Download Impedance and Equivalent Circuits

Impedance Concepts and
Thévenin Equivalent Systems
Learning Objectives
Electrical impedance
Description of 1-D waves in a fluid
equation of motion/constitutive equation
plane
l
waves - pulses
l andd harmonic
h
i waves
frequency and wavelength
acoustic impedance
Thévenin equivalent electrical circuits
Concept of Impedance
basic circuit elements
V(t)
I(t)
resistor
it
V (t ) = R I (t )
capacitor
dV (t )
C
= I (t )
dt
i d
inductor
dI ( t )
V (t ) = L
dt
For alternating (harmonic) voltages and currents
I (t )
= I 0 exp ( −iω t )
resistor
capacitor
inductor
general
impedance
V ( t ) = V0 exp ( −iω t )
V0 = R I 0
I0
V0 =
−iω C
V0 = −iω L I 0
V0 = Z (ω ) I 0
electrical impedance Z = Ze (ohms)
1-D plane wave traveling wave in a fluid
p(x, t ) … pressure in the fluid
ρ … fluid density
vx (x, t) … velocity
dy
p
dz
dx
∂p
p + dx
∂x
∑ F = ma
∂vx
∂p ⎞
⎛
− ⎜ p + dx ⎟ dydz + pdydz = ρ dxdydz
∂x ⎠
∂t
⎝
Equation of motion
∂vx
∂p
−
=ρ
∂x
∂t
Constitutive equation
∂u x
p = −K
∂x
dux
dx
1-D wave
equation
i
K … compressibility
ibili off the
h fluid
fl id
ux
… displacement
∂u x ΔV
=
= relative change in volume
∂x
V
∂ 2 ( ∂u x / ∂x )
∂2 p
− 2 =ρ
∂x
∂t 2
ρ ∂2 p
=
K ∂t 2
∂2 p 1 ∂2 p
− 2 2 =0
2
∂x
c f ∂t
cf =
K
ρ
wave speed in the
fl id
fluid
1-D plane wave solutions
p = f ( t − x / c f ) + g (t + x / c f )
f
g
x
Harmonic waves (or Fourier transforms of transient waves)
p = A ( f ) exp ⎡⎣ −2π if ( t − x / c f ) ⎤⎦ + B ( f ) exp ⎡⎣ −2π if ( t + x / c f ) ⎤⎦
Fourier transform pair
+∞
V ( f ) = ∫ v(t ) exp(2πi f t )dt
−∞
+∞
v(t ) = ∫ V ( f ) exp(−22πi f t )df
−∞
Let
then
V ( f ) = V% ( f ) exp ⎡⎣ 2π if ( x / c f ) ⎤⎦
v (t ) =
+∞
∫ V% ( f ) exp ⎡⎣2π if ( x / c
−∞
f
− t ) ⎤⎦ df
1-D plane wave traveling in the +x-direction
Thus v(t) is a superposition of plane waves
Thus,
Now, let
u = (t − x / c f
)
+∞
Then
% ( f ) exp [ −2π ifu ] df = v% ( u ) = v% ( t − x / c
V
f
∫
)
−∞
where
v% ( t ) ↔ V% ( f )
so
v% ( t − x / c f ) =
+∞
∫ V% ( f ) expp ⎡⎣2π iff ( x / c
−∞
f
− t ) ⎦⎤ dff
Thus, we can always synthesize a traveling plane wave pulse
with
i h a superposition
i i off harmonic
h
i plane
l
waves.
