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Sinusoidal Steady-State
responseresponse
EE3301
Kamran Kiasaleh
Learning Objectives
1. Be able to obtain the steady-state response of RLC circuits (in all forms) to a sinusoidal input
2. Be able to represent currents and voltages in “Phasor” format
3. Be able to obtain circuit impedance and admittance. 3. Be able to obtain circuit impedance and admittance.
4. Be able to obtain Thevenin and Norton equivalent circuits for steady-state sinusoidal circuits
5. Be able to write mesh, node, KVL, and KCL equations for sinusoidal steady-state circuit
6. Be able to conduct steady-state sinusoidal analysis of circuits with transformers
A sinusoidal voltage
What are the key parameters of a sinusoid
T;period(sec); f0 =1
T; frequency(Hz)
ω 0 = 2πf
v t( ) =Vm cos ω 0t +θ( )( ) m 0( )
Vrms =1
Tv t( )
0
T∫2
dt = lim T → ∞
1
Tv t( )
0
T∫2
dt
P =Vrms
2
R;
Vrms =Vm
2
Only for sinusoidal
signals
Phase shift (does not change the
frequency)-moves signal in time!
What is the rms of a triangular wave?
We have to average i2
Irms =IP
3
How do we assess the response of circuits to a sinusoidal signal (direct substitution)
1. First, write the differential equation that relates the desired output to the input
2. Ignore all initial conditions (this includes switches, etc.). All initial energies (initial conditions) are assumed to have dissipated in the resistive part of the circuit. circuit.
3. Assume that the response (in this case, the particular response) is also sinusoidal with different amplitude and phase, but the same frequency (linear circuit)
4. Plug in the proposed response in the differential equation and solve for the unknown amplitude and phase
An RL circuit excited by a sinusoidal voltage
source.
Differential Equation
1. Using KVL,
Ldi
dt+ Ri =V =Vm cos ωt + φ( )
This will disappear
Ldt
+ Ri =V =Vm cos ωt + φ( )
i t( ) = Im cos ωt + φ −θ( )− Im cos φ −θ( )e−R
Lt
i 0+( )= 0 = i 0−( )iss t( ) = Im cos ωt + φ −θ( )
Steady State Response
1. Using KVL,
Ldi
dt+ Ri =V =Vm cos ωt + φ( )
i t( ) = Im cos ωt + φ −θ( )m
−LImω sin ωt + φ −θ( )+ RIm cos ωt + φ −θ( )=
Vm cos ωt + φ( )RIm sin θ( )− LImω cos θ( ){ }sin ωt + φ( )+
RIm cos θ( )+ LImω sin θ( ){ }cos ωt + φ( )=
Vm cos ωt + φ( )
Steady State Response
RIm sin θ( )− LImω cos θ( ){ }sin ωt + φ( )+
RIm cos θ( )+ LImω sin θ( ){ }cos ωt + φ( )=
V cos ωt + φ( )
Is there an
easier and
more
intuitive way
to get this?
Vm cos ωt + φ( )
RIm sin θ( )− LImω cos θ( ) = 0⇒ θ = tan−1 ωLR
RIm cos θ( )+ LImω sin θ( ) =Vm ⇒ Im =Vm
ω 2L2 + R2
Phasors
1. Phasors are actually vector representation of sinusoidal signals
2. They suppress the element of time (if you know phase and amplitude, you can reconstruct the signal assuming a known frequency)
3. The length of the vector is the amplitude of the signal 3. The length of the vector is the amplitude of the signal (fixed) and the direction of the phasor at t=0 is the phase of the sinusoid
4. We can combine (add and subtract) phasors using vector addition
5. Sinusoidal signals may be related to phasors by observing the projection of the vector onto x and y axis
Asin θ t( )( )
Acos θ t( )( )
A
θ t( ) = ωt + φ
θ t( )
Acos θ t( )( )
e± jθ = cos θ( )± j sin θ( )Ae± jθ = Acos θ( )± jAsin θ( )Ae± j ωt+φ( ) = Acos ωt + φ( )± jAsin ωt + φ( )
How to add two phasors (22.43
degrees is the phase difference)
Can we use phasors to represent current and voltages of a passive devices?
