Đang tải... (xem toàn văn)
phân tích mạch trong Miền S của mạch điện tử phân tích mạch trong Miền S của mạch điện tử phân tích mạch trong Miền S của mạch điện tử phân tích mạch trong Miền S của mạch điện tử phân tích mạch trong Miền S của mạch điện tử phân tích mạch trong Miền S của mạch điện tử phân tích mạch trong Miền S của mạch điện tử phân tích mạch trong Miền S của mạch điện tử phân tích mạch trong Miền S của mạch điện tử phân tích mạch trong Miền S của mạch điện tử phân tích mạch trong Miền S của mạch điện tử phân tích mạch trong Miền S của mạch điện tử
S-Domain Analysis s-Domain Circuit Analysis Time domain (t domain) Linear Circuit Differential equation Complex frequency domain (s domain) Laplace Transform L Laplace Transform L Classical techniques Response waveform Transformed Circuit Algebraic equation Algebraic techniques Inverse Transform L-1 Response transform Kirchhoff’s Laws in s-Domain t domain s domain i2 (t ) Kirchhoff’s current law (KCL) i1 (t ) i3 (t ) i4 (t ) i1 (t ) + i2 (t ) − i3 (t ) + i4 (t ) = I1 ( s ) + I ( s ) − I ( s ) + I ( s ) = + v2 (t ) − Kirchhoff’s voltage law (KVL) − v1 (t ) + v2 (t ) + v3 (t ) = + v4 (t ) − + + + v1 (t ) v3 (t ) v5 (t ) − − − − V1 ( s ) + V2 ( s ) + V3 ( s ) = Signal Sources in s Domain t domain s domain i (t ) + v(t ) v(t ) = vS (t ) i (t ) = depends _ on circuit vS (t ) L V (s ) V ( s ) = VS ( s ) I ( s ) = depends on circuit I (s ) _ + _ i (t ) = iS (t ) v(t ) v(t ) = depends + on circuit VS (s ) _ i (t ) Current Source: Voltage Source: + _ + Voltage Source: I (s ) _ iS (t ) L V (s ) + Current Source: I S (s) I ( s ) = VS ( s ) V ( s ) = depends on circuit Time and s-Domain Element Models Impedance and Voltage Source for Initial Conditions Resistor: vR (t ) = RiR (t ) Inductor: diL (t ) vL (t ) = L dt Capacitor: iR (t ) s-Domain I R (s ) + + vR (t ) R L _ VR (s ) R _ iL (t ) I L (s ) + vL (t ) _ L L iC (t ) + t vC (t ) = ∫ iC (τ )dτ vC (t ) C _ + vC (0) Ls VL (s ) _ _ + I C (s ) L VC (s ) _ VR ( s ) = RI R ( s ) LiL (0) VL ( s ) = LsI L ( s ) − LiL (0) Capacitor: + C Resistor: Inductor: + _ + Time Domain Cs VC ( s ) = I C ( s) + Cs vC (0) vC( ) s s Impedance and Voltage Source for Initial Conditions • Impedance Z(s) Z ( s) = voltage transform current transform with all initial conditions set to zero • Impedance of the three passive elements Z R ( s) = VR ( s ) =R I R (s) VL ( s ) = Ls I L (s) with iL (0) = VC ( s ) = I C ( s ) Cs with vC ( ) = Z L ( s) = Z C (s) = Time and s-Domain Element Models Admittance and Current Source for Initial Conditions Time Domain Resistor: iR (t ) = vR (t ) R Inductor: iR (t ) + + vR (t ) iC (t ) = C dvC (t ) dt R L _ VR (s ) Resistor: R iL (t ) I L (s ) + L VL (s ) Ls _ iC (t ) I C (s ) + vC (t ) _ Inductor: + L + C L VC (s ) _ Cs VR ( s ) R I R (s) = _ t iL (t ) = ∫ vL (τ ) dτ vL (t ) L _ + iL (0) Capacitor: s-Domain I R (s ) I L (s) = i L ( 0) s VL ( s ) + Ls i L( ) s Capacitor: I C ( s ) = CsVC ( s ) − CvC (0) CvC (0) Admittance and Current Source for Initial Conditions • Admittance Y(s) Y ( s) = current transform = voltage transform Z ( s) with all initial conditions set to zero • Admittance of the three passive elements YR ( s ) = I R ( s) = VR ( s ) R YL ( s ) = I L ( s) = VL ( s ) Ls with iL (0) = YC ( s ) = I C ( s) = Cs VC ( s ) with vC( ) = Example: Solve for Current Waveform i(t) R L i (t ) L VA s + VR (s ) − _ + _ + V Au (t ) R I (s ) Ls _ + + VL (s ) LiL (0) _ VA By KVL: − + VR ( s ) + VL ( s ) = s Resistor: VR ( s ) = RI ( s ) Inductor: VL ( s ) = LsI ( s ) − LiL (0) V − A + RI ( s) + LsI ( s) − LiL (0) = s VA L iL (0) + I (s) = s ( s + R L) s + R L VA R VA R i L ( 0) = − + s s+R L s+R L R − t V A V A − RL t Inverse Transform: i (t ) = − e + iL (0)e L u (t ) R R forced response natural response Series Equivalence and Voltage Division I1 ( s ) I (s ) Rest of Circuit + V1 ( s ) − Z1 + V (s) − V1 ( s ) = Z1 ( s ) I1 ( s ) = Z1 ( s ) I ( s ) + Z2 V2 ( s) I ( s ) − V2 ( s ) = Z ( s ) I ( s ) = Z ( s ) I ( s ) KVL: V ( s ) = V1 ( s ) + V2 ( s ) = ( Z1 ( s ) + Z ( s )) I ( s ) I (s ) Rest of Circuit + V (s) Z EQ − Z EQ = Z1 + Z Z EQ ( s ) = Z1 ( s ) + Z ( s ) V1 ( s) = Z1 ( s ) V (s) Z EQ ( s ) V2 ( s ) = Z (s) V (s) Z EQ ( s) Parallel Equivalence and Current Division I1 ( s ) = Y1 ( s )V ( s ) I (s ) Rest of Circuit + I1 ( s ) I (s) V (s) Y Y2 − I ( s ) = Y2 ( s )V ( s ) KCL: I ( s ) = I1 ( s ) + I ( s ) = (Y1 ( s ) + Y2 ( s ))V ( s ) I (s ) Rest of Circuit + V (s) Y EQ − YEQ = Y1 + Y2 YEQ ( s ) = Y1 ( s ) + Y2 ( s ) I1 ( s ) = Y1 ( s ) I ( s) YEQ ( s ) I (s) = Y2 ( s) I (s) YEQ ( s ) Example: Equivalence Impedance and Admittance A L v1 (t ) R _ + B L A V1 ( s ) B Z EQ ( s ) = Ls + Z EQ1 ( s) = Ls + Ls Z EQ ZR EQ1 Z EQ Inductor current = at t = + capacitor voltage = C v2 (t ) Find equivalent impedance at A and B _ Solve for v2(t) RCs + 1 = + Cs = YEQ1 ( s ) = Z EQ1 ( s ) R R Z EQ1 RLCs + Ls + R + = RCs + 1 V2 ( s ) Z EQ1 ( s ) Cs _ V2 ( s ) = V1 ( s ) Z EQ R = V1 ( s) RCLs + Ls + R R RCs + _ + General Techniques for s-Domain Circuit Analysis • Node Voltage Analysis (in s-domain) – – – – Use Kirchhoff’s Current Law (KCL) Get equations of node voltages Use current sources for initial conditions Voltage source current source • Mesh Current Analysis (in s-domain) – – – – Use Kirchhoff’s Voltage Law (KVL) Get equations of currents in the mesh Use voltage sources for initial conditions Current source voltage source (Works only for “Planar” circuits) Formulating Node-Voltage Equations Step 0: Transform the circuit into the s domain using current sources to represent capacitor and inductor initial conditions Step 1: Select a reference node Identify a node voltage at each of the non-reference nodes and a current with every element in the circuit Step 2: Write KCL connection constraints in terms of the element currents at the non-reference nodes Step 3: Use the element admittances and the fundamental property of node voltages to express the element currents in terms of the node voltages Step 4: Substitute the device constraints from Step into the KCL connection constraints from Step and arrange the resulting equations in a standard form Example: Formulating Node-Voltage Equations L iS (t ) R C Step 0: Transform the circuit into the s domain using current sources to represent capacitor and inductor initial conditions Step 1: Identify N-1=2 node voltages and a current with each element t domain VA (s ) I ( s) I S (s) Reference node Step 2: Apply KCL at nodes A and B: iL (0) − − I1 ( s ) − I ( s ) = I s Node A : ( ) L S s iL (0) i (0) + I1 ( s ) − I ( s ) = Node B : CvC (0) + L s s Step 3: Express element equations in terms of node Ls VB (s ) voltages I1 ( s ) I ( s ) I1 ( s ) = YL ( s )[VA ( s ) − VB ( s )] = [VA ( s ) − VB ( s )] Ls R Cs CvC (0) I ( s ) = YR ( s)VA ( s ) = GVA ( s) where G = R s domain I ( s ) = YC ( s )VB ( s) = CsVB ( s ) Formulating Node-Voltage Equations (Cont’d) Step 2: Apply KCL at nodes A and B: iL (0) − I1 ( s ) − I ( s ) = Node A : I S ( s ) − s i (0) + I1 ( s ) − I ( s ) = Node B : CvC (0) + L s Step 3: Express element equations in terms of node