Coaxial Architecture Notes What is a cable network / system? What are the components and functions of coaxial cable network / system architecture? Components Functions

Tree-and-Branch Architecture Express Trunk More about… Tree and Branch Reference Section, Page 4–1 Notes Trunk lines, feeder lines, and drop cables in a traditional RF plant resemble a tree’s limbs and branches. The trunk provides a transport system through numerous amplifiers from the headend(s) to customers. What are the advantages and disadvantages of tree-and-branch architectures? Advantages Disadvantages

Reference Section, Page 4–2 HFC Architecture More about… HFC Architecture Reference Section, Page 4–2 Tap node HUBS Businesses amp in building Customer Homes Notes What are the components and functions of HFC cable system architecture? Components Functions

Rural Network Other cable outputs Fiber Optic Cable RF Active Device (Node) Optics RX/TX Headend Active Device Notes Head end consolidation, expanded bandwidths, and the demand for increased reliability led to the node concept. One node for __________ homes passed. (how many) Using optical fiber, the node could be as far as ______ miles from the signal source (assume 1310 nm lasers). (how many) The signal quality at the node was virtually equal to that at the headend. It is not usual for customers to receive signals through any more than _____ amplifiers after the node in a 550 MHz environment _____ amplifiers after the node in a 750 MHz environment

Near Passive Network Two-Way Tap Four-Way Tap Eight-Way Tap 11 14 26 29 Headend or Node Optics RX/TX 29 24 21 17 Two-Way Tap LPI Four-Way Tap Notes No subscriber receives signal through any more than _____ amplifiers after the node (how many) One node for __________ homes passed. Greater reliability and access than rural type due to less return paths. In contrast to traditional coaxial cable architecture, HFC architecture includes many more head ends which are used to transmit signals to smaller nodes. Therefore, a different set of signals can be sent to each node, serving as few as 500 subscribers. Eight-Way Tap LPS Directional Coupler Splitter

Passive Network Node Two Way Tap Four Way Tap Slope Equalizer 29 26 23 20 17 17 11 Headend or Node Node Slope Equalizer Two Way Tap Four Way Tap 4 Notes All fiber optic to node converts to RF and goes through taps. One node for __________ homes passed. (how many) No subscriber receives signal through any more than _____ amplifiers after the node Benefits offered by an HFC architecture:

Sample Headend 5008ET2 Public switch Com21 Analog Video 55-550 MHz HCX comController 2 Way RF splitter RF splitter 430MHz DWDD 2RRX 1 DT 815 Amp out Hub EDFA Return TX DWDM HUB 2 3 1) Analog or Video Channels. Also the Digital Channel line up can be added from 550 to 750 2) Combined input of the Modem and Telephony Downstream Telemetry Carriers. 3) The return Data from the modems and SIU’s. From here the signals will be routed to the internet service or the Telephone POP. The next slide is of a combining network done wrong.

Connector Parts Body Boot Ferrule Notes The ST (straight tip) connector is popular for single- and multi-mode fiber connections and is considered a good connector with an average loss of about 0.5 dB. Parts of optical fiber connectors: Ferrule - center portion of connector that contains and aligns the optical fiber Body - designed to screw on, twist lock, or snap in order to make the connection Strain-relief boot - relieves tension on the optical fiber

Types of Connectors Biconic FC ST SC D4 Notes Four types of optical fiber connectors: Biconic - Older style connector, has high loss, approx. 1 dB FC (ferruled connector) - Popular for single-mode fiber, also called FC-PC. Provides low loss, approx. 0.4 dB SC (snap connector) - Modular design, loss under 0.5 is common in single-mode applications D4 - primarily used in single-mode operation

Optical Loss Example Total Optical Loss = 0.83dB Notes Determining Optical Loss: (1550 nm example) Step 1: Multiply each significant length of fiber by the attenuation rate for that specific fiber (provided by thecable manufacturer). Step 2: Add the results of step 1 to the sum of the losses for all connectors and splices. Total Optical Loss = 0.83dB

Coaxial Plant Design and Operation Optical Transmitters and Receivers Introduction Module 6 introduces the basic equipment of a fiber-optic system, their components, and their functionality. This module also discusses the units used in fiber-optic systems to measure optical power and includes a discussion on determining power budget. Outcome At the end of this module, participants will be able to identify the basic components of a fiber-optic system as well as define the units of measure used for calculating and measuring optical power.

