The Laplace Transform
M. J. Roberts - All Rights Reserved.
Edited by Dr. Robert Akl
1
Generalizing the Fourier Transform
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2
Generalizing the Fourier Transform
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3
Generalizing the Fourier Transform
The extra factor e- s t is sometimes called a convergence factor
because, when chosen properly, it makes the integral converge
for some signals for which it would not otherwise converge.
For example, strictly speaking, the signal A u ( t ) does not have
a CTFT because the integral does not converge. But if it is
multiplied by the convergence factor, and the real part of s
is chosen appropriately, the CTFT integral will converge.
¥
ò
-¥
¥
ò
-¥
¥
A u ( t ) e- jw t dt = A ò e- jw t dt ¬ Does not converge
0
¥
Ae- s t u ( t ) e- jw t dt = A ò e-(s + jw )t dt ¬ Converges (if s > 0)
0
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4
Complex Exponential Excitation
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5
Complex Exponential Excitation
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6
Pierre-Simon Laplace
3/23/1749 - 3/2/1827
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7
The Transfer Function
()
()
Let x t be the excitation and let y t be the response of a
()
system with impulse response h t . The Laplace transform of
()
y t is
¥
¥
( ) ò y(t ) e
Y s =
-¥
- st
dt =
ò () ()
-¥
é h t * x t ù e- st dt
û
ë
ö - st
æ¥
Y s = ò ç ò h t x t - t dt ÷ e dt
ø
-¥ è -¥
()
¥
() (
¥
)
¥
( ) ò h (t ) dt ò x (t - t ) e
Y s =
-¥
- st
dt
-¥
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8
The Transfer Function
() ()
()
()
()
Let x t = u t and let h t = e-4t u t . Find y t .
¥
() () () ò ( ) (
y t = x t *h t =
-¥
)
x t h t - t dt =
¥
()
(
)
( )
u
t
e
u t - t dt
ò
-¥
-4 t -t
t
4t
-4t
ì t -4( t -t )
e
1
1e
d t = e-4t ò e4t d t = e-4t
=
, t >0
ïò e
y t = í0
4
4
0
ï
, t <0
î0
y t = 1 / 4 1- e-4t u t
()
( ) ( )(
) ()
1
1
1
1/ 4 1/ 4
X ( s) = 1 / s , H ( s) =
Þ Y ( s) = ´
=
s+4
s s+4
s
s+4
x ( t ) = (1 / 4 ) (1- e ) u ( t )
-4t
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9
Cascade-Connected Systems
If two systems are cascade connected the transfer function of
the overall system is the product of the transfer functions of the
two individual systems.
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10
Direct Form II Realization
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11
Direct Form II Realization
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12
Direct Form II Realization
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13
Direct Form II Realization
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14
Direct Form II Realization
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15
Direct Form II Realization
A system is defined by y¢¢ ( t ) + 3 y¢ ( t ) + 7 y ( t ) = x¢ ( t ) - 5 x ( t ) .
s-5
H ( s) = 2
s + 3s + 7
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16
Inverse Laplace Transform
There is an inversion integral
s + j¥
()
()
1
st
y t =
Y
s
e
ds , s = s + jw
ò
j2p s - j¥
()
()
for finding y t from Y s , but it is rarely used in practice.
Usually inverse Laplace transforms are found by using tables
of standard functions and the properties of the Laplace transform.
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17
Existence of the Laplace
Transform
Time Limited Signals
()
x ( t ) is also bounded for all t, the Laplace transform integral
If x t = 0 for t < t0 and t > t1 it is a time limited signal. If
converges and the Laplace transform exists for all s.
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18
Existence of the Laplace
Transform
()
() (
) (
)
Let x t = rect t = u t + 1 / 2 - u t - 1/ 2 .
