Post on 26-Apr-2018
Laplace Transform • Review
• Slides borrowed from Evans, U. of Texas @ Austin
Zero-State Response • Linear constant coefficient differential equation
Input x(t) and output Zero-state response: all initial conditions are zero Laplace transform both sides of differential equation with
all initial conditions being zero and solve for Y(s)/X(s)
)()()( tytyty statezeroinputzero −− +=
( ) ( )sYty ↔
( ) ( )sYstydtd rr
r
↔
0)0()()()('
=
=+−y
txtyty
( ) ( )sXtx ↔
( ) ( )sXstxdtd kk
k
↔
11
)()()(
)()()(
+==
=+
ssXsYsH
sXsYssY
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
( ) ( )( ) ( )
( ) ( )( )
Ts
Ts
esXsYsH
esXsYTtxty
−
−
==
=
−=
Transfer Function Examples • Laplace transform
• Assume input x(t) and output y(t) are causal • Ideal delay of T seconds
Initial conditions (initial voltages in delay buffer) are zero
( ) ( )∫∞ −−
=0
dtetxsX ts
Tx(t) y(t)
( )
( ) ( ) ( )( ) ( )
( )s
sXsYsH
sXsxsXssY
txdtdty
==
=−=
=
− )(0
)(( ) ( )
( ) ( ) ( )
( )ssX
sYsH
ys
sXs
sY
dxtyt
1)()(
0110
==
+=
=
−
∫ −ττ
Transfer Function Examples • Ideal integrator with
y(0-) = 0 • Ideal differentiator
with x(0-) = 0
( )•dtdx(t) y(t) ( )dt
t
0∫ −•
x(t) y(t)
Cascaded Systems • Assume input x(t) and output y(t) are causal
• Integrator first, then differentiator
• Differentiator first, then integrator
• Common transfer functions A constant (finite impulse response) A polynomial (finite impulse response) Ratio of two polynomials (infinite impulse response)
1/s s X(s)/s
( )∫ −
tdx
0ττx(t)
X(s)
x(t)
X(s)
s 1/s s X(s)
)(txdtd
x(t)
X(s)
x(t)
X(s)
Block Diagrams H(s) X(s) Y(s)
H1(s) + H2(s) X(s) Y(s) H1(s)
X(s) Y(s) H2(s)
Σ =
H1(s) X(s) Y(s) H2(s) H1(s)H2(s) X(s) Y(s) = W(s)
G(s) 1 + G(s)H(s) X(s) Y(s) G(s) X(s) Y(s)
H(s)
Σ - =
E(s)
19 - 8
Cascade and Parallel Connections • Cascade
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)
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) èY(s)/X(s) = H1(s) + H2(s)
H1(s) X(s) Y(s) H2(s) H2(s) X(s) Y(s) H1(s) ⇔
H1(s) + H2(s) X(s) Y(s) H1(s)
X(s) Y(s) H2(s)
Σ =
19 - 9
Feedback Connection • Governing equations
• Combining equations
( ) ( ) ( ) ( )( ) ( ) ( )sEsGsY
sYsHsFsE
=
−=
( ) ( ) ( ) ( ) ( )[ ]( ) ( ) ( ) ( ) ( ) ( )
( ) ( )( ) ( )
( )sFsHsG
sGsY
sFsGsYsHsGsY
sYsHsFsGsY
1
+=
=+
−=
• What happens if H(s) is a constant K? Choice of K controls all
poles in transfer function
G(s) 1 + G(s)H(s) F(s) Y(s) G(s) F(s) Y(s)
H(s)
Σ - =
E(s)
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
⎟⎠
⎞⎜⎝
⎛+
=⎟⎠
⎞⎜⎝
⎛+−
⎟⎠
⎞⎜⎝
⎛−
=⎟⎠
⎞⎜⎝
⎛
−
−=
11
11
11
11)( 2 ss
sss
ssH
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
• y(t) = H(s) e s t for a particular value of s
• Recall definition of frequency response:
( ) ( )
( ) ( )
( )( )
sH
sts
ts
ts
dehe
deh
ethty
∫∫
∞
∞−
−
∞
∞−
−
=
=
∗=
ττ
ττ
τ
τ
h(t) y(t) est
h(t) y(t) ej 2π f t
( ) ( )
( ) ( )
( )( )
fH
fjtfj
tfj
tfj
dehe
deh
ethty
∫∫
∞
∞−
∞
∞−
−
=
=
∗=
ττ
ττ
τππ
τπ
π
2 2
2
2
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
• What about h(t) = u(t)? We cannot convert H(s) to a frequency response However, this system has a frequency response
• What about h(t) = δ(t)?
