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Sinusoidal Steady State Response of Linear Circuits

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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 - R L + 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 dvc iC= (1.2) dt And thus the voltages vR and vL are given by dvc vR ==iR RC (1.3) dt di d 2vc vL ==L LC (1.4) dt dt 2 Substituting Equations (1.3) and (1.4) into Equation (1.1) we obtain d 2vc R dvc 11 ++vc =Vs (1.5) dt 2 L dt LC LC The solution to equation (1.5) is the linear combination of the homogeneous and the particular solution vc =vcph+vc The particular solution is 6.071/22.071 Spring 2006, Chaniotakis and Cory 1 vcp = Vs (1.6) And the homogeneous solution satisfies the equation dv2 c R dvc 1 hh+ +vc =0 (1.7) dt 2 L dt LC h Assuming a homogeneous solution is of the form Aest and by substituting into Equation (1.7) we obtain the characteristic equation R 1 ss2 + +=0 (1.8) L LC By defining R α = : Damping rate (1.9) 2L And 1 ω = : Natural frequency (1.10) ο LC The characteristic equation becomes 22 ss+ 2αω+ο =0 (1.11) The roots of the characteristic equation are 22 s1=−α + α−ωο (1.12) 22 s2 =−α − α−ωο (1.13) And the homogeneous solution becomes s1ts2t vch =+A1e A2e (1.14) The total solution now becomes s1ts2t vc =+Vs A1e +A2e (1.15) 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(0t = ) initial conditions of the system vc(t = 0) and . dt 22 The value of the term α −ωο 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 s12tst vc =+Vs A12e +A e 22 22 3. α < ωο . α −=ωωοj ο−α In this case the roots s1 and s2 are complex 22 22 numbers: sj1,=−α + ωαο− s2=−α−jωο−α. 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 • The frequency ω = (rad/sec) is called the natural frequency of the system ο LC or the resonant frequency. R • The parameter α = is called the damping rate and its value in relation to ω 2L ο determines the behavior of the response o α = ωο : Critically Damped o α > ωο : Over Damped o α < ωο : Under Damped L • The quantity 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 dv2 c 1 + vc = 0 (1.16) dt 2 LC Assuming a solution of the form Aest the characteristic equation is 22 s + ωο = 0 (1.17) 1 Where ω = ο LC The two roots are s1= + jωο (1.18) s2= − jωο (1.19) And the solution is a linear combination of A1es1t and A2es2t vc(t)=+A1e jtωo A2e− jωο t (1.20) By using Euler’s relation Equation (1.20) may also be written as vc(t)= B1cos()ωοt + B2sin(ωοt) (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 dvc(0t = ) For vc(0t ==)Vo and for = 0 (no current flowing in the circuit initially) we dt have from Equation (1.20) AA12+ =Vo (1.22) And jAωo12− jAωo= 0 (1.23) Which give Vo AA12== (1.24) 2 And the solution becomes Vo vc()t =+()e jtωωo e− jοt 2 (1.25) = Vo cos(ωot) The current flowing in the circuit is dvc iC= dt (1.26) =−CVoωοοsin(ω t) 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 (1.27) =−Vocos(ωot) 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. iC(t) iR(t) iL(t) + Is R 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. Establish the initial conditions for the system 2. Determine the equation that describes the system characteristics 3. Solve the equation 4. 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(0t = )= 0 Initial Conditions:⎨ (1.28) ⎩ vt(0= )= 0 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 diL vL= (1.30) dt And the currents iR and iC are vLdiL iR == (1.31) R Rdt dv d 2iL iC ==C LC (1.32) dt dt 2 Combining Equations (1.29), (1.31), and (1.32) we obtain di2 L 11diL 1 ++iL =Is (1.33) dt 2 RC dt LC LC The solution to equation (1.33) is a superposition of the particular and the homogeneous solutions. iL()t = iLp()t + iLh(t) (1.34) The particular solution is iLp ()t = Is (1.35) The homogeneous solution satisfies the equation di2 L 1diL 1 hh+ +iL =0 (1.36) dt 2 RC dt LC h By assuming a solution of the form Aest we obtain the characteristic equation 11 ss2 + +=0 (1.37) RCLC Be defining the following parameters 6.071/22.071 Spring 2006, Chaniotakis and Cory 10 1 ω ≡ : Resonant frequency (1.38) ο LC And 1 α = : Damping rate (1.39) 2RC The characteristic equation becomes 22 ss+ 2αω+ο =0 (1.40) The two roots of this equation are 22 s1=−α + α−ωο (1.41) 22 s2 =−α − α−ωο (1.42) The homogeneous solution is a linear combination of es1t and es2t s1ts2t iLh ()t =+A1e A2e (1.43) And the general solution becomes s1ts2t iL()t =+Is A1e +A2e (1.44) 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 22 22 α −=ωωοj ο−α In this case the roots s1 and s2 are complex numbers: 22 22 sj1,=−α + ωαο− s2=−α−jωο−α. System exhibits oscillatory behavior Let’s investigate the under damped case, α < ωο , in more detail. 22 22 For α < ωο , α −=ωωοοj−α≡jωd the solution is −α t jtωd− jtωd iL()t =+Is eN ( A1e +A2e ) (1.45) Decaying Oscillatory ± jtωd By using Euler’s identity e=±cosωdtjsinωdt, the solution becomes −α t iL()t =+Is eN (K1cosωdt +K2sinωdt) (1.46) Decaying Oscillatory Now we can determine the constants K1 and K2 by applying the initial conditions iL(0t = )=⇒0Is +K1 =0 (1.47) ⇒=KI1 −s diL = 0(⇒−αωKK12+ 0+ d )=0 dt t=0 (1.48) −α ⇒=KI2 s ωd And the solution is 6.071/22.071 Spring 2006, Chaniotakis and Cory 12 ⎡ ⎤ ⎢ ⎥ −α t ⎛⎞α iL()t =−Is ⎢1 eN ⎜cosωdt +sinωdt ⎟⎥ (1.49) ω ⎢ Decaying ⎝⎠ d ⎥ ⎣⎢ Oscillatory ⎦⎥ 22⎛⎞−1B2 By using the trigonometric identity Bt12cos +=Bsin tB1+B2cos⎜⎟t−tan the ⎝⎠B1 solution becomes ωο −α t ⎛⎞−1 α iL(t) =−Is Is e cos⎜ωd t −tan ⎟ (1.50) ωdd⎝⎠ω 22 Recall that ωd ≡ωο −α and thus ωd is always smaller than ωo Let’s now investigate the important limiting case: As R →∞ , α << ω0 22 −1 α −αt ωd ≡−ωαο ≈ωο and tan ≈ 0, e ≈ 1 ωο And the solution reduces to iL(t) = Is − Is cosωot which corresponds to the response of the circuit 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.
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