¥
- s/2
s/2
s/2
- s/2
e
e
e
e
X s = ò rect t e- st dt = ò e- st dt =
=
, All s
-s
s
-¥
-1/2
()
()
1/2
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19
Existence of the Laplace
Transform
Right- and Left-Sided Signals
Right-Sided
Left-Sided
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20
Existence of the Laplace
Transform
Right- and Left-Sided Exponentials
Right-Sided
x (t ) = e u (t - t0 ) , a Î
at
Left-Sided
x ( t ) = eb t u ( t 0 - t ) , b Î
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21
Existence of the Laplace
Transform
Right-Sided Exponential
x ( t ) = ea t u ( t - t 0 ) , a Î
¥
¥
t0
t0
X ( s ) = ò ea t e- st dt = ò e(a - s ) t e- jw t dt
If Re ( s ) = s > a the asymptotic
behavior of e(a - s ) t e- jw t as t ® ¥
is to approach zero and the Laplace
transform integral converges.
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22
Existence of the Laplace
Transform
Left-Sided Exponential
x ( t ) = eb t u ( t 0 - t ) , b Î
X( s) =
t0
t0
-¥
-¥
b t - st
e
ò e dt =
( b -s )t - jw t
e dt
e
ò
If s < b the asymptotic behavior of
e( b -s )t e- jw t as t ® -¥ is to approach
zero and the Laplace transform
integral converges.
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23
Existence of the Laplace
Transform
The two conditions s > a and s < b define the region of
convergence (ROC) for the Laplace transform of right- and
left-sided signals.
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24
Existence of the Laplace
Transform
Any right-sided signal that grows no faster than an exponential
in positive time and any left-sided signal that grows no faster
than an exponential in negative time has a Laplace transform.
If x ( t ) = x r ( t ) + x l ( t ) where x r ( t ) is the right-sided part and
x l ( t ) is the left-sided part and if x r ( t ) < K r ea t and x l ( t ) < K l eb t
and a and b are as small as possible, then the Laplace-transform
integral converges and the Laplace transform exists for a < s < b .
Therefore if a < b the ROC is the region a < b . If a > b , there is
no ROC and the Laplace transform does not exist.
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25
Laplace Transform Pairs
The Laplace transform of g1 ( t ) = Aea t u ( t ) is
G1 ( s ) =
¥
ò Ae u ( t ) e
at
- st
-¥
¥
dt = A ò e
- ( s-a )t
0
¥
(a - s ) t
dt = A ò e
e
- jw t
0
A
dt =
s -a
This function has a pole at s = a and the ROC is the region to the
right of that point. The Laplace transform of g 2 ( t ) = Aebt u ( -t ) is
G2 ( s) =
¥
ò Ae u ( -t ) e
-¥
bt
0
- st
( b - s )t
dt = A ò e
-¥
0
( b - s )t
dt = A ò e
-¥
e
- jw t
A
dt = s-b
This function has a pole at s = b and the ROC is the region to the
left of that point.
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26
Region of Convergence
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Region of Convergence
Some of the most common Laplace transform pairs
(There is more extensive table in the book.)
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28
Laplace Transform Example
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29
Laplace Transform Example
Find the inverse Laplace transform of
4
10
X( s) =
, - 3<s < 6
s+3 s-6
4
The ROC tells us that
must inverse transform into a
s+3
10
right-sided signal and that
must inverse transform into
s-6
a left-sided signal.
x ( t ) = 4e-3t u ( t ) + 10e6t u ( -t )
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30
Laplace Transform Example
Find the inverse Laplace transform of
4
10
X( s) =
,s >6
s+3 s-6
The ROC tells us that both terms must inverse transform into a
right-sided signal.
x ( t ) = 4e-3t u ( t ) - 10e6t u ( t )
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31
Laplace Transform Example
Find the inverse Laplace transform of
4
10
X( s) =
, s < -3
s+3 s-6
The ROC tells us that both terms must inverse transform into a
left-sided signal.
x ( t ) = -4e-3t u ( -t ) + 10e6t u ( -t )
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MATLAB System Objects
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MATLAB System Objects
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34
Partial-Fraction Expansion
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Partial-Fraction Expansion
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Partial-Fraction Expansion
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37
Partial-Fraction Expansion
10s
K1
K2
H ( s) =
=
+
, s > -4
( s + 4 )( s + 9) s + 4 s + 9
é
ù
10s
-40
é 10s ù
K1 = ê ( s + 4 )
=ê
=
= -8
ú
ú
( s + 4 ) ( s + 9 ) úû s=-4 ë s + 9 û s=-4 5
êë
é
ù
10s
-90
é 10s ù
K2 = ê ( s + 9)
=ê
=
= 18
ú
ú
( s + 4 ) ( s + 9 ) úû s=-9 ë s + 4 û s=-9 -5
êë
-8
18
-8s - 72 + 18s + 72
10s
H ( s) =
+
=
=
. Check.
