Transmission over Coaxial Cable Notes for EE456 University of - - PDF document

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 signals

• ver a medium of some sort to a receiver. The difference between the two systems is that one

embeds an analog message signal in the analog radio frequency signal while the other embeds digital information. For example commercial AM radio embeds an analog message signal into its radio frequency signal by amplitude modulating a carrier. Therefore, commercial AM is an analog communication system. On the other hand the radio frequency of a wireless router is modulated with digital data. Therefore, a computer linked to a router via WiFi is a digital communication system. Communications systems, whether analog or digital, use coaxial cable, or some other conduit like a micro strip transmission line, to get the radio frequency signal from one point to another. For example, a coaxial cable could be used in a wireless system to get the radio frequency (RF) signal from the high power amplifier to an antenna. Coaxial cable is also commonly used to connect the RF signal to test equipment while debugging or testing a communications system. Perhaps the best example is the role of coaxial cable in a cable TV distribution system, where it carries the radio signal from the transmitter to the receiver. In the notes to follow the properties and behavior of coaxial cables will be explored with a view to give the students a very good understanding at a macroscopic level. The detailed transmission line model that used to derive the transmission line equations is clearly presented, but the model is not analysed with calculus to obtain the equations. The resulting equations of this somewhat tedious analysis are just stated. These notes attempt to interpret and give meaning to the resulting equation 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 used to divert some of the power flowing inside a coaxial cable to a test instrument or a device such as a 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 electrical signals can be modelled as a transmission line. For example very long power lines are modelled as transmission lines as are very short mircrostrip lines, which is simply a track on a printed circuit board 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 conductor is called the sheath and the inner conductor is called the center wire. The material used for the solid insulator has a low dielectric constant and very low dielectric losses. An illustration of a coaxial cable 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 as the cascade of an infinite number of cable segments of length dx with each segment being modelled with lumped circuit elements. Each of the conductors is modelled as an inductor with inductance ℓ in series with a resistor with resistance r. As the conductors are in proximity to each other they are linked by a capacitor with capacitance c. The insulator separating the conductors is certain to

2 MODELLING A COAXIAL CABLE 3

v(t,0) x distance in meters center conductor sheath v(t,x) g r I(t,x+dx) v(t,x+dx) r r g r c c dx x x+dx I(t,0) I(t,0) I(t,x) I(t,x+dx) I(t,x) ℓ ℓ ℓ ℓ

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 case the 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 indicated in Figure 1, the values of the elements must also be infinitesimal. The infinitesimal values are ℓ = (L/2)dx, r = (R/2)dx, c = Cdx and g = Gdx, where L, R, C and G are the inductance per meter, resistance per meter, capacitance per meter and conductance per meter, respectively. I.e. L and R are the inductance and resistance of 1 meter of cable measured from the source end with the load end shorted and C and G are the capacitance and conductance of 1 meter of cable measured from the source end with the load end open. The conductance, G, models the losses in the dielectric. The loss per meter in the model is the square of the voltage across the dielectric insulator times G. Many if not all of the molecules in a dielectric with a relative dielectric constant greater than 1 are dipoles (or become dipoles in the 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, no mater how slight, bumps/rubs other molecule causing them to vibrate and increase their kinetic

• energy. This molecular kinetic energy is also referred to as heat. Since energy is conserved, the

kinetic energy (i.e. heat) comes from the energy in the electric field. Each time the polarity of the electric field changes, the molecules are rotated and energy is converted from the electric field to

• heat. Therefore, the power converted from the electric field to heat is proportional to the frequency
• f the alternating electric field. This means the conductance G is the function of frequency given

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

2 MODELLING A COAXIAL CABLE 4 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 results

• f the analysis for a finite length cable that is terminated with a load impedance ZL are presented

in 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 of time and position on the cable. For purposes of analysis the voltage of the center conductor w.r.t the sheath a distance x from the voltage source is denoted v(t, x) as illustrated in Figure 1. The current in both the center conductor and sheath at a distance of x from the source is similarly denoted I(t, x). Perhaps unnecessary, but it is pointed out the current I(t, x) differs from that of I(t, x + dx) due to the leakage paths through the capacitor with capacitance Cdx and the resistor with conductance Gdx. Using calculus as well as Kirchoff’s voltage and current laws produces the transmission line equations: −∂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 is

a constant with units volts, F is a constant with units Hz and φo is a constant with units

