* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Download Laplace Transforms I..
Survey
Document related concepts
Schmitt trigger wikipedia , lookup
Resistive opto-isolator wikipedia , lookup
Operational amplifier wikipedia , lookup
Lumped element model wikipedia , lookup
Switched-mode power supply wikipedia , lookup
Regenerative circuit wikipedia , lookup
Immunity-aware programming wikipedia , lookup
Topology (electrical circuits) wikipedia , lookup
Valve RF amplifier wikipedia , lookup
Flexible electronics wikipedia , lookup
Electronic engineering wikipedia , lookup
Integrated circuit wikipedia , lookup
Rectiverter wikipedia , lookup
Zobel network wikipedia , lookup
Opto-isolator wikipedia , lookup
RLC circuit wikipedia , lookup
Two-port network wikipedia , lookup
Transcript
Applications of Laplace Transforms Instructor: Chia-Ming Tsai Electronics Engineering National Chiao Tung University Hsinchu, Taiwan, R.O.C. Contents • • • • • • • Introduction Circuit Element Models Circuit Analysis Transfer Functions State Variables Network Stability Summary Introduction • To learn how easy it is to work with circuits in the s domain • To learn the concept of modeling circuits in the s domain • To learn the concept of transfer function in the s domain • To learn how to apply the state variable method for analyzing linear systems with multiple inputs and multiple outputs • To learn how the Laplace transform can be used in stability analysis Circuit Element Models • Steps in applying the Laplace transform: – Transform the circuit from the time domain to the s domain (a new step to be discussed later) – Solve the circuit using circuit analysis technique (nodal/mesh analysis, source transformation, etc.) – Take the inverse Laplace transform of the solution and thus obtain the solution in the time domain s-Domain Models for R and L For a resistor, v(t ) Ri (t ) Lv(t ) LRi (t ) V ( s ) RI ( s ) Time domain For an inductor, di (t ) v(t ) L dt V ( s) L sI ( s) i (0 ) sLI ( s) Li(0 ) 1 i (0 ) or I ( s ) V ( s) sL s s domain s domain s-Domain Model for C For a capacitor, dv(t ) i (t ) C dt I ( s) C sV ( s) v(0 ) sCV ( s) Cv(0 ) 1 v (0 ) or V ( s ) I ( s) sC s Time domain s domain s domain Summary For inductor: For capacitor: V ( s ) sLI ( s ) Li (0 ) I ( s ) sCV ( s ) Cv(0 ) 1 i (0 ) I ( s) V ( s) sL s 1 v (0 ) V (s) I ( s) sC s Summary • Impedance in the s domain – Z(s)=V(s)/I(s) • Admittance in the s domain – Y(s)=1/Z(s)=V(s)/I(s) Element Z(s) Resistor R Inductor sL Capacitor 1/sC *Assuming zero initial conditions Example 1 (1) Transforma tion to the s domain : 1 u (t ) s Z (1 H) sL s 1 1 3 Z ( 3 F) sC s (2) Applying mesh analysis : 1 3 3 s 1 s I1 s I 2 0 3 I1 s 5 3 I 2 0 s s 3 I2 3 s 8s 2 18s 3 Vo ( s ) sI 2 2 s 8s 18 3 2 2 ( s 4) 2 ( 2 ) 2 3 4t vo (t ) e sin 2t t 0 2 Example 2 vo (0) 5 V Example 2 (Cont’d) Applying nodal analysis at node a : 10 ( s 1) Vo Vo Vo 2 0.5 10 10 10 s 25s 35 A B Vo ( s 1)( s 2) s 1 s 2 Applying the residue method, A ( s 1)Vo ( s ) | 10 B ( s 2)Vo ( s ) | 15 s 1 s 2 vo(t) (10e t 15e 2t )u (t ) Example 3 io (0) I 0 V0 I ( s )( R sL ) LI 0 0 s LI 0 V0 I (s) R sL s ( R sL ) V0 R I 0 V0 R s sR L V0 t / V0 R i (t ) I 0 e , R R L Circuit Analysis • Operators (derivatives and integrals) into simple multipliers of s and 1/s • Use algebra to solve the circuit equations • All of the circuit