Topics Overview Optical Transmitters Optical Receivers Units of Optical Power Power Budget

Optical Transmitter and Receiver Input Electrical Signal Reproduced Electrical Signal Optical Fiber Receiver Optical Signal Transmitter Notes What are the three main components of a fiber optic system, and how do they function? Component Function How is a fiber optic system similar to a modem?

Reference Section, Page 6-1 Optical Transmitter More about… Optical Transmitters Reference Section, Page 6-1 Drive Level Control Drive Level Test Point RF Input Optical Fiber Optical Connector Laser Notes What are the components of an optical transmitter, and how do they function? Component Function Optical transmitters are available in analog and digital versions. They are characterized by their stability and noise levels, as well as how they interact with fiber to determine total link quality. The amount of wavelength light transmitted determines the signal attenuation.

Laser Performance Curve Laser Drive Levels Bias Current Modulated Optical Output Threshold Input RF Optical Output Power Laser Performance Curve Clipped Output Match items on the left with their descriptions on the right. _______ Threshold current a. steady state current applied to the laser that yields a steady state light output _______ RF input b. distortion of the desired signal due to nonlinear performance and improper setup _______ Modulated optical output c. 1310 nm and 1550 nm _______ Clipped output d. minimum current required to produce optical output _______ Bias current e. signal variation modulated onto the optical carrier _______ Common laser wavelengths f. signal of the channels being transmitted, at an amplitude set by the drive level control Name two types of solid state lasers and their key features:

Reference Section, Page 6-2 Optical Receiver More about… Optical Receivers Reference Section, Page 6-2 Bias Voltage Pre Amp Post RF Out Test Point Fiber Optical Connector Photo Detector Notes Optical receivers recover light wave signals transmitted through the optical system and convert (demodulate) the modulated light back to the intended original signal. Match items on the left with their descriptions on the right. _______ Bias voltage a. output port for the RF signal _______ Pre Amp b. detects the optical signal _______ Post Amp c. boosts the electrical level of the received signal _______ RF input connector d. used during setup and troubleshooting _______ RF output connector e. provides electrical energy for the process of optical to electrical conversion _______ RF test point connector f. boosts signal level beyond that of the pre amp _______ Photo connector (photodiode) g. brings optical signal into the receiver

Units of Optical Power Notes In the cable industry, which optical power unit measures are used for measuring optical power, and which unit measure is used for measuring electrical power? Absolute optical power measurements can be made in either a logarithmic scale (dBm) or a linear scale (Watts). Define dBm: Define mW: What tool is used to measure absolute optical power in mW or dBm? Which tools are used to measure power loss in dB?

Optical Power Equations dBm = 10 log mW mW = inverse log (dBm/10) Notes

+/-10dB Optical Power Table Optical Power (dBm) Optical Power (mW) 30 1,000 20 100 10 10 0 1 -10 0.1 -20 0.01 -30 0.001 Notes The tables in this and the next slide depict the simple relationships which help convert between mW and dBm. Note that large mW numbers can be converted to smaller dBm numbers. Adding 10 dB is the same as multiplying mW by ________. Subtracting __________ is the same as dividing mW by 10. For 30 dBm, the mW decimal place has been shifted ______ places to the right, yielding __________ mW. For -30 dBm, the mW decimal place has been shifted ______ places to the left, yielding __________ mW.

Reference Section, Page 6–3 Power Budget Formula More about… Power Budget Reference Section, Page 6–3 P b = T p - Rin where P b = the Power Budget T p = output Power of the Transmitter and R in = required input to the receiver Notes Why is determining and evaluating the optical power budget important? The optical power budget is calculated by subtracting the ________________________ from the ___________________________.

