Chapter3 laplace

Signals & Systems
Chapter 3

The Laplace Transform

INC212 Signals and Systems : 2 / 2554

Laplace Transform of
unit-step Function
∞
∞ X (σ + jω ) = ∫ e −(σ + jω ) t dt
X (ω ) = ∫ x(t )e − j ωt
dt 0
−∞ 1
∞
X (σ + jω ) = − [e −(σ + jω ) t ]tt =∞
=0
σ + jω
X (ω ) = ∫ 1⋅ e − jωt dt
0 1
X (σ + jω ) = − [ 0 − e − (σ + j ω ) 0 ]
σ + jω
∞
X (ω ) = ∫ e e −σt − jωt
dt X (σ + jω ) =
1
−∞
σ + jω
∞
X (ω ) = ∫ e −(σ + jω ) t dt σ + jω → s
0
1
X ( s) =
s

INC212 Signals and Systems : 2 / 2554 Chapter 3 The Laplace Transform

Laplace Transform of Signals
∞
X ( s ) = ∫ x(t )e dt ; s = σ + jω
− st
−∞

One-side transform x(t ) = 0; t < 0

∞
X ( s ) = ∫ x(t )e dt − st
0


Laplace Transform of Signals
∞ 1, t ≥ 0
X ( s ) = ∫ x(t )e dt − st
x(t ) = u (t ) = 
0 0, t < 0
∞ ∞
L[u (t )] = ∫ u (t )e dt − st
= ∫ (1)e − st dt
−∞ 0
− st ∞
e 1
=− = −0 − (− ), s > 0
s 0
s
1
∴ L[u (t )] = , s > 0
s


Relationship between the FT and ℒ T
One-side transform or Forward transform
x(t ) = 0; t < 0 ∴ s = jω ; σ = 0

∞ ∞
X (ω ) = ∫ x(t )e − jω t
dt X ( s ) = ∫ x(t )e − st dt
0 0

X (ω ) = X ( s ) s = jω
x(t ) ↔ X ( s )
X ( s) = L[ x(t )] ; x(t ) = L−1[ X ( s )]


Common ℒ T Pairs


Region of Convergence (ROC)
 g1 (t ) = Aeαt u (t ); α > 0

∞ ∞ ∞
G1 ( s ) = ∫
−∞
Aeαt u (t )e − st dt = A∫ e −( s −α )t dt = A∫ e (α −σ )t e − jωt dt
0 0

A
∴ G1 ( s ) = σ = Re( s ) > α
s −α


 g 2 (t ) = Ae −αt u (−t ) = g1 (−t ); α > 0

∞ 0 0
G2 ( s ) = ∫
−∞
Ae −αt u (−t )e − st dt = A ∫ e −( s +α )t dt = A ∫ e −(σ +α )t e − jωt dt
−∞ −∞

A
∴ G2 ( s) = = G1 (− s ) σ = Re( s ) < −α
s +α


 Example:
x(t ) = e − t u (t ) + e −2t u (t )

L ∞ ∞
x(t ) ↔ X ( s ) = ∫ [e u (t ) + e u (t )]e dt = A∫ [e −( s +1) t + e −( s + 2) t ]dt
−t − 2t − st

−∞ 0

1 1
∴ X (s) = + σ > −1
s +1 s + 2


 Example:
x(t ) = e −t u (t ) + e 2t u (−t )

L1 L 1
−t
e u (t ) ↔ σ > −1 e u (−t ) ↔ −
2t
σ <2
s +1 s−2

L 1 1
−t
∴ e u (t ) + e u ( −t ) ↔
2t
− −1 < σ < 2
s +1 s − 2


Properties of the ℒT
 Linearity

L L
x(t ) ↔ X ( s ) and v(t ) ↔ V ( s )

L
ax(t ) + bv(t ) ↔ aX ( s ) + bV ( s )


 Example: Linearity L[u (t ) + e − t u (t )]

1 −t 1
u (t ) ↔ and e u (t ) ↔
s s +1
−t 1 1
u (t ) + e u (t ) ↔ +
s s +1
−t 2s + 1
u (t ) + e u (t ) ↔
s ( s + 1)


 Right Shift in Time

x(t ) ↔ X ( s )
− cs
x(t − c)u (t − c) ↔ e X (s)


