- By
**koto** - Follow User

- 139 Views
- Uploaded on

Download Presentation
## PowerPoint Slideshow about ' D x' - koto

**An Image/Link below is provided (as is) to download presentation**

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.

- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -

Presentation Transcript

LettingΔx tend to zero:

Equations 3 & 4 are called the "Telegrapher's Equations"and are important because they govern the variation of I and V as a function of distance along the line (x) and time (t).

But we can substitute for I/t from one of the Telegrapher's Equations [Equation (4)]:

Rearranging:

Using similar steps it can also be shown that:

The above two equations govern the voltage and current along a general transmission line.

between these two

equations and . . .

. . . the wave equations

obtained from

Maxwell's Equations

in Prof. Murray's section:

Consider the special case of a lossless transmission line, i.e. one for which R = 0 and G = 0. The previous equations reduce to:

Equations 9 and 10 describe voltage and current WAVES

travelling down the transmission line.

Any function, F, which has t+x/v or t-x/v as the variable is a solution to these equations, v being a constant with the dimensions of velocity (i.e. units of m/s):

Solution to the wave equations for a lossless transmission line.

Note that F is a function of x AND t.

It is not surprising that F can be any function since we can propagate sine waves, square waves, triangular waves and waves of arbitrary shape down e.g. a coaxial cable.

To verify that F(tx/v) is a solution we need to put F into the wave equation and check that

If we do that for F(t-x/v), we find:

Therefore F(t-x/v) is a solution of the wave equations for the lossless transmission line (Equations 9 and 10) if

1/v2 = LC (or v = 1/√LC)

Similarly you can show that F(t+x/v) is a solution if

1/v2 = LC (or v = 1/√LC)

F(t-x/v)

F(t+x/v)

The “-” and “+” signs in the variable show which direction the wave is moving in.

“-” indicates the wave is moving to the right (+ve x direction)

“+” indicates the wave is moving to the left (-ve x direction)

Recalling that the velocity of a wave, v, is given by v = fl, then f/v = 1/l. Hence:

Putting w = 2p f and b = 2p / l (where b is termed the

PHASE CONSTANT):

V(x,t) = V1cos[wt - bx]

b is termed the phase constant because bx gives the phase of the wave at the point x.

Part 1 - Introduction and Basics

Lecture Topics

1. General definition

Practical definition

Types of transmission line: TE, TM, TEM modes

TEM wave equation - equivalent circuit approach

2. The "Telegrapher's Equations"

Solution for lossless transmission lines: F(t±x/v)

Simplest case of F(t±x/v)

3. Direction of travel of cos/sin (ωt ±x) waves

Phase velocity of a wave on a transmission line

General transmission line: attenuation

2.3 Properties of a wave on a transmission line.

A wave of frequency 4.5 MHz and phase constant 0.123 rad/m propagates down a lossless transmission line of length 500 m and with μr=1. Find:

(i) the wavelength

(ii) the velocity of the wave

(iii) the phase difference between the phasor voltages at

the two ends of the line

(iv) the time required for a reference point on the wave

to travel down the line

(v) the relative permittivity of the dielectric in the line

Direction of travel of cos/sin(wt +bx) waves

This figure shows how the function V=V1 sin (wt + bx) changes with time: a point, P, of constant phase moves to the LEFT with time.

P

l

V V1

0

-V1

T = period

t = 0 (a)

t=T/4 (b)

t= T/2 (c)

p 2p 3p 4p bx

P

V V1

0

-V1

p 2p 3p 4p bx

P

V V1

0

-V1

p 2p 3p 4p bx

V=V1 sin (wt + bx)

Direction of travel of cos/sin(wt +bx) waves

This figure shows how the function V=V1 sin (wt - bx) changes with time: a point, P, of constant constant phase moves to the RIGHT with time.

P

l

V V1

0

-V1

T = period

t = 0 (a)

t=T/4 (b)

t= T/2 (c)

p 2p 3p 4p bx

P

V V1

0

-V1

p 2p 3p 4p bx

P

V V1

0

-V1

p 2p 3p 4p bx

V=V1 sin (wt - bx)

V

x

Phase Velocity of a Wave on a Transmission Line

V=V1 sin (wt - bx)

The PHASE VELOCITY of a wave is defined as the velocity of a point of constant phase. For a point of constant phase, V is constant, hence

wt+ bx = constant

To find the velocity of this constant phase point we must obtain x/t, so differentiate the above w.r.t time, t:

w +b x/t = 0

\ PHASE VELOCITY = x/t = +w/b = + v

For a lossless transmission line with inductance per unit length L and capacitance per unit length C:

Example 2.1 shows that the velocity of a TEM wave on a lossless transmission line is the same as that of an EM wave in the dielectric medium of the line.

General Transmission Line length L and capacitance per unit length C: [R g 0, G g 0]

Until now we have dealt with the simplest type of transmission line, i.e. a lossless line (R=0 and G=0)

…what happens if the line isn’t lossless?

