Transcript Complex Numbers - Mar Dionysius College, Pazhanji

Complex Numbers
Definition
• A complex number z is a number of the form
x  jy
where j   1
• x is the real part and y the imaginary part, written as x = Re
z, y = Im z.
• j is called the imaginary unit
• If x = 0, then z = jy is a pure imaginary number.
• The complex conjugate of a complex number, z = x + jy,
denoted by z* , is given by
z* = x – jy.
• Two complex numbers are equal if and only if their real
parts are equal and their imaginary parts are equal.
1 August 2006
Slide 2
Complex Plane
A complex number can be plotted on a plane with two
perpendicular coordinate axes
 The horizontal x-axis, called the real axis
 The vertical y-axis, called the imaginary axis
y
P
z = x + iy
O
The complex plane
1 August 2006
x
Represent z = x + jy
geometrically as the point P(x,y)
in the x-y plane, or as the
vector OP from the origin to
P(x,y).
x-y plane is also known
as the complex plane.
Slide 3
Polar Coordinates
With x  r cos  ,
y  r sin 
z takes the polar form:
z  r (cos  j sin  )
r is called the absolute value or modulus or
magnitude of z and is denoted by |z|.
z  r  x 2  y 2  zz *
*
zz
 ( x  jy )( x  jy )
Note that :
 x2  y2
1 August 2006
Slide 4
Im
P
y
z = x + iy
|z
|=
r
θ
x
O
Re
Complex plane, polar form of a complex number
Geometrically, |z| is the distance of the point z from the
origin while θ is the directed angle from the positive xaxis to OP in the above figure.
From the figure,
1 August 2006
 y
  tan  
x
1
Slide 5
θ is called the argument of z and is denoted by arg z. Thus,
 y
  arg z  tan  
 x
1
z0
For z = 0, θ is undefined.
A complex number z ≠ 0 has infinitely many possible
arguments, each one differing from the rest by some
multiple of 2π. In fact, arg z is actually
 y
  tan    2n , n  0,1,2,...
 x
1
The value of θ that lies in the interval (-π, π] is called the
principle argument of z (≠ 0) and is denoted by Arg z.
1 August 2006
Slide 6
Euler Formula – an alternate polar form
The polar form of a complex number can be rewritten as :
z  r (cos  j sin  )  x  jy
 re j
This leads to the complex exponential function :
e z  e x  jy  e x e jy
 e x cos y  j sin y 

Further leads to :
1 August 2006

1 j
cos  e  e  j
2
1 j
sin  
e  e  j
2j


Slide 7
In mathematics terms,  is referred to as the argument of z
and it can be positive or negative.
In engineering terms,  is generally referred to as phase of z
and it can be positive or negative. It is denoted as z
The magnitude of z is the same both in Mathematics and
engineering, although in engineering, there are also
different interpretations depending on what physical system
one is referring to. Magnitudes are always > 0.
The application of complex numbers in engineering will
be dealt with later.
1 August 2006
Slide 8
Im
z1
x
r1
z1  r1e
j1
+1
-2
r2
z 2  r2 e  j2
x
z2
1 August 2006
Re
r1 , r2 , 1 ,  2  0
Slide 9
Example 1
A complex number, z = 1 + j , has a magnitude
| z | (12  12 )  2
and argument :
1


z  tan    2n    2n  rad
1
4

1
Hence its principal argument is :
Arg z   / 4
rad
Hence in polar form :



z  2  cos  j sin   2e
4
4

1 August 2006
j

4
Slide 10
Example 2
A complex number, z = 1 - j , has a magnitude
| z | (12  12 )  2
and argument :
 1
 

z  tan    2n     2n  rad
 1 
 4

1
Hence its principal argument is : Arg z  

4
rad
Hence in polar form :
z  2e
j

4



 2  cos  j sin 
4
4

In what way does the polar form help in manipulating
complex numbers?
1 August 2006
Slide 11
Other Examples
What about z1=0+j, z2=0-j, z3=2+j0, z4=-2?
z1  0  j1
 1e j 0.5
 10.5
z3  2  j0
 2e
 20
j0
1 August 2006
z 2  0  j1
 1e  j 0.5
 1  0.5
z 4  2  j 0
 2e  j
 2  
Slide 12
Im
●
z4 = -2
z1 = + j
0.5
●
z3 = 2
●
Re
● z2 = - j
1 August 2006
Slide 13
Arithmetic Operations in Polar Form
• The representation of z by its real and
imaginary parts is useful for addition and
subtraction.
• For multiplication and division, representation
by the polar form has apparent geometric
meaning.
1 August 2006
Slide 14
Suppose we have 2 complex numbers, z1 and z2 given by :
z1  x1  jy1  r1e
j1
z 2  x2  jy 2  r2 e
 j 2
z1  z 2  x1  jy1   x2  jy 2 
 x1  x2   j  y1  y 2 

z1 z 2  r1e
j1
 r1 r2 e
magnitudes multiply!
1 August 2006
r e
 j 2
2

j (1  (  2 ))
Easier with normal
form than polar form
Easier with polar form
than normal form
phases add!
Slide 15
For a complex number z2 ≠ 0,
j1
z1 r1e
r1 j (1 (  2 )) r1 j (1  2 )

 e
 e
j 2
z 2 r2 e
r2
r2
magnitudes divide!
z1
r1

z2
r2
1 August 2006
phases subtract!
z  1  ( 2 )  1   2
Slide 16
A common engineering problem involving
complex numbers
Given the transfer function model :
20
H (s) 
s 1
Generally, this is a frequency response model if s is
taken to be s  j .
In Engineering, you are often required to plot the
frequency response with respect to the frequency, .
1 August 2006
Slide 17
20e j 0
For a start : H ( s  0) 
 200
j0
1e
Let’s calculate H(s) at s=j10.
20
H ( j10) 

j10  1
H ( j10) 
20
20e j 0
101e
 10 
j tan1  
 1 
 2  20 log10 2 dB  5.98dB
101
H ( j10)  0  tan1 (10)  1.47 rad  84.30
Im
84.30
Re
2
x 2e  i1.47
1 August 2006
Slide 18
Let’s calculate H(s) at s=j1.
20
H ( j1) 

j1  1
H ( j1) 
20
20e j 0
2e
1
j tan1  
1
 14.142  20log10 14.142dB  23dB
2
H ( j1)  0  tan1 (1)  0.7854rad  450
Im
84.30
 450
Re
2
x H (i10)
H (i1)
x
1 August 2006
Slide 19
What happens when the frequency tends to infinity?
H ( s) s  j
H ( s) s  j
20

 0 ?
j  1
20
20


j tan1 
j  1 e
 0  900
When the frequency tends to infinity, H(s) tends to zero in
magnitude and the phase tends to -900!
1 August 2006
Slide 20
Polar Plot of H(s) showing the magnitude and phase of H(s)
1 August 2006
Slide 21
Frequency response of the system
Alternate view of the magnitude and phase of H(s)
1 August 2006
Slide 22