Sinusoidal Steady State Response of Linear Circuits

The RLC Circuit. Transient Response Series RLC circuit The circuit shown on Figure 1 is called the series RLC circuit. We will analyze this circuit in order to determine its transient characteristics once the switch S is closed. S + vR - + vL - L R Vs C + vc - Figure 1 The equation that describes the response of the system is obtained by applying KVL around the mesh vR + vL + vc = Vs (1.1) The current flowing in the circuit is i=C dvc dt (1.2) And thus the voltages vR and vL are given by vR = iR = RC vL = L dvc dt di d 2 vc = LC 2 dt dt (1.3) (1.4) Substituting Equations (1.3) and (1.4) into Equation (1.1) we obtain d 2 vc R dvc 1 1 + + vc = Vs 2 dt L dt LC LC (1.5) The solution to equation (1.5) is the linear combination of the homogeneous and the particular solution vc = vc p + vch The particular solution is 6.071/22.071 Spring 2006, Chaniotakis and Cory 1 vc p = Vs (1.6) And the homogeneous solution satisfies the equation d 2 vch R dvch 1 + + vch = 0 2 dt L dt LC (1.7) Assuming a homogeneous solution is of the form Ae st and by substituting into Equation (1.7) we obtain the characteristic equation s2 + R 1 =0 s+ L LC (1.8) By defining R : 2L α= Damping rate (1.9) And ωο = 1 : Natural frequency LC (1.10) The characteristic equation becomes s 2 + 2α s + ωο2 = 0 (1.11) The roots of the characteristic equation are s1 = −α + α 2 − ωο2 (1.12) s 2 = −α − α 2 − ωο2 (1.13) And the homogeneous solution becomes vch = A1e s1t + A2 e s 2t (1.14) vc = Vs + A1e s1t + A2 e s 2 t (1.15) The total solution now becomes 6.071/22.071 Spring 2006, Chaniotakis and Cory 2 The parameters A1 and A2 are constants and can be determined by the application of the dvc(t = 0) initial conditions of the system vc (t = 0) and . dt The value of the term α 2 − ωο2 determines the behavior of the response. Three types of responses are possible: 1. α = ωο then s1 and s2 are equal and real numbers: no oscillatory behavior Critically Damped System 2. α > ωο . Here s1 and s2 are real numbers but are unequal: no oscillatory behavior Over Damped System vc = Vs + A1e s1t + A2 e s 2 t 3. α < ωο . α 2 − ωο2 = j ωο2 − α 2 In this case the roots s1 and s2 are complex numbers: s1 = −α + j ωο2 − α 2 , s 2 = −α − j ωο2 − α 2 . System exhibits oscillatory behavior Under Damped System Important observations for the series RLC circuit. • • • As the resistance increases the value of α increases and the system is driven towards an over damped response. 1 (rad/sec) is called the natural frequency of the system LC or the resonant frequency. The frequency ωο = R is called the damping rate and its value in relation to ωο 2L determines the behavior of the response o α = ωο : Critically Damped The parameter α = o α > ωο : o α < ωο : • The quantity Over Damped Under Damped L has units of resistance C 6.071/22.071 Spring 2006, Chaniotakis and Cory 3 Figure 2 shows the response of the series RLC circuit with L=47mH, C=47nF and for three different values of R corresponding to the under damped, critically damped and over damped case. We will construct this circuit in the laboratory and examine its behavior in more detail. (a) Under Damped. R=500Ω (b) Critically Damped. R=2000 Ω (c) Over Damped. R=4000 Ω Figure 2 6.071/22.071 Spring 2006, Chaniotakis and Cory 4 The LC circuit. In the limit R → 0 the RLC circuit reduces to the lossless LC circuit shown on Figure 3. S + vL - L C + vc - Figure 3 The equation that describes the response of this circuit is d 2 vc 1 + vc = 0 dt 2 LC (1.16) Assuming a solution of the form Ae st the characteristic equation is Where ωο = s 2 + ωο2 = 0 (1.17) s1 = + jωο (1.18) s 2 = − jωο (1.19) 1 LC The two roots are And the solution is a linear combination of A1e s1t and A2e s 2t vc(t ) = A1e jωot + A2e − jωο t (1.20) By using Euler’s relation Equation (1.20) may also be written as vc(t ) = B1cos(ωο t ) + B 2sin(ωο t ) (1.21) The constants A1, A2 or B1, B2 are determined from the initial conditions of the system. 6.071/22.071 Spring 2006, Chaniotakis and Cory 5 For vc(t = 0) = Vo and for dvc (t = 0) = 0 (no current flowing in the circuit initially) we dt have from Equation (1.20) A1 + A2 = Vo (1.22) jωo A1 − jωo A2 = 0 (1.23) And Which give A1 = A2 = Vo 2 (1.24) And the solution becomes Vo jωot − jωο t e +e 2 = Vo cos(ωot ) vc(t ) = ( ) (1.25) The current flowing in the circuit is dvc dt = −CVoωο sin(ωο t ) i=C (1.26) And the voltage across the inductor is easily determined from KVL or from the element di relation of the inductor vL = L dt vL = −vc = −Vo cos(ωot ) (1.27) Figure 4 shows the plots of vc (t ), vL(t ), and i (t ) . Note the 180 degree phase difference between vc(t) and vL(t) and the 90 degree phase difference between vL(t) and i(t). Figure 5 shows a plot of the energy in the capacitor and the inductor as a function of time. Note that the energy is exchanged between the capacitor and the inductor in this lossless system 6.071/22.071 Spring 2006, Chaniotakis and Cory 6 (a) Voltage across the capacitor (b) Voltage across the inductor (c) Current flowing in the circuit Figure 4 6.071/22.071 Spring 2006, Chaniotakis and Cory 7 (a) Energy stored in the capacitor (b) Energy stored in the inductor Figure 5 6.071/22.071 Spring 2006, Chaniotakis and Cory 8 Parallel RLC Circuit The RLC circuit shown on Figure 6 is called the parallel RLC circuit. It is driven by the DC current source Is whose time evolution is shown on Figure 7. iR(t) R Is iC(t) iL(t) L C + v - Figure 6 Is 0 t Figure 7 Our goal is to determine the current iL(t) and the voltage v(t) for t>0. We proceed as follows: 1. 2. 3. 4. Establish the initial conditions for the system Determine the equation that describes the system characteristics Solve the equation Distinguish the operating characteristics as a function of the circuit element parameters. Since the current Is was zero prior to t=0 the initial conditions are: ⎧iL(t = 0) = 0 Initial Conditions: ⎨ ⎩ v(t = 0) = 0 (1.28) By applying KCl at the indicated node we obtain 6.071/22.071 Spring 2006, Chaniotakis and Cory 9 Is = iR + iL + iC (1.29) The voltage across the elements is given by d iL dt (1.30) v L d iL = R R dt (1.31) dv d 2iL = LC 2 dt dt (1.32) v=L And the currents iR and iC are iR = iC = C Combining Equations (1.29), (1.31), and (1.32) we obtain d 2iL 1 d iL 1 1 + + iL = Is 2 dt RC dt LC LC (1.33) The solution to equation (1.33) is a superposition of the particular and the homogeneous solutions. iL (t ) = iL p (t ) + iLh (t ) (1.34) iL p (t ) = Is (1.35) The particular solution is The homogeneous solution satisfies the equation d 2iLh 1 d iLh 1 + + iLh = 0 2 dt RC dt LC (1.36) By assuming a solution of the form Ae st we obtain the characteristic equation s2 + 1 1 =0 s+ RC LC (1.37) Be defining the following parameters 6.071/22.071 Spring 2006, Chaniotakis and Cory 10 1 : Resonant frequency LC ωο ≡ (1.38) And α= 1 : Damping rate 2RC (1.39) The characteristic equation becomes s 2 + 2α s + ωο2 = 0 (1.40) s1 = −α + α 2 − ωο2 (1.41) s 2 = −α − α 2 − ωο2 (1.42) The two roots of this equation are The homogeneous solution is a linear combination of e s1t and e s 2 t iLh (t ) = A1e s1t + A2 e s 2 t (1.43) iL(t ) = Is + A1e s1t + A2 e s 2 t (1.44) And the general solution becomes The constants A1 and A2 may be determined by using the initial conditions. Let’s now proceed by looking at the physical significance of the parameters α and ωο . 6.071/22.071 Spring 2006, Chaniotakis and Cory 11 The form of the roots s1 and s2 depend on the values of α and ωο . The following three cases are possible. 