theorems and relationships developed for dc circuits are perfectly valid in the s domain Example 1 10 Vs s i ( 0) 1 A v ( 0) 5 V V1 Vs V1 0 i (0) V1 v(0) s 0 10 3 5s s 1 (0.1s ) 40 5s 35 30 V1 ( s 1)( s 2) s 1 s 2 v1 (t ) 35e t 30e 2t u (t ) Example 2 Solved example 1 by using superposit ion : 30 30 30 V1 ( s 1)( s 2) s 1 s 2 v1 (t ) 30e t 30e 2t u (t ) 10 10 10 V2 ( s 1)( s 2) s 1 s 2 v2 (t ) 10e t 10e 2t u (t ) 5s 5 10 V3 ( s 1)( s 2) s 1 s 2 v3 (t ) 5e t 10e 2t u (t ) v(t ) v1 (t ) v2 (t ) v3 (t ) 35e (30 10 5)e t (30 10 10)e 2t u (t ) t 30e 2t u (t ) Example 3 Assume that no initial energy is stored. (a)Find Vo(s) using Thevenin’s theorem. (b)Find vo(0+) and vo() by apply the initial- and final-value theorems. (c) Determine vo(t). =10u(t) Example 3: (a) 10 50 Voc VTh 5 s s To find Z Th , use Z Th Voc I sc Applying nodal analysis : 10 (V1 2 I x ) 0 V1 0 0 s 5 2s V and I x 1 2s 100 V1 50 V1 , Ix 2s 3 2 s s (2 s 3) V 50 s Z Th oc 2s 3 I sc 50 s (2 s 3) Vo 5 5 125 50 VTh 5 Z Th 5 2 s 3 s s ( s 4) Example 3: (b), (c) Solution (b) : Solution (c) : 125 s ( s 4) The initial - value theorem gives 125 vo (0) lim sVo ( s ) lim 0 s s s 4 The final - value theorem gives 125 A B Vo ( s ) s ( s 4) s s 4 Applying the residue method, Vo ( s ) 125 125 vo () lim sVo ( s ) lim s 0 s 0 s 4 4 125 A sVo ( s )|s 0 4 31.25 125 B ( s 4)Vo ( s )| 31.25 s 4 4 vo (t ) 31.25(1 e 4t )u (t ) Transfer Functions • The transfer function H(s) is the ratio of the output response Y(s) to the input excitation X(s), assuming all initial conditions are zero. Y ( s) H (s) , Y ( s) X ( s) H ( s) X ( s) If x(t ) (t ) , X ( s ) 1 Thus Y ( s ) H ( s ) or y (t ) h(t ) It implies h(t ) : the unit impulse response of the network H ( s ) : the Laplace transform of h(t ) Transfer Functions (Cont’d) • Two ways to find H(s) – Assume an input and find the output – Assume an output and find the input (the ladder method: Ohm’s law + KCL) • Four kinds of transfer functions Vo ( s) I o ( s) H ( s) Voltage gain , H ( s) Current gain Vi ( s) I i ( s) V ( s) H ( s) Impedance I ( s) I ( s) , H ( s) Admittance V ( s) Example 1 If y (t ) 10e t cos 4t u (t ) when x(t ) e t u (t ). Find the transfer function and its impulse response. Solution : 1 10( s 1) X (s) , Y ( s) 2 2 s 1 ( s 1) 4 Y (s) 10( s 1) 4 H ( s) 10 40 2 2 2 2 X ( s) ( s 1) 4 ( s 1) 4 2 h(t ) 10 (t ) 40e sin 4t u (t ) t Example 2 Find H(s)=V0(s)/I0(s). Solution : By current division, ( s 4) I 0 I2 ( s 4) (2 1 2s) 2( s 4) I 0 V0 2 I 2 s 6 1 2s V0 ( s ) 4 s ( s 4) H ( s) 2 I 0 ( s ) 2 s 12 s 1 Example 2 (The Ladder Method) Let V0 1 V, By Ohm' s law, I 2 V0 2 1 2 1 1 4s 1 V1 I 2 2 1 2s 4s 4s V1 4s 1 I1 s4 4 s ( s 4) Applying KCL gives 2 s 2 12 s 1 I 0 I1 I 2 4 s ( s 4) V0 1 4s ( s 4) H (s) 2 I 0 I 0 2s 12 s 1 Example 3 Find (a) H(s) = Vo/Vi, (b) the impulse response, (c) the response when vi(t) = u(t) V, (d) the response when vi(t) = 8cos2t V. Example 3: (a), (b) Solution : By voltage division, 1 Vo Vab s 1 1 || ( s 1) Vab Vi 1 1 || ( s 1) ( s 1) ( s 2) Vi 1 ( s 1) ( s 2) s 1 Vi 2s 3 Vi 1 s 1 Vo Vi s 1 2s 3 2s 3 Vo 1 1 1 H ( s) Vi 2s 3 2 s 3 2 h(t ) 0.5e 3t 2u (t ) Example 3: (c), (d) Sol : (c) Sol : (d) 1 vi (t ) u (t ) Vi ( s ) s Vo ( s ) H ( s )Vi ( s ) 8s vi (t ) 8 cos 2t Vi ( s ) 2 s 4 Vo ( s ) H ( s )Vi ( s ) 4s A Bs C A B 2 3 3 2 s 4 3 3 s s s s 4 s 2 s s 2 2 2 2 24 24 64 1 1 A , B , C A , B 25 25 25 3 3 1 24 1 s 4 2 3t 2 vo (t ) 1 e u (t ) V Vo ( s ) 2 2 3 25 s 3 2 s 4 3 s 4 24 3t 2 4 vo (t ) e cos 2t sin 2t u (t ) V 25 3 1 State Variables • The state variables are those variables which, if known, allow all other system parameters to be determined by using only algebraic equations. • In an electric circuit, the state variables are the inductor current and the capacitor voltage since they collectively describe the energy state of the system. State Variable Method z1 (t ) z (t ) z(t ) 2 zm (t ) The state equation can be arranged as x Ax Bz where x1 (t ) x (t ) the state vector x (t ) 2 representi ng n state variable s xn (t ) y1 (t ) y (t ) 2 y(t ) ) t ( y p dx1 (t ) dt x1 (t ) dx (t ) dt x (t ) 2 x 2 dxn (t ) dt xn (t ) x Ax Bz y Cx Dz State Variable Method (Cont’d) x Ax Bz y Cx Dz Assuming zero initial conditions Y (s) H (s) C ( sI A) 1 B D Z (s) A system matrix B input coupling matrix and applying the Laplace transform , where C output matrix sX( s ) AX( s ) BZ( s ) D feedforwad matrix ( sI A)X( s ) BZ( s ) X( s ) ( sI A) 1 BZ( s ) I : the identity matrix Y( s ) CX( s ) DZ(s) C(sI A) 1 BZ( s ) DZ(s) In most cases, D 0 . So the degree of the numerator of H ( s ) is less than the degree of the denominato r of H ( s ). H ( s ) C ( sI A) 1 B How to Apply State Variable Method • Steps to apply the state variable method to circuit analysis: – Select the inductor current i and capacitor voltage v as the state variables (define vector x, z) – Apply KCL and KVL to obtain a set of first-order differential equations (find matrix A, B) – Obtain the output equation and put the final result in state-space representaion (find matrix C) – H(s)=C(sI-A)-1B Network Stability • A circuit is stable if its impulse response h(t) is bounded as t approaches ; it is unstable if h(t) grows without bound as t approaches . • Two requirements for stability – Degree of N(s) < Degree of D(s) N (s) R( s) n n 1 If H ( s) k n s k n 1s k1s k0 D( s ) D( s ) lim h(t ) t if n 1 – All the poles must lie in the left half of the s plane e pi t e ( i ji )t e it 0 only if i 0 Network Stability (Cont’d) • A circuit is stable when all the poles of its transfer function H(s) lie in the left half of the s plane. • Circuits composed of passive elements (R, L, and C) and independent sources either are stable or have poles with zero real parts. • Active circuits or passive circuits with controlled sources can supply energy, and can be unstable. Example 1 Vo 1 sC H ( s) Vs R sL 1 sC 1L 2 s s R L 1 LC p1, 2 2 02 R 2 L where 1 0 LC For R, L, C 0 : 0 (stable) For R 0 : 0 (unstable) Example 2 Applying mesh analysis gives Find k for a stable circuit. 1 I2 Vi R sC I1 sC 0 R 1 I I1 kI 0 2 1 sC sC 1 1 R I1 V i sC sC 0 k 1 R 1 I 2 sC sC The determinan t is 2 1 1 1 R k sC sC sC sR 2C 2 R k sC Let 0 , the single pole is given as k 2R R 2C For stable operation, p k 2R p 2 0 RC k 2R Summary • The methodology of circuit analysis using Laplace transform – Convert each element to its s-domain model – Obtain the s-domin solution – Apply the inverse Laplace transform to obtain the t-domain solution Summary • The transfer function H(s) of a network is the Laplace transform of the impulse response h(t) Y ( s) H (s) , Y ( s) X ( s) H ( s) X ( s) • A circuit is stable when all the poles of its transfer function H(s) lie in the left half of the s plane.