Download Review Lecture 5 • First-order circuit — The source-free R-C/R-L circuit

Review Lecture 5
• First-order circuit
— The source-free R-C/R-L circuit
— Step response of an RC/RL circuit
v(t )  v()  [v(t0 )  v()]e

t t 0

The time constant  = RC
The final capacitor
voltage v()
The initial capacitor
voltage v(t0)
Voltage/current-division
Lecture 7 AC Circuits
Sinusoids and Phasors
Contents
• Sinusoids
—Amplitude, angular/cyclic frequency, argument,
periodic function
• Phasors
—Rectangular/polar/exponential form, Euler’s identity,
and time-domain/phasor-domain representation
• Phasor relationships for circuit elements
— R, L, and C
• Impedance and admittance
— resistance and reactance,conductance and
susceptance
• Kirchhoff’s laws in the frequency domain
• Impedance combinations
Sinusoids
• Periodic functions
— f ( x  nT )  f ( x)
cos(x  n2 )  cos(x)
• Sinusoidal: sine or cosine function
—Time-varying (t) and argument, x, of cos(x)
—Amplitude Vm
phase
—Frequency:
v(t )  Vm cos(t   )
• angular  (rad/s) and cyclic f (Hz):   2f
argument
—Period T (s): T  2 T  1f
cos t  nT      cost     nT   cos(t   ) T  2
—use cosine with positive amplitude V  0

sin(t )  cos(t  )
2
m
 cos(t )  cos(t   )
Phase
• Phase angle between two signals
cos(x+/3)
leads
cos(x)
by /3
cos(x-/3)
lags
— Same frequency: 
Amplitude may vary
— Phase difference:
• Express them in the same form
– cosine and Vm > 0
v1  Vm1 cos(t  1 )
v2  Vm 2 cos(t  2 )
• Phase difference :   1  2 at any time t
— out of/in phase
— lead/lag
5cos(x+1)
In phase :   1  2  0
cos(x +2)
For a given , Vm and  are
important quantities.
cos(x)
by /3
Example:
Calculate the phase angle between v1 = −10 cos(ωt + /3)
and v2 = 12 sin(ωt + /6). State which sinusoid is leading.


2
(1) Same form: v1  10 cos(t  3 )  10 cos(t  3   )  10 cos(t  3 )
v2  12 sin(t 
(2) Compare:
v1  10 cos(t 

6
)  10 cos(t 


2
 
)  10 cos(t   )
3
3 3
v1 lags v2 by /3

 )  10 cos(t  )
6 2
3
Phasors
A phasor is a complex number that represents the amplitude (Vm)
and phase () of a sinusoid.
• Complex numbers - three
representations
— Rectangular: z  x  jy
z  r
— Polar:
j
— Exponential: z  re
• Phasor representaion sinusoid v(t)

— real part:
v(t )  Vm cos t   
Rectangular
Èxponential
Vm cost     j sin t   
 Vm e j t    Vm e e jt
 Re Vm cost     j sin t   

 Re Ve jt

V  Vm e j
ejωt is implicitly present
Sinor: rotating phasor
Sinor: Vm e e jt on the complex plane
• v(t) is the projection of the sinor on the real axis.
A circle of radius Vm
• The value of the sinor at time t = 0 is the phasor V of v(t).
A complex number:
magnitude and direction - vector
Projection
counterclockwise
One circle, 2,
Period T = time/circle = 2/
Phasor diagram
• Diagram/complex number
• Magnitude and phase
V  Vm e j
ejωt is implicitly present
Time domain and phasor(frequency) domain
To represent signals v(t), i(t):
• Time domain: v(t )  Vm cos(t   )
—Time dependent
—Always real
• Phasor (freqency) V  V e j
m
domain:
—Time independent
— Generally complex
—  is constant.
—Circuit response depends
on 
Example: Given i1(t) = 4 cos(ωt +
/6) and i2(t) = 5 sin(ωt − /3), find
their sum.
i1 (t )  I 1 , i2 (t )  I 2
j

I 1  4e , I 2  5e
6
i2 (t )  5 cos(t 
I 2  5e
j
 ...