• radians. 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 the

transients that result from turning on the source have died out. The analysis assumes the source 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 and R = 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 a phasor and characterize the sinusoid by the complex amplitude of the phasor at t = 0. A phasor is a complex function of time of the form Aej(2πFt+φ) = Aejφej2πFt. It is a vector in the complex plane that has a constant magnitude, which is A, and spins with a constant angular velocity, which is 2πF in units of radians/second. Its complex amplitude at time t = 0, which is Aejφ, is usually referred to as its complex amplitude without stating that it is the amplitude at t = 0. The real part of the phasor is the cosine A cos(2πFt + φ). The magnitude of the complex amplitude, which is A, is the amplitude of the cosine and the angle of the complex amplitude, which is φ, is the phase

• f the cosine at time t = 0. The frequency of the phasor, which is its angular velocity, is not a

parameter in the complex amplitude.

2 MODELLING A COAXIAL CABLE 5

real imaginary phasor at time spinning rate position of complex amplitude v(x) = Axejφx φ(x) = ∠ v(x) = 2πF rad/sec Ax cos(φx) Ax sin(φx) 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 conductor w.r.t. to sheath at any point on the cable will also be a sinusoid. The sinusoidal voltage at a distance

• f x from the source will have the form v(t, x) = Ax cos(2πFt + φx). It is easy to understand that

the amplitude and phase of the sinusoid will depend the distance from the source, which is x, so the amplitude and phase are subscripted with an x. The phasor representing v(t, x) for x = x1 is the 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 position

• f one or more phasors at time t = 0 with the understanding that all phasors in the diagram are

spinning with the same angular velocity. This particular diagram has only one phasor. The real part 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 y co-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 their

• sum. Of course the sinusoids must have the same frequency otherwise they can not be represented
• n 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 vector sum is illustrated in Figure 3. Vectors v1 and v2 are the complex amplitudes of the two sinusoids being 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 to follow in spots. To avoid the effort in following the development of the equations the results are presented without derivation. To better understand the final results the analysis is broken into steps 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.

2 MODELLING A COAXIAL CABLE 6 real imaginary 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 x from 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 = ℜ{γ} and β = ℑ{γ}. 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 the cable and the factor e−jβx provides the phase shift experienced by that same voltage in travelling the same distance. That phase shift is −βx. For practical coaxial cables where the frequency of the sinusoidal source is within the operational range 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)

2 MODELLING A COAXIAL CABLE 7

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 commercially available coaxial cables. It provides a reasonable approximation and certainly predicts the trend of how attenuation increases with frequency. The delay a sinusoidal voltage experiences as it travels along a cable is related to the phase shift it experiences. For example, a sinusoid that experiences a delay of τ can be expressed as sin( 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 has it 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 sinusoidal voltage as it travels along the cable depends on its frequency, but its velocity of propagation does not. The values for L and C do not significantly depend on frequency and can be calculated using the geometry of the cross-section of the coaxial cable along with the real parts of the relative permeability and relative permittivity of the dielectric material used as the insulator between the center conductor and the sheath. For the parameters given in Figure 4 the values of L and C are given by L = µrµo 2π ln(D d ) and C = 2πǫrǫ0 ln( D

d ) ,

(9)

2 MODELLING A COAXIAL CABLE 8 where µr and ǫr are the real parts of the relative permeability and relative permittivity of the insulator 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 also be derived from the transmission line equations. The complex amplitude of the current is found to be 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 frequencies near F = 0. The complex constant Zo relates the complex amplitudes of voltage and current in the same way 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 the
• perating 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 accurate approximation 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.

