impedance concepts and thévenin equivalent systemse_m.350/imped_thv 6.pdf · impedance concepts...
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Impedance Concepts andThévenin Equivalent Systems
Learning Objectives
Electrical impedance
Description of 1-D waves in a fluidequation of motion/constitutive equationl l d h iplane waves - pulses and harmonic waves
frequency and wavelengthacoustic impedance
Thévenin equivalent electrical circuits
Concept of Impedance
basic circuit elements
i t
V(t)I(t)
( ) ( )V t R I tresistor
capacitor
( ) ( )V t R I t=
( )( )dV tC I t=capacitor
i d
( )C I tdt
=
( ) ( )dI tV t Linductor ( ) ( )V t L
dt=
For alternating (harmonic) voltages and currents
( ) ( )0 expV t V i tω= −( )I t
resistor
( )( )0 expI i tω= −
0 0V R I=
capacitor0
0IVi Cω
=−
inductor
i Cω
0 0V i L Iω= −
generalimpedance
( )0 0V Z Iω=impedance
electrical impedance Z = Ze (ohms)
1-D plane wave traveling wave in a fluid
p(x, t ) … pressure in the fluidρ … fluid density
ppp dxx
∂+
∂dy
vx (x, t) … velocity
x∂
dxdz
m=∑F axvpp dx dydz pdydz dxdydz
x tρ ∂∂⎛ ⎞− + + =⎜ ⎟∂ ∂⎝ ⎠
xvpx t
ρ ∂∂− =
∂ ∂Equation of motion
x t∂ ∂
Constitutive equationibili f h fl id
xup Kx
∂= −
∂xu
K … compressibility of the fluid
… displacement
duxxu V
x V∂ Δ
=∂
= relative change in volume
( )22 /xu xp ρ∂ ∂ ∂∂
− =dx
x V∂
2 2
2
2
x tp
K
ρ
ρ
=∂ ∂
∂=
∂ 2K t∂2 2
2 2 2
1 0p p∂ ∂− =
∂ ∂1-D wave
if
Kcρ
= wave speed in thefl id2 2 2
fx c t∂ ∂equation ρ fluid
1-D plane wave solutions
( )/ ( / )f fp f t x c g t x c= − + +f g
x
Harmonic waves (or Fourier transforms of transient waves)
( ) ( ) ( ) ( )exp 2 / exp 2 /f fp A f if t x c B f if t x cπ π⎡ ⎤ ⎡ ⎤= − − + − +⎣ ⎦ ⎣ ⎦⎣ ⎦ ⎣ ⎦
V f( ) = v t( )exp 2πi f t( )dt+∞
∫
Fourier transform pair
V f( ) = v t( )exp 2πi f t( )dt−∞∫
v t( ) = V f( )exp 2πi f t( )df+∞
∫
( ) ( ) ( )2 /V f V f if⎡ ⎤%
v t( ) = V f( )exp −2πi f t( )df−∞∫
Let ( ) ( ) ( )exp 2 / fV f V f if x cπ⎡ ⎤= ⎣ ⎦
( ) ( ) ( )exp 2 /v t V f if x c t dfπ+∞
⎡ ⎤= −⎣ ⎦∫ %
Let
then1-D plane wave traveling in the +x-direction
( ) ( ) ( )exp 2 / fv t V f if x c t dfπ−∞
⎡ ⎤= ⎣ ⎦∫then
Thus v(t) is a superposition of plane wavesThus, v(t) is a superposition of plane waves
( )/ fu t x c= −Now, let
( ) [ ] ( ) ( )exp 2 / fV f ifu df v u v t x cπ+∞
−∞
− = = −∫ % % %Then
where ( ) ( )v t V f↔ %%
soso
( ) ( ) ( )/ exp 2 /f fv t x c V f if x c t dfπ+∞
⎡ ⎤− = −⎣ ⎦∫ %% ( ) ( ) ( )pf ff f f−∞
⎣ ⎦∫
Thus, we can always synthesize a traveling plane wave pulse i h i i f h i lwith a superposition of harmonic plane waves.
