Transmission over Coaxial Cable Notes for EE456 · PDF fileTransmission over Coaxial Cable Notes for EE456 University of Saskatchewan ... To analyze these transmission line equations:

Transmission over Coaxial Cable

Notes for EE456

University of Saskatchewan

Created Nov. 7, 2016

by Eric Salt

revised Nov 10, 2016: Implemented revisisons and suggestions

made by Prof. Ha Nguyen

revised Nov 11, 2016: Changed the title

revised Nov 24, 2016: Added frequency depency to equations

as suggested by Quang Nguyen and Brian

Bersheid. Also reworded several paragraphs

to make them more understandable.

revised Nov 25, 2016: Corrected typo and expanded a few explanations

Nov 28, 2016 to xxxx: revisions str in progress to

Correct low frequency model of R and Z_o.

Add examples.

Improve wording.

1

TRANSMISSION OVER COAXIAL CABLE 2

1 Introduction

Both analog and digital communications systems transmit bandpass analog radio frequency signalsover a medium of some sort to a receiver. The difference between the two systems is that oneembeds an analog message signal in the analog radio frequency signal while the other embedsdigital information. For example commercial AM radio embeds an analog message signal into itsradio frequency signal by amplitude modulating a carrier. Therefore, commercial AM is an analogcommunication system. On the other hand the radio frequency of a wireless router is modulatedwith digital data. Therefore, a computer linked to a router via WiFi is a digital communicationsystem.

Communications systems, whether analog or digital, use coaxial cable, or some other conduitlike a micro strip transmission line, to get the radio frequency signal from one point to another. Forexample, a coaxial cable could be used in a wireless system to get the radio frequency (RF) signalfrom the high power amplifier to an antenna. Coaxial cable is also commonly used to connect theRF signal to test equipment while debugging or testing a communications system. Perhaps the bestexample is the role of coaxial cable in a cable TV distribution system, where it carries the radiosignal from the transmitter to the receiver.

In the notes to follow the properties and behavior of coaxial cables will be explored with a viewto give the students a very good understanding at a macroscopic level. The detailed transmissionline model that used to derive the transmission line equations is clearly presented, but the modelis not analysed with calculus to obtain the equations. The resulting equations of this somewhattedious analysis are just stated. These notes attempt to interpret and give meaning to the resultingequation so they can be applied appropriately with confidence.

The notes go on to explain how devices like directional couplers work and how they can be usedto divert some of the power flowing inside a coaxial cable to a test instrument or a device such asa cable modem.

2 Modelling a Coaxial Cable

2.1 The Basic Model

A coaxial cable is modelled as a transmission line. Any pair of wires used to carry electricalsignals can be modelled as a transmission line. For example very long power lines are modelled astransmission lines as are very short mircrostrip lines, which is simply a track on a printed circuitboard above a ground plane on the other side of the board.

A coaxial cable is two concentric conductors separated by a solid insulator. The outer conductoris called the sheath and the inner conductor is called the center wire. The material used for the solidinsulator has a low dielectric constant and very low dielectric losses. An illustration of a coaxialcable is given at the top of Figure 1.

The transmission line model for a coaxial cable is given in Figure 1. The cable is viewed asthe cascade of an infinite number of cable segments of length dx with each segment being modelledwith lumped circuit elements. Each of the conductors is modelled as an inductor with inductanceℓ in series with a resistor with resistance r. As the conductors are in proximity to each other theyare linked by a capacitor with capacitance c. The insulator separating the conductors is certain to

2 MODELLING A COAXIAL CABLE 3

v(t,0)

xdistance in meters

center conductor

sheath

v(t,x) g

r I(t,x+dx)

v(t,x+dx)

r

r

g

r

cc

dxx x+dx

I(t,0)

I(t,0)

I(t,x)

I(t,x+dx)I(t,x)

ℓ ℓ

ℓℓ

Figure 1: A transmission line model of a coaxial cable

suffer dielectric losses so the link between the two conductors must include a resistor. In this casethe resistor is specified in term of conductance g, which has units of siemens, which is 1/ohm.

Since the length of the segment represented by the lumped elements is infinitesimal, as indicatedin Figure 1, the values of the elements must also be infinitesimal. The infinitesimal values areℓ = (L/2)dx, r = (R/2)dx, c = Cdx and g = Gdx, where L, R, C and G are the inductance permeter, resistance per meter, capacitance per meter and conductance per meter, respectively. I.e. Land R are the inductance and resistance of 1 meter of cable measured from the source end with theload end shorted and C and G are the capacitance and conductance of 1 meter of cable measuredfrom the source end with the load end open.

The conductance, G, models the losses in the dielectric. The loss per meter in the model isthe square of the voltage across the dielectric insulator times G. Many if not all of the moleculesin a dielectric with a relative dielectric constant greater than 1 are dipoles (or become dipoles inthe presence of an electric field). An electric field in a dielectric places a torque on the molecules(the dipoles) causing them to rotate in a direction that reduces the electric field. This rotation, nomater how slight, bumps/rubs other molecule causing them to vibrate and increase their kineticenergy. This molecular kinetic energy is also referred to as heat. Since energy is conserved, thekinetic energy (i.e. heat) comes from the energy in the electric field. Each time the polarity of theelectric field changes, the molecules are rotated and energy is converted from the electric field toheat. Therefore, the power converted from the electric field to heat is proportional to the frequencyof the alternating electric field. This means the conductance G is the function of frequency givenby F G′ where G′ is a constant.

Surprisingly, the resistance per unit length, i.e. R, is also a function of frequency. The reasonfor this has to do with a phenomenon called “skin effect”. At high frequencies most of the currentin a conductor flows near the outside of the conductor reducing it effective cross sectional area. Incoaxial cables this phenomenon starts at a frequency of about 0.1 MHz in coaxial cables. Above


this frequency the resistance per unit length is modeled as R = RDC +√FR′, where RDC and R′

are constants with units Ω/m and Ω/(m√Hz), respectively.

2.2 Analysis for a Cable of Infinite Length

In this subsection the results of the analysis for a cable of infinite length are discussed. The resultsof the analysis for a finite length cable that is terminated with a load impedance ZL are presentedin the next subsection.

The cable is driven at one end by a time dependent voltage source denoted v(t, 0) in Figure 1.This source voltage propagates down the cable creating a distribution in voltage as a function oftime and position on the cable. For purposes of analysis the voltage of the center conductor w.r.tthe sheath a distance x from the voltage source is denoted v(t, x) as illustrated in Figure 1. Thecurrent in both the center conductor and sheath at a distance of x from the source is similarlydenoted I(t, x). Perhaps unnecessary, but it is pointed out the current I(t, x) differs from that ofI(t, x + dx) due to the leakage paths through the capacitor with capacitance Cdx and the resistorwith conductance Gdx.