2 MODELLING A COAXIAL CABLE 9

V(0) real imaginary 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 be

frequencies below about 0.1 MHz, three of the four terms in Zo get smaller, but the fourth term, which is R approaches RDC. The effect on Zo is for it to increase magnitude and no longer remain real as the frequency decreases to zero. Having equations for the complex amplitudes of voltage and current as a function of x is only

• f value if they can be used in the design and analysis of systems that use coaxial cables. Normally

coaxial cables are used in communication systems to channel power from one circuit or device to another, e.g. from a transmitter to an antenna. To utilize the equations for the complex amplitudes

• f 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 operating range, 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 does not represent a practical situation. The situation being analysed here is a cable of infinite length driven by a voltage source. Power flows from the source into the cable and flows down the cable at a velocity of vp, but never reaches a load. Figure 5 shows the scenario just mentioned as well as the signals relevant to power flow calculation. The voltage source, which is shown at the top of Figure 5, drives the center conductor of the

• cable. 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 of the 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

2 MODELLING A COAXIAL CABLE 10 horizontal arrow under I(0). The average power delivered by the voltage source to the cable is Pave = 1 T T v(t, 0)I(t, 0)dt = 1 T T

• 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

• v(0)
• I(0)
• cos2( 2πFt + ∠ v(0) )dt

= 1 T T

• 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 complex amplitudes of the voltage and current, i.e. v(x) and I(x), are shown in the phasor diagram on the left-center of Figure 5. The relationship between the phasors and the time waveforms is shown on the right-center of Figure 5. The instantaneous power, which is v(t, x)I(t, x), as well as the average power are plotted on the bottom right of Figure 5. From the plot of the time waveforms v(t, x) and I(t, x) it clear that their phase shifts, which are the angles of v(x) and I(x), while dependent on x, 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 section at x does not depend on the angles of v(x) and I(x). It also shows the average power that passes through 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 any

x will be in the direction of propagation of the sinusoidal voltage that causes it when that sinusoidal voltage 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

2 MODELLING A COAXIAL CABLE 11 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. Since I(t, x) = v(t, x)/Ro for all x, the direction of the current flow when the voltage is positive is the same 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 for a given voltage is inversely proportional to the characteristic impedance. For example, cables in cable TV network have a 75 Ω characteristic impedance, if they had been 50 Ω cables lower voltages could have been used to deliver the same power.

Example:

Suppose the infinite length line of Figure 1 has a characteristic impedance 75∠0 Ω (cable is used in a CATV 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 is vp = 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 5

the voltage and current are expressed in terms of their complex amplitudes. Find v(0) and I(0) as shown in Figure 5. Also find the average power using v(0) and I(0).

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

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∠ 0 Ω I(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.

2 MODELLING A COAXIAL CABLE 12

• 3. The average power is the instantaneous power averaged over one period of the sinusoid. Since

F = 106 Hz, T = 1/F = 10−6 s. Pave = 1 T T 0.75 cos2(2π × 106 t + π/3) W dt = 1 T T 0.75 2

• cos(0) + cos(4π × 106 t + 2π/3)
• W dt

= 0.375 W.

• 4. The complex amplitude of the voltage applied to the cable is v(0) = 7.5∠ π/3 V. The complex

amplitude of the current entering the cable is I(0) = 0.1∠ π/3 A. The average power entering the cable is Pave(0) =

• v(0)
• I(0)
• 2

= 7.5 V 0.1 A 2 = 0.375 W.

• 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)(100 m)

=

• v(0)
• × 0.368

= 2.76 V. Now find β x. From the paragraph that precedes (7) delay or travel time is related to phase shift by τ = −θ/(2πF). In this case θ = β x. That same paragraph also explains that the velocity of propagation is the ratio of distance travelled to travel time. Therefore, τ is also equal 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 × 100 m 2 × 108 m/s = −π radians Therefore v(100 m) =

• v(100 m)
• ∠ (∠v(0) + β x) V

= 2.76∠(π/3 − π) V = 2.76∠ − 2π/3 V

2 MODELLING A COAXIAL CABLE 13

distance in meters rep 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(100 m) = v(100 m) Zo = 2.76∠ − 2π/3 V 75∠0 Ω = 0.0368∠ − 2π/3 A The average power can now be calculated and is Pave(100 m) =

• v(100 m)
• I(100 m)
• 2

= (2.76 V)(0.0368 A) 2 = 0.0505 W.

