Analysis of LRC with Sinusoidal Sources
ECE2040 – Summer 2007
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 Vs (t) or the
current Is (t)) is of the form
Vs (t) = Vm cos ωt
then forced response vo (t) of an RC circuit or RL circuit is of the form:
vf = A cos ωt + B sin ωt
√
2
2
+B
we get:
If we multiply the right side of the above equation by √A
A2 +B 2
√
B
A
cos ωt + √
sin ωt
vf = A2 + B 2 √
A2 + B 2
A2 + B 2
Now, if we define θ = tan−1 (B/A), we can say that:
sin θ = √
B
+ B2
cos θ = √
A2
A
+ B2
A2
So
vf = cos θ cos ωt + sin θ sin ωt
By the trigonometric identity:
cos α − β = cos α cos β + sin α sin β
we get:
vf =
√
A2 + 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
Analysis of LRC with Sinusoidal Sources
ECE2040 – Summer 2007
Now consider the circuit below where
vs (t) = Vm cos ωt
and we want to find the forced response if (t). This analysis is found in section 10.4 of Dorf and
Svoboda.
R
+
vs(t)
L
i(t)
−
Using KVL:
di
+ Ri = Vm cos ωt
dt
From chapter 8, we know that the forced response if (t) will be of the form:
L
if (t) = A cos ωt + B sin ωt
and:
di
= −ωA sin ωt + ωB cos ωt
dt
thus:
L(−ωA sin ωt + ωB cos ωt) + R(A cos ωt + B sin ωt) = Vm cos ωt
We can find A and B by chosing a value for ωt that makes sin ωt = 0 and cos ωt = 1.
ωLB + RA = Vb
and a value for ωt that makes sin ωt = 1 and cos ωt = 0.
−ωLA + RB = 0;
Solving for A and B we get:
A=
B=
R2
RVm
+ ω 2L2
ωLVm
R2 + ω 2L2
2
Analysis of LRC with Sinusoidal Sources
ECE2040 – Summer 2007
From equation 1, we know that:
So we need to find
√
vf = A cos ωt + B sin ωt =
A2 + B 2 :
A2 + B 2 =
√
A2 + B 2 (cos ωt − β)
ω 2 L2 Vm2
R2 Vm2
+
(R2 + ω 2L2 )(R2 + ω 2L2 ) (R2 + ω 2 L2 )(R2 + ω 2 L2 )
A2 + B 2 =
√
Vm2 (R2 + ω 2L2 )
(R2 + ω 2 L2 )(R2 + ω 2 L2 )
A2 + B 2 = √
Vm
R2 + ω 2 L2
So finally:
i(t) =
√
A2 + B 2 cos(ωt − β) =
Vm
cos(ωt − β)
Z
(2)
where:
Z=
√
R2 + ω 2L2
ωL
R
Equation 2 above is the solution for the forced response to an LR circuit with sinusoidal forcing
function, and should be memorized.
β = tan−1 (B/A) = tan−1
3
Analysis of LRC with Sinusoidal Sources
ECE2040 – Summer 2007
We now turn out attention to the RC circuit shown below, where vs (t) = Vm cos ωt. 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:
C
dv
dt
KVL around the loop gives:
Vs (t) = Vm cos ωt = RC
dv
+ vo (t)
dt
As before, assume the response vo (t) is of the form:
vo (t) = A cos ωt + B sin ωt
so substituting above for vo (t) and
dv
dt
we get:
Vm cos ωt = RCBω cos ωt − RCAω sin ωt + A cos ωt + B sin ωt
Again choosing appropriate values for cos ωt and sin ωt, we get:
B = RCAω
A = Vm − RCBω
Solving for A and B we get:
Vm
1 + R2 C 2 ω 2
RCωVm
B=
1 + R2 C 2 ω 2
A=
Again finding
√
A2 + B 2 =
A2 + B 2 :
Vm2 + R2 C 2 ω 2Vm2
Vm2 (1 + R2 C 2 ω 2)
Vm2
=
=
(1 + R2 C 2 ω 2 )(1 + R2 C 2 ω 2)
(1 + R2 C 2 ω 2 )(1 + R2 C 2 ω 2)
1 + R2 C 2 ω 2
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Analysis of LRC with Sinusoidal Sources
√
ECE2040 – Summer 2007
A2 + B 2 = √
Vm
1 + R2 C 2 ω 2
Finally, the forced response vo (t) is:
vo (t) =
Vm
cos(ωt − β)
P
(3)
where
P =
√
1 + R2 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
Analysis of LRC with Sinusoidal Sources
ECE2040 – Summer 2007
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 vs (t) is again Vm cos ωt.