Using calculus as well as Kirchoff’s voltage and current laws produces the transmission lineequations:

−∂v(t, x)

∂x= R× I(t, x) + L

∂I(t, x)

∂x

−∂I(t, x)

∂x= G× v(t, x) + C

∂v(t, x)

∂x

To analyze these transmission line equations:

1. The input voltage must be a sinusoid of the form v(t, 0) = Ao cos(2πFt + φo), where Ao isa constant with units volts, F is a constant with units Hz and φo is a constant with unitsradians. However, without loss of generality φo can be taken to be zero and the input can be,and will be, v(t, 0) = Ao cos(2πFt).

2. The system must be in steady state. The analysis is done under the assumptions all thetransients that result from turning on the source have died out. The analysis assumes thesource is turned on at time t = −∞ and the system has reached steady state by time t = 0.

Under the conditions given above G and R, while frequency dependent (recall G = G′F andR = RDC +R′

√F ), can be treated as constants since the variable in the differential equation is x.

In steady state analysis it is common practise to represent a sinusoidal function of time by aphasor and characterize the sinusoid by the complex amplitude of the phasor at t = 0. A phasoris a complex function of time of the form Aej(2πFt+φ) = Aejφej2πFt. It is a vector in the complexplane that has a constant magnitude, which is A, and spins with a constant angular velocity, whichis 2πF in units of radians/second. Its complex amplitude at time t = 0, which is Aejφ, is usuallyreferred to as its complex amplitude without stating that it is the amplitude at t = 0. The realpart of the phasor is the cosine A cos(2πFt+ φ). The magnitude of the complex amplitude, whichis A, is the amplitude of the cosine and the angle of the complex amplitude, which is φ, is the phaseof the cosine at time t = 0. The frequency of the phasor, which is its angular velocity, is not aparameter in the complex amplitude.


real

imag

inar

y phasor at time

spinning rate

position of

complex amplitudev(x) = Axe

jφx

φ(x) = ∠ v(x)

= 2πF rad/sec

Ax cos(φx)

Axsin(φ

x)

t = 0

Figure 2: Phasor Diagram that Illustrates the Complex Amplitude of a Phasor

If the input to the cable is a sinusoid, then in steady state, the voltage of the center conductorw.r.t. to sheath at any point on the cable will also be a sinusoid. The sinusoidal voltage at a distanceof x from the source will have the form v(t, x) = Ax cos(2πFt + φx). It is easy to understand thatthe amplitude and phase of the sinusoid will depend the distance from the source, which is x, sothe amplitude and phase are subscripted with an x. The phasor representing v(t, x) for x = x1 isthe complex number v(x)

∣

∣

x=x1

= v(x1), where v(x) is a complex function of the real variable x.The concept of a phasor is illustrated in Figure 2, which is referred to as a phasor diagram. A

phasor is a vector that spins with an angular velocity of 2πF . A phasor diagram shows the positionof one or more phasors at time t = 0 with the understanding that all phasors in the diagram arespinning with the same angular velocity. This particular diagram has only one phasor. The realpart of the phasor ( the x co-ordinate of the tip of the phasor) is the sinusoid Ax cos(2πFt + φx),which at t = 0 is Ax cos(φx) as illustrated in Figure 2. The imaginary part of the phasor ( the yco-ordinate of the tip of the phasor) is the sinusoid Ax sin(2πFt+ φx), which at t = 0 is Ax sin(φx)as illustrated in Figure 2.

Phasor diagrams provide insight into how the amplitudes and phases of two sinusoids affect theirsum. Of course the sinusoids must have the same frequency otherwise they can not be representedon the same phasor diagram. The complex amplitude of the sinusoid that is the sum of two sinu-soids is given by the vector addition of the complex amplitudes of the two sinusoids. The vectorsum is illustrated in Figure 3. Vectors v1 and v2 are the complex amplitudes of the two sinusoidsbeing summed while vsum is the complex amplitude of the resultant sinusoid.

The analysis of the transmission line model given in Figure 1 is tedious and a bit difficult tofollow in spots. To avoid the effort in following the development of the equations the results arepresented without derivation. To better understand the final results the analysis is broken intosteps with intermediate results. The first step is the analysis of an infinitely long transmission line.The line is made infinitely long to avoid the complexity of reflections from the end of the line.


realim

agin

ary

v1

v2

vsum = v1 + v2

φsum = ∠ vsum

Figure 3: Phasor Diagram that Illustrates the Sum of two Sinusoids

The complex amplitude of the voltage measured from center wire to sheath at a distance of xfrom the source, as illustrated in Figure 1, is found to be

v(x) = v(0)e−γx, (1)

where v(0) is the complex amplitude of the sinusoid applied at the source end and

γ = 2πF√LC

√

(j +R

2πFL)(j +

G

2πFC). (2)

Since γ is a complex number it is conventional to express it as γ = α + jβ, where α = ℜγ andβ = ℑγ. Then e−γx can be expressed as e−γx = e−αxe−jβx. This has

v(x) = v(0)e−αxe−jβx, (3)

The factor e−αx provides the attenuation a voltage experienced in travelling a distance x along thecable and the factor e−jβx provides the phase shift experienced by that same voltage in travellingthe same distance. That phase shift is −βx.

For practical coaxial cables where the frequency of the sinusoidal source is within the operationalrange of the cable, R/(2πFL) = (RDC +R′

√F ) / (2πFL) << 1 and G/(2πFC) = G′/(2πC) << 1.

This allows both α and β to be approximated. First, the parameter α can be approximated by

α ≈√LC

2

(

R

L+

G

C

)

in units of nepers/meter

α ≈√LC

2

(

RDC +R′√F

L+

G′F

C

)

in units of nepers/meter

α ≈ αDC + α1

√F + α2F in units of nepers/meter (4)

where αDC, α1 an α2 are constants with units nepers/m, nepers/(m√Hz) and nepers/(mHz). The

same two inequalities support the following approximation for β:

β ≈ 2πF√LC in units of radians/meter. (5)


d

D

µr, ǫr

Figure 4: Parameters that characterize a coaxial cable

With these approximations, the equation for v(x) can be expressed as

v(x) ≈ v(0)e−(α1

√F+α2F )x e−j2πFx

√LC . (6)

It must be pointed out that (4) does not accurately predict the attenuation of all commerciallyavailable coaxial cables. It provides a reasonable approximation and certainly predicts the trend ofhow attenuation increases with frequency.

