School of Electrical and Computer Engineering ECE2040 Dr. George F. Riley Summer 2007, GT Lorraine Analysis of LRC with Sinusoidal Sources

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1 School of Electrical and Computer Engineering ECE2040 Dr. George F. Riley Summer 2007, GT Lorraine Analysis of LRC with Sinusoidal Sources In chapter 8 in the textbook, we determined that if the forcing function (the voltage V s (t) or the current I s (t)) is of the form V s (t) = cosωt then forced response v o (t) of an RC circuit or RL circuit is of the form: v f = A cosωt B sin ωt If we multiply the right side of the above equation by A 2 B 2 A we get: 2 B 2 v f = ( ) A A 2 B 2 A2 B cosωt B 2 A2 B sin ωt 2 Now, if we define θ = tan 1 (B/A), we can say that: So sin θ = B A2 B 2 cosθ = A A2 B 2 By the trigonometric identity: we get: v f = cosθ cosωt sin θ sin ωt cosα β = cosαcosβ sin α sin β v f = A 2 B 2 cos(ωt θ) (1) The above result will be used in the next set of equation. The previous analysis is found in section 10.3 in Dorf and Svoboda. 1

2 Now consider the circuit below where v s (t) = cosωt and we want to find the forced response i f (t). This analysis is found in section 10.4 of Dorf and Svoboda. R vs(t) L i(t) Using KVL: L di dt Ri = cosωt From chapter 8, we know that the forced response i f (t) will be of the form: and: thus: di dt i f (t) = A cosωt B sin ωt = ωa sin ωt ωb cosωt L(ωA sin ωt ωb cosωt) R(A cosωt B sin ωt) = cosωt We can find A and B by chosing a value for ωt that makes sin ωt = 0 and cosωt = 1. ωlb RA = V b and a value for ωt that makes sin ωt = 1 and cos ωt = 0. Solving for A and B we get: ωla RB = 0; A = B = R R 2 ω 2 L 2 ωl R 2 ω 2 L 2 2

3 From equation 1, we know that: So we need to find A 2 B 2 : So finally: A 2 B 2 = v f = A cosωt B sin ωt = A 2 B 2 (cosωt β) R 2 V 2 m (R 2 ω 2 L 2 )(R 2 ω 2 L 2 ) ω 2 L 2 V 2 m (R 2 ω 2 L 2 )(R 2 ω 2 L 2 ) A 2 B 2 = V 2 m (R2 ω 2 L 2 ) (R 2 ω 2 L 2 )(R 2 ω 2 L 2 ) A2 B 2 = R2 ω 2 L 2 where: i(t) = A 2 B 2 cos(ωt β) = Z cos(ωt β) (2) Z = R 2 ω 2 L 2 β = tan 1 (B/A) = tan 1 ωl R Equation 2 above is the solution for the forced response to an LR circuit with sinusoidal forcing function, and should be memorized. 3

4 We now turn out attention to the RC circuit shown below, where v s (t) = cosωt. This analysis is not in the textbook, but is inportant and should be studied and understood. R vs(t) C vo(t) We know that the current is the same across all elements of this circuit, and that the current through a capacitor is: KVL around the loop gives: C dv dt V s (t) = cosωt = RC dv dt v o(t) As before, assume the response v o (t) is of the form: so substituting above for v o (t) and dv dt v o (t) = A cosωt B sin ωt we get: cosωt = RCBω cosωt RCAω sin ωt A cosωt B sin ωt Again choosing appropriate values for cosωt and sin ωt, we get: B = RCAω Solving for A and B we get: Again finding A 2 B 2 : A = RCBω A = 1 R 2 C 2 ω 2 B = RCω 1 R 2 C 2 ω 2 A 2 B 2 = V 2 m R2 C 2 ω 2 V 2 m (1 R 2 C 2 ω 2 )(1 R 2 C 2 ω 2 ) = V 2 m (1 R2 C 2 ω 2 ) (1 R 2 C 2 ω 2 )(1 R 2 C 2 ω 2 ) = V 2 m 1 R 2 C 2 ω 2 4

5 Finally, the forced response v o (t) is: A2 B 2 = 1 R2 C 2 ω 2 where v o (t) = P cos(ωt β) (3) P = 1 R 2 C 2 ω 2, β = tan 1 (B/A) = tan 1 (RCω) Equation 3 above is the response to an RC circuit with a sinusoidal forcing function and should be memorized. 5

6 Next, we will analyze an RL circuit with a complex exponential forcing function. The circuit below is the same as the LR circuit used previously, and the forcing function v s (t) is again cos ωt. However, in this case we will see that the analysis is a bit simpler. R vs(t) L i(t) We start by observing that: v s (t) = cos ωt = Re { e jωt} Where the Re notation indicates the real part of the complex variable. For the remainder of this analysis, we will drop the Re notation, and simply take the real part of the computed forced response when we are done. Using KVL: v s (t) = L di dt Ri We know that the response to an exponential forcing function is of the form: So: Solving for A: where: i o (t) = Ae jωt v s (t) = e jωt = AjωLe jωt RAe jωt = Ae jωt (jωl R) A = R jωl = Z ejβ Z = R 2 ω 2 L 2, β = tan 1 ωl R So the response is: { } Vm i o (t) = Re Z ejβ e jωt = Z ejωtβ = cos(ωt β) Z which matches the result earlier in equation 2. We could do a similar analysis for an RC circuit using the same technique, which would give the same results as in equation 3. 6

7 Now we look at using the concept of Complex Phasors to solve LR and RC circuit responses with sinusoidal forcing functions. In the previous analysis, we found that the term e jωt occurred often, and was unchanged from the forcing function to the respose function. For ease of notation, we use the notation: V = Re { e jβ e jωt} = β The boldface V indicates the value is a Phasor, which has a magnitude and a phase angle. Phasors also have an e jωt term which is not written but assumed present. To see how this is useful, consider the circuit below with v s (t) = cos(ωt φ). R vs(t) L i(t) Assume the response i(t) is of the form: v s (t) = cos(ωt φ) = Re { e j(ωtφ)} By KVL: i(t) = I m cos(ωt β) = Re { I m e jωtβ} v s (t) = L di dt Ri e j(ωtφ) = (jωli m RI m )e j(ωtβ) To convert to phasor notation, first remove the e jωt from every term: e jφ = (jωli m RI m )e jβ Thus in phasor notation, we have: V s = e jφ, I = I m e jβ (jωl R)I = V s I = V s jωl R 7

8 If we let φ = 10, ω = 100rad/s, R = 200Ω, L = 2H, we get: V s I = jωl R = j = = Converting back to time domain representation, we get: V s i(t) = 283 cos(100t 35 ) As another example, use Phasor notation to solve for v o (T) in the figure below, where i(t) = 10 cosωt. i(t) R C vo(t) By KCL: i(t) = 10 cosωt = I = 10 0 i(t) = v o(t) R C dv dt i(t) = 10Re { e jωt} v o (t) = Re { e jωtβ} R ej(ωtβ) jωc e j(ωtβ) = 10e jωt Suppress the e jωt term to convert to phasor notation: ( 1 R jωc ( 1 R jωc Let R = 1Ω, C = 10mF, ω = 100, we get: ) e jβ = 10e j0 ) V = I (1 j1)v = I, V = I 1 j1, V = Converting back to time domain representation, we get: = v o (t) = 10 2 cos(100t 45 ) 8