various forms of a harmonic plane wave traveling in the +x-direction
( )( )
exp 2 / 2
exp 2 /
fp A ifx c ift
A ifx c i t
π π
π ω
= −
= ( )( )( )
exp 2 /
exp
2 /
fA ifx c i t
A ikx i t
A i i t
π ω
ω
λ
= −
= −/ 2/
2 / /f
fk c
k c f
ω πω
λ π
==
= =( )exp 2 /A ix i tπ λ ω= − 2 / /fk c fλ π= =
f ... frequency (cycles/sec)
ω …frequency (rad/sec)
k … wave number (rad/length)
λ wavelength (length/cycle)λ …wavelength (length/cycle)
( )exp 2 / 2fp A ifx c iftπ π= −
magnitude of the pressure at a fixed x versus time, t:
t
T… period (sec/cycle)
2 2ifT iπ π=
1f f ( l / )fT
= frequency (cycles/sec)
2 fω π= circular frequency (rad/sec)2 fω π
rad/cycle
q y ( )
( )exp 2 / 2fp A ifx c iftπ π= −
i d f h fi d i dimagnitude of the pressure at a fixed time, t, versus distance, x:
x
λ ... wavelength (length/cycle)
1xf
λ= spatial frequency (cycles/length)
22 xk f ππλ
= = spatial circular frequency( b ) ( d/l h)λ
rad/cycle(wave number) (rad/length)
( )exp 2 / 2fp A ifx c iftπ π= −( )f
magnitude of the pressure at a fixed time, t, versus distance, x:
x
λ ... wavelength (length/cycle)
2 / 2fi f c iπ λ π=
fλ ff cλ=fundamental relationship that shows how the f f l h i li ifrequency of a plane harmonic traveling wave is related to its wavelength
Example:Example:
consider a 5MHz plane wave traveling in water ( c =1500m/sec).What is the wave length?What is the wave length?
6
6
1.5 10 / sec/
mml
λ ×= 65 10 / sec
0.3 /cycle
mm cycle×
=
F l li i h di iFor a plane pressure wave traveling in the x-direction
( )/p f t x c= −( )/ fp f t x c= −
/ fu t x c= −Let
( ) 1df up u fx du x c
∂ ∂ ′= = −∂ ∂
then
1f
x
f
x du x c
v ft cρ
∂ ∂
∂ ′=∂
xvpx t
ρ ∂∂− =
∂ ∂1 1
f
xf f f f
df f pv f t duc c du c c
ρ
ρ ρ ρ ρ′= ∂ = = =∫ ∫
x t∂ ∂
f f f f
f xp c vρ=x
S … area
specific acoustic impedanceof a plane wave (pressure/velocity)
afz cρ=
p (p y)
f x
a
F pS c v Sρ= =a
xF Z v=
aZ c Sρ=acoustic impedanceof a plane wave
fZ c Sρ= of a plane wave(force/velocity)
F l (h i ) ( )aF ZFor more general (harmonic) waves ( )aF Z vω=
Thévenin equivalent circuit (harmonic voltages and currents)
R1 C1I
Vi R2 V
L1R3
I
V ( ) Zeq(ω) V
I
Vs(ω) Zeq(ω) V
Determining the Thévenin equivalent sourceDetermining the Thévenin equivalent source
Vs(ω) Zeq(ω) V0 open-circuit voltage
0sV V=
D t i i th Thé i i l t i dDetermining the Thévenin equivalent impedance es L eq
L L
V V Z I
V R I
− =
=
Vs(ω) Zeq(ω) RL VL
1sVZ R⎛ ⎞⎜ ⎟1s
eq LL
Z RV
= −⎜ ⎟⎝ ⎠
Example: find the Thévenin equivalent circuit forExample: find the Thévenin equivalent circuit forthe following circuit
R
Vi C
R
Vi C 0VI open-circuit voltageC 0I p g
0iV V IRIV
− =
=0Vi Cω
=−
eliminating I, we findg ,
0 1iVV
i RCω=
− Thévenin equivalent source1 i RCω
R
Vi CI1RLI2 VL
( )1 2I I−( )1 21 2i L L L L
I IV V I R V I R V
i Cω− = = =
−
iVeliminating I1 , I2 gives ( )1 /
iL
L
VVi RC R Rω
=− +
( )1 /i RC R RV ω⎧ ⎫+⎛ ⎞ ⎪ ⎪so ( )( )
0 1 /1 1
1
/
Leq L L
L
i RC R RVZ R RV i RC
R R R
ωω
⎧ ⎫− +⎛ ⎞ ⎪ ⎪= − = −⎨ ⎬⎜ ⎟ −⎪ ⎪⎝ ⎠ ⎩ ⎭⎧ ⎫⎪ ⎪
( ) ( )/
1 1L
LR R RR
i RC i RCω ω⎧ ⎫⎪ ⎪= =⎨ ⎬− −⎪ ⎪⎩ ⎭
R
Vi(ω) Ci( ) C
( ) ( ) RZ( ) ( )
1i
S
VV
i RCω
ωω
=−
( )1eqZ
i RCω
ω=
−
In many EE books the equivalent impedance is obtained by simply shorting out (removing) the source andby simply shorting out (removing) the source and examining the ratio V/I at the output port:
R I
VI1
C
( )1
1
V R I IIV
= −( )V R I i CVω= +
1Vi Cω
=− ( )1
eq
V i RC RIV RZ
ω− =
= =1eq I i RCω−
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