The delay a sinusoidal voltage experiences as it travels along a cable is related to the phaseshift it experiences. For example, a sinusoid that experiences a delay of τ can be expressed assin( 2πF (t− τ) ) or as sin(2πFt+ θ). The phase shift θ is related to delay by τ = −θ/(2πF ). From(5) the phase shift experienced by a sinusoidal voltage travelling a distance x along the cable is−2πFx

√LC, the delay it experiences is x

√LC. Since the distance travelled divided by the travel

time is velocity, the velocity of propagation is

vp =x

x√LC

=1√LC

(7)

Often vp is listed in the data sheets for coaxial cable as a percentage of the speed of light which hasit

vp = 100× 1√LC × 3.00× 108 m/s

in % of speed of light, (8)

where L and C are inductance and capacitance in H/m and F/m, respectively.It must be emphasized that since R and G are functions of frequency α is a function of frequency

as well, but β is not. The important implication is the attenuation experienced by a sinusoidalvoltage as it travels along the cable depends on its frequency, but its velocity of propagation doesnot.

The values for L and C do not significantly depend on frequency and can be calculated usingthe geometry of the cross-section of the coaxial cable along with the real parts of the relativepermeability and relative permittivity of the dielectric material used as the insulator between thecenter conductor and the sheath. For the parameters given in Figure 4 the values of L and C aregiven by

L =µrµo

2πln(

D

d) and C =

2πǫrǫ0

ln(Dd), (9)


where µr and ǫr are the real parts of the relative permeability and relative permittivity of theinsulator used in the cable and

µo = 1.257× 10−6 H/m and ǫo = 8.854× 10−12 F/m. (10)

Using theses values for L and C along with the fact that the speed of light is given by 1/√µoǫo has

the velocity of propagation equal to

Vp =100√µrǫr

in % of the speed of light, (11)

It is pointed out that µr = 1, or very nearly so, for the insulating materials used in coaxial cable.

Now attention is turned from voltage to current. The current on the transmission line can alsobe derived from the transmission line equations. The complex amplitude of the current is found tobe

I(x) =v(x)

Zo

(12)

where Zo is given by

Zo =

√

R + j2πFL

G+ j2πFC. (13)

Since R = RDC + R′√F << 2π F L and G = G′ F << 2π F C over the operating range of the

cable, Zo can be approximated by

Zo ≈√

L

C. (14)

It is pointed out that as F → 0, Zo →√

RDC/0. That is to say that as F → 0, |Zo| → ∞. Since Zo

is constant over the operating range of the cable, the operating range can not include frequenciesnear F = 0.

The complex constant Zo relates the complex amplitudes of voltage and current in the sameway the resistance relates voltage and current in Ohm’s law. Such a quantity is called impedance.The impedance Zo is called the characteristic impedance of the cable.

Careful observation of the equation for Zo shows:

1. Zo has units of ohms.

2. Zo does not depend on x yet it relates I(t, x) to v(t, x) for any value of x.

3. In general Zo may be complex, but coaxial cables are designed so that Zo is real over theoperating frequency range.

4. Since, in the frequency range of normal operation, R = R′√F << 2πFL and G = G′ F <<

2πFC, Zo is a very weak function of frequency. Ignoring R and G produces the quite accurateapproximation Zo =

√

L/C, which shows Zo is independent of frequency.

5. Over the frequency range where Zo =√

L/C it is real and could be symbolized as Ro.


V(0)

real

imag

inar

y

t

t

|I(x)| |v(x)|

∠ v(x) = ∠ I(x)

∠ v(x) = ∠ I(x)

I(t, x)v(t, x)

v(t, x)I(t, x)Pave

I(0)

x

I(x)

I(x)

v(x)

v(x)

Figure 5: Power flow in a cable

6. At low frequencies, which for many of the more commonly used coaxial cables would befrequencies below about 0.1 MHz, three of the four terms in Zo get smaller, but the fourthterm, which is R approaches RDC . The effect on Zo is for it to increase magnitude and nolonger remain real as the frequency decreases to zero.

Having equations for the complex amplitudes of voltage and current as a function of x is onlyof value if they can be used in the design and analysis of systems that use coaxial cables. Normallycoaxial cables are used in communication systems to channel power from one circuit or device toanother, e.g. from a transmitter to an antenna. To utilize the equations for the complex amplitudesof voltage and current they need to be related to the power flowing through the cable.

Since the characteristic impedances of cables used in industry are real over their useful operatingrange, only the case where Zo is real, i.e. Zo = Ro, will be considered going forward.

At this first step in the analysis of transmission lines the cable does not have a load so doesnot represent a practical situation. The situation being analysed here is a cable of infinite lengthdriven by a voltage source. Power flows from the source into the cable and flows down the cable ata velocity of vp, but never reaches a load. Figure 5 shows the scenario just mentioned as well as thesignals relevant to power flow calculation.

The voltage source, which is shown at the top of Figure 5, drives the center conductor of thecable. Therefore, the complex amplitude of the current passing through the source is I(0) = v(0)/Zo.Since Zo is real, i.e. Zo = Ro, I(0) has the same angle as v(0). This means current flows out ofthe positive terminal of the voltage source whenever the positive terminal has a positive voltage.The direction the current flows while the positive terminal has a positive voltage is indicated the


horizontal arrow under I(0).The average power delivered by the voltage source to the cable is

Pave =1

T

∫ T

0

v(t, 0)I(t, 0)dt

=1

T

∫ T

0

∣

∣v(0)∣

∣ cos( 2πFt+ ∠ v(0) )∣

∣I(0)∣

∣ cos( 2πFt+ ∠ I(0) )dt, (15)

where T is the period of the sinusoid, i.e. T = 1/F .Since in this case the characteristic impedance of the cable is Ro, the complex amplitude of the

current is I(0) = v(0)/Ro and since Ro is real, ∠ I(0) = ∠ v(0). Substituting the angle for I(0) into(15) has

Pave =1

T

∫ T

0

∣

∣v(0)∣

∣

∣

∣I(0)∣

∣ cos2( 2πFt+ ∠ v(0) )dt

=1

T

∫ T

0

∣

∣v(0)∣

∣

∣

∣I(0)∣

∣

2(1 + cos( 4πFt+ 2∠ v(0) ) )dt

=

∣

∣v(0)∣

∣

∣

∣I(0)∣

∣

2(16)

Since I(0) = v(0)/Ro the average power is also given by

Pave =

∣

∣v(0)∣

∣

2

2Ro

.

The power injected into the cable by the source travels down the cable with a velocity of vp.Along the way some of the power is lost as heat due to cable parameters R and G being non-zero.The power that reaches x is

Pave =∣

∣v(x)∣

∣

∣

∣I(x)∣

∣/2.