various forms of a harmonic plane wave
traveling in the +x-direction
p = A exp ( 2π ifx / c f − 2π ift )
= A exp ( 2π ifx / c f − iωt )
f = ω / 2π
= A exp ( ikx − iωt )
k = ω / cf
= A exp ( 2π ix
i / λ − iωt )
λ = 2π / k = c f / f
f ... frequency (cycles/sec)
ω …frequency (rad/sec)
k … wave number (rad/length)
λ …wavelength
wavelength (length/cycle)
p = A exp ( 2π ifx / c f − 2π ift )
magnitude of the pressure at a fixed x versus time, t:
t
T… period (sec/cycle)
2π ifT = 2π i
f =
1
T
ω = 2π f
rad/cycle
f
frequency
( l / )
(cycles/sec)
circular frequency
q
y ((rad/sec))
p = A exp ( 2π ifx / c f − 2π ift )
magnitude
i d off the
h pressure at a fixed
fi d time,
i
t, versus distance,
di
x:
x
λ ... wavelength (length/cycle)
fx =
1
λ
k = 2π f x =
rad/cycle
spatial frequency (cycles/length)
2π
λ
spatial circular frequency
(
(wave
number)
b ) (rad/length)
( d/l
h)
p = A exp ( 2π ifx / c f − 2π ift )
magnitude of the pressure at a fixed time, t, versus distance, x:
x
λ ... wavelength (length/cycle)
2π i f λ / c f = 2π i
f λ = cf
fundamental relationship that shows how the
f
frequency
off a plane
l
harmonic
h
i traveling
li wave is
i
related to its wavelength
Example:
consider a 5MHz plane wave traveling in water ( c =1500m/sec).
What is the wave length?
1.5 × 106 mm / sec
λ=
5 × 106 cycle
l / sec
= 0.3mm / cycle
F a plane
For
l
pressure wave traveling
li in
i the
h x-direction
di i
p = f (t − x / c f
Let
)
u = t − x / cf
then
∂vx
∂p
−
=ρ
∂x
∂t
∂p df ( u ) ∂u
1
=
=−
f′
∂x
du ∂x
cf
∂vx
1
=
f′
∂t
ρc f
1
vx =
ρc f
∫
1
f ′∂t =
ρc f
df
f
p
∫ du du = ρ c f = ρ c f
p = ρ c f vx
za = ρc f
S … area
x
specific acoustic impedance
of a pplane wave (p
(pressure/velocity)
y)
F = pS = ρ c f vx S
F = Z a vx
Z a = ρc f S
acoustic impedance
of a plane wave
(force/velocity)
F more generall (harmonic)
For
(h
i ) waves F = Z a (ω ) v
Thévenin equivalent circuit (harmonic voltages and currents)
R1
Vi
C1
R2
R3
V
L1
I
Vs(ω)
( )
Zeq(ω)
I
V
Determining the Thévenin equivalent source
Vs(ω)
Zeq(ω)
V0 open-circuit voltage
Vs = V0
D t
Determining
i i the
th Thévenin
Thé i equivalent
i l t impedance
i
d
Vs − VL = Z eqe I
VL = RL I
Vs(ω)
Zeq(ω)
RL
VL
⎛ Vs
⎞
Z eq = RL ⎜ − 1⎟
⎝ VL ⎠
Example: find the Thévenin equivalent circuit for
the following circuit
R
Vi
C
R
Vi
I
V0
C
open-circuit
p
voltage
g
Vi − V0 = IR
V0 =
I
−iω C
eliminatingg I,, we find
Vi
V0 =
1 − iω RC
Thévenin equivalent source
R
Vi
I1
Vi − VL = I1 R
C
RL
I2
VL = I 2 RL
VL
VL
I1 − I 2 )
(
=
−iω C
Vi
eliminating I1 , I2 gives VL =
(1 − iω RC ) + R / RL
so
⎧⎪ (1 − iω RC ) + R / RL ⎫⎪
⎛ V0 ⎞
Z eq = RL ⎜ − 1⎟ = RL ⎨
− 1⎬
(1 − iω RC )
⎪⎩
⎝ VL ⎠
⎭⎪
⎧⎪ R / RL ⎫⎪
R
= RL ⎨
=
⎬
⎪⎩ (1 − iω RC ) ⎪⎭ (1 − iω RC )
R
Vi((ω))
Vi (ω )
VS (ω ) =
1 − iω RC
C
Z eq (ω ) =
R
1 − iω RC
In many EE books the equivalent impedance is obtained
by simply shorting out (removing) the source and
examining the ratio V/I at the output port:
R
I
I1
V
C
V = R ( I − I1 )
I
V= 1
−iωC
V = R ( I + iωCV )
V (1 − iω RC ) = RI
V
R
Z eq = =
I 1 − iω RC