the polarity marks to make the mutual inductance M positive. PDF Created with deskPDF PDF Writer - Trial :: http://www.docudesk.comEE 422G Notes: Chapter 6 Instructor: Cheung Page 6-6Example : Apply mesh analysis to the following circuit Using Laplace Transform )()(1)()()()0()(1))0()(()()()()(sRIsICssILssRIsvssICissILsVsVsVsVCRCL++=+++−=++=− Define ‘Generalized Resistors’ (Impedances) )()()()( sZsIsVLssZLLL=⇒= )()()(1)( sZsIsVCssZCCC=⇒= Both capacitor and inductor behave exactly like a resistor! RsZsZsVsIsRIsIsZsIsZsVCLsCLs++=⇒++=⇒)()()()()()()()()()( Everything we know about resistive network can be applied to dynamic network in Laplace domain: analysismesh and analysis Nodalcircuit EquivalentKCL and KVLionsuperpositLaw Ohms dGeneralize PDF Created with deskPDF PDF Writer - Trial :: http://www.docudesk.comEE 422G Notes: Chapter 6 Instructor: Cheung Page 6-73. Laplace transform models of circuit elements. What if the initial conditions are not zero? 1) Capacitor Alternatively, you can also represent it as an impedance and a parallel current source (Norton equivalent circuit) BE VERY CAREFUL ABOUT THE POLARITY OF VOLTAGE SOURCE AND THE DIRECTION OF CURRENT SOURCE!! ZC Cv(0-) I(s) + - V(s) PDF Created with deskPDF PDF Writer - Trial :: http://www.docudesk.comEE 422G Notes: Chapter 6 Instructor: Cheung Page 6-82) Inductor Alternatively, you can also represent it as an impedance and a parallel current source (Norton equivalent circuit) 3) Resistor V(s) = RI(s) 4) Voltage and Current Sources (Don’t forget to apply Laplace Transform on them) 5) Op-Amp : same ideal model assumption ZL i(0-)/s I(s) + - V(s) PDF Created with deskPDF PDF Writer - Trial :: http://www.docudesk.comEE 422G Notes: Chapter 6 Instructor: Cheung Page 6-96) Mutual Inductance (Transformers) ⇓ Laplace transform model: Obtain it by using inductance model PDF Created with deskPDF PDF Writer - Trial :: http://www.docudesk.comEE 422G Notes: Chapter 6 Instructor: Cheung Page 6-10Example: Find Complex Norton Equivalent circuit given 0)0( =−cv Solution 1) Compute the Short-Circuit Current scsIsI =)( Straightforward to see: )(2)( sIsIS−= To compute I(s), apply mesh analysis on the left loop: 32)(2)(31)()(3)()31(1)(3 1)(1+−=−=⇒+=⇒+=+=+Ω×=ssIsIssIsIsssIsssIsIss No need to do inverse Laplace transform as the equivalent circuit is in the s-domain. PDF Created with deskPDF PDF Writer - Trial :: http://www.docudesk.comEE 422G Notes: Chapter 6 Instructor: Cheung Page 6-112) Find the equivalent impedance sZ Normally, we can just kill all the independent sources and combine the impedances (using resistive combination rules). However, as there is a dependent source, we need to drive it with a test voltage: )(2)()()(sIsVsIsVZtesttesttests== Mesh analysis on the left loop: 0)()(3)() 1(0)(3 1)(=⇒−=Ω⇒=+Ω×sIsIssIssIsI So we got an interesting