Sinusoidal Steady State Response of Linear Circuits

Sinusoidal Steady State Response of Linear Circuits The circuit shown on Figure 1 is driven by a sinusoidal voltage source vs(t) of the form vs (t ) = vo cos(ωt ) i(t) (1.1) R + vR (t) + vc(t) vs (t) C - Figure 1. Series RC circuit driven by a sinusoidal forcing function Our goal is to determine the voltages vc(t) and the current i(t) which will completely characterize the “Steady State” response of the circuit. The equation that describes the behavior of this circuit is obtained by applying KVL around the mesh. vR (t ) + vc (t ) = vs (t ) (1.2) Using the current voltage relationship of the resistor and the capacitor, Equation (1.2) becomes RC dvc (t ) + vc (t ) = vo cos(ωt ) dt (1.3) Note that the coefficient RC has the unit of time. (Ohm)(Farad) → seconds Before proceeding with the solution of this differential equation let’s explore its physical significance. This linear circuit is driven (forced) by an independent sinusoidal source, vs(t). We may view its response (its effect on the circuit) as the superposition of the response with the source set equal to zero (source-free or natural response (vch(t)) ) and the forced response (vcp(t)). vc (t ) = vch (t ) + vcp (t ) 6.071/22.071 Spring 2006, Chaniotakis and Cory (1.4) 1 Schematically the superposition is shown on Figure 2(a) and (b). R R + + vch (t) C vcp (t) vs (t) C - - (a) (b) Figure 2 In mathematical language we call these two responses the homogeneous solution (vch(t)) ) and the particular solution (vcp(t)) of the equation that characterizes the system (Equation (1.3) in our case) The homogeneous solution corresponds to the differential equation RC dvch (t ) + vch (t ) = 0 dt (1.5) And the particular solution to the equation RC dvcp (t ) dt + vcp (t ) = vo cos(ωt ) (1.6) The homogeneous solution (or the natural response of the system) has the form ⎡ −t ⎤ vch (t ) = B exp ⎢ ⎣ RC ⎥⎦ (1.7) The particular solution (or the forced response of the system) is the cosine function of amplitude A, frequency ω, and phase φ. vcp (t ) = A cos(ω t + φ ) (1.8) And so the general form of the system response is ⎡ −t ⎤ vc (t ) = A cos(ωt + φ ) + B exp ⎢ ⎣ RC ⎥⎦ 6.071/22.071 Spring 2006, Chaniotakis and Cory (1.9) 2 At this time let’s recall the problem statement which says that we are interested in obtaining the Steady State response of the system. This is equivalent to saying that the source vs(t) was connected to the system long time ago and all transient phenomena are gone. In order to be more precise, if the time t  RC then the exponential term in Equation (1.9) would go to zero. In this case the observable and thus the important response of the system is the Steady State response which is given by vc (t ) = A cos(ωt + φ ) (1.10) We may now proceed to determine the details of the solution by calculating the amplitude A and the phase φ. To do this we substitute the form of the solution (Equation (1.10) ) into the differential Equation (1.3). Before we make that substitution lets use the trigonometric identities to expand Equation (1.10). vc (t ) = A cos(ωt + φ ) = A cos φ cos(ωt ) − A sin φ sin(ωt ) (1.11) And the corresponding derivative dvc = − Aω cos φ sin(ωt ) − Aω sin φ cos(ωt ) dt (1.12) Substituting Equations (1.11) and (1.12) into Equation (1.3) we have − Aω cos φ sin(ωt ) − Aω sin φ cos(ωt ) + v 1 [ A cos φ cos(ωt ) − A sin φ sin(ωt )] = 0 cos(ωt ) RC RC (1.13) Collecting the coefficients of cos(ωt ) and sin(ω t ) we have the following equations for the unknowns A and φ. A sin φ = 0 RC (1.14) v A cos φ = 0 RC RC (1.15) − Aω cos φ − − Aω sin φ + These equations are independent and thus they may be solved simultaneously for A and φ. 6.071/22.071 Spring 2006, Chaniotakis and Cory 3 From Equation (1.14) we obtain the phase φ = arctan ( − RC ω ) (1.16) and A becomes A= vo cos φ − ω RC sin φ (1.17) Therefore, the Steady State response of the system is vc (t ) = vo cos (ωt + φ ) cos φ − ω RC sin φ (1.18) where φ = arctan ( −ω RC ) Figure 3 shows a plot of the phase and the amplitude ratio A / vo as a function of the dimensionless parameter ω RC . Observe the frequency selectivity of this system. It passes “low” frequencies while it attenuates “higher” frequencies. 