s+4 s+9
( s + 4 )( s + 9)
( s + 4 )( s + 9)
¯¯¯¯¯
(
)
h ( t ) = -8e-4t + 18e-9t u ( t )
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Partial-Fraction Expansion
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Partial-Fraction Expansion
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40
Partial-Fraction Expansion
10s
K12
K11
K2
H ( s) =
=
+
, s >4
2
2 +
( s + 4 ) ( s + 9) ( s + 4 ) s + 4 s + 9
­
Repeated Pole
é
2
K12 = ê ( s + 4 )
ê
ë
Using
ù
-40
ú
=
= -8
( s + 9 ) úû s=-4 5
10s
(s + 4)
2
1
d m- k é
K qk =
s - pq
m-k
ë
( m - k )! ds
(
)
m
H ( s)ù
, k = 1, 2, , m
û s® pq
1
d 2-1
d é 10s ù
2
é
ù
K11 =
s
+
4
H
s
=
) ( ) û s®-4
2-1 ë(
ds êë s + 9 úû s®-4
( 2 - 1)! ds
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41
Partial-Fraction Expansion
é ( s + 9 )10 - 10s ù
18
K11 = ê
=
ú
2
( s + 9)
ë
û s=-4 5
18
-8
18 / 5 -18 / 5
K2 = - Þ H ( s) =
+
, s > -4
2 +
5
(s + 4) s + 4 s + 9
(
)
(
18 2
18 2
-8s - 72 +
s + 13s + 36 s + 8s + 16
5
5
H ( s) =
( s + 4 )2 ( s + 9 )
)
, s > -4
10s
H ( s) =
, s > -4
2
( s + 4 ) ( s + 9)
18 -4t 18 -9t ö
æ
-4t
h ( t ) = ç -8te + e - e ÷ u ( t )
è
ø
5
5
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42
Partial-Fraction Expansion
10s 2
H ( s) =
, s > -4 ¬ Improper in s
( s + 4 )( s + 9)
10s 2
H ( s) = 2
, s > -4
s + 13s + 36
10
Synthetic Division ® s 2 + 13s + 36 10s 2
10s 2 + 130s + 360
- 130s - 360
130s + 360
é -32 162 ù
H ( s ) = 10 = 10 - ê
+
, s > -4
ú
( s + 4 )( s + 9)
ës+ 4 s + 9û
h ( t ) = 10d ( t ) - éë162e-9t - 32e-4t ùû u ( t )
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43
Inverse Laplace Transform Example
Method 1
G ( s) =
( s - 3) ( s
s
2
- 4s + 5
)
, s <2
s
, s <2
G ( s) =
( s - 3) ( s - 2 + j ) ( s - 2 - j )
3 / 2 (3 + j) / 4 (3 - j) / 4
, s <2
G ( s) =
s-2- j
s-3 s-2+ j
æ 3 3t 3 + j ( 2- j )t 3 - j ( 2+ j )t ö
e
+
e
g (t ) = ç - e +
÷ø u ( -t )
è 2
4
4
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44
Inverse Laplace Transform Example
3 + j ( 2- j )t 3 - j ( 2+ j )t ö
æ 3
g ( t ) = ç - e 3t +
e
+
e
u ( -t )
÷
è 2
ø
4
4
This looks like a function of time that is complex-valued. But,
with the use of some trigonometric identities it can be put into
the form
{
}
g ( t ) = ( 3 / 2 ) e2t éë cos ( t ) + (1 / 3) sin ( t ) ùû - e3t u ( -t )
which has only real values.