 Example: Right Shift in Time
1, 0 ≤ t < c
x(t ) = 
0, all other t

x(t ) = u (t ) − u (t − c)
− cs − cs
1 e 1− e
u (t ) − u (t − c) ↔ − =
s s s


Properties of the ℒ T


Computation of the Inverse ℒ T

1 c + j∞
2π j ∫c − j∞
x(t ) = X ( s )e st ds


The Inverse ℒ T using
Partial-Fraction Expansion
B( s)
X ( s) =
A( s )
Let p1, p2, …, pN
B ( s ) = bM s M + bM −1s M −1 +  + b1s + b0 denote the roots
A( s ) = a N s N + a N −1s N −1 +  + a1s + a0 of the equation
A( s ) = 0
The pi for i = 1, 2,
A( s ) = a N ( s − p1 )( s − p2 )  ( s − p N )
…,N are called the B( s)
X (s) =
poles of X(s) a N ( s − p1 )( s − p2 )  ( s − p N )


Distinct Poles
c1 c2 cN
X ( s) = + ++
s − p1 s − p2 s − pN
ci = [( s − pi ) X ( s )]s = pi , i = 1,2, , N

x(t ) = c1e p1t
+ c2 e p2t
+  + cN e pN t
, t≥0


Example: Distinct Poles
s+2 ci = [( s − pi ) X ( s )]s = pi , i = 1,2, , N
X ( s) = 3
s + 4 s 2 + 3s s+2 2
A( s ) = s + 4 s + 3s = s ( s + 1)( s + 3) c1 = [ sX ( s )]s =0 = ( s + 1)( s + 3)
3 2 =
3
s =0
A( s ) = 0 = s ( s + 1)( s + 3)
s+2 1
p1 = 0, p2 = −1, p3 = −3 c2 = [( s + 1) X ( s )]s = −1 = =
s ( s + 3) s = −1 − 2
c1 c2 c3
X ( s) = + + s+2 −1
s − 0 s − (−1) s − (−3) c3 = [( s + 3) X ( s )]s = −3 = =
s ( s + 1) s = −3 6
c c c
X ( s) = 1 + 2 + 3 2 1 − t 1 − 3t
s s +1 s + 3 x(t ) = − e − e , t ≥ 0
3 2 6


Distinct Poles with 2 or More
Poles Complex
c1 c1 c3 cN
X (s) = + + ++
s − p1 s − p1 s − p3 s − pN

x(t ) = c1e p1t
+ c1e p1t
+ c3e p3t +  + c N e p N t
σt
c1e p1t
+ c1e p1t
= 2 c1 e cos(ωt + ∠c1 )
σt
x(t ) = 2 c1 e cos(ωt + ∠c1 ) + c3e p3 t
+  + cN e pN t


Example: Distinct Poles with 2 or
More Poles Complex
ci = [( s − pi ) X ( s)]s = p , i = 1,2,, N
s − 2s + 1
2 i

X (s) = 3 s 2 − 2s + 1
s + 3s 2 + 4 s + 2 c1 = [( s + 1 − j ) X ( s )]s =−1+ j =
( s + 1 + j )( s + 1) s =−1+ j
A( s ) = s 3 + 3s 2 + 4 s + 2
= ( s + 1 − j )( s + 1 + j )( s + 1) −3
c1 = + j2
2
p1 = −1 + j , p2 = −1 − j , p3 = −1
9 5
c1 = +4 = ;
X ( s) =
c1
+
c1
+
c3 4 2
s − (−1 + j ) s − (−1 − j ) s − (−1) −4
∠c1 = 180° + tan −1 = 126.87°
c1 c1 c3 3
X ( s) = + +
s +1− j s +1+ j s +1 s 2 − 2s + 1
c3 = [( s + 1) X ( s )]s = −1 = 2 =4
s + 2 s + 2 s = −1

x(t ) = 5e −t cos(t + 126.87°) + 4e − t , t ≥ 0


Repeated Poles
B( s) c c2 cr c cN
X (s) = X ( s) = 1 + ++ + r +1 +  +
A( s ) s − p1 ( s − p1 ) 2 ( s − p1 ) r s − pr +1 s − pN
ci = [( s − pi ) X ( s )]s = pi , i = r + 1, r + 2, , N
cr = [( s − p1 ) r X ( s )]s = p1
1d 
i = 1,2,  , r − 1 i = 1; cr −1 = [( s − p1 ) r X ( s )]
1!  ds
  s = p1
1  di 
cr − i =  i [( s − p1 ) X ( s )]
r
1  d2 
i!  ds  s = p1 i = 2; cr − 2 =  2 [( s − p1 ) r X ( s )]
2!  ds  s= p 1

x(t ) = c1e p1t + c2te p1t +  + cr t r −1e p1t + cr +1e pr +1t +  + c N e p N t , t ≥ 0