A cross section of the cable at position x is illustrated at the top of Figure 5. The complexamplitudes of the voltage and current, i.e. v(x) and I(x), are shown in the phasor diagram on theleft-center of Figure 5. The relationship between the phasors and the time waveforms is shown onthe right-center of Figure 5. The instantaneous power, which is v(t, x)I(t, x), as well as the averagepower are plotted on the bottom right of Figure 5. From the plot of the time waveforms v(t, x) andI(t, x) it clear that their phase shifts, which are the angles of v(x) and I(x), while dependent onx, are equal to each other for all x. The plot on the bottom right shows that the average power,which is obviously the time average of the instantaneous power, that flows through the cross sectionat x does not depend on the angles of v(x) and I(x). It also shows the average power that passesthrough the cross section at x is Pave =

∣

∣v(x)∣

∣

∣

∣I(x)∣

∣ / 2.

NB: Going forward it will be very important to know that the current flow in the cable at anyx will be in the direction of propagation of the sinusoidal voltage that causes it when that sinusoidalvoltage is positive. That is to say, at any time while the sinusoidal voltage at x, i.e. v(t, x), is pos-itive the sinusoidal current at x caused by v(t, x), i.e. I(t, x), flows in the direction of propagation.The explanation for this begins with the current that flows through the voltage source. At any


instant the voltage source in Figure 5 is positive, current, which is given by I(t, 0) = v(t, 0)/Ro,flows out of the positive terminal and into the cable, which is in the direction of propagation. SinceI(t, x) = v(t, x)/Ro for all x, the direction of the current flow when the voltage is positive is thesame for all values of x.

When it come to power flow the characteristic impedance is a measure of the size of the pipe.It determines the power that flows down the cable for a given voltage source. The power flows fora given voltage is inversely proportional to the characteristic impedance. For example, cables incable TV network have a 75Ω characteristic impedance, if they had been 50Ω cables lower voltagescould have been used to deliver the same power.

Example:Suppose the infinite length line of Figure 1 has a characteristic impedance 75∠0Ω (cable is used in aCATV application) at a frequency of F = 106 Hz, which is in the normal operating range of the cable.A voltage source drives the cable as shown in Figure 1 with v(t, 0) = 7.5 cos(2π × 106 t + π/3)V.The attenuation constant for F = 106 Hz is α = 0.01 nepers/m. The velocity of propagation isvp = 2× 108 m/s.

1. Find I(t, 0) as defined in Figure 1. I.e. find the current entering the cable.

2. Find the instantaneous power that enters the cable, which is a function of time.

3. Find the average power that enters the cable.

4. Figure 5 shows the same infinite length cable that was shown in Figure 1. However, in Figure 5the voltage and current are expressed in terms of their complex amplitudes. Find v(0) andI(0) as shown in Figure 5. Also find the average power using v(0) and I(0).

5. Find v(100m), I(100m) and the average power crossing the cross section plane at x = 100m.

Solution

1. From (12) I(x) = v(x)/Zo. Therefore I(0) = v(0)/Zo. Therefore

I(t, 0) =7.5 cos(2π × 106 t+ π/3− ∠Zo) V

|Zo|

Since Zo = 75∠ 0ΩI(t, 0) = 0.1 cos(2π × 106 t+ π/3)A.

2. The instantaneous power entering the cable at time t is v(t, 0)I(t, 0). Therefore

Pinstantaneous = 7.5 cos(2π × 106 t+ π/3)V × 0.1 cos(2π × 106 t+ π/3)A

= 0.75 cos2(2π × 106 t+ π/3)W.


3. The average power is the instantaneous power averaged over one period of the sinusoid. SinceF = 106 Hz, T = 1/F = 10−6 s.

Pave =1

T

∫ T

0

0.75 cos2(2π × 106 t+ π/3)W dt

=1

T

∫ T

0

0.75

2

(

cos(0) + cos(4π × 106 t+ 2π/3))

W dt

= 0.375W.

4. The complex amplitude of the voltage applied to the cable is v(0) = 7.5∠π/3V. The complexamplitude of the current entering the cable is I(0) = 0.1∠π/3A. The average power enteringthe cable is

Pave(0) =

∣

∣v(0)∣

∣

∣

∣I(0)∣

∣

2

=7.5V 0.1A

2= 0.375W.

5. From (3) v(x) = v(0) e−αxe−jβx. First find the∣

∣v(x)∣

∣.

∣

∣v(x)∣

∣ =∣

∣v(0)∣

∣ e−αx

=∣

∣v(0)∣

∣ e−(0.01 nepers/m)(100m)

=∣

∣v(0)∣

∣× 0.368

= 2.76V.

Now find β x. From the paragraph that precedes (7) delay or travel time is related to phaseshift by τ = −θ/(2πF ). In this case θ = β x. That same paragraph also explains that thevelocity of propagation is the ratio of distance travelled to travel time. Therefore, τ is alsoequal to x/vp. Equating −θ/(2πF ) to x/vp and solving for θ has

β x = θ =−2πFx

vp

=−2π radians/cycle× 106 cycles/s× 100m

2× 108 m/s

= −π radians

Therefore

v(100m) =∣

∣v(100m)∣

∣∠ (∠v(0) + β x) V

= 2.76∠(π/3− π) V

= 2.76∠− 2π/3V


distance in metersPSfrag replacemen

xL

IL(t)

IL(t)

vL(t)vs(t)

Is(t)

Is(t)

Zs

Figure 6: A Terminated Finite Length Transmission Line

Now find I(x).

I(100m) =v(100m)

Zo

=2.76∠− 2π/3V

75∠0Ω= 0.0368∠− 2π/3A

The average power can now be calculated and is

Pave(100m) =

∣

∣v(100m)∣

∣

∣

∣I(100m)∣

∣

2

=(2.76V)(0.0368A)

2= 0.0505W.

End Example

2.3 Analysis for a Terminated Finite Length Cable

This subsection discusses the voltage and current along a finite length cable that is driven by avoltage source that has output impedance Zs and is terminated with a load (i.e. a circuit) that hasan impedance of ZL. This situation is illustrated in Figure 6, where the output impedance of thevoltage source is shown as a resistor, but it could have a capacitive or inductive component, andthe load is a resistor in parallel with a capacitor making the load impedance for the circuit shownZL = RL + 1/(j2πFCL).

The presence of the load places boundary conditions on the transmission line equations andmakes the analysis more complicated than for the case of the infinite length line. When the boundaryconditions are included the analysis shows that when the forward propagating voltage reached theend of the cable where the load is connected some of it is reflected and propagates in the oppositedirection, which is back toward source.

In preparation for analysing a cable where voltages can be reflected and thereby have twovoltages propagating in different directions it is necessary to modify the notation to include thedirection of propagation. A sinusoidal voltage propagating in the forward direction, which is fromsource to load, as well as it complex amplitude will be superscripted with a + and a sinusoidalvoltage propagating in the reverse direction, which is from load to source, as well as its complexamplitude will be superscripted with a −. This means v+(x) is the complex amplitude for the


forward propagating sinusoidal voltage v+(t, x) and v−(x) is the complex amplitude for the reversepropagating sinusoidal voltage v−(t, x).

