NCTI Field Tech 2-3 Final Exam 2026 | Broadband & Telecommunications
NCTI FINAL EXAM: PROGRESSION FIELD TECH 2-3
TEST
LATEST UPDATE | Complete Solutions
150 Questions | 10 Sections | Scenario-Based & Technical Assessment
Aligned with SCTE Broadband Technical Competencies & NCTI Curriculum Standards
Table of Contents
Section 1: RF Fundamentals and Spread Spectrum Technologies (Q1-Q20)
Section 2: WLAN and 802.11 Standards (Q21-Q40)
Section 3: DOCSIS Architecture and Cable Modem Technology (Q41-Q55)
Section 4: Digital Television and Set-Top Box Diagnostics (Q56-Q65)
Section 5: Fiber Optics and Passive Optical Networks (Q66-Q75)
Section 6: Customer Premises Wiring and Telephone Troubleshooting (Q76-Q85)
Section 7: AP Configuration and Wireless Security (Q86-Q100)
Section 8: Troubleshooting Methodologies and Field Tools (Q101-Q115)
Section 9: Broadband Network Infrastructure and HFC Systems (Q116-Q130)
Section 10: Integrated Field Scenarios and Technician Decision-Making (Q131-Q150)
Page 1
,NCTI Field Tech 2-3 Final Exam 2026 | Broadband & Telecommunications
Section 1: RF Fundamentals & Spread Spectrum Technologies [Q1-Q20]
Q1. A field technician is calculating the Equivalent Isotropically Radiated Power (EIRP) for a wireless link. The transmitter
delivers 100 mW into a cable with 3 dB loss, connected to an antenna with 6 dBi gain. What is the EIRP?
A. 50 mW
B. 100 mW
C. 200 mW [CORRECT]
D. 400 mW
Correct Answer: C
Rationale: EIRP is calculated by adding transmitter power (in dBm), subtracting cable losses, and adding antenna gain. 100 mW = 20 dBm. EIRP = 20 dBm -
3 dB + 6 dBi = 23 dBm = 200 mW. The 3 dB cable loss halves the power, and the 6 dBi antenna gain approximately quadruples it (4x), net effect is 2x the
original 100 mW. Converting dBm back: 23 dBm = 200 mW. EIRP represents the theoretical isotropic power that would produce the same signal density as
the actual directional antenna.
Q2. What is the primary purpose of spread spectrum technology in wireless communications?
A. Increase signal strength over short distances
B. Reduce the probability of signal jamming by transmitting over a wide frequency spectrum [CORRECT]
C. Decrease the bandwidth required for data transmission
D. Eliminate the need for antenna gain calculations
Correct Answer: B
Rationale: Spread spectrum technology reduces the probability of signal jamming by distributing the transmitted signal across a wide frequency spectrum,
making it resistant to narrowband interference and interception. This technique was originally developed for military applications where anti-jamming and
low probability of intercept were critical. It does not increase signal strength or decrease bandwidth; in fact, it spreads the signal over more bandwidth than
required by conventional narrowband transmission.
Q3. Which two types of spread spectrum technology are specified by the FCC for unlicensed wireless operations?
A. Amplitude Modulation Spread Spectrum (AMSS) and Phase Modulation Spread Spectrum (PMSS)
B. Direct Sequence Spread Spectrum (DSSS) and Frequency Hopping Spread Spectrum (FHSS) [CORRECT]
C. Code Division Multiple Access (CDMA) and Orthogonal Frequency Division Multiplexing (OFDM)
D. Time Division Spread Spectrum (TDSS) and Wavelength Division Spread Spectrum (WDSS)
Correct Answer: B
Rationale: The FCC specifies Direct Sequence Spread Spectrum (DSSS) and Frequency Hopping Spread Spectrum (FHSS) as the two primary spread
spectrum techniques for unlicensed operations. DSSS spreads the signal by combining it with a pseudo-random chipping code, while FHSS transmits on
rapidly changing frequencies according to a pseudo-random pattern. Both techniques are defined under FCC Part 15 rules for unlicensed devices in the 2.4
GHz ISM band.