Optical Node The next slide shows a block diagram of a 4 by 1 node System RF out Return Path Alignment - The photo above depicts the four port Diamond Net node. When dealing with reverse path alignment the main items to focus on are - Input levels to the device (whether optical or RF) - Output levels of the device. Main components which make up the node are Optical receivers, Optical transmitter, RF amplifier and optional plug in devices (such as transponder , return signal switcher, etc.) Express the need for the forward to be correct. The reverse cannot work if the forward is bad. The next slide shows a block diagram of a 4 by 1 node NOR NRT

Optical Node Operation Post Amp H L Node/Amplifier Block Diagram -20 dB Rf TP 20 dBmV Post Amp H L Light to RF Converter Pre Amp Splitter Post Amp Attenuator Level Control Tilt Generator H L RF to Light Converter to RF Post Amp H L Combiner

Forward Optical Receivers/NOR’S RF Gain adjust 9A pad Optical monitoring T/P -30 down Test Point Optical alarm

Diamond Net RF Module

Coaxial Plant Design and Operation Amplifier Technology Introduction Module 8 reviews several types of amplifier technology, their functionalities, advantages, and disadvantages. Outcome By understanding the characteristics of each amplifier type, technicians can effectively maintain and troubleshoot a system that is reliable for their customers.

Topics Semiconductor Configurations in CATV Single - Ended Amplifier Push - Pull Amplifier Parallel - Hybrid Amplifier Notes

Semiconductor Amplifier Configurations More about… Amplifiers Reference Section, Page8–1 Semiconductor Amplifier Configurations Notes Cable television’s distribution systems use amplifier stations in their trunk and feeder lines to amplify signals. The modules that amplify the signals are called ______________. The most important parts of the amplifying module are the _____________________. Amplifier stations used in trunk lines are called either _______________________ or ___________________________. Amplifier stations used in feeder lines are called ____________________________. CATV network systems have serial connections of amplifiers where the output of one feeds another. The industry refers to these connected amplifiers as a ___________. What are the most important parts of amplifying equipment, and why? What four amplifier configurations have been used in the CATV industry?

Single - Ended Amplifier 2nd Harmonic plus Noise Notes The single-ended transistor amplifier circuit remains the basic building block for the more complex amplifying circuits used today. This slide depicts a single-ended amplifier stage of an RF broadband circuit. It is more complex and requires a higher quality transistor than an ordinary single-ended circuit for amplifying audio. In the circuit shown above: Gain = Voutput / Vinput = 80Vpk pk / 2V pk pk = 40 Output signal is 180° out of phase with the input Input is an unmodulated 50 MHz carrier signal Transistor powered by power supply labeled as VCC. The peak to peak magnitude of V out can never exceed VCC. More complex, requiring a higher quality transistor than an ordinary single - ended circuit for amplifying audio. What are the benefits and drawbacks of a single-ended amplifier?

Push - Pull Amplifier Notes A push-pull amplifier is a hybrid which uses two single-ended transistor circuits in which one single ended circuit amplifies only during the positive half wave of the input signal and the other amplifies only during the negative part of the cycle. This slide depicts two single-ended transistor stages running in parallel but fed out-of-phase from an inductive splitter. Summarize how the push-pull hybrid functions: What advantages does a push-pull hybrid amplifier provide over a single-ended amplifier?

Parallel - Hybrid Amplifier Push Pull Stage Pin Pout Push Pull Stage Notes The parallel-hybrid amplifier is the most common amplifier configuration used today. It consists of two push-pull stages connected in parallel and fed in phase from an inductive splitter. Each output from the splitter goes to a separate push-pull stage. The outputs of the two push-pull stages are then re-combined into a single output signal. Summarize how the parallel hybrid amplifier functions: What advantages does a parallel hybrid amplifier provide over a single-ended amplifier and a push-pull hybrid amplifier? ADVANTAGES: High Gain and Reduced Distortions

Coaxial Plant Design and Operation Amplifier Configurations Introduction This module introduces amplifier configurations used in a cable network, discussing their functionality and common characteristics. Understanding the use and proper utilization of amplifiers in a cable network helps to maintain current network infrastructure and optimizes the delivery of services to subscribers. Outcome By understanding amplifier configurations and their uses, technicians will be able to troubleshoot network problems, ensuring a successful and reliable cable system.

Objectives Describe the most common amplifier configurations and discuss their usage. Identify the components for each of the amplifier configuration and explain their functions and importance. Notes

Forward Amplifier Characteristics Diamond Line 2 Amp has two equal or three unequal high level RF distribution outputs. Two Post amplifiers are used one for the main output and one for the secondary outputs. A plug-in port selector allows the use of one or both of the secondary ports with the use of a jumper or a coupler. The Diamond Line 2 Amp is a distribution amplifier similar in function to the Philips GNA series Global Network Amplifier.DO NOT CONFUSE WITH A GNA AMPLIFIER.

Amplifier Output Tilt 11dB of tilt @ 750 MHz Notes More about… Balancing System Amplifiers Reference Section, Page14–1 11dB of tilt @ 750 MHz Notes Why is it important to properly balance a system? What is tilt, and why is it important when balancing a cable network system? This slide depicts a typical tilt for a 750 MHz system. The tilt is typically calculated from typically 54 MHz to 750 MHz or channels 2 to 116. The highest channel must be an analog carrier as digital is 10 dB below analog.

Attenuator Function 20 dBmV 20 dBmV 10 dBmV 10 dBmV 0 dBmV 0 dBmV 50 MHz 750 MHz 50 MHz 750 MHz Notes Attenuators, also referred to as ___________, help balance an amplifier in the following ways: They reduce energy levels of all ___________________ in the spectrum equally. They compensate for the length of ______________ (normally a short span). When placed at the input of an amplifier, they directly affect _______________ ___________________________ (C/N). They are used to set the gain block input level in case the interamplifier losses are __________________________ amplifier gain.

Equalizer Function Effect of Cable Combined Results 20 dBmV 10 dBmV 0 dBmV 10 dBmV 50 MHz 750 MHz Combined Results 0 dB 50 MHz 750 MHz 10 dB Effect of Equalizer Notes Equalizers help balance an amplifier in the following ways: They reduce the energy level of ______ frequency signals more than ______ frequency signals. They compensate for the _______________ of cable. Equalizer response pattern compliments the response pattern of the cable to produce a flat broadband output signal. Equalizers should not be used to create tilt. Instead, you should use ___________________________ to perform this function. What are the consequences of choosing attenuators and equalizers incorrectly? 20 dB 50 MHz 750 MHz

Response Equalizer Examples of Peak to Valley Responses. With Response Equalizers Installed Notes Response equalizers, also called _____________________, help balance an amplifier in the following ways: They flatten the ______________________________ of cascaded amplifiers, compensating for system signature imperfections. They help to correct undesirable _________ or _________ in the forward response of cascaded amplifiers by attenuating a portion of the _______________ while minimizing the inserting bandwidth to be “bumped.” They create “bumps” or __________ to equalize peaks or valleys within narrow frequency ranges. Response equalizers are available in many different configurations/ frequency ranges. They may be shaped and controlled to provide the exact response necessary for your network requirements. Response equalizers are not to be used to cover-up a system problem. These are available in either bumps or traps

Equalizer Selection = Interstage Eq Set for Secondary Eq Set desired Tilt @ Output Secondary Eq Set Additional Tilt 50 MHz 750 MHz 20 dB 12 dB 62E750/11 11 dB 8 dB Notes How does a technician determine the need for a response equalizer in a particular amplifier in the field? If a cascade appears to need response equalizers more frequently than once in every third amplifier, what might you infer about your cable system? In network amplifiers with an output tilt of greater than 8 dB the additional tilt must be accomplished at the secondary equalizer position. The input equalizer must be chosen so that the spectrum at the input to the pre-amp is flat. As you will see later, this is important to establishing C/N.

Amplifier Block Diagram with ALSC / AGC Post Amp High Low DC TP Shorting Stub to One Secondary or DC 4-8-or 12 Pad Input Atten EQ Pre Manual Gain Adj. Response Equalizer Interstage Slope Eq. ALSC Optional Plug In Inter Stage Dist Atten. Notes What is the importance of AGC and ALSC when balancing systems? This slide illustrates the basic downstream structure and components of amplifier stations. Common downstream structure consists of several cascaded gain stages and circuits that compensate for the loss in the preceding section of cable.

Forward Functional Block Diagram Forward Amplifier Characteristics Forward Functional Block Diagram Input / Output Test Point -20 dB direction couplers.They can be used as sweep points for return signals. Input / Output Diplex Filter These filters route the forward and return signals to there proper locations. Input Attenuator The 9-A-WC attenuator reduces RF Input levels to meet design requirements. Input Equalizer 7-2E-WC Input equalizer compensates for the effects of cable preceding the Amplifier. Pre-Amplifier A push pull hybrid provides gain and sets the noise figure (C/N) of the Amp. Alignment Attenuator Sets the Amps. Overall gain.Should not be changed. Interstage Attenuator Reduces the overall gain of the Amp. ALSC Used to overcome changes in signal level .Caused by changes in temperature. Response Equalizer An EDB series equalizer,when needed will compensate for variations in the system. Interstage Equalizer 7-2E-WC Equalizer is factory installed for the system designed tilt,at the output. Interstage Amp. Power Doubling hybrid,further amps.the RF signals. Distribution Equalizer 7-2E-WC Equalizer will provide additional slope to all outputs.(jumper from the factory) Main output alignment attenuator Fine tunes the RF level at the main output per amplifier specs. (DO NOT CHANGE) Main output attenuator 9A attenuator reduces the signal level to the output of the Amp. (Shipped with a zero pad) Main output post Amp Power doubling hybrid that sets the final output level. (MAIN ONLY) Output diplex filter Separates the forward signals from the return. Output T/P -20db directional coupler T/P.Will measure both forward and reverse signals. Secondary output alignment Atten. Fine tunes the RF level at the secondary output ports. Secondary output Atten. 9A attenuator reduces the signal level to the secondary output ports. Secondary output post Amp. Power doubling hybrid.Provides final amplification to the secondary outputs. Output port selector Used to route signals to secondary outputs.Available in dc-4,dc-8 or dc-12. Secondary output diplex filters Separates the forward from the reverse signals. Secondary output T/P’s. -20db directional couplers T/P.Will measure both forward and reverse signals.

Forward Sweep SYSTEM AMPLIFIER METER SWEEP GEAR Notes System sweeping, also known as _________________________________ or ___________________________________, helps to identify signal quality problems that occur between the band edges. Routine, periodic sweep testing ensures that customers receive consistently high-quality services. Sweep testing also helps technicians detect other problems, such as . . . Name the three types of sweep systems:

Sweep System Requirements More about… Sweep System Requirements Reference Section, Page14–3 Fiber Optic Interconnect Sweep Transmitter Headend Combiner Fiber Transmitter Node AMP 1 Amp 4 Amp 3 Amp 2 Reference Notes What is the goal of sweeping? Alignment within each cascade requires a reference point. What part of the system is used to create a reference point? Each amplifier is aligned from ___________________________ to _____________________________ with the amplifier reference point. For this reason, the reference point must be as _____________ as possible. Amp 6 Amp 5 *The remaining amplifiers in the cascade are compared to the reference.

Reference Section, Page 14–5 Raw Sweep More about… Sweep Test Equipment Reference Section, Page 14–5 AFTER BEFORE Notes What equipment is used to sweep a system, and how does it function? Equipment Function Both devices rapidly sweep through a succession of frequencies; hence the term “sweep testing.” The display on the sweep receiver depicts the frequency-selective impact of the span and the amplifier compared to some reference. If needed, a _____________________________ may be selected to adjust the amplifier’s response so that the output more closely resembles the reference.