Example: Repeated Poles
5s − 1 c1 c2 c
X (s) = 3 X (s) = + + 3
s − 3s − 2 s + 1 ( s + 1) 2 s − 2

1d   d 5s − 1  −9
c1 = [( s + 1) 2 X ( s )] = [ ] = = −1
1!  ds
  s = −1  ds s − 2  s = −1 ( s − 2)
2
s = −1

5s − 1
c2 = [( s + 1) 2 X ( s )]s = −1 = =2
s − 2 s = −1
5s − 1
c3 = [( s − 2) X ( s )]s = 2 = =1
( s + 1) 2 s =2

x(t ) = −e − t + 2te − t + e 2t , t ≥ 0


Case when M ≥ N
B(s)
X (s) =
A( s ) R( s)
X ( s) = Q( s) +
B( s ) = bM s M + bM −1s M −1 +  + b1s + b0 A( s )
A( s ) = a N s N + a N −1s N −1 +  + a1s + a0
R( s)
X ( s ) = Q( s ) + V ( s ), V ( s ) =
A( s )
Q(s)
A( s ) B( s ) x(t ) = q (t ) + v(t )
A( s ) * Q ( s ) dN
q(t ); δ (t ) ↔ s N
R( s) dt N
v(t ); v(t ) ↔ V ( s )


Example: Case when M ≥ N
s 3 + 2s − 4 20 s − 12
X (s) = 2 = s−4+ 2
s + 4s − 2 s + 4s − 2
X ( s) = Q( s) + V ( s)
Q( s) = s − 4
d
q(t ) = δ (t ) − 4δ (t )
dt
20s − 12 20.6145 0.6145
V ( s) = 2 = −
s + 4 s − 2 s + 4.4495 s − 0.4495
v(t ) = 20.6145e − 4.4495t − 0.6145e 0.4495t , t ≥ 0
d
x(t ) = δ (t ) − 4δ (t ) + 20.6145e − 4.4495t − 0.6145e 0.4495t , t ≥ 0
dt


Transform of the I/O
Differential Equation
 First-Order Case
y (0 − ) b
Y ( s) = + X (s)
dy (t ) s+a s+a
+ ay (t ) = bx(t ) b
dt Y ( s) = X (s)
sY ( s ) − y (0 − ) + aY ( s ) = bX ( s ) s+a
b
( s + a )Y ( s ) = y (0 − ) + bX ( s ) H (s) =
s+a
Y ( s) = H ( s) X ( s)

H(s) Transfer Function (TF) of the system


 Example: First-Order Case

dy (t ) 1 1
+ y (t ) = x(t )
dt RC RC

y (0 − ) 1 RC
Y ( s) = +
s + 1 RC ( s + 1 RC ) s
y (0 − ) 1 RC
Y (s) = + X (s) y (0 − ) 1 1 RC
s + 1 RC s + 1 RC Y ( s) = + −
s + 1 RC s s + 1 RC
y (t ) = y (0 − )e −(1 RC ) t + 1 − e −(1 RC ) t , t ≥ 0

 Second-Order Case
d 2 y (t ) dy (t ) dx(t )
2
+ a1 + a0 y (t ) = b1 + b0 x(t )
dt dt dt
[ ]
s 2Y ( s ) − y (0 − ) s − y (0 − ) + a1 sY ( s ) − y (0 − ) + a0Y ( s ) = b1sX ( s ) + b0 X ( s )

y (0 − ) s + y (0 − ) + a1 y (0 − )
 b1s + b0
Y (s) = + 2 X ( s)
s + a1s + a0
2
s + a1s + a0

If initial condition = 0 :
b1s + b0 b1s + b0
Y (s) = 2 X ( s) ; H ( s) = 2
s + a1s + a0 s + a1s + a0


