Transcript
BME 301
5-Complex Circuits
1
Letrsquos Connect a Series RC circuit
C
R
+
--VIN(t)
I(t)
+ VR - + VC -
Since this is a Series Circuit let s use KVL and we get( ) where from Ohms Law and
1 from Gauss Law
1( ) +
To solve this differential equation lets assume tha
IN R C R
C
IN
V t V V V IR
V IdtC
V t IR IdtC
= + =
=
=
int
int
5
5
5 5 5
5
t ( ) since this is a linear system then ( ) must
also have the form where A is unknownSubstituting for and ( ) we get
1 + 5
1( ) ( )1 155 5
tIN
t
IN
t t t
t
V t Ve I tAe
I V t
Ve Ae R AeC
V VV A R A I t eC R R
C C
=
=
= + rArr = rArr =+ +
2
Do we always have to solve Differential EquationThere is a simpler way
But first some assumptionsLets assume that most signals are of the form ( )
( ) ( )Then the current through a resistor is ( ) and ( )
Then the cur
st
st st
st
V t AeV t Ae V t AeI t R
AeR R I tR
=
= = = =
( ) ( ) 1rent through a capacitor is ( ) and ( )
1 1 1 ( )Then the current through an inductor is ( ) ( ) and 1( )
If we define
stst st
st
stst st
st
dV t d V t AeI t C C Ae sCAedt dt I t sCAe sC
V t AeI t V t dt Ae dt Ae sLL L sL I t Ae
sL
= = = = =
= = = = =int int
( ) as the impedance then the impedance of ( )
( )a resistor ( )
( ) 1a capacitor and( )( )an inductor ( )
V tI t
V t RI tV tI t sCV t sLI t
=
=
=3
Do we always have to solve Differential Equation continued
There is a simpler wayNow lets assume that most signals are either sinusoids or related to sinusoidsSo ( ) cos( ) however from Euler s formula we know that
cos( ) and that cos( )2
j j j
V t A te e et
θ θ
ω
θ ωminus
=
+= = where is the imaginary number 1
2So to make things simpler lets concentrate on signals which have the form ( )Then the impedance of
( )a resistor ( )
( ) 1a capacitor ( )
t j t
j t
e j
V t Ve
V t RI tV tI t
ω ω
ω
minus+minus
=
=
= and
( )an inductor ( )
j CV t j LI t
ω
ω=
4
Do we always have to solve Differential Equation continued
5
C
R
+
--VIN(t)
I(t)
+ VR - + VC -
Since this is a Series Circuit let s use KVL and use impedances to solve for the current we get
( ) ( ) (t) where ( ) ( ) and 1 1( ) = ( )
1 1( ) ( ) + ( ) ( ) ( )
IN R C R
C
IN
V t V t V V t I t R
V t Idt I tC j C
V t I t R I t R I tj C j C
ω
ω ω
= + =
=
= = +
int
This is just an arithmetic equation but with complex numberssince
( )( ) 1( )
Before we go any further lets review complex numbers
INV tI tR
j Cω
=+
6
Complex Numbers
bull Complex numbers What are theybull What is the solution to this equation
ax2+bx+c=0bull This is a second order equation whose solution
is
aacbbx 2
42
21minusplusmnminus=
7
What is the solution to
1 x2+4x+3=0
31224
244
212164
23444 2
21
minusminus=plusmnminus=plusmnminus=
minusplusmnminus=timesminusplusmnminus=x
8
What is the solution to
2 x2+4x+5=0
244
220164
25444 2
21
minusplusmnminus=
minusplusmnminus=timesminusplusmnminus=x
9
What is the Square Root of a Negative Number
bull We define the square root of a negative number as an imaginary number
bull We define
bull Then our solution becomesans)mathematicfor ( engineersfor 1 i j rArrminus
12122
242
44 2
442
201642
5444 2
21
jjjj
x
minusminus+minus=plusmnminus
=plusmnminus
=minusplusmnminus
=
minusplusmnminus=
timesminusplusmnminus=
10
The Complex Planebull z = x+jy is a complex number where
x = Rez is the real part of zy = Imz is the imaginary part of z