( ) { } 0Refor 1>= s
ssH
( ) ( ) 1 allfor 1 freq =⇒= fHssH
( ) ( )fjs
sHfHπ2freq
==
Frequency Selectivity in Filters • Lowpass filter
• Highpass filter
• Bandpass filter
• Bandstop filter
Linear time-invariant filters are BIBO stable
f
|Hfreq(f)|
f
|Hfreq(f)|
f
|Hfreq(f)| 1
f
|Hfreq(f)|
1
Passive Circuit Elements • Laplace transforms
with zero-valued initial conditions
• Capacitor
• Inductor
• Resistor
( )
( ) ( )
( ) ( )( )
sLsIsVsH
sIsLsVdtdiLtv
==
=
=
( )
( ) ( )
( ) ( )
( ) ( )( ) sCsIsVsH
sIsC
sV
sVsCsIdtdvCti
1
1
==
=
=
=
( ) ( )( ) ( )
( ) ( )( )
RsIsVsH
sIRsVtiRtv
==
=
=
+
– v(t)
+
– v(t)
+
– v(t)
First-Order RC Lowpass Filter
x(t) y(t) ++
C
R
X(s) Y(s) ++
R
sC 1
Time domain
Laplace domain CRs
CRsXsY
sX
sCR
sCsY
sIsC
sY
sCR
sXsI
1
1
)()(
)(
1
1
)(
)(
1)(
1)()(
+=
+=
=
+=
i(t)
I(s)
Plot Bode plot – Amplitude & Phase?
First-Order RC Highpass Filter
x(t) y(t) ++
R
C
X(s) Y(s) ++
sC 1
Time domain
Laplace domain
CRs
ssXsY
sX
sCR
RsY
sIRsYsC
R
sXsI
1)(
)(
)(
1)(
)( )(
1)()(
+=
+=
=
+=
i(t)
RI(s) Frequency response is
also an example of a notch filter Plot Bode plot – Amplitude & Phase?
Passive Circuit Elements • Laplace transforms
with non-zero initial conditions
• Capacitor
( )
( ) ( ) ( )[ ]( ) ( )
( ) ( )⎥⎦
⎤⎢⎣
⎡−=
−=
−=
=
−
−
−
sisIsL
iLsIsL
isIsLsVdtdiLtV
0
0
0
• Inductor
( )
( ) ( ) ( )[ ]
( ) ( ) ( )
( ) ( )[ ]−
−
−
+=
+=
−=
=
0
1
0
1
0
vCsIsC
svsI
sCsV
vsVsCsIdtdvCti
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)
+ _
y(t) +
_
+ _ vx(t)
Operational Amplifier Circuit • Assuming that Vx(s) = 0,
• How to realize a gain of –1? • How to realize a gain of 10?