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 a voltage source that has output impedance Zs and is terminated with a load (i.e. a circuit) that has an impedance of ZL. This situation is illustrated in Figure 6, where the output impedance of the voltage source is shown as a resistor, but it could have a capacitive or inductive component, and the load is a resistor in parallel with a capacitor making the load impedance for the circuit shown ZL = RL + 1/(j2πFCL). The presence of the load places boundary conditions on the transmission line equations and makes the analysis more complicated than for the case of the infinite length line. When the boundary conditions are included the analysis shows that when the forward propagating voltage reached the end of the cable where the load is connected some of it is reflected and propagates in the opposite direction, which is back toward source. In preparation for analysing a cable where voltages can be reflected and thereby have two voltages propagating in different directions it is necessary to modify the notation to include the direction of propagation. A sinusoidal voltage propagating in the forward direction, which is from source to load, as well as it complex amplitude will be superscripted with a + and a sinusoidal voltage propagating in the reverse direction, which is from load to source, as well as its complex amplitude will be superscripted with a −. This means v+(x) is the complex amplitude for the

2 MODELLING A COAXIAL CABLE 14 forward propagating sinusoidal voltage v+(t, x) and v−(x) is the complex amplitude for the reverse propagating sinusoidal voltage v−(t, x). Since the current generated by a propagating voltage will flow in the direction of propagation while that voltage is positive, the current generated by v−(t, x) while it is positive will be in the

• pposite direction to the current generated by v+(t, x) while it is positive. For this reason it is

necessary to symbolize current in a way that makes it clear which direction the voltage that generates it 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 positive current is arbitrarily defined as the direction of the current generated while a forward propagating sinusoidal voltage is positive. With this definition of positive current I+(t, x) = v+(t, x)/Ro and I−(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 from the source is v+(x) + v−(x) then the complex amplitude of the current in the center at the same point 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 the voltage across load and the current though the load can now be calculated. Trusting that including boundary conditions in the analysis of the transmission line equations (not done here) indicates that some of the forward propagating voltage gets reflected at the load, the amount of voltage reflected at the load can be found by balancing the currents at the node where the load is attached to the center conductor of the coaxial cable. Defining the direction of positive current through the load as being from the center conductor to sheath as shown in Figure 6, the complex amplitude of the current through the load must be IL = 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 complex amplitudes of the currents generated by v+(xL), which is the complex amplitude of the forward propagating voltage, and v−(xL), which is the complex amplitude of the voltage reflected from the load, respectively. The voltage across the load is the voltage of the center conductor w.r.t. the sheath at end of the cable, i.e. at x = XL. Therefore, the complex amplitude of the sinusoidal voltage across the load 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)

2 MODELLING A COAXIAL CABLE 15 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 of

• F. ΓL relates the amplitude and phase of the reflected sinusoidal voltage to those of the forward

propagating sinusoidal voltage. The reflected sinusoidal voltage at x = xL has

• ΓL
• times the

amplitude of the forward propagating sinusoidal voltage and is phase shifted by ∠ ΓL w.r.t. the forward travelling voltage. A few load impedances and their associated reflection coefficients are given below. The values for 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 voltage

is reflected.

• 2. For a short circuit the load impedance is ZL = 0 and ΓL = −1. In this case the reflected

voltage has the same amplitude as the forward propagating voltage, but is 180 degrees out of phase.

• 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 amplitude

as 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 propagating voltage 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 the sinusoidal voltage across the load and vs is the complex amplitude of the sinusoidal voltage of the

• source. This involves finding v+(0) in terms of vs and v−(0), where v+(0) is the forward propagating

2 MODELLING A COAXIAL CABLE 16 voltage voltage in the cable at x = 0 and v−(0) is what is left of the reverse propagating voltage reflected 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 node where the source resistor connects to the center conductor of the cable are balanced. The resulting equation 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 for v+(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 for vL 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 5 equations with 5 unknowns. Solving these equations for vL produces vL = 1 + ΓL 1 − ΓsΓL e−γxL Zo Zs + Zo vs. The other parameter that is of value is the input impedance at the source end of a terminated finite length cable. The input impedance of an infinite length cable is Zo, but the reflection from the load end of a finite length cable changes the input impedance. The input impedance is defined as the ratio of the voltage of center conductor w.r.t. the sheath divided by the current flowing in the 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 the source, 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 as

