EEE598D: Analog Filter & Signal
Processing Circuits
Instructor:
Dr. Hongjiang Song
Department of Electrical Engineering
Arizona State University
Dr. Hongjiang Song, Arizona State University
Thursday January 17, 2002
Today: Continuous-Time Filter Fundamental
• Fourier Transformation
• Laplace Transformation
• Transfer Function (TF)
Dr. Hongjiang Song, Arizona State University
RLC Network and Response
• The traditional question:
For a given input vi(t) signal and a RLC network,
what is the output signal vo(t)?
Vi
L
R
Vo
C
d 2 vo (t )
dvo (t )
+ RC
+ vo (t ) = vi (t )
LC
2
dt
dt
vo (t ) |t =0 = vo (0)
(1.1)
dvo (t )
dv (0)
|t =0 = o
dt
dt
• The solution:
It is a differential equation.
– Solvable but not trivial!!
Dr. Hongjiang Song, Arizona State University
Steady State Response
• Sometimes we only interest in the steady
state responses of the form:
vi (t ) = Vi e jωt and
vo (t ) = Vo e
(1.2)
jω t
• Equation (1.1) can be simplified as:
(1 − LCω 2 + jωRC )Vo = Vi
or
Vo
1
H ( jω ) =
=
Vi 1 − LCω 2 + jωRC
Dr. Hongjiang Song, Arizona State University
(1.3)
Frequency Response
• The magnitude (gain) response of the
network is
1
H ( jω ) =
(1 − LCω ) + ( RCω )
2 2
• We have:
H ( jω ) ω → 0 = 1
H ( jω ) ω →∞ = 1 / ω 2 |ω →∞ = 0
H ( jω ) ω →1/
LC
Dr. Hongjiang Song, Arizona State University
1 L
=
≡Q
R C
2
(1.5)
Gain Response
Q≡
Dr. Hongjiang Song, Arizona State University
ω LC
1 L
R C
Frequency Response
• The phase response of the network is
RCω
θ ( jω ) = tan (
)
2
(1 − LCω )
−1
• We have:
θ ( jω ) |ω →0 = 0
θ ( jω ) |ω →1/
LC
= −π / 2
θ ( jω ) |ω →∞ = −π
Dr. Hongjiang Song, Arizona State University
(1.6)
Phase Response
ω LC
Dr. Hongjiang Song, Arizona State University
General Network Responses
• Question 1: How is the network response to
a general input (non-sinusoidal) signal?
• Question 2: What is the network’s transient
response (non-steady-state response)?
Dr. Hongjiang Song, Arizona State University
Answer to Q1: Fourier Transformation
• A generic input signal can be expressed by
combination of sinusoidal signal using
Fourier transformation
Dr. Hongjiang Song, Arizona State University
The Fourier Transformation
• The Fourier Transform pair is defined as
∞
F ( jω ) =
∫
f (t )e − jωt dt
−∞
1
f (t ) =
2π
Spectrum of f(t)
Dr. Hongjiang Song, Arizona State University
∞
j ωt
(
)
ω
F
j
e
dt
∫
−∞
(1.7)
Useful Fourier Transform Pairs
(1.8)
Dr. Hongjiang Song, Arizona State University
Useful Fourier Transform Pairs
(1.9)
Dr. Hongjiang Song, Arizona State University
Response by Fourier Transformation
• Apply Fourier Transform to (1.1), we have
− LCω 2Vo ( jω ) + jωRCVo ( jω ) + Vo ( jω ) = Vi ( jω )
Vo ( jω )
1
⇒ H ( jω ) ≡
=
Vi ( jω ) 1 − LCω 2 + jωRC
Dr. Hongjiang Song, Arizona State University
Answer to Q2: Laplace Transformation
• The transient response of the network can
be calculated by using the Lapalce
Transformation
Dr. Hongjiang Song, Arizona State University
Laplace Transformation
• The (one side) Laplace Transform is
defined as:
∞
F ( s ) ≡ L( f (t )) ≡ ∫ f (t )e − st dt
0
σ + j∞
1
− st
f (t ) = L ( F ( s )) =
F ( s )e dt
∫
2πj σ − j∞
−1
Dr. Hongjiang Song, Arizona State University
(1.10)
Useful Laplace Transform Pairs
(1.11)
Dr. Hongjiang Song, Arizona State University
Useful Laplace Transform Pairs
(1.12)
Dr. Hongjiang Song, Arizona State University
General Continuous-Time Linear Filter
• A general continuous-time (CT) linear filter
can be expressed as a differential equation:
k
n
m
k
d vi (t )
d vo (t )
bk
= ∑ ak
∑
k
k
dt
dt
k =0
k =0
k
k
d vo (t )
d vo (0)
|t =0 =
k
k
dt
dt
Dr. Hongjiang Song, Arizona State University
(1.13)
k = 0,1,...n − 1
Filter Response using Laplace Transform
• The response of the above filter can be
solved using Laplace Transform:
m
d k vo (t )
d k vi (t )
L(∑ bk k ) = L(∑ ak k )
dt
dt
k =0
k =0
n
m
Vo ( s ) =
∑a s
k =0
n
(1.14)
k
k
k
b
s
∑k
Vi ( s ) + T ( s )
Initial condition
k =0
m
k
a
s
∑ k
vo (t ) = L−1[( k =n0
∑b s
k =0
