Download Laplace Transformations

Laplace Transformations
Dr. Holbert
February 25, 2008
Lect10
EEE 202
1
Sinusoids
• Period: T = 2p/ω = 1/f
– Time necessary to go through one cycle
• Frequency: f = 1/T
– Cycles per second (Hertz, Hz)
• Angular frequency: ω = 2p f
– Radians per second
• Amplitude: A
– For example, could be volts or amps
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Example Sinusoid
• What is the amplitude, period, frequency,
and angular (radian) frequency of this
sinusoid?
8
6
4
2
0
-2 0
0.01
0.02
0.03
0.04
0.05
-4
-6
-8
Time (sec)
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Phasors
• A phasor is a complex number that
represents the magnitude and phase
angle of a sinusoidal signal:
X M cost   
X  X M 
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Complex Numbers
imaginary
axis
• x is the real part
• y is the imaginary
part
• z is the
magnitude
y

x
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real
axis
• θ is the phase
angle
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5
More Complex Numbers
• Polar coordinates: A = z  θ
• Rectangular coordinates: A = x + jy
• Polar ↔ rectangular conversion relations:
y  z sin 
x  z cos
z x y
2
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  tan
2
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1
y
x
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Arithmetic with Complex
Numbers
• To compute phasor voltages and currents, we
need to be able to perform basic computations
with complex numbers
– Addition
– Subtraction
– Multiplication
– Division
• Appendix A has a review of complex numbers
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Complex Number
Addition and Subtraction
• Addition is most easily performed in rectangular
coordinates:
A = x + jy
B = z + jw
A + B = (x + z) + j(y + w)
• Subtraction is also most easily performed in
rectangular coordinates:
A - B = (x - z) + j(y - w)
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Complex Number
Multiplication and Division
• Multiplication is most easily performed in polar
coordinates:
A = AM  
B = BM  f
A  B = (AM  BM)  (  f)
• Division is also most easily performed in polar
coordinates:
A / B = (AM / BM)  (  f)
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Are You a Technology “Have”?
• There is a good chance that your
calculator will convert from rectangular to
polar, and from polar to rectangular
• Convert to polar: 3 + j4 and –3 – j4
• Convert to rectangular: 2 45 & –2 45
• Add: 3 30 + 5 20
• Determine the complex conjugate of the
phasor: A = z  θ
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Singularity Functions
• Singularity functions are either not finite or
don't have finite derivatives everywhere
• The two singularity functions of interest
are
1. unit step function, u(t)
and its construct: the gate function
2. delta or unit impulse function, (t)
and its construct: the sampling function
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Unit Step Function, u(t)
• The unit step function, u(t)
– Mathematical definition
0
u (t )  
1
t0
t 0
– Graphical illustration
u(t)
1
t
0
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Extensions of Unit Step Function
• A more general unit
step function is u(t-a)
0
u (t  a)  
1
1
ta
ta
0
t
a
• The gate function can
be constructed from u(t)
– a rectangular pulse that
starts at t= and ends at
t=+T : u(t-) – u(t--T)
– like an on/off switch
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1
0

+T
t
13
Delta or Unit Impulse Function, (t)
• The delta or unit impulse function, (t)
– Mathematical definition (non-purist version)
t  t0
0
 (t  t 0 )  
1
t  t0
– Graphical illustration
(t)
1
0
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t0
t
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Extensions of the Delta Function
• An important property of the unit impulse
function is its sampling property
– Mathematical definition (non-purist version)
t  t0
 0
f (t )  (t  t 0 )  
 f (t 0 )
t  t0
f(t)
f(t) (t-t0)
0
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t0
t
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Laplace Transform
• Applications of the Laplace transform
– solve differential equations (both ordinary and partial)
– application to RLC circuit analysis
• Laplace transform converts differential
equations in the time domain to algebraic
equations in the frequency domain, thus three
important processes:
1. transformation from the time to frequency domain
2. manipulate the algebraic equations to form a solution
3. inverse transformation from the frequency to time
domain (we’ll wait to the next class time for this)
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Definition of Laplace Transform
• Definition of the unilateral (one-sided) Laplace
transform

L f t   Fs    f t  e
 st
dt
0
where s=+j is the complex frequency, and
f(t)=0 for t<0
• The inverse Laplace transform requires a course
in complex variable analysis (e.g., MAT 461)
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Laplace Transform Pairs
Most of the time
we find Laplace
transforms from
tables like Table
5.1 rather than
using the
defining integral
(t)
F(s)
δ(t)
1
u(t) {a constant}
1
s
e–at
t
t e–at
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1
sa
1
s2
1
s  a  2
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Some
Laplace
Transform
Properties
Property
(t)
F(s)
Scaling
A (t)
A F(s)
Linearity
1(t) ± 2(t)
F1(s) ± F2(s)
(a·t)
Time Scaling
a0
(t–t0) u(t–t0)
e–s·t0 F(s) t00
Frequency Shifting
e–a·t (t)
F(s+a)
Time Domain
Differentiation
d f (t )
dt
s F(s) – (0)
Time Shifting (delay)
Frequency Domain
Differentiation
t (t)

Time Domain
Integration
Convolution
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1 s
F 
a a
t
0

t
0
f ( ) d
f1 ( ) f 2 (t   ) d
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 d F (s)
ds
1
F ( s)
s
F1(s) F2(s)
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Class Examples
• Drill Problem P5-2 (solve using both the
defining Laplace integral, and the table of
integrals with appropriate properties)
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