1. α = ωο : Critically Damped System. s1 and s2 are equal and real numbers: no oscillatory behavior 2. α > ωο : Over Damped System Here s1 and s2 are real numbers but are unequal: no oscillatory behavior 3. α < ωο : Under Damped System α 2 − ωο2 = j ωο2 − α 2 In this case the roots s1 and s2 are complex numbers: s1 = −α + j ωο2 − α 2 , s 2 = −α − j ωο2 − α 2 . System exhibits oscillatory behavior Let’s investigate the under damped case, α < ωο , in more detail. For α < ωο , α 2 − ωο2 = j ωο2 − α 2 ≡ jωd the solution is −α t iL(t ) = Is + eN (A1e jωd t + A2e− jωd t ) Decaying (1.45) Oscillatory By using Euler’s identity e ± jωd t = cos ωd t ± j sin ωd t , the solution becomes −α t iL(t ) = Is + eN (K1 cos ωd t + K 2 sin ωd t ) Decaying (1.46) Oscillatory Now we can determine the constants K1 and K 2 by applying the initial conditions iL(t = 0) = 0 ⇒ Is + K1 = 0 ⇒ K1 = − Is diL = 0 ⇒ −α K1 + (0 + K 2ωd ) = 0 dt t =0 ⇒ K2 = −α ωd (1.47) (1.48) Is And the solution is 6.071/22.071 Spring 2006, Chaniotakis and Cory 12 ⎡ ⎤ ⎢ ⎞⎥ α −α t ⎛ iL(t ) = Is ⎢1 − eN sin ωd t ⎟ ⎥ ⎜ cos ωd t + ωd ⎢ Decaying ⎝ ⎠⎥ ⎢⎣ ⎥⎦ Oscillatory (1.49) ⎛ B ⎞ By using the trigonometric identity B1 cos t + B2 sin t = B12 + B22 cos ⎜ t − tan −1 2 ⎟ the B1 ⎠ ⎝ solution becomes iL(t ) = Is − Is ⎛ ωο −α t α ⎞ e cos ⎜ ωd t − tan −1 ⎟ ωd ωd ⎠ ⎝ (1.50) Recall that ωd ≡ ωο2 − α 2 and thus ωd is always smaller than ωo Let’s now investigate the important limiting case: As R → ∞ , α << ω0 α ≈ 0 , e −α t ≈ 1 ωο And the solution reduces to iL(t ) = Is − Is cos ωot which corresponds to the response of the circuit ωd ≡ ωο2 − α 2 ≈ ωο and tan −1 Is L C The plot of iL(t) is shown on Figure 8 for C=47nF, L=47mH, Is=5A and for R=20kΩ and 8kΩ, The dotted lines indicate the decaying characteristics of the response. For convenience and easy visualization the plot is presented in the normalized time ωο t / π . Note that the peak current through the inductor is greater than the supply current Is. 6.071/22.071 Spring 2006, Chaniotakis and Cory 13 (a) For R=8kΩ (b) For R=20kΩ Figure 8. 6.071/22.071 Spring 2006, Chaniotakis and Cory 14 (a) R=20kΩ 6.071/22.071 Spring 2006, Chaniotakis and Cory 15 (b) R=8kΩ Figure 9 The energy stored in the inductor and the capacitor is shown on Figure 10. Figure 10. Energy as a function of time Figure 11 shows the plot of the response corresponding to the case where α << ω0 . This shows the persistent oscillation for the current iL(t) with frequency ω0 . 6.071/22.071 Spring 2006, Chaniotakis and Cory 16 Figure 11 6.071/22.071 Spring 2006, Chaniotakis and Cory 17 The Critically Damped Response. When α = ωο the two roots of the characteristic equation are equal s1=s2=s. And our assumed solution becomes iL(t ) = A1e st + A2e st = A3e st (1.51) Now we have only one arbitrary constant. This is a problem for our second order system since our two initial conditions can not be satisfied. The problem stems from an incorrect assumption for the solution for this special case. For α = ωο the differential equation of the homogeneous problem becomes d 2iLh d iLh + 2α + α 2iLh = 0 2 dt dt (1.52) The solution of this equation is1 iL (t ) = A1te −α t + A2 e −α t (1.53) Which is a linear combination of the exponential term and an exponential term multiplied by t. d ⎛ di d 2i di ⎞ ⎛ di ⎞ + 2α + α 2i = 0 may be rewritten as ⎜ + α i ⎟ + α ⎜ + α i ⎟ = 0 , by 2 dt ⎝ dt dt dt ⎠ ⎝ dt ⎠ dξ di defining ξ = + α i the equation becomes + αξ = 0 whose solution is ξ = K1e−α t . Therefore dt dt di d eα t + eα tα i = K1 which may be written as (eα t i ) = K1 . By integration we obtain the solution dt dt −α t −α t i = K1te + K 2 e 1 The equation 6.071/22.071 Spring 2006, Chaniotakis and Cory 18 Summary of RLC transient response ωο α Critically Damped Series 1 ωο = LC R α= 2L Parallel 1 ωο = LC 1 α= 2RC α = ωο Response: A1te −α t + A2 e −α t α < ωο Under Damped Response: eN ( K1 cos ωd t + K 2 sin ωd t )  Decaying −α t Oscillatory Where ωd ≡ ωο2 − α 2 Over Damped α > ωο Response: A1e s1t + A2 e s 2t Where s1, 2 = −α ± α 2 − ωο2 6.071/22.071 Spring 2006, Chaniotakis and Cory 19