Since the current generated by a propagating voltage will flow in the direction of propagationwhile that voltage is positive, the current generated by v−(t, x) while it is positive will be in theopposite direction to the current generated by v+(t, x) while it is positive. For this reason it isnecessary to symbolize current in a way that makes it clear which direction the voltage that generatesit is propagating. Current generated by a forward propagating voltage will be superscripted with a+ and current generated by a reverse propagating voltage will be superscripted by a −.

For purposes of analysis a convention for the direction of positive current flow must be chosen.Of course there are only two directions, but either could be chosen. The direction of positivecurrent is arbitrarily defined as the direction of the current generated while a forward propagatingsinusoidal voltage is positive. With this definition of positive current I+(t, x) = v+(t, x)/Ro andI−(t, x) = −v−(t, x)/Ro. This also means I+(x) = v+(x)/Ro and I−(x) = −v−(x)/Ro. Therefore,if the complex amplitude of the voltage of the center conductor w.r.t. the sheath a distance x fromthe source is v+(x) + v−(x) then the complex amplitude of the current in the center at the samepoint is I+(x) + I−(x) = v+(x)/Zo − v−(x)/Zo in the direction from the source to the load.

Having modified the notation and defined the direction of positive current, the equations for thevoltage across load and the current though the load can now be calculated. Trusting that includingboundary conditions in the analysis of the transmission line equations (not done here) indicates thatsome of the forward propagating voltage gets reflected at the load, the amount of voltage reflectedat the load can be found by balancing the currents at the node where the load is attached to thecenter conductor of the coaxial cable.

Defining the direction of positive current through the load as being from the center conductorto sheath as shown in Figure 6, the complex amplitude of the current through the load must beIL = I+(xL) + I−(xL), where the IL is the complex amplitude of the current though the load, xL

is the length of cable between the source and the load and I+(xL) and I−(xL) are the complexamplitudes of the currents generated by v+(xL), which is the complex amplitude of the forwardpropagating voltage, and v−(xL), which is the complex amplitude of the voltage reflected from theload, respectively.

The voltage across the load is the voltage of the center conductor w.r.t. the sheath at end ofthe cable, i.e. at x = XL. Therefore, the complex amplitude of the sinusoidal voltage across theload is VL = v+(xL) + v−(xL).

Since the complex amplitude of the current through the load is given by

IL =VL

ZL

=v+(xL) + v−(xL)

ZL

(17)

and it is also given by

IL = I+(xL) + I−(xL)

=v+(xL)− v−(xL)

Zo

, (18)


the left hand sides of (17) and (18) must be equal to each other, which means

v+(xL) + v−(xL)

ZL

=v+(xL)− v−(xL)

Zo

.

Rearranging the equation above to isolate v−(xL) has

v−(xL) =ZL − Zo

ZL + Zo

v+(xL).

To make the equation above more compact the complex constant preceding v+(xL) is symbolized

ΓL =ZL − Zo

ZL + Zo

(19)

and is referred to as the reflection coefficient for the load. Using ΓL, the equation becomes

v−(xL) = ΓL v+(xL). (20)

It is pointed out that ΓL depends on F through ZL, but is a constant for any specific value ofF . ΓL relates the amplitude and phase of the reflected sinusoidal voltage to those of the forwardpropagating sinusoidal voltage. The reflected sinusoidal voltage at x = xL has

∣

∣ΓL

∣

∣ times theamplitude of the forward propagating sinusoidal voltage and is phase shifted by ∠ΓL w.r.t. theforward travelling voltage.

A few load impedances and their associated reflection coefficients are given below. The valuesfor the reflection coefficients can be easily verified using (19).

1. For an open circuit the load impedance is ZL = ∞ and ΓL = 1. This means all of the voltageis reflected.

2. For a short circuit the load impedance is ZL = 0 and ΓL = −1. In this case the reflectedvoltage has the same amplitude as the forward propagating voltage, but is 180 degrees out ofphase.

3. For a load that is either an inductor or a capacitor the load impedance is purely imaginary.Then

ΓL =jχ−Ro

jχ+Ro

= 1∠ 2arctan(−Ro, χ), (21)

where the load impedance is jχ and has value jχ = j2π F LL for an inductor and jχ =−j/(2π F CL) for a capacitor. Since

∣

∣ΓL

∣

∣ = 1, the reflected voltage has the same amplitudeas the forward propagating voltage, but is 2 arctan(−Ro, χ) radians out of phase.

4. For a load that has an impedance equal to the characteristic impedance of the cable, i.e.ZL = Zo, ΓL = 0. This means no voltage is reflected and the entire forward propagatingvoltage is absorbed by the load.

The next step in the objective to find vL in terms of vs, where vL is the complex amplitude of thesinusoidal voltage across the load and vs is the complex amplitude of the sinusoidal voltage of thesource. This involves finding v+(0) in terms of vs and v−(0), where v+(0) is the forward propagating


voltage voltage in the cable at x = 0 and v−(0) is what is left of the reverse propagating voltagereflected from the load after it has travelled back to x = 0.

The analysis parallels that of finding v−(xL) in terms of v+(xL). The currents at the nodewhere the source resistor connects to the center conductor of the cable are balanced. The resultingequation is

vs − (v+(0) + v−(0))

Zs

=v+(0)− v−(0)

Zo

Rearranging the equation above has

v+(0) =Zo

Zo + Zs

vs +Zs − Zo

Zs + Zo

v−(0).

As was done at the load end, the constant preceding v−(0) is symbolized

Γs =Zs − Zo

Zs + Zo

(22)

and is referred to as the reflection coefficient for the source. Using this constant in the equation forv+(0) has

v+(0) =Zo

Zo + Zs

vs + Γsv−(0). (23)

At this point there are two equations, which are (20) and (23), and four unknowns. To solve forvL three more equations are required. Two of them are obtained using (1) and they are v+(xL) =v+(0)e−γxL and v−(xL) = v−(0)eγxL . The third is vL = v+(xL) + v−(xL). This provides a set of 5equations with 5 unknowns. Solving these equations for vL produces

vL =1 + ΓL

1− ΓsΓL

e−γxLZo

Zs + Zo

vs.