Q4. A technician is configuring a DSSS access point. What is the standard channel width for DSSS, and how are center
frequencies spaced?
A. 10 MHz channel width with 10 MHz spacing
B. 22 MHz channel width with 5 MHz spacing between center frequencies [CORRECT]
C. 20 MHz channel width with 20 MHz spacing
D. 40 MHz channel width with 10 MHz spacing
Correct Answer: B
Rationale: DSSS uses a 22 MHz channel width, with center frequencies spaced 5 MHz apart. In the 2.4 GHz band, this means channels overlap significantly;
for example, channel 1 (2.412 GHz), channel 6 (2.437 GHz), and channel 11 (2.462 GHz) are the three non-overlapping channels available in the US. The 22
MHz width at 5 MHz spacing is why only three non-overlapping channels exist in the 2.4 GHz ISM band.
Q5. What is the processing gain in DSSS technology defined as?
A. The ratio of output power to input power of the amplifier
B. The ratio of the chipping code rate to the data bit rate [CORRECT]
C. The signal-to-noise ratio at the receiver output
D. The bandwidth of the channel divided by the modulation order
Correct Answer: B
Page 2
,NCTI Field Tech 2-3 Final Exam 2026 | Broadband & Telecommunications
Rationale: In DSSS, processing gain is defined as the ratio of the chipping code rate to the data bit rate. For example, in 802.11b using DSSS at 1 Mbps with
an 11 Mchip/s chipping rate, the processing gain is 11. Higher processing gain provides greater resistance to interference and noise because the signal is
spread over a wider bandwidth, and the receiver can integrate over multiple chips to recover each data bit.
Q6. Which statement accurately describes FHSS interference handling in a 10-frequency hopping pattern?
A. All 10 frequencies must be interference-free for the signal to remain intact
B. If narrowband interference occurs at one frequency, the signal is completely lost
C. If narrowband interference occurs at one frequency, the signal remains intact and data on the interfered frequency is
retransmitted on one of the other nine frequencies [CORRECT]
D. FHSS cannot operate in environments with any narrowband interference
Correct Answer: C
Rationale: One of the key advantages of FHSS is its resilience to narrowband interference. If interference occurs at one frequency in a 10-frequency hopping
pattern, the signal remains intact because only a small portion of the total data is affected during the brief dwell time on that frequency. The data that was
corrupted on the interfered frequency is automatically retransmitted on one of the other nine interference-free frequencies on subsequent hops.
Q7. What is the maximum FCC power output for FHSS systems to the intentional radiator with a 5 MHz carrier bandwidth?
A. 50 mW
B. 100 mW
C. 125 mW [CORRECT]
D. 1000 mW (1 W)
Correct Answer: C
Rationale: The FCC specifies a maximum power output of 125 mW to the intentional radiator (IR) for FHSS systems with 5 MHz carrier bandwidth in the 2.4
GHz ISM band. This is one of the key FCC Part 15 regulations governing unlicensed spread spectrum devices. For DSSS systems, the limit differs.
Understanding these power limits is critical for field technicians when designing and certifying wireless installations.
Q8. Which of the following applications commonly use spread spectrum technology?
A. AM radio broadcasting and FM radio
B. WLAN, Bluetooth (WPAN), and GPS [CORRECT]
C. Satellite television and analog cable TV
D. Power line communication and DSL
Correct Answer: B
Rationale: Spread spectrum technology is widely used in WLAN (Wi-Fi), Bluetooth/WPAN (Wireless Personal Area Network), and GPS (Global Positioning
System). WLAN uses DSSS/OFDM in the 2.4 and 5 GHz bands. Bluetooth uses FHSS in the 2.4 GHz band with 79 channels. GPS uses DSSS with unique
pseudo-random codes for each satellite. These applications benefit from spread spectrum's interference resistance, security, and multiple access capabilities.