e−2αxLe−j2βxL. Since practical cables are lossy, α is greater than 0. This means e−2αxL → 0 and 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 is short and βxL = π/2, then e−j2βxL = −1 and e−2γxL ≈ −1. In this case the numerator in (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 cable is short and βxL = π/2 as it was in the previous case, then again e−j2βxL = −1 and e−2γxL ≈ −1. However, in this case ΓL = −1 so the denominator in (24) becomes very small 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 the input signal is measured by the ratio of signal power to noise power, i.e. SNR, where the signal power 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 signal power from one device to another. For example the modulation/demodulation scheme is designed to maximize the efficiency of transporting power across a specific channel, which, for example, could be 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 in a large number of applications, but only one is addressed in this section. It is the distribution of signal power from the head end of a CATV system to the modems in the customer premises over coaxial 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 nuclear reactions, energy can only be converted from one form to another. The term “power delivered to a load” 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, and similarly 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 flow through it. The difference is the current is phase aligned with the voltage when that element is a resistor 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”.

3 POWER FLOW IN COAXIAL CABLES 18 the current through an element is in quadrature to the sinusoidal voltage across it, the product is a sinusoid that has no DC value. However, if the current was phase shifted to be aligned with the voltage then the product would have a DC value and that DC value is referred to a imaginary power. The power delivered to the resistive elements in a load, which is sometimes referred to as real power to better distinguish it from the concocted imaginary power, is the time average value of the product of the voltage across the load and the component of current that is phase aligned with the voltage. Coaxial cable transports power from a source to a load in the form of a forward propagating electromagnetic wave. The underlying physics of the propagating electromagnetic wave can be loosely viewed as a bucket brigade consisting of the capacitors and inductors illustrated in Figure 1 that take energy from the source and pass it along the line to the load. At the load end of the cable some of the power in the forward propagating electromagnetic wave is converted to heat by the load and the rest of the power is reflected back toward the source as a reverse propagating electromagnetic wave. Again using the loose physical analogy, the energy in the reverse propagating electromagnetic 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 impedance

• ZL. Both ways are straight forward, but one way leads to a more useful equation. The way that

finds the less useful equation expresses the power in terms of the voltage across ZL, i.e. vL, and the current through ZL, i.e. IL. To get the equation for power delivered the current is separated into quadrature components with one of the components being phase aligned with the voltage across the load. In doing this the amplitude of the component of current phase aligned with the voltage is

• vL

ZL

• 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 must be the difference between the forward propagating and reverse propagating power at x = xL. This has 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 and

confidence in the equations. First suppose that ZL = ∞. Then ΓL = 1 and both equations indicate no power is absorbed by the load. This confirms what is obvious since there is no load. Next suppose that ZL = 0. If ZL = 0 the load voltage is zero regardless of the current through it and no power is delivered to the load. Since VL = 0 the first equation indicates no power is delivered to the load

3 POWER FLOW IN COAXIAL CABLES 19 and since ΓL = −1 the second equation indicates no power is delivered to the load. Next suppose that the load is a capacitor. Since a capacitor does not convert electrical power to heat obviously no power is delivered to the load. This is verified by both equations since cos(∠ZL) = cos(π/2) = 0 and

• 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 imprecise components the input impedance will vary from device to device and will not exactly equal the characteristic impedance. The quality of these devices are measured by how closely their input impedances match the characteristic impedance. This quality is measured through a characteristic called return loss. Return loss is the ratio of power carried to the load by the forward propagating voltage to the power carried away from the load by the reverse propagating voltage. Return loss is usually 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 RL for return loss creates some ambiguity since it also represents the product of R and L, which are resistance and inductance per unit length. It will be clear from the context in which RL is used when 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 coaxial cable 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 of the 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 modems it 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 cable in the forward direction to a subscriber. A directional coupler is a four-port device converted to a three port device by terminating the fourth port inside the device. An ideal three port directional coupler is illustrated in Figure 8. The diagram in the top left corner of Figure 8 shows three external ports named: input port, thought port and coupled port. The through port is also called the output port. The diagram shows one internal port that is terminated in a 50Ω resistor. This usually implies the device is designed for use on cables with a characteristic impedance of 50 Ω, but sometimes the device is designed in a way that the internal port must be terminated in an impedance