The other parameter that is of value is the input impedance at the source end of a terminatedfinite length cable. The input impedance of an infinite length cable is Zo, but the reflection fromthe load end of a finite length cable changes the input impedance. The input impedance is definedas the ratio of the voltage of center conductor w.r.t. the sheath divided by the current flowing inthe center conductor in the forward direction. Therefore,

Zin =v+(0) + v−(0)

I+(0) + I−(0)

=v+(0) + v−(0)

(v+(0)− v−(0))/Zo

=v+(0) + v−(0)

v+(0)− v−(0)Zo

Since v−(0) results from v+(0) travelling to the load, being reflected and travelling back to thesource, v−(0) = (e−γxL ΓL e

−γxL) v+(0). Substituting this into the equation above yields

Zin =

(

1 + ΓLe−2γxL

1− ΓLe−2γxL

)

Zo. (24)

Several observations can be made from (24):

3 POWER FLOW IN COAXIAL CABLES 17

1. Zin depends on the length of cable through the term e−2γxL , which could be expressed ase−2αxLe−j2βxL . Since practical cables are lossy, α is greater than 0. This means e−2αxL → 0and Zin → Zo for very long cables regardless of the load.

2. For short cables or cables of modest length Zin depends on the load through ΓL.

(a) If ZL = Zo then ΓL = 0 and Zin = Zo regardless of the length of the cable.

(b) If ZL = ∞, Zin can still be very small. For ZL = ∞, ΓL = 1. However if the cable isshort and βxL = π/2, then e−j2βxL = −1 and e−2γxL ≈ −1. In this case the numeratorin (24) becomes very small and Zin → 0.

(c) If ZL = 0, Zin can still be very large. For ZL = 0, ΓL = −1. However, if the cableis short and βxL = π/2 as it was in the previous case, then again e−j2βxL = −1 ande−2γxL ≈ −1. However, in this case ΓL = −1 so the denominator in (24) becomes verysmall and Zin → ∞.

3 Power Flow in Coaxial Cables

3.1 Introduction

The receivers in communication systems are designed to extract information from the signal pre-sented to their input. For reasons that are beyond the scope of this analysis the quality of theinput signal is measured by the ratio of signal power to noise power, i.e. SNR, where the signalpower is that delivered to1 the real component of the input impedance. For this reason communica-tion systems are designed with the primary goal of maximizing the efficiency of transporting signalpower from one device to another. For example the modulation/demodulation scheme is designedto maximize the efficiency of transporting power across a specific channel, which, for example, couldbe twisted-pair, optical fiber, coaxial cable, or wireless (over-the-air).

In this section we look at transporting signal power over coaxial cable. Coaxial cable is used ina large number of applications, but only one is addressed in this section. It is the distribution ofsignal power from the head end of a CATV system to the modems in the customer premises overcoaxial cable.

3.2 Concept of Power and Energy

Energy can not be created or destroyed, except perhaps in nuclear reactions where it can be con-verted to or created from mass. Setting aside the mass-energy conversion that happens in nuclearreactions, energy can only be converted from one form to another. The term “power delivered to aload” means “power converted to heat by the resistive elements on the load”.

A capacitive or inductive element can not accept power over a long term. A capacitor, andsimilarly an inductor, accepts energy while it charges, but then expels that energy when it discharges.A voltage across any element, whether it is a resistor, capacitor or inductor causes a current to flowthrough it. The difference is the current is phase aligned with the voltage when that element is aresistor and is in quadrature with the voltage when that element is an inductor or capacitor. When

1It will be explained later that “delivered to” means “converted to heat by”.


the current through an element is in quadrature to the sinusoidal voltage across it, the productis a sinusoid that has no DC value. However, if the current was phase shifted to be aligned withthe voltage then the product would have a DC value and that DC value is referred to a imaginarypower.

The power delivered to the resistive elements in a load, which is sometimes referred to as realpower to better distinguish it from the concocted imaginary power, is the time average value of theproduct of the voltage across the load and the component of current that is phase aligned with thevoltage.

Coaxial cable transports power from a source to a load in the form of a forward propagatingelectromagnetic wave. The underlying physics of the propagating electromagnetic wave can beloosely viewed as a bucket brigade consisting of the capacitors and inductors illustrated in Figure 1that take energy from the source and pass it along the line to the load. At the load end of thecable some of the power in the forward propagating electromagnetic wave is converted to heatby the load and the rest of the power is reflected back toward the source as a reverse propagatingelectromagnetic wave. Again using the loose physical analogy, the energy in the reverse propagatingelectromagnetic wave is carried back to the source by the same capacitive and inductive elements,but they need to be viewed as a separate bucket brigade that does not interfere with the first.

There are a couple of ways to calculate the power converted to heat by a load with impedanceZL. Both ways are straight forward, but one way leads to a more useful equation. The way thatfinds the less useful equation expresses the power in terms of the voltage across ZL, i.e. vL, and thecurrent through ZL, i.e. IL. To get the equation for power delivered the current is separated intoquadrature components with one of the components being phase aligned with the voltage acrossthe load. In doing this the amplitude of the component of current phase aligned with the voltage

is∣

∣

∣

vLZL

∣

∣

∣cos(∠ZL). Therefore the power converted to heat by the load is

PL =

∣

∣vL∣

∣

2

2∣

∣ZL

∣

∣

cos(∠ZL).

The second way is use the conservation of power. The power converted to heat by the load mustbe the difference between the forward propagating and reverse propagating power at x = xL. Thishas

PL =

∣

∣v+(xL)∣

∣

2

2∣

∣Zo

∣

∣

−∣

∣v−(xL)∣

∣

2

2∣

∣Zo

∣

∣

=

∣

∣v+(xL)∣

∣

2

2∣

∣Zo

∣

∣

−∣

∣ΓL v+(xL)

∣

∣

2

2∣

∣Zo

∣

∣

=(

1−∣

∣ΓL

∣

∣

2)

∣

∣v+(xL)∣

∣

2

2∣

∣Zo

∣

∣

At this point it is worth doing a few quick checks to elevate both physical understanding andconfidence in the equations. First suppose that ZL = ∞. Then ΓL = 1 and both equations indicateno power is absorbed by the load. This confirms what is obvious since there is no load. Next supposethat ZL = 0. If ZL = 0 the load voltage is zero regardless of the current through it and no poweris delivered to the load. Since VL = 0 the first equation indicates no power is delivered to the load


and since ΓL = −1 the second equation indicates no power is delivered to the load. Next supposethat the load is a capacitor. Since a capacitor does not convert electrical power to heat obviouslyno power is delivered to the load. This is verified by both equations since cos(∠ZL) = cos(π/2) = 0and

∣

∣ΓL

∣

∣ = 1.In CATV applications all of the devices that are connected to the coaxial cable are designed to

have an input impedance equal to characteristic impedance. Of course, due to the use of imprecisecomponents the input impedance will vary from device to device and will not exactly equal thecharacteristic impedance. The quality of these devices are measured by how closely their inputimpedances match the characteristic impedance. This quality is measured through a characteristiccalled return loss. Return loss is the ratio of power carried to the load by the forward propagatingvoltage to the power carried away from the load by the reverse propagating voltage. Return loss isusually specified in dB and is given by

RL = 10 log(P+(xL)

P−(xL)

)

= 10 log(

∣

∣v+(xL)∣

∣

2/∣

∣Zo

∣

∣

∣

∣v−(xL)∣

∣

2/∣

∣Zo

∣

∣

)

= 10 log(

∣

∣v+(xL)∣

∣

2

∣

∣ΓL v+(xL)∣

∣

2

)

= −10 log(∣

∣ΓL

∣

∣

2)

= −20 log(∣

∣ΓL

∣

∣

)

, (25)

where RL is return loss in dB for the load located at x = xL. Note that using the symbol RLfor return loss creates some ambiguity since it also represents the product of R and L, which areresistance and inductance per unit length. It will be clear from the context in which RL is usedwhen it means return loss.