Dr. Hongjiang Song, Arizona State University
k
k
)Vi ( s )] + L−1[T ( s )]
(1.15)
Example of Laplace Transform
• Look at filter response of the above filter
for given initial condition
Vi
L
R
Vo
C
dvo (t )
d 2 vo (t )
− jωt
+ RC
+ vo (t ) = Ae
LC
2
dt
dt
(1.16)
vo (t ) |t =0 = 0
dvo (t )
|t =0 = 0
dt
Dr. Hongjiang Song, Arizona State University
Example of Laplace Transform
• Using (1.5), we have
• Let
Vm
1
Vo ( s ) =
( LCs 2 + RCs + 1) ( s + jω )
R
α≡
2L
R 2
1
β≡
−( )
LC 2 L
(1.17)
(1.18)
• We have
Vo ( s ) =
Vm
1
( s + α + jβ )( s + α − jβ ) ( s + jω )
Dr. Hongjiang Song, Arizona State University
(1.19)
Example of Laplace Transform
• With partial-fraction expansion, we have
A
B
C
(1.20)
+
+
Vo ( s ) =
( s + jω ) ( s + α + jβ ) ( s + α − jβ )
• where
Vm
| s = jω
A= 2
s LC + sRC + 1
Vm
|s =α + jβ
B=
( s + α − jβ )( s + jω )
Vm
|s =α − jβ
C=
( s + α + jβ )( s + jω )
Dr. Hongjiang Song, Arizona State University
(1.21)
Example of Laplace Transform
• Using (1.3) the response can be solved as:
vi (t ) = Ae
− jωt
+ Be
−αt − jβt
+ Ce
−αt + jβt
(1.22)
transient response
(→ 0 as t → ∞)
Vo ( s )
A=
|s = jω steady-state response
Vi ( s )
Dr. Hongjiang Song, Arizona State University
Transfer Function
• For steady-state response, the s-domain
ratio of the output signal to input signal is
known as the transfer function (TF) of the
filter:
V (s)
H ( s) =
o
Vi ( s )
(1.23)
• For the given example above, TF is given
as:
Vo ( s )
1
= 2
H ( s) =
Vi ( s ) s LC + sRC + 1
Dr. Hongjiang Song, Arizona State University
(1.24)
S-Domain Representation of Operators
• By using the Laplace Transform, we have:
Integrator:
t
1
y (t ) = ∫ x(τ )dτ ⇒ Y ( s ) = X ( s ) ⇒
s
to
∫ dt
⇔
1 (1.25)
s
Differentiator:
dx(t )
⇒ Y ( s ) = sX ( s ) ⇒
y (t ) =
dt
Dr. Hongjiang Song, Arizona State University
d
dt
⇔ s
(1.26)
Transfer Function
• Generally, for a filter described the
following linear differential equation:
k
n
m
k
d vo (t )
d vi (t )
bk
= ∑ ak
∑
k
k
dt
dt
k =0
k =0
(1.27)
• The transfer function can be calculated
using Laplace Transform as:
m
H (s) =
k
a
s
∑ k
k =0
n
k
b
s
∑k
k =0
Dr. Hongjiang Song, Arizona State University
am s m + am −1s m −1 + ... + a1s + a0
=
(1.28)
n
n −1
bn s + bn −1s + ... + b1s + b0
S-Domain Impedance of Components
• By Laplace Transform, we have:
Resistor:
v(t ) = Ri (t ) ⇒ V ( s ) = RI ( s )
⇒
ZR = R
Capacitor:
dv(t ) I (t )
1
=
⇒ V (s) =
I (s)
dt
C
sC
⇒
1
(1.29)
ZC =
sC
⇒
Z L = sL
Inductor:
di (t )
v(t ) = L
⇒ V ( s ) = sLI ( s )
dt
Dr. Hongjiang Song, Arizona State University
Calculation of Transfer Function
• The transfer function of a linear circuit can
be calculated by using s-domain impedance
through Kirchhoff’s law:
Vi
sL
R
Vo
1/sC
Vo ( s )
ZC
1 / sC
=
=
Vi ( s ) Z L + Z R + Z C sL + R + 1 / sC
1
= 2
s LC + sRC + 1
Dr. Hongjiang Song, Arizona State University
(1.30)
System Model of CT Filters
• Linear Systems
constant
constant
K (a1 x1 (t ) + a2 x(t )) = a1 K ( x1 (t )) + a2 K ( x2 (t ))
• Time- (shift-) Invariant Systems
if y (t ) = K ( x(t ))
Then y (t − to ) = K ( x(t − to ))
Dr. Hongjiang Song, Arizona State University
S-Domain Representation of Linear TimeInvariant (LTI) Systems
∞
L{ y (t )} = L{K ( x(t ))} = L{K ( ∫ x(ζ )δ (t − ζ )dζ )}
−∞
∞
= L{ ∫ x(ζ ) K (δ (t − ζ ))dζ } = L{x(t )} ⋅ L{K (δ (t ))}
−∞
Y ( s ) ≡ L{ y (t )}
Let:
X ( s ) ≡ L{x(t )}
Then:
h(t ) ≡ K (δ (t ))
H ( s ) ≡ L{h(t )}
Y (s)
= H (s)
X ( s)
Transfer Function (TF) of system
Impulse response of system
Remark: Continuous-time LTI System can be completely determined
by its impulse response.
Dr. Hongjiang Song, Arizona State University
Steady State Response of ContinuousTime LTI Systems
x(t ) = Au (t )e jωt
∞
⇒ X ( s ) = A∫ e −( s − jω ) t dt =
0
Y (s) = H (s) ⋅ X (s) =
A
s − jω
H ( s ) H ( jω )
=
+ Other Terms
s − j ω s − jω
y (t ) t →∞ = Au (t ) H ( jω )e jωt + yh (t ) = H ( jω ) x (t )
Approach zero for stable system
Steady State Response of the system
Dr. Hongjiang Song, Arizona State University