+ _
Y(s) +
_
+ _ Vx(s)
Zf(s)
Z(s)
+ _ F(s)
I(s) ( ) ( ) ( )
( ) ( )( )
( )( )( )
( )
( ) ( )( )
( )( )sZsZ
sFsYsH
sFsZsZ
sY
sZsFsI
sZsIsY
f
f
f
−==
−=
=
−= H(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
210
1
10log is defined as Decibel GaindBPGP
=
22 2
2 2 2 110 10 10 102
11 1 2
1
2 2 210 10 10 2 1
1 1 1
10log 10log 10log 10log
20log 10log 20log if
dB
VP R V RG
VP V RR
V R V R RV R V
⎛ ⎞= = = +⎜ ⎟
⎝ ⎠
= − = =
2 210 10
1 1
10log 20logdBP VGP V
= =
DECIBEL SCALE
210 10
1
20log 20logdBVH HV
= =
Magnitude H Decibel Value HdB
0.001 -60 0.01 -40 0.1 -20 0.5 -6
1/√2 -3 1 0 √2 3 2 6 10 20 20 26 100 40
1 2 1 2
11 2
2
log log log
Some Properties of Logar
log log log
log loglog
ith s
1 0
m
n
PP P P
P P PP
P n P
= +
⎛ ⎞= −⎜ ⎟
⎝ ⎠
=
=
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.
Typical Sound Levels and Their Decibel Levels.
• Express transfer function in Standard form.
• Express the magnitude and phase responses.
• 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
( )( )200( )2 10jH
j jω
ωω ω
=+ +
10STANDARD FORM ( )(1 2)(1 10)
jHj j
ωω
ω ω=
+ +
10
-1 -1
10 10 1020log 10 20log 20log 1 2 20log 1
90 tan ( 2)
10
tan ( 10)dbH j j j
φ
ω
ω
ω
ω
ω
= ° −
= +
−
− + − +
EXAMPLE 1 Construct Bode plots for ( )( )
200( )2 10jH
j jω
ωω ω
=+ +
10 10 10 1020log 10 20log 20log 1 2 20log 1 10dbH j j jω ω ω= + − + − +
-1 -190 tan ( 2) tan ( 10)φ ω ω= °− −
1020log 2 6dB=
26 dB
X X
XX
EXAMPLE 2 Continued: Let us calculate |H| and φ at ω=50 rad/sec graphically.
26 dB
10(50) (10) 20log (50 /10) 26 20 0.7 26 14 12H H dB= − = − × = − =
10 10 10(50) 90 45 log (1/ 0.2) 90 log (20 /1) 45 log (50 / 20)90 45 0.7 90 1.3 45 0.4 90 31.5 117 18 76.5
φ = °− °× − °× − °×
= °− °× − °× − °× = °− °− °− ° = − °
ω=50
ω=50
EXAMPLE 1 Construct Bode plots for ( )210( )5
jHj j
ωω
ω ω
+=
+
2
0.4(1STANDARD F
10)( )(1 5)
ORMjH
j jω
ωω ω
+=
+
10 10 10 1020log 0.4 20log 1 10 20log 40log 1 5dbH j j jω ω ω= + + − − +
-1 -10 tan ( 10) 90 2 tan ( 5)φ ω ω= °+ − °−
• Express transfer function in Standard form.
• Express the magnitude and phase responses.
• 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 ( )210( )5
jHj j
ωω
ω ω
+=
+
10
10
10
10
20 log 0.420log 1 10
20log
40log 1 5
dbHj
j
j
ω
ω
ω
= +
+ −
−
+
-1
-1
0 tan ( 10)90 2 tan ( 5)
φ ω
ω
= °+
− °−
X O
X O
PRACTICE PROBLEM 4 Obtain the transfer function for the Bode plot given.
2 2
2
2
A zero at 0.5,
A pole at 1,
Two poles at 10,
1 0.5 (1 0.5)(0.5 ) 200( 0.5)( )(1 1)(1 10) (1 100)(1 )(10 ) ( 1)( 10)
1 0.51
1 11
(1 10)
j
j
j j sHj j j j s s
j
ω
ω
ω
ω ωω
ω
ω
ω
ω ω
ω
ω
=
=
=
+ + += = =
+ + + + + +
+
+
+