3.3 Distribution of Signal Power in CATV networks

In principle signal power is distributed to the subscribers of a CATV service using one long coaxialcable with devices referred to as “taps” inserted to extract some power from the cable for subscribers.A simplified distribution system is shown in Figure 7. The critical element in the distribution ofthe signal is the device referred to in the industry as a “tap”, but it is really a directional coupler.To understand how signal power is transported between the head end and the subscriber modemsit is necessary to understand the operation of a directional coupler, which is a so called tap.

A directional coupler is a device used to redirect some of the power flowing through the cablein the forward direction to a subscriber. A directional coupler is a four-port device converted to athree port device by terminating the fourth port inside the device. An ideal three port directionalcoupler is illustrated in Figure 8. The diagram in the top left corner of Figure 8 shows threeexternal ports named: input port, thought port and coupled port. The through port is also calledthe output port. The diagram shows one internal port that is terminated in a 50Ω resistor. Thisusually implies the device is designed for use on cables with a characteristic impedance of 50Ω, butsometimes the device is designed in a way that the internal port must be terminated in an impedance


modemsubscriber

modemsubscriber

modemsubscriber

HeadEnd

coupledport

taps( i.e. directional couplers )

drop cablethrough

port

port

main cable

input

Figure 7: Simplified distribution system for CATV

50Ω

50Ω

50Ω

V1

(c) voltage applied to coupled port

50Ω

50Ω50Ω

V1

(b) voltage applied to ouput port

50Ω

50Ω

50Ω

(a) voltage applied to input port

V1

50Ω

coupledport

outputport

also calledthe through

port

inputport

internally terminated

power flow map and port definitionfor a directional coupler

Figure 8: Power flow in an ideal directional coupler


larger than the characteristic impedance. The two most common characteristic impedances are 50Ω(commonly used to connect lab equipment) and 75Ω ( used in CATV networks ). A directionalcoupler is usually drawn on a schematic as a three port device with the internal port not shown atall.

The diagram in the top left corner of Figure 8 also shows the paths over which power can flow.The large two-headed arrow between the input and output ports indicates that a voltage applied toeither the input or output port will cause power to flow through the device and produce a voltageat the other port. The little two-headed arrow between the input port and coupled port indicatesthat if a voltage is applied at one of those ports power will flow to the other. The double-headedarrow between the input and coupled ports is drawn thinner to indicate less power flows in thispath than the path indicated with the large two-headed arrow. This means if a voltage is appliedto the input port most of the power flows over the big arrow path to the output port and less powerflows over the small arrow path to the coupled port.

Notice there is no path for power to flow between the output and coupled ports. It is the absenceof this path that makes the device a directional coupler.

It must be emphasized that a voltage source can be applied to any of the external ports, includingthe port called the output port. This does not fit with normal naming convention where it is notpermissible to apply a voltage to the output of a device. However, for a directional coupler it isperfectly normal to apply a voltage source to its output port. To avoid the confusion the outputport probably should have been called something else. Perhaps it is for this reason the output portis sometimes referred to as the through port.

Another noteworthy point is directional couplers are designed for use on cables with a specificcharacteristic impedance. The coaxial cables and therefore the directional couplers used in labora-tories usually have characteristic impedances of Zo = 50 Ω since the source impedances of signalgenerators and spectrum analysers are generally Zo = 50 Ω. This means if two of the three portsare terminated with Zo = 50 Ω, the impedance looking into the third port is also Zo = 50 Ω.

Theoretically, a directional coupler is lossless, which means all the power entering the input portleaves through the other two ports. However, practical directional couplers have internal thermalloss and some of the power is converted to heat in the device. Therefore, the power entering thedirectional coupler through the input port is equal to the sum of three powers: the power leaving asheat and the power leaving through the output and coupled ports. The power equation becomes:

Pinput = Poutput + Pcoupled + Pthermal

A directional coupler is designed to transfer a pre-specified fraction of the power it receives inits input port to the load connected to its coupled port, when the impedance of that load equalsthe characteristic impedance. That fraction of power is specified as the coupled loss in dB. It isgiven by

coupled_loss = 10 log(Pinput/Pcoupled ).

For example, if a directional coupler is specified to have a 20 dB coupled loss, then the powertransferred from its input port to the cable or termination resistor connected to its coupled portwill be 20 dB below the power flowing through its input port. In linear terms the power flowingthrough the coupled port to its load will be Pcoupled = Pinput × 10−20/10 = Pinput/100. This is 1% ofthe input power.


m

n

m

n

input port

coupled port isolated port

output orthrough port

(internal port)

Figure 9: One of many possible schematic diagrams for a broad-band CATV directional coupler

The power that flows through the output port to its load is the power left after the powerconverted heat and the power leaving through the coupled port are subtracted. The loss in thetransfer of power from the input port to the output port is referred to as insertion loss. Insertionloss, in units of dB, is given by

insertion_loss = 10 log(Pinput/Poutput ).

Since thermal loss is implementation specific, the insertion loss is specified in the data sheet for thedirectional coupler.

The limitation of practical directional couplers is not the thermal loss, but rather the powerthat leaks between the output and coupled ports. Ideally there is no leakage and the coupled portis completely isolated from the output port. The leakage, measured in dB, is the loss in dB that asignal applied to the output port experiences in traversing the leakage paths to the coupled port.Ideally this leakage, which is referred to as isolation, is infinite, but in practice it is not. For thedirectional coupler to be useful the isolation must be much greater that the coupled loss. Forexample, if the isolation was equal to the coupled loss, the same power would reach the couple portwhether it entered the input port or output port. In this case the device would have no directivity.The difference between the isolation in dB and coupled loss in dB is called the directivity of thedirectional coupler. A directivity of 30 dB or more is considered quite good.

