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−