Low Level Signal Telemetry Notes Low level signal telemetry, the tool of choice for today’s hybrid fiber-coaxial (HFC) networks, has the following characteristics: Produces a test signal whose level is in the _______________________, 10 dB or more below the level of the video carriers. Sweeps between active carriers and causes little ___________________ to pictures. A _________________________________ can be used to receive the sweep readings rather than a calibrated sweep receiver. Works only on ____________ cascades. Resolution is sometimes insufficient. In some cases, some carriers must be ______________________ to conduct a sweep test. Can be used over ____________________________________. Why can’t older high-level sweep testing be done over fiber-optic cable?

Coaxial Plant Design and Operation Frequency Response Specifics Introduction This module explores the ways in which hardware, network integrity, and environmental factors affect system flatness and service delivery. In a perfect broadband network, signals delivered to and from subscribers would remain at constant relative energy levels across the broadband spectrum, yielding high service quality at all frequencies. However, components in the network (signal sources, signal processors, etc.) as well as other factors such as the environment, aging equipment, and inadequate work practices affect signal levels, compromising system flatness and leading to signal loss. Outcome By understanding how components in the broadband network affect system flatness, participants will be able to maintain a high quality system and troubleshoot it effectively, thus minimizing outage times and keeping the system within FCC specifications.

Non Linear Cable Loss Characteristics 550MHz 50 MHz Signal Level Ideal Response Cable Kinks Z Mismatch Non Linear Notes Ideally, the response of the network would be to pass all desired frequencies at the same level of energy, without affecting frequencies out of the range of the filter. Uniform signal quality (consistent amplitude) would produce service quality at all frequencies, making the system flat or linear. The combined contributions of cable, connectors, etc. result in something less than a flat response, or non linear loss. Causes of non linear loss: Manmade Environmental Cable system components

Hardware Points of Concern More about… Hardware issues Reference Section, Page 13–1 Connector Tap Cable Amplifier Cable Cable Connector Notes What might be some hardware issues that would affect the overall cable’s signal performance? What is component signature, and how does it affect signal linearity in general? How does each type of hardware listed below affect system flatness? Amplifiers Connectors Passives

Frequency Characteristics Notes

Peak to Valley Notes What is a peak? What is a valley? What is peak-to-valley deviation?

Impedance Mismatch Notes What is impedance mismatch? What are the most common causes of impedance mismatch?

RF Suckout Notes What are the probable causes of suckout? How does suckout affect a cable system’s signal flatness? How would a technician fix a suckout in a cable system?

Low End Loss Notes Low end loss appears as the opposite of a typical cable response in that the high end channels may appear normal on a TV screen while low end channels appear noisy or not at all. What symptom(s) of low end loss distinguish it from suckout? What are the most common causes of low end loss? What must a technician do to correct low end loss?

Correcting the Characteristics of an Amplifier Signature Notes Every amplifier from any manufacturer has its own innate frequency response characteristics, also known as ____________________________, that manifest themselves as resonant peaks distorting the signal flatness as the signal moves from amplifier to amplifier. What can be done to compensate for amplifier signatures which compromise overall signal performance?

Correcting the Characteristics of an Amplifier Signature Notes: If my amplifier cascade specification is N/2+1@750MHz and I have 6 amplifiers in cascade, what is the total peak to valley in dB that can be accepted. Each division is 2dB. 6/2+1 = 4dB. Is the sweep picture above within this specification? No P-V is 5dB. If not what can be done to get it into spec?Insert a 750 bump. 54 750

Before

After

Forward Amplifier Characteristics

Forward Amplifier Characteristics Automatic Gain Control(AGC series plug-in) Optional Thermal Control (TGSC plug-in) Optional Thru Power Plug Power Supply +24 VDC Test Point Surge Arrestor AC Return Amplifier Module LER series plug-in Return RF Input/ H Pre-Amp Atten. Equalizer -30 dB Resistive Upstream Downstream Output Post-Amp Output/ Input

