Download Lecture 11 - kmutt-inc

INC 112 Basic Circuit Analysis
Week 9
Force Response of
a Sinusoidal Input and
Phasor Concept
Forced Response of Sinusoidal
Input
In this part of the course, we will focus on finding the force response
of a sinusoidal input.
• Start oscillate from stop
input
displacement
Period that have transient
• Have oscillated for a long time
input
displacement
We will only be interested in this case for force response (not count the transient)
Input
Phase shift
Amplitude
Output
Theory
Force response of a sinusoidal input is also a sinusoidal signal
with the same frequency but with different amplitude and phase shift.
v2(t) Sine wave
v1(t) Sine wave
i(t) Sine wave
+
2Ω
5sin(3t+π/3)
AC
-
3H
vL(t) Sine wave
i(t)
R
+
sin(t)
AC
C
-
What is the relationship between sin(t) and i(t) ?
sin(t)
i(t)
Phase shift
i(t)
+
AC
10sin(2πt + π/4)
2Ω
Find i(t)
-
v(t )  i (t ) R
10 sin( 2t   / 4)  i(t )  2
10 sin( 2t   / 4)
i (t ) 
2
i (t )  5 sin( 2t   / 4)
Note: Only amplitude changes, frequency and phase still remain the same.
i(t)
+
Asin(ωt)
AC
L
Find i(t)
-
from
di (t )
v(t )  L
dt
1
1
i (t )   v(t )dt   A sin tdt
L
L
A
A   cos t 
  sin tdt  

L
L  
A
A


( cos t ) 
(sin t  )
L
L
2
i(t)
i(t)
+
+
Asin(ωt)
AC
L
Asin(ωt)
AC
R
-
-
A

i (t ) 
(sin t  )
L
2
A
i (t )  (sin t )
R
ωL เรียก ความต้ านทานเสมือน (impedance)
Phase shift -90
Phasor Diagram of an inductor
Phasor Diagram of a resistor
v
v
i
i
Power = (vi cosθ)/2 = 0
Power = (vi cosθ)/2 = vi/2
Note: No power consumed in inductors
i lags v
i(t)
+
Asin(ωt)
AC
C
Find i(t)
-
dv(t )
d ( A sin t )
i (t )  C
C
dt
dt
 AC (cos t )
A


sin( t  )
2
 1 


 C 
ความต้ านทานเสมือน (impedance)
Phase shift +90
Phasor Diagram of a capacitor
Phasor Diagram of a resistor
i
v
v
i
Power = (vi cosθ)/2 = 0
Power = (vi cosθ)/2 = vi/2
Note: No power consumed in capacitors
i leads v
Kirchhoff's Law with AC Circuit
KCL,KVL still hold.
vR
i
vR (t )  3sin( t )
i(t)
v(t)
R
+
v(t)
AC
i
C
-
v(t )  3 sin( t )  4 sin( t  90o )
 5 sin( t  53.13 )
o
vC
vC (t )  4 sin( t  90o )
v(t )  5 sin( t  53.13o )
 5 sin( t ) cos(53.13 )  5 cos(t ) sin( 53.13 )
 3 sin( t )  4 cos(t )
o
o
 3 sin( t )  4 sin( t  90 )
o
This is similar to adding vectors.
Therefore, we will represent sine voltage with a vector.
3
5
4
Vector Quantity
Complex numbers can be viewed as vectors where
X-axis represents real parts
Y-axis represents imaginary parts
There are two ways to represent complex numbers.
• Cartesian form 3+j4
• Polar form
5∟53o
Operation add, subtract, multiply, division?
Complex Number Forms
(Rectangular, Polar Form)
a  jb
b
r  
r  a b
2
r
θ
a
2
b
  arctan  
a
a  r cos 
b  r sin 
Interchange Rectangular, Polar form
jω
s = 4 + j3
3
σ
4
Rectangular form:
Polar form
4 + j3
magnitude=5, angle = 37
บวก ลบ คูณ หาร vector ??
Rectangular form
Add, Subtraction
(4  j 3)  (1  j )  5  j 4
Polar form
Multiplication
537  2  12
 5  2(37  12 )  1025
Division
537  2  12
5
 49
2
Note: Impedance depends on frequency and R,L,C values
Cartesian form
Polar form
a  jb
cd
c  magnitude  a 2  b 2

b
d  phase  tan  
a
1
Example: Find impedance in form of polar value for ω = 1/3 rad/sec
3H
1Ω
1
R  jL  1  j  3   1  j  245
3
Rules that can be used in
Phasor Analysis
• Ohm’s law
• KVL/KCL
• Nodal, Mesh Analysis
• Superposition
• Thevenin / Norton
Example
+ vR(t) +
AC
-
1Ω
2sin(t/3)
i(t)
+
vL(t)
-
Find i(t), vR(t), vL(t)
3H
V
20

I


2


45
Z total
245
Phasor form
VR  IR  2  45 1  2  45


VL  IZ L  I  jL  2  45 190  245




i (t )  2 sin( t / 3  )
4
I  2  45
VR  2  45
VL  245



vR (t )  2 sin( t / 3  )
4

vL (t )  2 sin( t / 3  )
4
V
I
In an RLC circuit with sinusoidal voltage/current source,
voltages and currents at all points are in sinusoidal wave form too
but with different amplitudes and phase shifts.
Summary of Procedures
• Change voltage/current sources in to phasor form
• Change R, L, C value into phasor form
R
L
C
R
jωL
1/jωC
• Use DC circuit analysis techniques normally, but the value of
voltage, current, and resistance can be complex numbers
• Change back to the time-domain form if the problem asks.
Example
+ vR(t) 2Ω
+
5sin(3t+π/3)
AC
-
i(t)
+
vL(t)
-
Find i(t), vL(t)
3H
i(t)
+
2
5∟60
AC
-
j9
i(t)
+
2
5∟60
AC
-
j9
Z total  2  j 9  9.2277.47 
V
560

I


0
.
54


17
.
47
Z total 9.2277.47
i (t )  0.54 sin( 3t  0.3)
VL  IZ L  I  j9  0.54  17.47  990  4.8872.53
vL (t )  4.88 sin( 3t  1.27)
VR  IR  I  2  0.54  17.47  20  1.08  17.47



Phasor Diagram
V
I  0.54  17.47
VL
VL  4.8872.53

VR  1.08  17.47
VR
Resistor consumes power
I
2
0
.
54
2
2
P  I rms
R
2
Inductor consumes no power P = 0