3
?
cosine and Im > 0
i (t )  I m cos(t   )


5
 )  5 cos(t  )
3 2
6
5
6
I 1  I 2  4e
 4(cos
j

j

6
 5e
j
5
6

5
5
 j sin )  5(cos
 j sin )
6
6
6
6
Derivative and integral in phasor domain
v(t )  Vm cos(t   )
V  Vm e j
dv (t )
d cos(t   )
 Vm
 Vm sin(t   )
dt
dt


 Vm cos(t    )  Vm cos(t      )
2
2
3
 Vm cos(t   
 2 )
2

 Vm cos(t    )
2
Vm e
 
j  
2

 Vm e j e
j
Example: Using the phasor approach,
determine the current i(t) in a circuit
described by the integrodifferential equation.
4i  8 idt  3
Frequency domain

2

j
8I
4I 
 3 jI  50e 3
j



 V  cos  j sin   jV
2
2

dv(t )
dt
jV
 v(t )
V
j
di

 50 cos(2t  )
dt
3
Useful in finding the
steady-state solution:
same frequency!
 =2
Phasor relationships for circuit elements
i leads v by /2
v and i in phase
v(t )  Vm cos(t   )
i (t )  I m cos(t   )
  I m e j
v(t )  RI m cos(t   )
V  RI m e j  RI
Ohm’ law holds
v leads i by /2
  I m e j
i (t )  I m cos(t   )
di (t )
v(t )  L
 LI m sin(t   )
dt

 LI m cos(t    )
2
V  jLI m e j  jLI
v has a phase
+/2
i (t )  C
dv (t )
 CVm sin(t   )
dt

 CVm cos(t    )
2
V  Vm e j
i has a phase
+/2
I  jCVm e j  jCV
V
I
jC
Impedance and admittance
opposition to the flow of sinusoidal current
V  RI
V  jLI
V
V  ZI
I
jC
Impedance
Z
V Phasor Voltage


I Phaseor Current
Admittance
Complex quantity, but
not a phasor
Z
1
j
C
Y
1 I
 S
Z V
More about Impedance…
1
j  , an open circuit
C
Inductor: Z  jL  0, a short circuit
Capacitor, Z  
 0

Capacitor, Z  
1
j  0, a short circuit
C
Circuit response
depends on the
frequency!
Inductor: Z  jL  , an open circuit
Z is a complex quantity
Rectangular form (x+jy)
Reactance
Z  R  jX
Resistance
1
j)
C
X  0, inductive/lagging reactance... (Z  jL)
X  0, capacitive/leading reactance... (Z  
1
Y   G  jB
Z
Conductance
Current leads voltage
Susceptance
Note: G does not always equal to 1/R
Kirchhoff’s laws in the frequency domain
Both KVL and KCL hold in the frequency domain
Time domain
v1  v2  ...  vn  0
vi (t )  Vmi cos(t  i )
vi  Re(Vmi e ji e jt )
Use the Re of a complex quantity
to represent the signal
Re(Vm1e j1 e jt )  Re(Vm 2e j2 e jt )  ...  Re(Vmn e jn e jt )  0
Re(Vm1e j1 e jt  Vm 2e j2 e jt  ...  Vmn e jn e jt )  0


Re (Vm1e j1  Vm 2e j2  ...  Vmn e jn )e jt  0
Frequency domain
Vm1e j1  Vm 2 e j2  ...  Vmn e jn  0
KVL
V1  V2  ...  Vn  0
KCL
I 1  I 2  ...  I n  0
e jt  0
Impedance combinations
Combination of Impedance is similar to resistance circuits.
KVL
V  V1  V2  ...  Vn
V  Z1I  Z 2I  ...  Zn I  Z1  Z 2  ...  Zn I
 Z eq I
Z eq  Z1  Z 2  ...  Z n
Voltage/current division holds.
Y- and -Y transformations as well
Example: Impedance
Find the input impedance of the circuit Assume that the circuit operates at ω = 50 rad/s.
Z1 = Impedance of the 2-mF capacitor
Z2 = Impedance of the 3- resistor in series with
the 10-mF capacitor
Z3 = Impedance of the 0.2-H inductor in series
with the 8-resistor
Z
1
j
C
Z  jL
Example: Solve the circuit
Determine vo(t) in the circuit
Step 1: Phasor-domain equivalent
Z1
Step 2: Voltage-division
Z2
Step 3: convert to time-domain
Appendix: Mathematics
• Complex numbers
• Trigonometry
z1  z 2   x1  x2   j  y1  y2 
sin  A  B   sin A cos B  cos A sin B
z1  z 2   x1  x2   j  y1  y2 
cos A  B   cos A cos B  sin A sin B
z1 z 2  r1r2 e j 1 2 
sin t      sin t
z1 r1 j 1 2 
 e
z 2 r2
cost      cos t


sin t     cos t
2

1 1  j
 e
z r


cos t     sin t
2

A cos t  B sin t  C cost   ,
where
C  A2  B 2 ,
  tan 1
z  re
j

2
,
z *  x  jy  re  j
B
A
1
j
j
e  j  cos   j sin 
Lecture 8 AC Circuits
Sinusoidal Steady-State Analysis