However, in this case we will see that the analysis is a bit simpler.
R
+
vs(t)
i(t)
L
−
We start by observing that:
vs (t) = Vm cos ωt = Re Vm ejωt
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:
di
+ Ri
dt
We know that the response to an exponential forcing function is of the form:
vs (t) = L
io (t) = Aejωt
So:
vs (t) = Vm ejωt = AjωLejωt + RAejωt = Aejωt (jωL + R)
Solving for A:
A=
Vm
Vm −jβ
=
e
R + jωL
Z
where:
Z=
√
R2 + ω 2 L2 , β = tan−1
ωL
R
So the response is:
Vm −jβ jωt
Vm jωt−β Vm
io (t) = Re
=
e e
e
=
cos(ωt − β)
Z
Z
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
Analysis of LRC with Sinusoidal Sources
ECE2040 – Summer 2007
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 ejωt occurred often,
and was unchanged from the forcing function to the respose function. For ease of notation, we use
the notation:
V = Re Vm ejβ ejωt = Vm ∠β
The boldface V indicates the value is a Phasor, which has a magnitude and a phase angle. Phasors
also have an ejωt term which is not written but assumed present. To see how this is useful, consider
the circuit below with vs (t) = Vm cos(ωt + φ).
R
+
vs(t)
L
i(t)
−
vs (t) = Vm cos(ωt + φ) = Re Vm ej(ωt+φ)
Assume the response i(t) is of the form:
By KVL:
i(t) = Im cos(ωt + β) = Re Im ejωt+β
vs (t) = L
di
+ Ri
dt
Vm ej(ωt+φ) = (jωLIm + RIm )ej(ωt+β)
To convert to phasor notation, first remove the ejωt from every term:
Vm ejφ = (jωLIm + RIm )ejβ
Vs = Vm ejφ , I = Im ejβ
Thus in phasor notation, we have:
(jωL + R)I = Vs
I=
Vs
jωL + R
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Analysis of LRC with Sinusoidal Sources
ECE2040 – Summer 2007
If we let φ = 10◦ , ω = 100rad/s, R = 200Ω, L = 2H, we get:
Vs
Vm ∠10◦
Vm ∠ − 35◦
Vs
=
=
=
jωL + R
j200 + 200
283∠45◦
283
Converting back to time domain representation, we get:
I=
Vm
cos(100t − 35◦ )
283
As another example, use Phasor notation to solve for vo (T ) in the figure below, where i(t) =
10 cos ωt.
i(t) =
+
i(t)
R
C
vo(t)
−
i(t) = 10 cos ωt = I = 10∠0◦
By KCL:
dv
vo (t)
+C
R dt
i(t) = 10Re ejωt
vo (t) = Vm Re ejωt+β
i(t) =
Vm j(ωt+β)
e
+ jωCVm ej(ωt+β) = 10ejωt
R
Suppress the ejωt term to convert to phasor notation:
1
◦
+ jωC Vm ejβ = 10ej0
R
1
+ jωC V = I
R
Let R = 1Ω, C = 10mF, ω = 100, we get:
10∠ − 45◦
I
10∠0◦
√
=
, V= √
1 + j1
2∠45◦
2
Converting back to time domain representation, we get:
(1 + j1)V = I, V =
10
vo (t) = √ cos(100t − 45◦)
2
8