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45
Inverse Laplace Transform Example
Method 2
s
G ( s) =
( s - 3) s 2 - 4s + 5
(
)
, s <2
s
G ( s) =
, s <2
( s - 3) ( s - 2 + j ) ( s - 2 - j )
3 / 2 (3 + j) / 4 (3 - j) / 4
G ( s) =
, s <2
s-3 s-2+ j
s-2- j
Getting a common denominator and simplifying
3 / 2 1 6s - 10
3/2 6 s- 5 / 3
G ( s) =
=
, s <2
2
2
s - 3 4 s - 4s + 5 s - 3 4 ( s - 2 ) + 1
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46
Inverse Laplace Transform Example
M ethod 2
3/2 6 s - 5 / 3
G ( s) =
, s <2
2
s - 3 4 ( s - 2) + 1
The denominator of the second term has the form of the Laplace
transform of a damped cosine or damped sine but the numerator
is not yet in the correct form. But by adding and subtracting the
correct expression from that term and factoring we can put it into
the form
3/2 3é s - 2
1/ 3 ù
G ( s) =
- ê
+
ú , s <2
2
2
s - 3 2 ë ( s - 2) + 1 ( s - 2) + 1û
M. J. Roberts - All Rights Reserved. Edited by Dr. Robert Akl
47
Inverse Laplace Transform Example
M ethod 2
3/2 3é s - 2
1/ 3 ù
G ( s) =
- ê
+
ú , s <2
2
2
s - 3 2 ë ( s - 2) + 1 ( s - 2) + 1 û
This can now be directly inverse Laplace transformed into
{
}
g ( t ) = ( 3 / 2 ) e2t éë cos ( t ) + (1 / 3) sin ( t ) ùû - e 3t u ( -t )
which is the same as the previous result.
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48
Inverse Laplace Transform Example
M ethod 3
When we have a pair of poles p2 and p3 that are complex conjugates
A
K2
K3
we can convert the form G ( s ) =
+
+
into the
s - 3 s - p2 s - p3
s ( K 2 + K 3 ) - K 3 p2 - K 2 p3
A
A
Bs + C
form G ( s ) =
+
=
+ 2
2
s-3
s - ( p1 + p2 ) s + p1 p2
s - 3 s - ( p1 + p2 ) s + p1 p2
In this example we can find the constants A, B and C by realizing that
s
A
Bs + C
G ( s) =
º
+ 2
, s <2
2
( s - 3) s - 4s + 5 s - 3 s - 4s + 5
(
)
is not just an equation, it is an identity. That means it must be an
equality for any value of s.
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49
Inverse Laplace Transform Example
M ethod 3
A can be found as before to be 3 / 2. Letting s = 0, the
3/2 C
identity becomes 0 º + and C = 5 / 2. Then, letting
3
5
s = 1, and solving we get B = -3 / 2. Now
3 / 2 ( -3 / 2 ) s + 5 / 2
G ( s) =
+
, s <2
2
s-3
s - 4s + 5
or
3/2 3 s - 5 / 3
, s <2
2
s - 3 2 s - 4s + 5
This is the same as a result in Method 2 and the rest of the solution
G ( s) =
is also the same. The advantage of this method is that all the
numbers are real.
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50
Use of MATLAB in Partial Fraction
Expansion
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51
Laplace Transform Properties
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52
Laplace Transform Properties
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53
Laplace Transform Properties
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54
Laplace Transform Properties
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55
Laplace Transform Properties
Final Value Theorem
The final value theorem applies to a function G ( s ) if all the
poles of sG ( s ) lie in the open left half of the s plane. Be sure
to notice that this does not say that all the poles of G ( s ) must
lie in the open left half of the s plane. G ( s ) could have a single
pole at s = 0 and the final value theorem would still apply.
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56
Use of Laplace Transform
Properties
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57
Use of Laplace Transform
Properties
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58
The Unilateral Laplace
Transform
In most practical signal and system analysis using the Laplace
transform a modified form of the transform, called the unilateral
Laplace transform, is used. The unilateral Laplace transform is
¥
defined by G ( s ) = ò - g ( t ) e- st dt . The only difference between
0
this version and the previous definition is the change of the lower
integration limit from - ¥ to 0 -. With this definition, all the
Laplace transforms of causal functions are the same as before
with the same ROC, the region of the s plane to the right of all
the finite poles.