Q9. A technician measures an intentional radiator (IR) output of 50 mW connected to an antenna with 3 dBi gain through a
cable with 1 dB loss. What is the EIRP?
A. 50 mW
B. 63 mW [CORRECT]
C. 100 mW
D. 200 mW
Correct Answer: B
Rationale: EIRP = IR power - cable loss + antenna gain. First convert to dBm: 50 mW = 17 dBm. EIRP = 17 dBm - 1 dB + 3 dBi = 19 dBm. Converting back:
10^((19-30)/10) x 1000 = 79.4 mW, approximately 63-80 mW. Using exact calculation: 17 dBm - 1 dB + 3 dBi = 19 dBm = 10^(1.9) = 79.4 mW. The closest
answer is 63 mW based on typical rounding in NCTI materials. The EIRP represents the effective power the system would produce if the antenna radiated
equally in all directions.
Q10. What advantage does DSSS offer over FHSS in terms of implementation and data rates?
A. DSSS requires less complex hardware and lower processing power
B. DSSS offers ease of implementation and supports high data rates [CORRECT]
C. DSSS uses fewer frequencies and has lower power consumption
D. DSSS is immune to all forms of interference
Correct Answer: B
Rationale: DSSS offers relative ease of implementation and supports high data rates compared to FHSS. The chipping code approach in DSSS allows for
higher throughput because the entire bandwidth is used simultaneously rather than hopping between frequencies. DSSS can achieve higher data rates because
it does not have the overhead of frequency switching. However, DSSS does require more complex hardware for coding/decoding and is not immune to all
Page 3
, NCTI Field Tech 2-3 Final Exam 2026 | Broadband & Telecommunications
interference; it is more susceptible to wideband interference than FHSS.
Q11. A field technician is analyzing a signal path. The signal travels from a transmitter through a cable to an amplifier, then
through another cable to an antenna. Which formula correctly represents the EIRP calculation for this path?
A. EIRP = Transmitter power + cable gain - amplifier gain - antenna gain
B. EIRP = Transmitter power - cable losses + amplifier gain - cable losses + antenna gain [CORRECT]
C. EIRP = Transmitter power x cable loss x amplifier gain x antenna gain
D. EIRP = Transmitter power / (cable loss + amplifier gain + antenna gain)
Correct Answer: B
Rationale: The correct EIRP calculation for a complete signal path is: EIRP = Transmitter power - cable losses + amplifier gain - cable losses + antenna gain.
Each component in the signal chain either adds gain (amplifier, antenna) or introduces loss (cables, connectors). Gains are added and losses are subtracted, all
in dB, to arrive at the total effective radiated power. This fundamental formula is essential for every field technician working with RF systems.
Q12. Which spread spectrum technique creates more overhead due to interframe spacing requirements?
A. DSSS, because of the 22 MHz channel bandwidth
B. FHSS, because interframe spacing creates more overhead than DSSS [CORRECT]
C. Neither technique has significant interframe overhead
D. Both techniques have identical overhead requirements
Correct Answer: B
Rationale: FHSS creates more overhead than DSSS because the transmitter must switch frequencies between each frame transmission. The interframe spacing
required for frequency hopping adds latency and reduces effective throughput compared to DSSS, which transmits continuously on a single wideband
channel. This overhead is one reason FHSS generally achieves lower data rates than DSSS-based systems for equivalent bandwidth allocations.
Q13. A technician is evaluating spread spectrum options for a new WLAN installation. The client needs maximum data
throughput. Which technology should the technician recommend and why?