Forward Amplifier Characteristics Input= 11/17.7dB 40dB gain + 43.1 dBmV 33.6 dBmV +48/35.5 +38.0/31.5 dBmV +32.5/29 dBmV 29dB -0.4 23dB -0.6dB 20dB -1.0dB 17dB -1.5 14dB2.0 11dB -3.5 300ft= 4.5dB@870 1.5dB@50 Each port has 15.5/12dBmV out. Needs a minimum of 8dBMv at 870MHz Each port has 19/6.5dBmV out From the tap port back to the RF amplifier there are the additional losses of the tap value and of the coax cable. The tap distribution system is designed based mainly on forward level requirements. Values are selected that keep forward signal levels within a controlled window. Taps near amplifiers have larger values, while taps farther from the amplifiers have lower values since the forward RF signals have been attenuated by the coax cable. The return signals see the same tap losses as the forward signals, so homes nearer to RF actives will have more return path loss than homes farther away. In this example, home A has a return path loss of 35dB from the tap port back to the closest RF active, while home B has only 13dB. Overall, there is a fairly large spread in return path loss from home to home. The next graph shows this distribution. The next slide shows how the losses will vary in the return. Cable Losses @870 MHz= 1.5dB/100ft @50 MHz=0.5dB/100ft

Network Operation and Maintenance Procedures Notes System design dictates the signal levels required at each active device to effectively transfer signal energy through the network system. These levels directly affect accuracy and service quality. It is therefore important that initial setup of an amplifier or a series of amplifiers is accomplished in accordance with design specifications. The two setup procedures used to construct and set up a cable network system are balancing and sweeping. Define balancing: Define sweeping: What equipment most efficiently tests frequency response in a cable network system?

Coaxial Plant Design and Operation Intermodulation Distortions Introduction This module explores how distortions in a cable network result in unwanted interference signals called composite triple beat (CTB), composite second order (CSO), and cross modulation distortions (XMOD). These types of distortions are the most annoying and therefore among the major causes of customer complaints. Understanding these distortions, their impact on picture quality, and how meeting analog performance measurements for CTB, CSO, and XMOD is a good starting point to achieving signal quality and reducing customer complaints. Outcome By understanding that signals produce and suffer from different types of distortion, and by knowing how to prevent and correct these problems, technicians will improve their ability to maintain good picture quality for their customers in their cable network.

Distortion Cause: Amplifiers 55.25MHz Fs Amp Fs+ Fs VCC 2 Fs 2nd Harmonic 110.5MHz Notes What is distortion, and how is it caused? The basic distortion is the __________________. Second and third order distortions are derived from the multiple of ______________________ we try to amplify. Example: Channel 2 is 55.25 MHz. Its first harmonic is ________________. Its second harmonic is ________________.

Intermodulation Distortion Notes Intermodulation distortions are caused when different _____________________ in the signal mix to produce _________ and _____________________ frequencies which did not exist in the signal. Unwanted signals interfere with other signals and degrade the desired signal’s content until quality can no longer be sustained. The unwanted signals are outside the analog channels that create them, causing the _________________________________________________________ and the _________________________________________________________ distortion products. In a cable system where every channel is occupied, these distortion products fall in the __________________ of another signal, causing distortions.

Distortions 2nd Order Distortion Cross Modulation Discrete Third Order More about… Intermodulation Distortions Reference Section, Page 16–1 Notes Define each of the following types of distortions: Discrete Second Order-- Composite Second Order (CSO)-- Discrete Third Order-- Composite Triple Beat (CTB)-- Cross Modulation Distortion (XMOD)--

Discrete Second Order Distortions 55.25 MHz 175.25 MHz 230.50 MHz 229.25 MHz A+B 121.25 MHz 120.00 MHz A-B B A Carrier 1 Carrier 2 Carrier 1 Carrier 2 Active Beat Products = Carrier 1 +/- Carrier 2 Beat Product Notes Example: Channel A is 55.25 MHz. Channel B is 175.25 MHz. The second order beats fall above Channel 25 or 229.25 MHz and below Channel 14 at 120 MHz. All second order beats are  0.75 MHz or  1.25 MHz above or below a standard video carrier.