0.0 -10.0 -20.0 -30.0 -40.0 -50.0 -60.0 -70.0 -80.0 -90.0 0 1 2 3 4 5 6 7 8 9 10 7 8 9 10 ωRC 1.0 0.8 0.6 0.4 0.2 0.0 0 1 2 3 4 5 6 ωRC Figure 3 6.071/22.071 Spring 2006, Chaniotakis and Cory 4 The voltage across the resistor, vR (t ) , may also be determined by calculating the current i(t) and multiplying it by R. i (t ) = C dvc (t ) vo Cω π⎞ ⎛ cos ⎜ ωt + φ + ⎟ = dt cos φ − ω RC sin φ 2⎠ ⎝ vR (t ) = i (t ) R = (1.19) voω RC π⎞ ⎛ cos ⎜ ωt + φ + ⎟ cos φ − ω RC sin φ 2⎠ ⎝ (1.20) Figure 4 shows the plot of amplitude ratio vR / vo as a function of the dimensionless parameter ω RC . Here we see the complementary behavior to that shown on Figure 3. Observe again the frequency selectivity of this system. If the output is thus taken across the resistor it passes “high” frequencies while it attenuates “lower” frequencies. 1.0 0.8 0.6 0.4 0.2 0.0 0 1 2 3 4 5 6 7 8 9 10 ωRC Figure 4. 6.071/22.071 Spring 2006, Chaniotakis and Cory 5 Now let’s look at a few frequencies of interest 1. For ω = 0 (dc signal) the phase φ = 0 and the amplitude A = vo . The voltage across the capacitor is constant. No current flows in the circuit The capacitor behaves as an open circuit. The circuit equivalent when ω = 0 is shown on Figure 5. R R C vo ≡ C vo Figure 5 2. For ω RC = 1 The phase φ = −450 and the amplitude A = vo v = o 1/ 2 + 1/ 2 2 R vo cos(ωt ) C vo 2 cos(ωt − π / 4) Here current is flowing and thus some power is dissipated in R. Also the capacitor stores some energy. 6.071/22.071 Spring 2006, Chaniotakis and Cory 6 3. For ω RC  1 The phase φ = −900 and the amplitude A = vo v = o →0 0 − ω RC (−1) ω RC As ω increases more power is dissipated in the resistor R. When ω → ∞ the capacitor acts as a short circuit and all power is dissipated in R. The circuit equivalent when ω → ∞ is shown on Figure 6. R vo cos(ωt ) R C ⎯⎯⎯ → ω →∞ vo cos(ωt ) C Figure 6 6.071/22.071 Spring 2006, Chaniotakis and Cory 7 Now let’s consider the RL circuit i(t) R + vL(t) vs (t) L - We would like to characterize this circuit by obtaining the current i(t) and the voltage vL(t). Apply KVL around the mesh vs (t ) = i (t ) R + L di (t ) dt (1.21) With vs(t) of the form vs (t ) = vo cos(ωt ) , Equation (1.21) becomes v di (t ) R + i (t ) = 0 cos(ωt ) dt L L The coefficient (1.22) L Henry is a time constant. → seconds R Ohm Here again we are interested in the behavior of the system for times that are long L compared to the time constant . In this scenario the only contribution to the solution is R again the one forced by the sinusoidal source voltage vs (t ) = vo cos(ωt ) . And the form of the forced response solution is i (t ) = A cos(ωt + φ ) = A cos φ cos(ωt ) − A sin φ sin(ωt ) (1.23) di (t ) = − Aω cos φ sin(ωt ) − Aω sin φ cos(ωt ) dt (1.24) Substituting back into the differential equation (1.22) we get 6.071/22.071 Spring 2006, Chaniotakis and Cory 8 − Aω cos φ sin(ωt ) − Aω sin φ cos(ωt ) + v R [ A cos φ cos(ωt ) − A sin φ sin(ωt )] = 0 cos(ωt ) L L (1.25) We now separate the cosine and sine functional forms of the solution since they are independent contributions. The coefficients of the sin(ω t ) terms are: − Aω cos φ − RA sin φ = 0 L (1.26) And the coefficients of the cos(ωt ) terms are: − Aω sin φ + v RA cos φ = 0 L L (1.27) Equation (1.26) gives us the expression for the phase φ. ⎛ ωL ⎞ ⎟ ⎝ R ⎠ φ = tan −1 ⎜ − (1.28) And the amplitude A becomes A= vo / R ωL cos φ − sin φ R (1.29) And the solution for i(t) becomes vo / R cos (ωt + φ ) L cos φ − ω sin φ R ⎛ ωL ⎞ Where the phase φ = tan −1 ⎜ − ⎟ ⎝ R ⎠ i (t ) = (1.30) Figure 7 (a) and (b) show the plots of the phase and the amplitude as a function of the ωL parameter . R 6.071/22.071 Spring 2006, Chaniotakis and Cory 9 0.0 -10.0 -20.0 -30.0 -40.0 -50.0 -60.0 -70.0 -80.0 -90.0 0 1 2 3 4 5 6 7 8 9 10 7 8 9 10 ωL/R (a) 1.0 0.8 0.6 0.4 0.2 0.0 0 1 2 3 4 5 6 ωL/R (b) Figure 7 6.071/22.071 Spring 2006, Chaniotakis and Cory 10 Let’s look at a few important frequencies as before 1. For ω = 0 (dc signal) the phase φ = 0 and the amplitude A = vo / R . The voltage across the inductor is zero and the current flowing in the circuit is vo / R . All power is dissipated in R. A magnetic field is generated in the inductor but it does not change over time (no change in the magnetic field no voltage) The inductor behaves as a short circuit as indicated on Figure 8. R L vo R vo /R ≡ L vo Figure 8 2. For ω L / R = 1 The phase φ = −450 and the amplitude of the current is A = vo / R 1 vo = 1/ 2 + 1/ 2 2 R R vo cos(ωt ) L i= 1 vo cos(ωt − π / 4) 2 R Here current is flowing and thus some power is dissipated in R. Also the inductor stores some energy. 