 Example: Second-Order Case
x(t) = u(t) so that X(s) = 1/s; initial cond. = 0
d 2 y (t ) dy (t )
2
+6 + 8 y (t ) = 2 x(t )
dt dt
2
H (s) = 2 2 1
s + 6s + 8 Y (s) = H (s) X (s) =
s 2 + 6s + 8 s
0.25 0.5 0.25
Y (s) = − +
s s+2 s+4
y (t ) = 0.25 − 0.5e − 2t + 0.25e − 4t , t ≥ 0


 Example: Second-Order Case y (0 − ) = 1
x(t) = u(t) with the initial condition
y (0 − ) = 2

s +8 2 1
Y ( s) = 2 + 2
s + 6s + 8 s + 6s + 8 s
s 2 + 8s + 2
=
s ( s 2 + 6 s + 8)
0.25 2.5 1.75
Y ( s) = + −
s s+2 s+4
y (t ) = 0.25 + 2.5e − 2t − 1.75e − 4t , t ≥ 0


 Nth-Order Case
d N y (t ) N −1 d i y (t ) M d i x(t ) C (s) B(s)
N
+ ∑ ai i
= ∑ bi i
; Y ( s) = + X (s)
dt i =0 dt i =0 dt A( s ) A( s )
B ( s ) = bM s M + bM −1s M −1 +  + b1s + b0 ; A( s ) = s N + a N −1s N −1 +  + a1s + a0
C ( s ) = y ( 0 − ) s + y ( 0 − ) + a1 y ( 0 − )


B( s) bM s M +  + b1s + b0
Y (s) = X (s) = N N −1
X (s)
A( s ) s + a N −1s +  + a1s + a0
bM s M +  + b1s + b0
H (s) = N
s + a N −1s N −1 +  + a1s + a0


Convolution Integral
t
y (t ) = h(t ) ∗ x(t ) = ∫ h(λ ) x(t − λ )dλ , t ≥ 0
0

Y ( s) = H ( s) X ( s)
h(t ) ↔ H ( s )
Y (s)
H (s) =
X (s)


 Example: Determining the TF
y (t ) = 2 − 3e −t + e −2t cos 2t , t ≥ 0 2 3 s+2
− +
2 3 s+2 s s + 1 ( s + 2) 2 + 4
Y (s) = − + H (s) =
s s + 1 ( s + 2) + 4 2 1
s +1
1 2( s + 1) ( s + 1)( s + 2)
X (s) = = −3+
s +1 s ( s + 2) 2 + 4
[2( s + 1) − 3s ][( s + 2) 2 + 4] + s ( s + 1)( s + 2)
=
s[( s + 2) 2 + 4]
s 2 + 2s + 16
= 3
s + 4 s 2 + 8s


 Finite-Dimensional Systems

bM s M + bM −1s M −1 +  + b1s + b0
H ( s) = N
s + a N −1s N −1 +  + a1s + a0
( s N + a N −1s N −1 +  + a1s + a0 )Y ( s ) = (bM s M + bM −1s M −1 +  + b1s + b0 ) X ( s )
d N y (t ) N −1 d i y (t ) M d i x(t )
N
+ ∑ ai i
= ∑ bi
dt i =0 dt i =0 dt i


 Poles and zeros of a Systems
bM s M + bM −1s M −1 +  + b1s + b0
H ( s) = N
s + a N −1s N −1 +  + a1s + a0
bM ( s − z1 )( s − z 2 )  ( s − z M )
H ( s) =
( s − p1 )( s − p2 )  ( s − p N )
zi : “zeros of H (s)” or “zeros of system”
pi : “poles of H (s)” or “poles of system”
N : “number of poles of system” or “order N of system”


 Example: Third-Order System
2 s 2 + 12 s + 20
H ( s) = 3
s + 6 s 2 + 10s + 8
2( s + 3 − j )( s + 3 + j )
H ( s) =
( s + 4)( s + 1 − j )( s + 1 + j )
z1 = −3 + j and z1 = −3 − j
p1 = −4, p 2 = −1 + j , p3 = −1 − j


Exercises
 Sketch the pole-zero plot and ROC for
these signals.
 x(t ) = e −8t u (t )
 x(t ) = e 3t cos(20πt )u (−t )
 x(t ) = e 2t u (−t ) − e −5t u (t )