bull We can define the complex plane and we can define 2 representations for a complex number
Rez
Imz
x
y
z = x+jy
(xy)
11
Rectangular Formbull Rectangular (or cartesian) form of a complex
number is given asz = x+jyx = Rez is the real part of zy = Imz is the imaginary part of z
Rez
Imz
Rectangular or Cartesian
x
y
z = x+jy
(xy)
12
Polar Formbull is a complex number wherebull r is the magnitude of zbull θ is the angle or argument of z (arg z)
Rez
Imz
Polar
x
y
z = r e jθ
(rθ)
θ
r
13
Relationships between the Polar and Rectangular Forms
z = x + jy = r e jθ
bull Relationship of Polar to the Rectangular Formx = Rez = r cos θy = Imz = r sin θ
bull Relationship of Rectangular to Polar Form
)arctan( and 22
xyyxr =+= θ
14
Addition of 2 complex numbersbull When two complex numbers are added it is best to use
the rectangular formbull The real part of the sum is the sum of the real parts and
imaginary part of the sum is the sum of the imaginary parts
bull Example z3 = z1 + z2
)()(
2121
2121
2211213
222111
yyjxxjyjyxx
jyxjyxzzzjyxzjyxz
+++=+++=
+++=+=+=+=
Re
Im
z1
z2
x1
y1
y2
x2
x1 + x2
y1 + y2
z3
15
Multiplication of 2 complex numbers
bull When two complex numbers are multiplied it is best to use the polar form
bull Example z3 = z1 x z2
bull We multiply the magnitudes and add the phase angles
)(21
)()(21
)(2
)(1213
)(22
)(11
2121
21
21
θθθθ
θθ
θθ
+==
times=times=
==
jjj
jj
jj
erreerrererzzz
erzerz
Re
Im
θ1
r1
θ2r2r3 = r1 r2
θ3= θ1 +θ2
16
Eulerrsquos Formulae jθ = cos θ + j sin θ
bull We can use Eulerrsquos Formula to define complex numbers
z = r e jθ = r cos θ + j r sin θ= x + j y
Rez
Imz
θ
1
Some shorthand
17
Some Examples
Rez
Imz
-1
Rez
Imz1
Rez
Imz
-1
j
- j
Rez
Imz1
118
Some Examples4 2
4
2
4 2 2 2 1 1
2 21) 2 2[cos( 4) sin( 4)]
1 12[cos( 4) sin( 4)] 2[ ] 2 22 2
2) 2 0 2
3) 2 2 0 2 2 (141 2) 2 (141 2) 2 058605862 2 ( 2) (0586) tan ( ) 153 tan (0
2
j j
j
j
j j
e ee j
j j j
e j
j j j j j
e e
π π
π
π
π π
π π
π π
minus
minus
minus minus minus
+
= minus + minus
= minus = minus = minus
= +
minus + + = minus minus = + minus = +
there4 + = + ang = ang 41)
Re
Im
Note 0785 03920692 2254 8π π
= = rArr
19
Some Examples1
32 2 tan ( 1)2 45 5 5 5 5( 1 ) 5 ( 1) 1 5 2
j j j je e j j e eπ π
π minus minus+ = minus = minus + = times minus + =
Re
Im
20
21
Complex Exponential Signalsbull Since we are dealing with sinusoid signal letrsquos look at
the complex exponential signal which is defined as
bull Note that it is defined in polar form wherendash the magnitude of z(t) is |z(t)| = A ndash the angle (or argument arg z(t) ) of z(t) = (ωot + θ)
bull Where ω is called the radian frequency and θ is the phase angle (phase shift)
( )( ) j tz t Ae ω θ+=
22
Complex Exponential Signalsbull Note that by using Eulerrsquos formula we can rewrite the
complex exponential signal in rectangular form as
bull Therefore real part is the cosine signal and imaginary part is a sine signal both of radial frequency ω and phase angle of θ
( )( )cos( ) sin( )
j tz t AeA t jA t
ω θ
ω θ ω θ
+== + + +
23
Plotting the waveform of a complex exponential signal
bull For an complex signal we plot the real part and the imaginary part separately
bull Example z(t) = 20e j(2π(40)t-04π) = 20e j(80πt-04π)
= 20 cos(80πt-04π) + j20 sin(80πt-04π)
-20
-15
-10
-5
0
5
10
15
20
-003 -002 -001 0 001 002 003 004
real part
-20
-15
-10
-5
0
5
10
15
20
-003 -002 -001 0 001 002 003 004
imaginary part