Download Sinusoidal Steady

Sinusoidal Steady-State
response
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.
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
1
f 0 = ; frequency(Hz)
T
T; period(sec);
ω 0 = 2πf
v (t ) = Vm cos(ω 0 t + θ )
1
Vrms =
T
2
Vrms
P=
;
R
Vm
Vrms =
2
∫
T
0
1
v (t ) dt = lim T →
∞ T
2
2
∫ v (t ) dt
T
0
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.
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,
di
L + Ri = V = Vm cos(ωt + φ )
dt
This will disappear
i(t ) = Im cos(ωt + φ − θ ) − Im cos(φ − θ )e
i(0
+
) = 0 = i(0 )
−
iss (t ) = Im cos(ωt + φ − θ )
R
− t
L
Steady State Response
1. Using KVL,
di
L + Ri = V = Vm cos(ωt + φ )
dt
i(t ) = Im cos(ωt + φ − θ )
−LImω sin(ωt + φ − θ ) + RIm cos(ωt + φ − θ ) =
Vm cos(ωt + φ )
{RI
{RI
m
sin(θ ) − LImω cos(θ )}sin(ωt + φ ) +
m
cos(θ ) + LImω sin(θ )}cos(ωt + φ ) =
Vm cos(ωt + φ )
Steady State Response
{RI
{RI
m
sin(θ ) − LImω cos(θ )}sin(ωt + φ ) +
m
cos(θ ) + LImω sin(θ )}cos(ωt + φ ) =
Vm cos(ωt + φ )
Is there an
easier and
more
intuitive way
to get this?
 ωL 
RIm sin(θ ) − LImω cos(θ ) = 0 ⇒ θ = tan  
 R
Vm
RIm cos(θ ) + LImω sin(θ ) = Vm ⇒ Im =
ω 2 L2 + R 2
−1
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
(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
θ (t ) = ωt + φ
Asin(θ (t ))
A
θ (t )
Acos(θ (t ))
e
± jθ
Ae
Ae
= cos(θ ) ± j sin(θ )
± jθ
= Acos(θ ) ± jAsin(θ )
± j (ωt +φ )
= Acos(ωt + φ ) ± jA sin(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
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 ) = RI sin(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
di
v=L
dt
i = I sin (ωt + φ )
v = I ω L cos (ωt + φ )
v = V sin (ωt + φ + 90 )
V
V = jω LI ⇒ = jω L
I
Inductor response
Capacitor
dv
i =C
dt
v = V sin(ωt + φ )
i = VωC cos(ωt + φ )
i = VωC sin(ωt + φ + 90)
i = I sin(ωt + φ )
V
1
I = VjωC ⇒ =
I
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
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 + jX
Reactance
KVL
Vab = (Z1 + Z 2 + ...+ Z n )I
Vab = ZI
Z = Z1 + Z 2 + ...+ Z n
Example 9.6.
The circuit at the 800 Hz
KCL
Vab = Z1I1 = Z 2 I2 = ... = Z n In
I = I1 + I2 + ...+ In
Vab = ZI
Z = Z1 || Z 2 || ... || Z n
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 Transformers
Mutual Inductance
Vs = (R1 + jωL1 + Z S )I1 − jωMI2
0 = − jωMI1 + (R2 + jωL2 + Z L )I2
Transformer as a 2-port network
Z11 = Z s + R1 + jωL1
Z 22 = Z L + R2 + jωL2
Z ab = R1 + jωL1 + Z r2
Z cd = R2 + jωL2 + Z r1
Self-impedance primary
Self-impedance secondary
ω 2M 2
Z r2 =
R2 + RL + j (ωL2 + X L )
ω 2M 2
Z r1 =
R1 + Rs + j (ωL1 + X s )
Z r2 ω 2 M 2
k1 = * =
Z 22 | Z 22 |2
Z r1 ω 2 M 2
k2 = * =
Z11 | Z11 |2
Reflected impedance
Scaling factor
How do you model an ideal
transformer?
Coefficient of coupling
k=
M
L1L2
L1  N1 
= 
L2  N 2 
2
Ideal transformer
V1 N1
=
V2 N 2
k =1
L1 →∞
I1 N 2
=
I2 N1
L2 →∞
L1  N1 
= 
L2  N 2 
2
No power
loss
V1I1 = V2 I2
 N1 
Z in =   Z L
 N2 
2
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