Directional couplers can be built in many different ways. At microwave frequencies directionalcouplers can be implemented on printed circuit boards by placing a track that is connected tothe coupled port in proximity to, but not touching, the track that connects the input and output.In CATV applications where the operating frequencies are below 1 GHz, directional coupler areusually implemented by two (sometimes three) tiny transformers. One arrangement of transformersthat implements a directional coupler is shown in Figure 9. Two pictures showing the inside of adirectional coupler manufactured by Mini Circuits is shown in Figure 10. The ruler in the pictureon the left is marked in centimeters so the box is about 3cm x 3cm x 1cm. The picture on the rightis a zoomed in view. It shows two very small transformers. The termination resistor for the isolatedport can not be seen in either picture as it is mounted on the bottom of the circuit board.

The directional couplers used on the main cable in a CATV system are custom chosen for eachsubscriber. The coupling loss is chosen to place the signal power at the input to the subscriber’s cablemodem within pre-specified limits. For example, consider the abbreviated CATV system shown inFigure 7. Suppose the head end transmits at a power of 40 dBmV for a 6 MHz channel and cable


Figure 10: An inside look at a direction coupler with 20 dB coupled loss manufactured by MiniCircuits


Table 1: Typical values for a set of CATV directional couplers (Zo = 75Ω) offered by one manu-facturer.

3 6 9 12 20 24 27 3016

3.6 2.5 1.4 0.8 0.8 0.7 0.7 0.7 0.7

30 26 28 30 32 33 25 38 38

insertion loss (dB)

isolation(dB)

loss (dB)coupled

modemsubscriber

HeadEnd

40dBmV

modemsubscriber

modemsubscriber

5dB

0.7dB

24dB5.3dBmV

0.3dBmV

5dB

5dB27dB

0.7dB

8dBmV

5dB

0.8dB

16dB

22.8dBmV

7.6dBmV

2.6dBmV

29.3dBmV 23.6dBmV28.6dBmV34.3dBmV

5dB 5dB

35dBmV

3dBmV

Figure 11: Schematic of example CATV system annotated to show coupling loss (bold blue),insertion loss (blue), cable loss (blue) and signal levels (black).

modems are designed to receive that signal at a power level between 0 and 5 dBmV. For ease ofcalculation, also suppose that that each cable segment has a loss of 5 dB. Once this informationis known the coupling loss in a directional couplers can be calculated. Then, a directional couplerwith a coupling loss close to the calculated value can be chosen from the ones that are commerciallyavailable.

Manufactures sell directional couplers in families that have the same basic design and are man-ufactured in the same way, but each member of the family has a different coupling loss. Typically aproduct family has coupling losses going from 3 dB to 30 dB in 3 dB steps. Table 1 shows a familyof directional couplers offered by one manufacturer.

The choice of directional couplers for system in the example currently under consideration isgiven in Figure 11. The losses, which have units of dB, are shown in blue font with the couplingloss being bold blue font and the signal levels, which have units dBmV, are shown in black font.The system is designed starting with the subscriber closest to the head end. The coupling loss ischosen to place the signal at the modem in the pre-specified range. After the directional coupler ischosen its insertion loss is known and the signal level at the input to the next directional couplercan be computed. It is then possible to chose the second directional coupler and so on.

The CATV network can also transport power from the subscriber’s modem to the head end.The transport loss between a modem and the head end is the same regardless of the direction. Forexample, if the most distant modem from the head end in Figure 11 transmits a 40 dBmV signal,that signal will travel down the drop cable to the coupled port of its directional coupler losing 5 dBof power along the way. The remaining power, which is 40− 5 = 35 dBmV, flows into the coupled


LPF

HP

F

LPF

HP

F

LPF

HP

F

HPF

LPF

head end

subscriber’s modems

Figure 12: Schematic showing how filters could be used to prevent the transmitter from saturatingthe LNA in the receiver

port and splits with most of the power being transferred to the isolated port terminated inside thedirectional coupler and some of the power being transported to the input port. The power leavingthrough the input port is 16 dB, i.e. coupling loss, below the power flowing into the coupled port.Therefore the power leaving the input port is 35−16 = 19 dBmV. The signal then travels down themain cable through three sections of cable and two directional couplers to the head end. Along theway is loses 16.4 dB power, which is comprised of 5 dB of cable loss, followed by 0.7 dB of insertionloss, followed by another 5 dB of cable loss, followed by another 0.7 dB of insertion loss and finallyfollowed by another 5 dB of cable loss. The signal level at the head end is 19− 16.4 = 2.6 dBmV.This is the same level received by cable modem when the head end transmitted 40 dBmV.

CATV systems support two way communication using frequency division. The spectrum ispartitioned into two bands: A low frequency band for transmitting from the cable modems tothe head end (called upstream transmission) and a high frequency band for transmitting from thehead end to the cable modems (called downstream transmission). In theory there is no problemat all with such a system, but in practice, even though the transmitted signal occupies a differentfrequency band than the received signal, it is has much more power than the received signal andwill saturate the receiver’s low noise amplifier.

In theory, the receiver’s low noise amplifier can be shielded from the high powered transmittedsignal by placing low-pass and high-pass filters on the inputs and outputs of the amplifiers asillustrated in Figure 12.

The difficulty in this solution is matching the impedance of the two filters to the characteristicimpedance of the cable. The impedance the cable sees is that of the input impedance of one filterin parallel with the output impedance of the other. It is very difficult if not impossible to designtwo filters so that the input impedance of one in parallel with the output impedance of the other isa real constant.

The solution is to use another directional coupler. This directional coupler is designed so thatthe coupled loss is exactly the same as the insertion loss, which is 3 dB plus half the thermal loss.Directional couplers designed to split power equally between two ports are called hybrid couplers.They are also referred to as 2-wire to 4-wire interfaces. A block diagram of a CATV system that


subscriber’s modems

HP

F

LPF

HP

F

LPF

HP

F

LPF

head end

HPF

LPF

hybidcoupler

2−wire side

4−wire side

leakagepath

Figure 13: Schematic diagram for a two way CATV system

supports two-way communication is given in Figure 13. The power propagating down the cabletoward the modem enters the 2-wire side of the hybrid coupler. The hybrid coupler splits the powerand delivers half to each of the two ports on the 4-wire side of the hybrid coupler. Obviously thepower delivered to the transmit port is consumed by the output resistance and therefore wasted.

The power entering the hybrid coupler from the transmit port on the 4-wire side is split withhalf that power being delivered to the cable and the other have being delivered to the internallyterminated port. In theory, the two ports on the 4-wire side are completely isolated, but in practicethere is a leakage path that couples the transmit port to the receive port. In practice the leakageis about 30 dB below the transmit level, but that by itself is not enough to prevent the LNA fromsaturating. Therefore, it is still necessary to precede the LNA with a filter.

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Transmission over Coaxial Cable Notes for EE456 · PDF fileTransmission over Coaxial Cable Notes for EE456 University of Saskatchewan ... To analyze these transmission line equations: