Laplace Transform
•  Review
•  Slides borrowed from Evans, U. of Texas @
Austin
Zero-State Response
•  Linear constant coefficient differential equation
Input x(t) and output y (t ) = y zero−input (t ) + y zero− state (t )
Zero-state response: all initial conditions are zero
y(t ) ↔ Y (s )
x(t ) ↔ X (s )
dr
r
(
)
y
t
↔
s
Y (s )
r
dt
dk
k
(
)
x
t
↔
s
X (s )
k
dt
Laplace transform both sides of differential equation with
all initial conditions being zero and solve for Y(s)/X(s)
y ' (t ) + y (t ) = x(t )
sY ( s ) + Y ( s ) = X ( s )
y (0 − ) = 0
H (s) =
Y ( s)
1
=
X (s) s + 1
Transfer Function
•  H(s) is called the transfer function because it
describes how input is transferred to the output
in a transform domain (s-domain in this case)
Y(s) = H(s) X(s)
y(t) = L-1{H(s) X(s)} = h(t) * x(t) ⇒ H(s) = L{h(t)}
•  Transfer function is Laplace transform of
impulse response
Transfer Function Examples
•  Laplace transform
∞
X (s ) = ∫ − x(t ) e − s t dt
0
•  Assume input x(t) and output y(t) are causal
•  Ideal delay of T seconds
Initial conditions (initial voltages in delay buffer) are zero
y (t ) = x(t − T )
Y (s ) = X (s ) e − s T
Y (s )
H (s ) =
= e−s T
X (s )
x(t)
T
y(t)
Transfer Function Examples
•  Ideal integrator with
y(0-) = 0
t
y (t ) = ∫ − x(τ )dτ
0
1
1
Y (s ) = X (s ) + y 0 −
s
s
Y (s) 1
H (s ) =
=
X (s) s
( )
x(t)
t
∫ (•) dt
0−
y(t)
•  Ideal differentiator
with x(0-) = 0
d
y (t ) = x(t )
dt
Y (s ) = s X (s ) − x 0 − = s X ( s )
Y (s )
H (s ) =
=s
X (s )
( )
x(t)
d
y(t)
(•)
dt
Cascaded Systems
•  Assume input x(t) and output y(t) are causal
x(t)
•  Integrator first,
then differentiator X(s)
•  Differentiator first, x(t)
then integrator
X(s)
t
1/s
s
∫ x(τ )dτ
0−
X(s)/s
d
x(t )
dt
s X(s)
s
1/s
•  Common transfer functions
A constant (finite impulse response)
A polynomial (finite impulse response)
Ratio of two polynomials (infinite impulse response)
x(t)
X(s)
x(t)
X(s)
Block Diagrams
X(s)
X(s)
H1(s)
W(s)
H(s)
Y(s)
H2(s)
Y(s)
=
X(s)
H1(s)H2(s)
Y(s)
Σ
Y(s)
=
X(s)
H1(s) + H2(s)
Y(s)
Y(s)
=
X(s)
G(s)
1 + G(s)H(s)
Y(s)
H1(s)
X(s)
H2(s)
X(s)
-
Σ
E(s)
G(s)
H(s)
Cascade and Parallel Connections
•  Cascade
X(s)
W(s) = H1(s) X(s)
Y(s) = H2(s)W(s)
Y(s) = H1(s) H2(s) X(s) ⇒ Y(s)/X(s) = H1(s)H2(s)
H1(s)
H2(s)
Y(s) ⇔ X(s)
H2(s)
H1(s)
Y(s)
One can switch the order of the cascade of two LTI
systems if both LTI systems compute to exact precision
•  Parallel Combination
Y(s) = H1(s)X(s) + H2(s)X(s) èY(s)/X(s) = H1(s) + H2(s)
H1(s)
X(s)
Σ
H2(s)
Y(s)
=
X(s)
H1(s) + H2(s)
Y(s)
19 - 8
Feedback Connection
•  Governing equations
E (s ) = F (s ) − H (s ) Y (s )
Y (s ) = G(s ) E (s )
•  What happens if H(s) is
a constant K?
