Sinusoids and Complex Numbers
The input to this circuit is the voltage source voltage, v s ( t ) .
The output is the voltage across the capacitor, v C ( t ) . The
input and output are the sinusoids shown below. We expect
v s ( t ) = A cos (ω t + θ ) and v C ( t ) = B cos (ω t + φ )
where A, B, ω, θ and φ are constants to be determined.
From the plot
2π
= 100 rad/s .
0.06283
The peak-to-peak amplitude of the source voltage is 50 V so A = 25 V.
The peak-to-peak amplitude of the capacitor voltage is 40 V so B = 20 V.
⎛ 24.15 ⎞
v s ( 0 ) = 24.15 V and v s ( t ) is decreasing at time t = 0 so θ = cos −1 ⎜
⎟ = 15° .
⎝ 25 ⎠
⎛ 18.54 ⎞
v C ( 0 ) = 18.54 V and v C ( t ) is increasing at time t = 0 so φ = − cos −1 ⎜
⎟ = −22° .
⎝ 20 ⎠
1) The period of both sinusoids is T = 62.83 ms so ω =
2)
3)
4)
5)
so
v s ( t ) = 25cos (100 t + 15° ) V and v C ( t ) = 20 cos (100 t − 22° ) V
Apply KVL in the circuit to get
v R ( t ) = v s ( t ) − v C ( t ) = 25cos (100 t + 15° ) − 20 cos (100 t − 22° )
First, we solve this equation using trigonometry:
cos (α ± β ) = cos (α ) cos ( β ) ∓ sin (α ) sin ( β )
Recall that
so
25cos (100 t + 15° ) = 25 ⎡⎣cos (100 t ) cos (15° ) − sin (100 t ) sin (15° ) ⎤⎦
= 25cos (15° ) cos (100 t ) − 25sin (15° ) sin (100 t )
= 24.15cos (100 t ) − 6.47 sin (100 t )
and
20 cos (100 t − 22° ) = 20 ⎡⎣cos (100 t ) cos ( 22° ) + sin (100 t ) sin ( 22° ) ⎤⎦
= 20 cos ( 22° ) cos (100 t ) + 20sin ( 22° ) sin (100 t )
= 18.54 cos (100 t ) + 7.49sin (100 t )
The KVL equation indicates that
v R ( t ) = v s ( t ) − v C ( t ) = 25cos (100 t + 15° ) − 20 cos (100 t − 22° )
= ⎡⎣ 24.15cos (100 t ) − 6.47 sin (100 t ) ⎤⎦ − ⎡⎣18.54 cos (100 t ) + 7.49sin (100 t ) ⎤⎦
= ( 24.15 − 18.54 ) cos (100 t ) + ( −6.47 − 7.49 ) sin (100 t )
= 5.61cos (100 t ) − 16.96sin (100 t )
We need another trigonometric identity. Consider
A cos (ω t ) + B sin (ω t )
⎡
A
cos (ω t ) +
= A2 + B 2 ⎢
2
2
+
A
B
⎣
⎤
sin (ω t ) ⎥
A2 + B 2
⎦
B
= A2 + B 2 ⎡⎣cos ( θ ) cos (ω t ) + sin ( θ ) sin (ω t ) ⎤⎦
sin ( θ ) =
cos ( θ ) =
B
2
A + B2
A
A2 + B 2
= A2 + B 2 ⎡⎣cos ( −θ ) cos (ω t ) − sin ( −θ ) sin (ω t ) ⎤⎦
= A2 + B 2 cos (ω t − θ )
⎛B⎞
where θ = tan −1 ⎜ ⎟
⎝ A⎠
Consequently
⎛
⎛ −16.96 ⎞ ⎞
v R ( t ) = 5.61cos (100 t ) − 16.96sin (100 t ) = 5.612 + 16.962 cos ⎜ 100 t − tan −1 ⎜
⎟⎟
⎝ 5.61 ⎠ ⎠
⎝
= 15cos (100 t − ( −68.1° ) )
= 15cos (100 t + 68.1° ) V
Second, we will solve the KVL equation using complex numbers. To do so we associate the
complex number A∠θ with the sinusoid A cos (ω t + θ ) . That is
A cos (ω t + θ ) ⇔
A∠θ
Using this association:
25cos (100 t + 15° ) − 20 cos (100 t − 22° ) ⇔ 25∠15° − 20∠ − 22° = ( 24.15 + j 6.47 ) − (18.54 − j 7.49 )
= ( 24.15 − 18.54 ) + j ( 6.47 + 7.49 )
= 5.61 + j 13.96 = 15∠68.1°
Then
15cos (100 t + 68.1° ) ⇔ 15∠68.1°
We have achieved the same result with considerably less effort.
i (t ) =
Check:
v R (t )
300
But also
(
i ( t ) = 25 × 10−6
)
=
15cos (100 t + 68.1° )
= 0.05cos (100 t + 68.1° )
300
d v C (t )
dt
(
) dtd ( 20 cos (100 t − 22°) )
= ( 25 ×10−6 ) ( 20 )(100 ) ( − sin (100 t − 22° ) )
= 25 × 10−6
= 0.05sin (100 t + 180° − 22° ) = 0.05sin (100 t + 158° )
= 0.05cos (100 t + 158° − 90° )
= 0.05cos (100 t + 68° )