CSO Beats in a 77 Channel System More about… CSO Reference Section, Page 16–1 COMPOSITE SECOND ORDER (CSO) CSO( Single Amp.) = CSO(Spec.) + 2*(Rated Output-Actual Output) 68 + 2*(46 - 48) = 68 + 2*( -2 ) = 68 + -4 = 50 100 200 300 400 500 550 10 20 30 40 Subtraction Beats: CSO -F1, -F2, -F3 Addition Beats CSO +F1, +F2, +F3 Frequency in MHz NUMBER OF BEATS 60 Notes Composite Second Order (CSO) distortions result from the addition and subtraction of all the carriers in the system take two at a time. These energy combinations fall in an unequal distribution around other carriers. Example: A composite second order (cascade) = 66 – 15Log N where N is the number of amps in a cascade. If N = 2, what is the composite second order? 64 dBc

Discrete Third Order Active Carrier 1 Carrier 3 Carrier 2 Beat Product Beat Product = Carrier 1 +/- Carrier 2 +/- Carrier 3 Notes Third order beat products are the result of processing three signals at a time. The output will contain amplified versions of the three inputs. There will also be measurable energy at frequencies which result from adding and/or subtracting the three. The output of the amplifier includes amplified equivalents of the input signals and the sum and difference frequency of each of the input signals three at a time. A 550 MHz.system contains a large number of beats. EXAMPLE: BEAT PRODUCT = CHANNEL 11 - CHANNEL 10 + CHANNEL 12: 199.25MHz – 193.25MHz + 205.25MHz = 211.25MHz CHANNEL 13 VIDEO CARRIER = 211.25MHz

Composite Triple Beat Distortions More about… CTB Reference Section, Page16–3 Notes Also known as composite third order distortions, composite triple beat distortions are output products. If the output increases, composite triple beat performance degrades. Composite triple beat products are signal-level related. For each change of 1 dB in the output there is a ________ change in composite triple beat performance. Doubling or halving the number of amplifiers will cause a _________ change in composite triple beat performance. CTB1Amp = CTBspec + 2(Rated Output [RO] - Actual Output [AO]) EXAMPLE: Calculate the CTB1Amp when CTB spec = 74 dB, RO = 48 dBmV@79 channels, and AO = 45 dBmV. As you increase or decrease the output level by 1 dBmV, CTB performance changes by ____________.

Cross Modulation Video Aural Channel A Un-modulated Carrier Channel B More about… XMOD Reference Section, Page16–4 Video Aural Channel A Un-modulated Carrier Channel B Channel A with Cross Modulation from Channel B Notes Cross modulation is a third-order effect that modulates the modulation within the distribution amplifiers themselves. It produces amplitude modulation on the dominant system signals (visual carriers). Cross modulation can only be measured on an ___________________ channel. In systems carrying 20 channels or more, cross modulation appears after ___________________________________________________. Viewing the same channel after a number of amplifiers reveals that the signal has been modulated. This modulation is the result of mixings of energy with the other channels, which are modulated at the output of the amplifiers. XMOD (single) = XMOD (spec.) + 2 (RO – AO) Example: Calculate the XMOD (single) when XMOD (spec.) = 71, Rated Output = 46, and Actual Output = 48. If Actual Output increases by 1, what is the XMOD?

CTB#Amps = CTB1Amp - 20log (#Amps) Composite Triple Beat Multiple Amplifiers CTB#Amps = CTB1Amp - 20log (#Amps) 2 Amplifiers CTB2Amps = 68- 20Log (2) CTB2Amps = 68- 20 x .3 CTB2Amps = 62 dBc This is true if all the amplifiers are identical. CTB is 20 Log because it is a voltage function. . Notes

Cross Modulation Calculation XMOD ( Cascade ) = XMOD ( Single ) - 20Log ( N ) where N is the number of amplifiers in cascade. Notes