Choice of K controls all
poles in transfer function
•  Combining equations
Y (s ) = G (s ) [ F (s ) − H (s ) Y (s ) ]
Y (s ) + G (s ) H (s ) Y (s ) = G (s ) F (s )
G (s )
Y (s ) =
F (s )
1 + G (s ) H (s )
F(s)
-
Σ
E(s)
G(s)
H(s)
Y(s)
=
F(s)
G(s)
1 + G(s)H(s)
Y(s)
19 - 9
External Stability Conditions
•  Bounded-input bounded-output stability
Zero-state response given by h(t) * x(t)
Two choices: BIBO stable or BIBO unstable
•  Remove common factors in transfer function H(s)
•  If all poles of H(s) in left-hand plane,
All terms in h(t) are decaying exponentials
h(t) is absolutely integrable and system is BIBO stable
•  Example: BIBO stable but asymptotically
unstable
⎛ s − 1 ⎞ ⎛ 1 ⎞⎛ s − 1 ⎞ ⎛ 1 ⎞
H ( s) = ⎜
⎟ = ⎜
⎟⎜
⎟ = ⎜
⎟
2
s
−
1
s
+
1
s
+
1
⎠⎝
⎠ ⎝
⎠
⎝ s − 1 ⎠ ⎝
Based on slide by Prof. Adnan Kavak
Internal Stability Conditions
•  Stability based on zero-input solution
•  Asymptotically stable if and only if
Characteristic roots are in left-hand plane (LHP)
Roots may be repeated or non-repeated
•  Unstable if and only if
(i) at least characteristic root in right-hand plane and/or
(ii) repeated characteristic roots are on imaginary axis
•  Marginally stable if and only if
There are no characteristic roots in right-hand plane and
Some non-repeated roots are on imaginary axis
Based on slide by Prof. Adnan Kavak
Frequency-Domain Interpretation
est
h(t)
y(t)
y (t ) = h(t )∗ e s t
∞
•  y(t) = H(s) e s t
for a particular value of s
= ∫ h(τ )e s (t −τ )dτ
−∞
=e
st
∞
−s τ
(
)
h
τ
e
∫−∞dτ
H (s )
•  Recall definition of
frequency response:
ej 2π f t
h(t)
y(t)
y (t ) = h(t )∗ e j 2π
∞
ft
= ∫ h(τ )e j 2π
f (t −τ )
−∞
=e
j 2π f t
dτ
∞
j 2π f τ
(
)
h
τ
e
∫−∞dτ
H(f
)
Frequency-Domain Interpretation
•  Generalized frequency: s = σ + j 2 π f
•  We may convert transfer function into
frequency response by if and only if region of
convergence of H(s) includes the imaginary axis
H freq ( f ) = H (s )
s = j 2πf
•  What about h(t) = u(t)?
1
H (s ) =
s
for Re{s} > 0
We cannot convert H(s) to a frequency response
However, this system has a frequency response
•  What about h(t) = δ(t)? H (s ) = 1 for all s ⇒ H freq ( f ) = 1
Frequency Selectivity in Filters
•  Lowpass filter
|Hfreq(f)|
•  Bandpass filter
|Hfreq(f)|
1
1
•  Highpass filter
f
|Hfreq(f)|
f
•  Bandstop filter
|Hfreq(f)|
f
Linear time-invariant filters are BIBO stable
f
Passive Circuit Elements
•  Laplace transforms
with zero-valued
initial conditions
•  Capacitor
+
dv
v(t)
i (t ) = C
dt
–
I (s ) = C s V (s )
1
(
)
V s =
I (s )
Cs
V (s ) 1
H (s ) =
=
I (s ) C s
•  Inductor
+
v(t)
di
v(t ) = L
dt
–
V (s ) = L s I (s )
V (s )
H (s ) =
=Ls
I (s )
•  Resistor
v(t ) = R i (t )
V (s ) = R I (s )
V (s )
H (s ) =
=R
I (s )
+
v(t)
–
First-Order RC Lowpass Filter
R
+
x(t)
C
i(t)
+
y(t)
Time domain
R
+
X(s)
I(s)
1
Cs
+
Y(s)
X ( s)
1
R+
Cs
1
Y (s) =
I ( s)
Cs
1
Y ( s) = C s X ( s)
1
R+
Cs
1
Y (s)
= RC
X (s) s + 1
RC
I ( s) =
Laplace domain
Plot Bode plot – Amplitude & Phase?
First-Order RC Highpass Filter
C
+
x(t)
i(t)
R
+
y(t)
Time domain
+
X(s)
1
Cs
I(s)
R
+
Y(s)
Laplace domain
Plot Bode plot – Amplitude & Phase?
X ( s)
1
R+
Cs
Y (s) = R I (s)
R
Y (s) =
X ( s)
1
R+
Cs
Y (s)
s
=
X (s) s + 1
RC
I ( s) =
Frequency response is
also an example of a
notch filter
Passive Circuit Elements
•  Laplace transforms
with non-zero initial
conditions
•  Capacitor
i (t ) = C
[
[
( )]
= L s I (s ) − L i (0 )
⎡
i (0 )⎤
= L s I (s ) −
V (s ) = L s I (s ) − i 0 −
−
( )]
v(0 )
I (s ) +
−
−
1
V (s ) =
Cs
s
1
=
I (s ) + C v 0 −
Cs
[
di
V (t ) = L
dt
−
dv
dt
I (s ) = C s V (s ) − v 0
•  Inductor
( )]
⎢
⎣
⎥
s ⎦
Operational Amplifier
•  Ideal case: model this nonlinear circuit as
linear and time-invariant
Input impedance is extremely high (considered infinite)
vx(t) is very small (considered zero)
+
vx(t) _
_
+
+
y(t)
_
Operational Amplifier Circuit
•  Assuming that Vx(s) = 0,
Y (s ) = − I (s ) Z f (s )
F (s )
Z (s )
F(s)
Z f (s )
Y (s ) = −
F (s )
+
_
Z (s )
Z f (s )
Y (s )
H (s ) =
=−
F (s )
Z (s )
H(s)
I (s ) =
Z(s)
•  How to realize a gain of –1?
