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Forced Response of Sinusoidal Input In this part of the course, we will focus on finding the force response of a sinusoidal input.
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INC 111 Basic Circuit Analysis
Week 11Force Response of
a Sinusoidal Input andPhasor Concept
Two Types of Analysis
• non-periodic electric source
• periodic electric source
(Transient response analysis of a step input)
(Steady state response analysis of a sinusoidal input)
Forced Response of Sinusoidal Input
In this part of the course, we will focus on finding the force responseof 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)
TheoryForce response of a sinusoidal input is also a sinusoidal signalwith the same frequency but with different amplitude and phase shift.
AC
+
-
5sin(3t+π/3)
i(t)
3H
2Ω
v2(t) Sine wave
Sine wave
v1(t) Sine wave
vL(t) Sine wave
Input
Output
Phase shift
Amplitude
AC
+
-
sin(t)
i(t)
RC
What is the relationship between sin(t) and i(t) ?
sin(t)
i(t)
Phase shift
AC
+
-10sin(2πt + π/4) 2Ω
i(t)
Find i(t)
)4/2sin(5)(2
)4/2sin(10)(
2)()4/2sin(10)()(
tti
tti
titRtitv
Note: Only amplitude changes, frequency and phase still remain the same.
R circuit
AC
+
-
Asin(ωt)
i(t)
L Find i(t)
)2
(sin)cos(
cossin
sin1)(1)(
)()(
tLAt
LA
tLAtdt
LA
tdtAL
dttvL
ti
dttdiLtvfrom
L circuit
AC
+
-
Asin(ωt) R
i(t)
AC
+
-
Asin(ωt)
i(t)
L
)2
(sin)(
tLAti )(sin)( t
RAti
ωL เรยก ความตานทานเสมอน (impedance) Phase shift -90
Phasor Diagram of an inductor
v
i
Phasor Diagram of a resistor
vi
Note: No power consumed in inductorsi lags v 90o
AC
+
-
Asin(ωt)
i(t)
C Find i(t)
)2
sin(1
)(cos
)sin()()(
t
C
AtCA
dttAdC
dttdvCti
ความตานทานเสมอน (impedance) Phase shift +90
C circuit
Phasor Diagram of a capacitor
v
i
Phasor Diagram of a resistor
vi
Note: No power consumed in capacitorsi leads v 90o
Kirchhoff's Law with AC Circuit
AC
+
-
v(t)
i(t)
RC
vRi
vC
i
)13.53sin(5
)90sin(4)sin(3)(o
o
t
tttv
v(t)
)sin(3)( ttvR
)90sin(4)( oC ttv
KCL,KVL still hold.
)90sin(4)sin(3
)cos(4)sin(3)13.53sin()cos(5)13.53cos()sin(5
)13.53sin(5)(
o
oo
o
tt
tttt
ttv
This is similar to adding vectors.Therefore, we will represent sine voltage with a vector.
4
3
5
Vector QuantityComplex numbers can be viewed as vectors whereX-axis represents real partsY-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?
Interchange Rectangular, Polar form
jba
a
b
r
θ
r
sincos
arctan
22
rbra
abbar
Complex Number Forms(Rectangular, Polar Form)
s = 4 + j3
jω
σ
3
4
Rectangular form: 4 + j3
Polar form magnitude=5, angle = 37
บวก ลบ คณ หาร vector ??
Rectangular form
Add, Subtraction
Polar form
45)1()34( jjj
Multiplication
2510)1237(25
122375
Division
4925
122375
ImpedanceCompare to ohm’s law, impedance is a ratio of V/I in when V and I is in the vector format.
AC
+
-
Asin(ωt)
i(t)
L
Inductor
LjL
LAA
IVZ
LAt
LAti
AtAtv
o
o
o
o
o
9090
0
90)2
(sin)(
0)(sin)(
AC
+
-
Asin(ωt)
i(t)
C
CjCCAA
IVZ
CAtCAti
AtAtv
oo
o
o
o
190
190
0
90)2
(sin)(
0)(sin)(
Capacitor
1Ω
3H
Example: Find impedance in form of polar value for ω = 1/3 rad/sec
Note: Impedance depends on frequency and R,L,C values
45213131 jjLjR
Rules that can be used inPhasor Analysis
• Ohm’s law
• KVL/KCL
• Nodal, Mesh Analysis
• Superposition
• Thevenin / Norton
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
AC
+
-
2sin(t/3)
i(t)
3H
1Ω
+ vR(t) -
+vL(t)
-
Find i(t), vR(t), vL(t)
45245202
totalZVI
4521452 IRVR
452901452 LjIIZV LL
Phasor form
)4
3/sin(2)(
)4
3/sin(2)(
)4
3/sin(2)(
ttv
ttv
tti
L
R
452 I
452 RV
452LV
V
I
Example
AC
+
-
5sin(3t+π/3)
i(t)
3H
2Ω
+ vR(t) -
+vL(t)
-
Find i(t), vL(t)
AC
+
-
5∟60
i(t)
j9
2
47.1754.047.7722.9
605
totalZVI
53.7288.490947.1754.09 jIIZV LL
AC
+
-
5∟60
i(t)
j9
247.7722.992 jZ total
)3.03sin(54.0)( tti
)27.13sin(88.4)( ttvL 47.1708.10247.1754.02 IIRVR
V
IVR
VL47.1754.0 I
53.7288.4 LV
47.1708.1 RV
Phasor Diagram