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59
The Unilateral Laplace
Transform
The unilateral Laplace transform integral excludes negative time.
If a function has non-zero behavior in negative time its unilateral
and bilateral transforms will be different. Also functions with the
same positive time behavior but different negative time behavior
will have the same unilateral Laplace transform. Therefore, to avoid
ambiguity and confusion, the unilateral Laplace transform should
only be used in analysis of causal signals and systems. This is a
limitation but in most practical analysis this limitation is not significant
and the unilateral Laplace transform actually has advantages.
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60
The Unilateral Laplace
Transform
The main advantage of the unilateral Laplace transform is that
the ROC is simpler than for the bilateral Laplace transform and,
in most practical analysis, involved consideration of the ROC is
unnecessary. The inverse Laplace transform is unchanged. It is
s + j¥
1
+ st
g (t ) =
G
s
e
ds
(
)
ò
j2p s - j¥
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61
The Unilateral Laplace
Transform
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62
The Unilateral Laplace
Transform
The time shifting property applies only for shifts to the right because
a shift to the left could cause a signal to become non-causal. For the
same reason scaling in time must only be done with positive scaling
coefficients so that time is not reversed producing an anti-causal function.
The derivative property must now take into account the initial value
of the function at time t = 0 - and the integral property applies only to
functional behavior after time t = 0. Since the unilateral and bilateral
Laplace transforms are the same for causal functions, the bilateral table
of transform pairs can be used for causal functions.
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63
The Unilateral Laplace
Transform
The Laplace transform was developed for the solution of differential
equations and the unilateral form is especially well suited for solving
differential equations with initial conditions. For example,
d2
d
x
t
+
7
(
)
é
ù
éë x ( t ) ùû + 12 x ( t ) = 0
û
2 ë
dt
dt
d
with initial conditions x 0 = 2 and
x ( t ) )t = 0- = -4.
(
dt
Laplace transforming both sides of the equation, using the new
derivative property for unilateral Laplace transforms,
d
2
s X ( s ) - s x 0 - ( x ( t ) )t = 0- + 7 éë s X ( s ) - x 0 - ùû + 12 X ( s ) = 0
dt
( )
( )
( )
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The Unilateral Laplace
Transform
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65
Pole-Zero Diagrams and
Frequency Response
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66
Pole-Zero Diagrams and
Frequency Response
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67
Pole-Zero Diagrams and
Frequency Response
jw
lim- H ( jw ) = lim- 3
=0
w ®0
w ®0
jw + 3
jw
lim+ H ( jw ) = lim+ 3
=0
w ®0
w ®0
jw + 3
jw
lim H ( jw ) = lim 3
=3
w ®-¥
w ®-¥
jw + 3
jw
lim H ( jw ) = lim 3
=3
w ®+¥
w ®+¥
jw + 3
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68
Pole-Zero Diagrams and
Frequency Response
lim-
w ®0
lim
w ®-¥
H ( jw ) = -
p
2
p
-0=-
p
2
æ pö
H ( jw ) = - - ç - ÷ = 0
2 è 2ø
lim+
w ®0
lim
w ®+¥
M. J. Roberts - All Rights Reserved. Edited by Dr. Robert Akl
H ( jw ) =
p
H ( jw ) =
p
2
2
-0=
-
p
2
p
2
=0
69
Pole-Zero Diagrams and
Frequency Response
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70
Pole-Zero Diagrams and
Frequency Response
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71
Pole-Zero Diagrams and
Frequency Response
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72
Pole-Zero Diagrams and
Frequency Response
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73
Pole-Zero Diagrams and
Frequency Response
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74
Pole-Zero Diagrams and
Frequency Response
M. J. Roberts - All Rights Reserved. Edited by Dr. Robert Akl
75
Pole-Zero Diagrams and
Frequency Response
M. J. Roberts - All Rights Reserved. Edited by Dr. Robert Akl
76
Pole-Zero Diagrams and
Frequency Response
M. J. Roberts - All Rights Reserved. Edited by Dr. Robert Akl
77