•  How to realize a gain of 10?
I(s)
Zf(s)
+ _
Vx(s) _
+
+
Y(s)
_
Differentiator
•  A differentiator amplifies high frequencies, e.g.
high-frequency components of noise:
H(s) = s for all values of s (see next slide)
Frequency response is H(f) = j 2 π f ⇒ | H( f ) |= 2 π | f |
•  Noise has equal amounts of low and high
frequencies up to a physical limit
•  A differentiator may amplify noise to drown
out a signal of interest
•  In analog circuit design, one would generally
use integrators instead of differentiators
DECIBEL SCALE
P2
GdB = 10 log10
P1
is defined as Decibel Gain
P2
V2
GdB = 10 log10 = 20 log10
P1
V1
V2 2
2
⎛ V2 ⎞
P2
R2
R
GdB = 10 log10
= 10 log10 2
= 10 log10 ⎜ ⎟ + 10 log10 1
V1
P1
R2
⎝ V1 ⎠
R1
= 20 log10
V2
R
V
− 10 log10 2 = 20 log10 2
V1
R1
V1
if
R2 = R1
DECIBEL SCALE
Ø  The DECIBEL value is a logarithmic measurement of the ratio of one variable to another of the same type.
Ø Decibel value has no dimension.
Ø  It is used for voltage, current and power gains.
Magnitude H
Decibel Value HdB
0.001
-60
0.01
-40
0.1
-20
0.5
-6
Some Properties of Logarithms
1/√2
-3
log P1 P2 = log P1 + log P2
1
0
√2
3
2
6
10
20
20
26
⎛ P1 ⎞
log ⎜ ⎟ = log P1 − log P2
⎝ P2 ⎠
log P n = n log P
log 1 = 0
100
40
H dB
V2
= 20 log10 H = 20 log10
V1
Typical Sound Levels and Their Decibel Levels.
EXAMPLE 1 Construct Bode plots for
H (ω ) =
200 jω
( jω + 2)( jω + 10)
•  Express transfer function in Standard form.
STANDARD FORM H (ω ) =
10 jω
(1 + jω 2)(1 + jω 10)
•  Express the magnitude and phase responses.
H db = 20log10 10 + 20log10 jω − 20log10 1 + jω 2 − 20log10 1 + jω 10
φ = 90° − tan -1 (ω 2) − tan -1 (ω 10)
•  Two corner frequencies at ω=2, 10 and a zero at the origin ω=0.
•  Sketch each term and add to find the total response.
EXAMPLE 1 Construct Bode plots for
H (ω ) =
200 jω
( jω + 2)( jω + 10)
H db = 20log10 10 + 20log10 jω − 20log10 1 + jω 2 − 20log10 1 + jω 10
26 dB
20log10 2 = 6dB
X
X
φ = 90° − tan -1 (ω 2) − tan -1 (ω 10)
X
X
EXAMPLE 2 Continued: Let us calculate |H| and φ at ω=50 rad/sec graphically.
H (50) = H (10) − 20log10 (50 /10) = 26 − 20 × 0.7 = 26 − 14 = 12 dB
φ (50) = 90° − 45°× log10 (1/ 0.2) − 90°× log10 (20 /1) − 45°× log10 (50 / 20)
= 90° − 45°× 0.7 − 90°× 1.3 − 45°× 0.4 = 90° − 31.5° − 117° − 18° = −76.5°
26 dB
ω=50
ω=50
EXAMPLE 1 Construct Bode plots for
H (ω ) =
jω + 10
jω ( jω + 5 )
2
•  Express transfer function in Standard form.
STANDARD FORM
0.4(1 + jω 10)
H (ω ) =
jω (1 + jω 5) 2
•  Express the magnitude and phase responses.
H db = 20log10 0.4 + 20log10 1 + jω 10 − 20log10 jω − 40log10 1 + jω 5
φ = 0° + tan -1 (ω 10) − 90° − 2 tan -1 (ω 5)
•  Two corner frequencies at ω=5, 10 and a zero at ω=10.
•  The pole at ω=5 is a double pole. The slope of the magnitude is -40
dB/decade and phase has slope -90 degree/decade.
•  Sketch each term and add to find the total response.
EXAMPLE 3 Construct Bode plots for
H (ω ) =
jω + 10
jω ( jω + 5 )
2
H db = 20 log10 0.4 +
20 log10 1 + jω 10 −
X
O
X
O
20 log10 jω −
40 log10 1 + jω 5
φ = 0° + tan -1 (ω 10)
-1
− 90° − 2 tan (ω 5)
PRACTICE PROBLEM 4 Obtain the transfer function for the Bode plot given.
A zero at ω = 0.5, 1 + jω 0.5
1
A pole at ω = 1,
1 + jω 1
Two poles at
ω = 10,
1
(1 + jω 10) 2
1 + jω 0.5
(1 0.5)(0.5 + jω )
200 ( s + 0.5)
H (ω ) =
=
=
(1 + jω 1)(1 + jω 10) 2 (1 100)(1 + jω )(10 + jω ) 2 ( s + 1)( s + 10) 2