Exercises
 Using the time-shifting property, find
the LT of these signals.
 x(t ) = u (t ) − u (t − 1)

 x(t ) = 3e −3(t − 2)u (t − 2)
 x(t ) = 3e −3t u (t − 2)
 x(t ) = 5 sin(π (t − 1))u (t − 1)


Exercises
 Find the inverse LT of these functions.
 24  20
X ( s) = X (s) =
s ( s + 8) s 2 + 4s + 3

 s2  s
X ( s) = 2 X ( s) = 2
s − 4s + 4 s + 4s + 4

 5  2s
X ( s) = 2 X ( s) = 2
s + 6 s + 73 s + 2 s + 13


Direct Construction of the TF
 RLC Circuits

dv(t ) 1
= i (t )
dt C
v(t ) = Ri (t ) di (t )
1 v(t ) = L
sV ( s ) − v(0) = I ( s ) dt
V ( s ) = RI ( s ) C
1 1 V ( s ) = LsI ( s ) − Li (0)
V (s) = I ( s ) + v(0)
Cs s


 Series and Parallel Connection

Z1 ( s ) Z 2 (s)
V1 ( s ) = V ( s) I1 ( s ) = I ( s)
Z1 ( s ) + Z 2 ( s ) Z1 ( s ) + Z 2 ( s )
Z 2 (s) Z1 ( s )
V2 ( s ) = V ( s) I 2 (s) = I ( s)
Z1 ( s ) + Z 2 ( s ) Z1 ( s ) + Z 2 ( s )


 Example: Series RLC Circuit

Output = VC(s) Output = VR(s)
1 Cs R
Vc ( s ) = X ( s) VR ( s ) = X (s)
Ls + R + (1 Cs ) Ls + R + (1 Cs )
1 LC ( R L) s
= 2 X ( s) = 2 X ( s)
s + ( R L) s + (1 LC ) s + ( R L) s + (1 LC )
1 LC ( R L) s
H ( s) = 2 H ( s) = 2
s + ( R L) s + (1 LC ) s + ( R L) s + (1 LC )


 Interconnections of Integrators


sQ1 ( s) = −4Q1 ( s ) + X ( s)
 Example: Q1 ( s) =
1
s+4
X (s)

sQ2 ( s ) = Q1 ( s) − 3Q2 ( s) + X ( s )
1
Q2 ( s ) = [Q1 ( s ) + X ( s )]
s+3
1  1 
=  + 1 X ( s )
s +3 s +4 
s+5
= X ( s)
( s + 3)( s + 4)
Y ( s ) = Q2 ( s ) + X ( s )
s+5
= X ( s) + X ( s)
( s + 3)( s + 4)
s 2 + 8s + 17
= X ( s)
( s + 3)( s + 4)
s 2 + 8s + 17 s 2 + 8s + 17
H ( s) = =
( s + 3)( s + 4) s 2 + 7 s + 12


TF of Block Diagrams
 Parallel Interconnection

Y ( s ) = Y1 ( s ) + Y2 ( s )
Y1 ( s ) = H1 ( s ) X ( s )
Y2 ( s ) = H 2 ( s ) X ( s )

Y ( s ) = H1 ( s ) X ( s ) + H 2 ( s) X ( s)
= ( H1 ( s ) + H 2 ( s )) X ( s )
H ( s) = H1 ( s) + H 2 ( s )


 Series Connection

Y1 ( s ) = H1 ( s ) X ( s )
Y2 ( s ) = H 2 ( s )Y1 ( s )
Y ( s ) = Y2 ( s ) = H 2 ( s ) H1 ( s ) X ( s )
H ( s ) = H 2 ( s ) H1 ( s ) = H1 ( s ) H 2 ( s )


 Feedback Connection Y ( s) = H1 ( s) X 1 ( s)
X 1 ( s ) = X ( s ) − Y2 ( s )
= X ( s ) − H 2 ( s )Y ( s )
Y ( s ) = H1 ( s )[ X ( s ) − H 2 ( s )Y ( s )]
H1 ( s )
Y (s) = X (s)
1 + H1 ( s) H 2 ( s)
H1 ( s)
H (s) =
1 + H1 ( s) H 2 ( s)
H1 ( s)
H (s) =
1 − H1 ( s ) H 2 ( s )


Chapter3 laplace

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Chapter3 laplace