voltages I1 ( s ) = YL ( s)[VA ( s) − VB ( s)] = [VA ( s ) − VB ( s )] Ls I ( s ) = YR ( s )VA ( s) = GVA ( s ) where G = R I ( s ) = YC ( s )VB ( s ) = CsVB ( s) Step 4: Substitute eqns in Step into eqns in Step and collect common terms to yield node-voltage eqns i (0) Node A : G + VA ( s ) − VB ( s ) = I S ( s ) − L s Ls Ls i (0) Node B : − VA ( s ) + + Cs VB ( s) = CvC (0) + L s Ls Ls Solving s-Domain Circuit Equations G + Ls − Ls • Circuit Determinant: ∆( s ) = − Ls Cs + Ls = (G + Ls )(Cs + Ls ) − (1 Ls ) GLCs + Cs + G = Ls Depends on circuit element parameters: L, C, G=1/R, not on driving force and initial conditions • Solve for node A using Cramer’s rule: − Ls I S ( s ) + i L ( 0) s ∆ A ( s) iL (0) s + CvC (0) Cs + Ls = VA ( s ) = ∆(s) ∆(s) ( LCs + 1) I S ( s) − LCsiL (0) + CvC (0) = + GLCs + Cs + G GLCs + Cs + G Zero State when initial condition sources are turned off Zero input when input sources are turned off Solving s-Domain Circuit Eqns (Cont’d) • Solve for node B using Cramer’s rule: G + Ls I S ( s ) − iL (0) s − Ls iL (0) s + CvC (0) ∆ B (s) = ∆( s) ∆( s) I S ( s) GLiL (0) + (GLs + 1)CvC (0) = + GLCs + Cs + G GLCs + Cs + G VB ( s ) = Zero State Zero input Network Functions Network function = Zero - state Response Transform Input Signal Transform • Driving-point function relates the voltage and current at a given pair of terminals called a port V (s) = Z ( s) = I ( s) Y ( s) • Transfer function relates an input and response at different ports in the circuit V (s) TV ( s ) = Voltage Transfer Function = V1 ( s ) I (s) TI ( s ) = Current Transfer Function = V1 I1 ( s ) I ( s) In TY ( s) = Transfer Admittance = V1 ( s) V (s) V1 TZ ( s) = Transfer Impedance = I1 ( s ) _ + _ + In I (s ) + Circuit in the zero-state V (s ) − Circuit Output in the V1 or I1 V or I zero-state 2 Input I2 + V2 − TV (s ) Out I1 TI (s ) In Out I2 TY (s ) + V2 − I1 TZ (s ) Out In Out Calculating Network Functions Z1 V1 ( s ) + Z2 V2 ( s) − I (s) I1 ( s ) Y1 Y2 • Driving-point impedance Z EQ ( s ) = Z1 ( s) + Z ( s ) • Voltage transfer function: Z (s) V2 ( s ) = V1 ( s ) Z1 ( s ) + Z ( s ) V (s) Z ( s) TV ( s) = = V1 ( s ) Z1 ( s ) + Z ( s ) • Driving-point admittance YEQ ( s ) = Y1 ( s ) + Y2 ( s) • Voltage transfer function: Y2 ( s ) I (s) = Y1 ( s) Y1 ( s) + Y2 ( s ) I (s) Y2 ( s ) TI ( s ) = = I1 ( s ) Y1 ( s ) + Y2 ( s) _ + Impulse Response and Step Response • Input-output relationship in s-domain Input Y ( s) = T ( s) X ( s) X (s ) • When input signal is an impulse x(t ) = δ (t ) Y ( s) = T ( s) ×1 = T ( s) T(s) Circuit Output Y (s ) – Impulse response equals network function – H(s) = impulse response transform – h(t) = impulse response waveform • When input signal is a step x(t ) = u (t ) – G(s) = step response transform – g(t) = step response waveform T (s) H (s) = G (s) = s s g ( s) = ∫ h(τ )dτ , t (=) means equal almost everywhere, dg (t ) excludes those points at which g(t) h(t )(=) dt has a discontinuity ... YL ( s) [VA ( s) − VB ( s) ] = [VA ( s ) − VB ( s )] Ls I ( s ) = YR ( s )VA ( s) = GVA ( s ) where G = R I ( s ) = YC ( s )VB ( s ) = CsVB ( s) Step 4: Substitute eqns in Step into eqns in Step... conditions • Solve for node A using Cramer s rule: − Ls I S ( s ) + i L ( 0) s ∆ A ( s) iL (0) s + CvC (0) Cs + Ls = VA ( s ) = ∆ (s) ∆ (s) ( LCs + 1) I S ( s) − LCsiL (0) + CvC (0) = + GLCs + Cs +... H (s) = impulse response transform – h(t) = impulse response waveform • When input signal is a step x(t ) = u (t ) – G (s) = step response transform – g(t) = step response waveform T (s) H (s)