Figure 1. Series RC circuit driven by a sinusoidal forcing ...

1 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( t ) ( ). i(t) R. + vR (t) - +. vs (t) vc(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 ) ( ). Using the current voltage relationship of the resistor and the capacitor, Equation ( ).

2 Becomes dvc (t ). RC + vc (t ) = vo cos( t ) ( ). dt 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 ) ( ). Spring 2006, Chaniotakis and Cory 1. Schematically the superposition is shown on Figure 2(a) and (b).

3 R R. + +. vch (t) C vs (t) vcp (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 ( ) in our case). The homogeneous solution corresponds to the differential equation dvch (t ). RC + vch (t ) = 0 ( ). dt And the particular solution to the equation dvcp (t ). RC + vcp (t ) = vo cos( t ) ( ). dt The homogeneous solution (or the natural response of the system) has the form t . vch (t ) = B exp ( ). RC . The particular solution (or the forced response of the system) is the cosine function of amplitude A, frequency , and phase.

4 Vcp (t ) = A cos( t + ) ( ). And so the general form of the system response is t . vc (t ) = A cos( t + ) + B exp ( ). RC . Spring 2006, Chaniotakis and Cory 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 ( ) 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( t + ) ( ).

5 We may now proceed to determine the details of the solution by calculating the amplitude A and the phase . To do this we substitute the form of the solution (Equation ( ) ). into the differential Equation ( ). Before we make that substitution lets use the trigonometric identities to expand Equation ( ). vc (t ) = A cos( t + ). ( ). = A cos cos( t ) A sin sin( t ). And the corresponding derivative dvc = A cos sin( t ) A sin cos( t ) ( ). dt Substituting Equations ( ) and ( ) into Equation ( ) we have A cos sin( t ) A sin cos( t ) +. 1 v ( ). [ A cos cos( t ) A sin sin( t )] = 0 cos( t ). RC RC. Collecting the coefficients of cos( t ) and sin( t ) we have the following equations for the unknowns A and.

6 A. A cos sin = 0 ( ). RC. A v A sin + cos = 0 ( ). RC RC. These equations are independent and thus they may be solved simultaneously for A and . Spring 2006, Chaniotakis and Cory 3. From Equation ( ) we obtain the phase = arctan ( RC ) ( ). and A becomes vo A= ( ). cos RC sin . Therefore, the Steady State response of the system is vo vc (t ) = cos ( t + ) ( ). cos RC sin . where = arctan ( RC ). Figure 3 shows a plot of the phase and the amplitude ratio A / vo as a function of the dimensionless parameter RC . Observe the frequency selectivity of this system. It passes low frequencies while it attenuates higher frequencies.

7 0 1 2 3 4 5 6 7 8 9 10. RC. 0 1 2 3 4 5 6 7 8 9 10. RC. Figure 3. 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. dvc (t ) vo C . i (t ) = C = cos t + + ( ). dt cos RC sin 2 . vo RC . vR (t ) = i (t ) R = cos t + + ( ). cos RC sin 2 . Figure 4 shows the plot of amplitude ratio vR / vo as a function of the dimensionless parameter 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.

8 0 1 2 3 4 5 6 7 8 9 10. RC. Figure 4. Spring 2006, Chaniotakis and Cory 5. Now let's look at a few frequencies of interest 1. For = 0 (dc signal) the phase = 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 = 0 is shown on Figure 5. R R.. vo C vo C. Figure 5. 2. For RC = 1. vo v The phase = 450 and the amplitude A = = o 1/ 2 + 1/ 2 2. R. vo vo cos( t ) C cos( t / 4). 2. Here current is flowing and thus some power is dissipated in R. Also the capacitor stores some energy. Spring 2006, Chaniotakis and Cory 6.

9 3. For RC 1. vo v The phase = 900 and the amplitude A = = o 0. 0 RC ( 1) RC. As increases more power is dissipated in the resistor R. When the capacitor acts as a short circuit and all power is dissipated in R. The circuit equivalent when is shown on Figure 6. R R. vo cos( t ) C .. vo cos( t ) C. Figure 6. Spring 2006, Chaniotakis and Cory 7. Now let's consider the RL circuit i(t) R. +. vs (t) vL(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 di (t ). vs (t ) = i (t ) R + L ( ). dt With vs(t) of the form vs (t ) = vo cos( t ) , Equation ( ) becomes di (t ) R v + i (t ) = 0 cos( t ) ( ).

10 Dt L L. L Henry The coefficient 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( t ) . And the form of the forced response solution is i (t ) = A cos( t + ). ( ). = A cos cos( t ) A sin sin( t ). di (t ). = A cos sin( t ) A sin cos( t ) ( ). dt Substituting back into the differential equation ( ) we get Spring 2006, Chaniotakis and Cory 8. A cos sin( t ) A sin cos( t ) +. R v ( ).