Δx

LDx

RDx

CDx

GDx

For a lossless line (R = 0 and G = 0) we know that the voltage is governed by the wave equation

with solution V = V1e jwt e jx

Lossless Line (R = 0 and G = 0)

General Line (R g 0, G g 0)

For a line which isn't lossless we saw that the voltage is governed by Equation 7:

By analogy with the solution for the wave equation, for the general line let's take as a trial solution to Equation 7:

V = V1e jwt e gx

= V1e(jwt+gx) (14)

but take γ to be complex rather than imaginary like j.

V = V voltage is governed by the wave equation1 e jwt e jbx Lossless Line ( is real)

V = V1 e jwt e gx General Line (g is complex)

Here we are assuming that the time variation is as before but the spatial variation is different and as yet unknown since g is unknown.

To check that our trial function is a solution we need to

put it in Equation 7 and check that L.H.S. = R.H.S.

For RHS: voltage is governed by the wave equation

Substituting for V/t and 2V/t2 in 7:

For LHS: voltage is governed by the wave equation

Comparing (16) and (17), we see that for our trial function to be a solution:

γ2V = (R + jwL)(G + jwC)V

=> g2= (R + jwL)(G + jwC) (18)

Thus V1 e(jwt+gx)is a solution to Equation 7 if:

g = + {(R + jwL)(G + jwC)}1/2 (19)

The to be a solution: + sign again indicates the direction the wave is

travelling in:

the solution with e-gxcorresponds to a forward

travelling wave (+x direction)

the solution with e+gxcorresponds to a backward

travelling wave (-x direction)

The general solution is

V = V1 e jwt e -gx +V2 e jwt e +gx

where V1 and V2 are independent arbitrary amplitudes

which depend on the circumstances.

Physically, this corresponds to two waves moving simultaneously in opposite directions on the transmission line:

V = V1 e jwt e -gx +V2 e jwt e +gx

− Backward (Reflected)

voltage wave

−Forward

voltagewave

-x

+x

V = V simultaneously in opposite directions on the transmission line:1 e jwt e jbx Lossless Line ( is real)

V = V1 e jwt e gx General Line (g is complex)

We need to check that g reduces to j when the line is

lossless – so put R = G = 0 in Equation 19:

For the lossless case

g = + {(R + jwL)(G + jwC)}1/2 (19)

g = +(j2w2LC)1/2

= + jw(LC)1/2

= + jw/v [since v = 1/(LC)1/2]

= + jb [since w/b = v]

Thus for the lossless case e simultaneously in opposite directions on the transmission line:+gx=e+ jbx

i.e. g is purely imaginary (γ = ±j).

In general,g is complex and has both real and imaginary

parts:

g = +{(R+jwL)(G+jwC)}1/2 = +(a + jb)

Hence:

e+gx = e+(a+jb)x=e+axe+jbx

The factor e+ax operates on the amplitude of the wave, decreasing it exponentially.

a is termed the ATTENUATION CONSTANT

e+ax gives the amplitude attenuation as the wave travels

e+jbxgives the phase change over distance x

g is termed the PROPAGATION CONSTANT

For a forward travelling wave: simultaneously in opposite directions on the transmission line:

V = V1 e jwt e-gx = V1 e-ax e j(wt-bx)

amplitude factor phase factor

time variation

Voltage

V1

x

Derive an expression for the phase velocity of a wave on a simultaneously in opposite directions on the transmission line:

general transmission line.

Example 3.1 - Phase velocity of a wave

Voltage

V1

P

x

Determine approximate expressions for simultaneously in opposite directions on the transmission line:a and b when w is large (i.e. at high frequencies) or when R and G are small.

Example 3.2 - High-frequency expressions for the attenuation and phase constants, a and b.

Example 3.3 - Calculation of simultaneously in opposite directions on the transmission line:a and b.

For a parallel wire transmission line the primary line constants at 3 kHz are

R = 6.74 Ω/km

L = 0.00352 H/km

G = 0.29x10-6 S/km

C = 0.0087x10-6 F/km

Find the attenuation and phase constants (a and b) and the phase velocity of the line at 3 kHz. Find also the distance at which the wave amplitude has decayed to 0.1 of its initial value.

Summary simultaneously in opposite directions on the transmission line:

q Direction of travel of cos/sin(wt+bx) waves

the + sign gives the DIRECTION OF TRAVEL for

the waves:

+ indicates the wave is travelling to the left (i.e. in

-x direction)

- indicates the wave is travelling to the right (i.e. in

+x direction)

q The PHASE VELOCITY of a wave on a transmission line is defined as the velocity of a point of constant phase.

Phase velocity = v = fl = w/b [ = 1/(LC)1/2 for a lossless line]

q simultaneously in opposite directions on the transmission line: Voltage on a general transmission line [R g 0, G g 0]

V = V1e(jwt+gx)

where g = +{(R+jwL)(G+jwC)}1/2 = +(a + jb)

g is termed the PROPAGATION CONSTANT

a is termed the ATTENUATION CONSTANT

b is the PHASE CONSTANT

V = V1e(jwt-gx) = V1e-axe j(wt-bx)

GENERAL SOLUTION is:

V = V1ejwte-gx + V2ejwte+gx

forward reflected

wave wave

Download Presentation

Connecting to Server..