3 POWER FLOW IN COAXIAL CABLES 20

modem subscriber modem subscriber modem subscriber Head End coupled port taps ( i.e. directional couplers ) drop cable through port port main cable input

Figure 7: Simplified distribution system for CATV

50Ω 50Ω 50Ω V1 (c) voltage applied to coupled port 50Ω 50Ω 50Ω V1 (b) voltage applied to ouput port 50Ω 50Ω 50Ω (a) voltage applied to input port V1 50Ω coupled port

• utput

port also called the through port input port internally terminated power flow map and port definition for a directional coupler

Figure 8: Power flow in an ideal directional coupler

3 POWER FLOW IN COAXIAL CABLES 21 larger than the characteristic impedance. The two most common characteristic impedances are 50 Ω (commonly used to connect lab equipment) and 75 Ω ( used in CATV networks ). A directional coupler is usually drawn on a schematic as a three port device with the internal port not shown at all. 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 to either the input or output port will cause power to flow through the device and produce a voltage at the other port. The little two-headed arrow between the input port and coupled port indicates that if a voltage is applied at one of those ports power will flow to the other. The double-headed arrow between the input and coupled ports is drawn thinner to indicate less power flows in this path than the path indicated with the large two-headed arrow. This means if a voltage is applied to the input port most of the power flows over the big arrow path to the output port and less power flows 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 absence

• f 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, including the port called the output port. This does not fit with normal naming convention where it is not permissible to apply a voltage to the output of a device. However, for a directional coupler it is perfectly normal to apply a voltage source to its output port. To avoid the confusion the output port probably should have been called something else. Perhaps it is for this reason the output port is sometimes referred to as the through port. Another noteworthy point is directional couplers are designed for use on cables with a specific characteristic 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 signal generators and spectrum analysers are generally Zo = 50 Ω. This means if two of the three ports are 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 port leaves through the other two ports. However, practical directional couplers have internal thermal loss and some of the power is converted to heat in the device. Therefore, the power entering the directional coupler through the input port is equal to the sum of three powers: the power leaving as heat 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 in its input port to the load connected to its coupled port, when the impedance of that load equals the characteristic impedance. That fraction of power is specified as the coupled loss in dB. It is given by coupled_loss = 10 log( Pinput/Pcoupled ). For example, if a directional coupler is specified to have a 20 dB coupled loss, then the power transferred from its input port to the cable or termination resistor connected to its coupled port will be 20 dB below the power flowing through its input port. In linear terms the power flowing through the coupled port to its load will be Pcoupled = Pinput × 10−20/10 = Pinput/100. This is 1% of the input power.

3 POWER FLOW IN COAXIAL CABLES 22

m n m n input port coupled port isolated port

• utput or

through 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 power converted heat and the power leaving through the coupled port are subtracted. The loss in the transfer of power from the input port to the output port is referred to as insertion loss. Insertion loss, 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 the directional coupler. The limitation of practical directional couplers is not the thermal loss, but rather the power that leaks between the output and coupled ports. Ideally there is no leakage and the coupled port is completely isolated from the output port. The leakage, measured in dB, is the loss in dB that a signal 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 the directional coupler to be useful the isolation must be much greater that the coupled loss. For example, if the isolation was equal to the coupled loss, the same power would reach the couple port whether 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 the directional coupler. A directivity of 30 dB or more is considered quite good. Directional couplers can be built in many different ways. At microwave frequencies directional couplers can be implemented on printed circuit boards by placing a track that is connected to the 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 are usually implemented by two (sometimes three) tiny transformers. One arrangement of transformers that implements a directional coupler is shown in Figure 9. Two pictures showing the inside of a directional coupler manufactured by Mini Circuits is shown in Figure 10. The ruler in the picture

• n the left is marked in centimeters so the box is about 3cm x 3cm x 1cm. The picture on the right

is a zoomed in view. It shows two very small transformers. The termination resistor for the isolated port 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 each

• subscriber. The coupling loss is chosen to place the signal power at the input to the subscriber’s cable

modem within pre-specified limits. For example, consider the abbreviated CATV system shown in Figure 7. Suppose the head end transmits at a power of 40 dBmV for a 6 MHz channel and cable

3 POWER FLOW IN COAXIAL CABLES 23 Figure 10: An inside look at a direction coupler with 20 dB coupled loss manufactured by Mini Circuits

3 POWER FLOW IN COAXIAL CABLES 24 Table 1: Typical values for a set of CATV directional couplers (Zo = 75 Ω) offered by one manu- facturer.