1. If we add (subtract) two sinusoidal signals,
the resulting phasors add (subtract)
2. KVL and KCL still applies for sinusoidal signals
3. This implies that we can apply KVL and KCL 3. This implies that we can apply KVL and KCL
for circuits using phasors assuming that we
have access to the relationship between
current and voltage of all components
4. Let us consider passive components
Resistors
v = Ri
V sin ωt + φ( )= RIsin ωt + φ( )⇒V = RI
V = RI
Figure 9.9 A plot showing that the
voltage and current at the terminals of
a resistor are in phase.
Inductor
( )sin
div L
dt
i I tω φ
=
= +( )( )
( )cos
sin 90
v I L t
v V t
VV j LI j L
I
ω ω φ
ω φ
ω ω
= +
= + +
= ⇒ =
Inductor response
Capacitor
i = Cdv
dt
v =V sin ωt + φ( )i =VωCcos ωt + φ( )i =VωCcos ωt + φ( )i =VωC sin ωt + φ + 90( )i = I sin ωt + φ( )
I =VjωC ⇒V
I=1
jωC
Capacitor response
What do previous observations imply
1. Capacitors, inductor, and resistors may be looked as having “Impedance”
2. Impedance of a resistor is real and is called resistance resistance
3. The impedance of a capacitors or an inductor is purely imaginary (is called Reactance)
4. The reactance is frequency dependent
5. Given KVL and KCL, we can treat R, L, and C as we would treat a simple resistors through
V = ZI
Z = impedance = R + jXReactance
KVL
Vab = Z1 + Z2 + ...+ Zn( )IVab = ZI
Z = Z1 + Z2 + ...+ Zn
Example 9.6.
The circuit at the 800 Hz
KCL
Vab = Z1I1 = Z2I2 = ...= ZnIn
I = I1 + I2 + ...+ In
Vab = ZI
Z = Z1 || Z2 || ... || Zn
A parallel circuit
Previous circuit at ω=200,000
We can use ∆∆∆∆-Y transformation
Example
Simplified circuit
source transformation
Thevenin equivalent
Norton equivalent
Example
Use of Thevenin to solve circuit
Compute output voltage (no
load)
Thevenin Impedance
Final circuit
Writing Node Equation
Mesh Current
Linear TransformersMutual Inductance
Vs = R1 + jωL1 + ZS( )I1 − jωMI2
0 = − jωMI1 + R2 + jωL2 + ZL( )I2
Transformer as a 2-port networkZ11 = Zs + R1 + jωL1Z22 = ZL + R2 + jωL2Zab = R1 + jωL1 + Zr2
Zcd = R2 + jωL2 + Zr1
=ω 2M 2
Self-impedance primary
Self-impedance secondary
Zr2 =ω 2M 2
R2 + RL + j ωL2 + XL( )
Zr1 =ω 2M 2
R1 + Rs + j ωL1 + X s( )
k1 =Zr2
Z22*
=ω 2M 2
| Z22 |2
k2 =Zr1
Z11*
=ω 2M 2
| Z11 |2
Reflected impedance
Scaling factor
How do you model an ideal
transformer?
M
Coefficient of coupling
k =M
L1L2
L1
L2=
N1
N2
2
Ideal transformer
k =1
L1 →∞
L →∞
V1
V2=N1
N2
I1 =N2
No power
loss
L2 →∞
L1
L2=
N1
N2
2
1
I2= 2
N1
V1I1 =V2I2
Zin =N1N2
2
ZL
Ideal Transformer
Input/Output relationship
Can we use phasors to solve a
circuit?
The complex number -7 – j3 = 7.62
-156.80°.
Example
The phasor diagram
Example
Impedance model
Phasor diagram for the circuit
addition of a capacitor
One more phasor
Final picture
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