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ECE 3144 Lecture 30
Dr. Rose Q. Hu
Electrical and Computer Engineering Department
Mississippi State University
1
Reminder from Lecture 29
• Representation of sinusoidal functions
Acos(t+)
• Steady state response of Sinusoid functions
v(t)=VM cost
i(t ) 
VM
R  L
2
2
2
cos(t  tan
1
L
R
i(t)
2
)
Complex Number
•
•
A complex number a+jb:
– a = Re(a+jb) is the real part of the complex number a+jb
– b=Im(a+jb) is the imaginary part of the complex number a+bj
Complex number relationships
a+bj = rej
r
•
a2  b2
  tan 1
b
a
a=rcos, b=rsin
Now we want to establish a relationship between time-varying sinusoidal
functions and complex numbers through Euler’s equation
e jt  exp( jt )  cos t  j sin t
cost = Re(ejt)
sint = Im(ejt)
3
Complex forcing function
• Think about applying a complex forcing function
v(t) = VMcos(t+) + jVMsin(t+ ) to a linear circuit network.
– We expect a complex forcing function to produce a complex response;
– The real part of the forcing function, VMcos(t+ ), will produce the real
part of the response IMcos(t+)
– The imaginary part of the forcing function, jVMsin(t+ ), will result in
the imaginary portion of the response jIMsin(t+)
– Based on the superposition, the response to the complex forcing function
v(t) = VMcos(t+ ) + jVMsin(t+ ) is i(t) = IMcos(t+) +
jIMsin(t+)
• Applying Euler’s identity
– v(t) = VMcos(t+) + jVMsin(t+  )= VMej(t+)
– i(t) =IMcos(t+) + jIMsin(t+) = IMej(t+)
– The complex source and response are illustrated as
v(t) = VMej( t+)
i(t) =IMej( t+)
4
Apply complex forcing function to the problem as shown in Lecture 29
Once gain, let us determine the current in the RL circuit. However, rather
than applying VMcost, we will apply VMejt as the input signal
v(t)=VM cost
The forced response of input VMejt will be of form
i(t) = Imej(t+) , where IM and  need to be decided.
i(t)
The KVL equation for the circuit is
L
di (t )
 Ri (t )  VM e jt
dt
Substituting i(t) = Imej(t+) into the above differential equation
d
( I M e j (t  ) )  VM e jt
dt
 jLI M e j (t  )  VM e jt
=>
RI M e j (t  )  L
RI M e j (t  )
=>
RI M e j  jLI M e j  VM
=>
VM
I M e j 
, converting the right side to exponential form =>
R  jL
I M e j 
VM
R  L
2
2 2
e j (  tan
1
(L / R ))
IM 
=>
VM
R 2   2 L2
and
   tan 1
L
R
Since the actual forcing function was VMcost, the actual response is the real part of the complex response
i (t )  I M cos(t   ) 
VM
R  L
2
2
2
cos(t  tan 1
L
R
)
5
•
•
•
•
•
•
Phasor
By introducing in complex forcing function, the differential equations becomes simple
algebraic equations with coefficients as complex numbers.
The actual response is the real part of the complex response from the complex forcing
function.
For the given example, the forcing function is v(t) = VMej t and steady sate response is i(t) =
Imej( t+) . The steady state response in the network has the same form and the same frequency
with the forcing function.
So from now on, we will simply drop the frequency ej t since each term in the equation will
contain it.
Thus every voltage and current can be fully described by a magnitude and phase. This
abbreviated complex representation is called a phasor.
For a voltage v(t)=VM cos(t+), a current i(t)=IM cos(t+),
– Their exponential notations are
v(t) = Re[VMei(t+ )], i(t) = Re[IMei(t+ )]
– Their phasor notations are
V  VM 
I  I M 
•
Phasor operation:
•
v(t) represents a voltage in time domain while phasor V represents the voltage in the frequency
domain.
A11
A
 1 1   2
A2  2
A2
Time Domain
Acos(t)
Asin(t)
Frequency Domain
A  
A    90o
6
Phasor relationship for a resistor R
The voltage-current relationship is known as
v(t) = Ri(t)
(1)
j (t  v )
Applying the complex voltage source VM e
results in the
j (t  )
complex current I M e
, the above equation becomes:
i
VM e j (t  v )  RI M e j (t i )
VM e j v  RI M e j i
=>
(2)
The above equation can be written in phasor form as
V=RI
Where
V  VM e j v  VM  v
(3)
I  I M e j i  I M  i
From equation (2) we see that v=I. Thus the current through and voltage
across a resistor are in phase.
7
Phasor relationship for an inductor L
The voltage-current relationship for an inductor is
di (t )
dt
v (t )  L
j (t  v )
Applying the complex voltage source VM e
results in the
j (t  )
complex current I M e
, the above equation becomes:
i
d
I M e j (t  i )
dt
=>
VM e j (t  v )  jLI M e j (t i )
=>
VM e j (t  v )  L
VM e j v  jLI M e j i
(1)
Equation (1) in phasor notation is
V = jLI
Note that
(2)
j  1e j 90  1900
o
Thus the voltage and current for an inductor are 90o out of phase. Actually I lags
V by 90o for an inductor.
8
Phasor relationship for a capacitor C
The voltage-current relationship for an inductor is
i (t )  C
dv (t )
dt
Applying the complex voltage source VM e j (t  ) results in the
complex current I M e j (t  ) , the above equation becomes:
v
i
I M e j (t  i )  C
d
VM e j (t  v )
dt
I M e j (t i )  jCVM e j (t  v )
I M e j i  jCVM e j v
=>
=>
(1)
Equation (1) in phasor notation is
I = jCV
Note that
(2)
j  1e j 90  1900
o
Thus the voltage and current for a capacitor are 90o out of phase. Actually V lags
I by 90o for a capacitor.
9
Homework for lecture 30
• Problems 7.5, 7.6, 7.7
• Due April 8
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