3 6 9 12 20 24 27 30 16 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

modem subscriber Head End 40dBmV modem subscriber modem subscriber 5dB

0.7dB

24dB 5.3dBmV 0.3dBmV 5dB 5dB 27dB

0.7dB

8dBmV 5dB

0.8dB

16dB 22.8dBmV 7.6dBmV 2.6dBmV 29.3dBmV 23.6dBmV 28.6dBmV 34.3dBmV 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 of calculation, also suppose that that each cable segment has a loss of 5 dB. Once this information is known the coupling loss in a directional couplers can be calculated. Then, a directional coupler with a coupling loss close to the calculated value can be chosen from the ones that are commercially available. 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 a product family has coupling losses going from 3 dB to 30 dB in 3 dB steps. Table 1 shows a family

• f directional couplers offered by one manufacturer.

The choice of directional couplers for system in the example currently under consideration is given in Figure 11. The losses, which have units of dB, are shown in blue font with the coupling loss 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 is chosen to place the signal at the modem in the pre-specified range. After the directional coupler is chosen its insertion loss is known and the signal level at the input to the next directional coupler can 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. For example, 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 dB

• f power along the way. The remaining power, which is 40 − 5 = 35 dBmV, flows into the coupled

3 POWER FLOW IN COAXIAL CABLES 25

LPF HPF LPF HPF LPF HPF

HPF LPF

head end subscriber’s modems

Figure 12: Schematic showing how filters could be used to prevent the transmitter from saturating the LNA in the receiver port and splits with most of the power being transferred to the isolated port terminated inside the directional coupler and some of the power being transported to the input port. The power leaving through 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 the main cable through three sections of cable and two directional couplers to the head end. Along the way is loses 16.4 dB power, which is comprised of 5 dB of cable loss, followed by 0.7 dB of insertion loss, followed by another 5 dB of cable loss, followed by another 0.7 dB of insertion loss and finally followed 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 is partitioned into two bands: A low frequency band for transmitting from the cable modems to the head end (called upstream transmission) and a high frequency band for transmitting from the head end to the cable modems (called downstream transmission). In theory there is no problem at all with such a system, but in practice, even though the transmitted signal occupies a different frequency band than the received signal, it is has much more power than the received signal and will saturate the receiver’s low noise amplifier. In theory, the receiver’s low noise amplifier can be shielded from the high powered transmitted signal by placing low-pass and high-pass filters on the inputs and outputs of the amplifiers as illustrated in Figure 12. The difficulty in this solution is matching the impedance of the two filters to the characteristic impedance of the cable. The impedance the cable sees is that of the input impedance of one filter in parallel with the output impedance of the other. It is very difficult if not impossible to design two filters so that the input impedance of one in parallel with the output impedance of the other is a real constant. The solution is to use another directional coupler. This directional coupler is designed so that the 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

3 POWER FLOW IN COAXIAL CABLES 26

subscriber’s modems

HPF LPF HPF LPF HPF LPF

head end

HPF LPF

hybid coupler 2−wire side 4−wire side leakage path

Figure 13: Schematic diagram for a two way CATV system supports two-way communication is given in Figure 13. The power propagating down the cable toward the modem enters the 2-wire side of the hybrid coupler. The hybrid coupler splits the power and delivers half to each of the two ports on the 4-wire side of the hybrid coupler. Obviously the power 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 with half that power being delivered to the cable and the other have being delivered to the internally terminated port. In theory, the two ports on the 4-wire side are completely isolated, but in practice there is a leakage path that couples the transmit port to the receive port. In practice the leakage is about 30 dB below the transmit level, but that by itself is not enough to prevent the LNA from

• saturating. Therefore, it is still necessary to precede the LNA with a filter.