6.071/22.071 Spring 2006, Chaniotakis and Cory 11 3. For ω L / R  1 The phase φ = −900 and the amplitude A = vo / R v /R = o →0 ωL ωL 0− (−1) R R As ω increases the current decreases and as ω → ∞ the inductor acts as an open circuit (see Figure 9). R vo cos(ωt ) R L ⎯⎯⎯ → ω →∞ vo cos(ωt ) L Figure 9 6.071/22.071 Spring 2006, Chaniotakis and Cory 12 Using the complex forcing function Our goal is to be able to analyze RC and RL circuits without having to every time employ the differential equation method, which can be cumbersome. If we draw upon our current understanding of RC and RL networks and the fact that they represent linear systems we will be able to considerably simplify the mathematical steps involved in the computation. This simplification will require the use of fundamental complex arithmetic and will in the end reduce the differential equations into simple algebraic equations. (indeed a simplification using complex numbers!) The linearity of the system implies that if we use a forcing function of the form vo cos (ωt + θ ) then the output will have the same frequency but with a different phase A cos (ωt + φ ) . Also if we were to scale the source by a factor k then the output will be scaled by the same factor. Figure 10 graphically demonstrates these two statements. vo cos(ωt + θ ) Linear syestem A cos(ωt + φ ) k A cos(ωt + θ ) Linear k A cos(ωt + φ ) syestem Figure 10 If the factor k is an imaginary number like j = −1 , then linearity still holds as we graphically demonstrate on Figure 11. Linear syestem j vo sin(ωt + θ ) j A sin(ωt + φ ) Figure 11 Therefore by superposition a forcing function of the form vo cos(ωt ) + jvo sin(ωt ) (1.31) Will produce a response of the form A cos(ω t + φ ) + jA sin(ωt + φ ) 6.071/22.071 Spring 2006, Chaniotakis and Cory (1.32) 13 By using Euler’s identity Equations (1.31) and (1.32), the forcing function and the corresponding response, become respectively vo e j ( ωt ) (1.33) and Ae j (ωt +φ ) (1.34) Therefore we could employ the complex form of the forcing function, proceed with the development of the solution and then extract the desirable part of the response depending on whether the forcing function was a cosine or a sine function. So what is the advantage of this method? 1. Very easy to perform algebra with the exponentials 2. Reduces differential equations to algebraic equations. Let’s explore this method by considering again the RL circuit analyzed previously and shown again on Figure 12. The source has the form vs (t ) = vo cos(ωt ) . i(t) R + vL(t) vs (t) vo cos(ωt ) L - Figure 12 Since vo cos(ωt ) is the real part of vo e jωt we will proceed with the analysis and at the end we will simply extract the real part of the solution. The equation characterizing the system is v di (t ) R + i (t ) = s dt L L (1.35) vo e jωt (1.36) If we use the complex source 6.071/22.071 Spring 2006, Chaniotakis and Cory 14 Then the corresponding complex response is Ae j (ωt +φ ) (1.37) Equation (1.35) becomes jω Ae jφ + A R jφ vo e = L L (1.38) which upon simplification becomes Ae jφ ( R + jω L ) = vo (1.39) Continue with more simplification we obtain Ae jφ = vo / R 1 + jω L / R (1.40) And by writing the complex number in polar form we have jφ Ae = vo / R 1+ e ω 2 L2 ⎛ ⎛ ωL ⎞⎞ j ⎜ tan −1 ⎜ − ⎟⎟ ⎝ R ⎠⎠ ⎝ (1.41) R2 Where the amplitude of the current is A= vo / R 1+ ω 2 L2 (1.42) R2 And the phase is ⎛ ωL ⎞ ⎟ ⎝ R ⎠ φ = tan −1 ⎜ − (1.43) The complete complex response of the system is 6.071/22.071 Spring 2006, Chaniotakis and Cory 15 vo / R 1+ ω L 2 2 e ⎛ ⎛ ωL ⎞⎞ j ⎜ tan −1 ⎜ − ⎟⎟ ⎝ R ⎠⎠ ⎝ e jωt (1.44) R2 Since our forcing source was the cosine function, all we need to do is extract the real part of the function given by Equation (1.44) which is i (t ) = ⎛ ⎛ ωL ⎞ ⎞ cos ⎜ ωt + tan −1 ⎜ − ⎟⎟ ⎝ R ⎠⎠ ⎝ ω L 1+ 2 R vo / R 2 2 (1.45) Which is the same as Equation (1.30). 6.071/22.071 Spring 2006, Chaniotakis and Cory 16