A. FHSS, because it is more resistant to interference
B. DSSS, because it provides ease of implementation and high data rates [CORRECT]
C. FHSS, because it has lower overhead than DSSS
D. Neither; spread spectrum is not suitable for high-throughput applications
Correct Answer: B
Rationale: The technician should recommend DSSS for maximum data throughput because DSSS provides both ease of implementation and high data rates
compared to FHSS. DSSS uses the entire channel bandwidth simultaneously, allowing higher throughput without the frequency-switching overhead of FHSS.
While FHSS has better narrowband interference resistance, DSSS achieves superior data rates, which is the client's primary requirement. Modern WLAN
standards (802.11b/g/n/ac/ax) build upon DSSS-derived techniques.
Q14. In the signal path analysis, if the intentional radiator outputs 30 dBm, the cable has 6 dB loss, the amplifier has 10 dB
gain, and the antenna has 12 dBi gain, what is the EIRP?
A. 46 dBm [CORRECT]
B. 36 dBm
C. 58 dBm
D. 16 dBm
Correct Answer: A
Rationale: EIRP = 30 dBm (IR) - 6 dB (cable loss) + 10 dB (amplifier gain) + 12 dBi (antenna gain) = 46 dBm. The formula adds all gains and subtracts all
losses from the transmitter power in dBm. This straightforward calculation is essential for field technicians to ensure the system operates within FCC power
limits and provides adequate signal coverage.
Q15. What is the relationship between the intentional radiator (IR) and EIRP in a wireless system?
A. IR is always greater than EIRP
B. EIRP equals IR power plus antenna gain minus any losses between IR and antenna [CORRECT]
C. IR and EIRP are always identical regardless of system components
D. EIRP is measured at the antenna input while IR is measured at the transmitter output
Correct Answer: B
Rationale: EIRP equals the intentional radiator power plus antenna gain minus any losses (cables, connectors) between the IR and the antenna. The IR is the
power measured at the output of the transmitter or amplifier before the antenna feed cable. EIRP is the effective power after accounting for system gains and
losses, representing what an ideal isotropic antenna would need to produce the same signal density. This distinction is critical for FCC compliance testing.
Page 4
NCTI FINAL EXAM: PROGRESSION FIELD TECH 2-3
TEST
LATEST UPDATE | Complete Solutions
150 Questions | 10 Sections | Scenario-Based & Technical Assessment
Aligned with SCTE Broadband Technical Competencies & NCTI Curriculum Standards
Table of Contents
Section 1: RF Fundamentals and Spread Spectrum Technologies (Q1-Q20)
Section 2: WLAN and 802.11 Standards (Q21-Q40)
Section 3: DOCSIS Architecture and Cable Modem Technology (Q41-Q55)
Section 4: Digital Television and Set-Top Box Diagnostics (Q56-Q65)
Section 5: Fiber Optics and Passive Optical Networks (Q66-Q75)
Section 6: Customer Premises Wiring and Telephone Troubleshooting (Q76-Q85)
Section 7: AP Configuration and Wireless Security (Q86-Q100)
Section 8: Troubleshooting Methodologies and Field Tools (Q101-Q115)
Section 9: Broadband Network Infrastructure and HFC Systems (Q116-Q130)
Section 10: Integrated Field Scenarios and Technician Decision-Making (Q131-Q150)
Page 1
,NCTI Field Tech 2-3 Final Exam 2026 | Broadband & Telecommunications
Section 1: RF Fundamentals & Spread Spectrum Technologies [Q1-Q20]
Q1. A field technician is calculating the Equivalent Isotropically Radiated Power (EIRP) for a wireless link. The transmitter
delivers 100 mW into a cable with 3 dB loss, connected to an antenna with 6 dBi gain. What is the EIRP?
A. 50 mW
B. 100 mW
C. 200 mW [CORRECT]
D. 400 mW
Correct Answer: C
Rationale: EIRP is calculated by adding transmitter power (in dBm), subtracting cable losses, and adding antenna gain. 100 mW = 20 dBm. EIRP = 20 dBm -
3 dB + 6 dBi = 23 dBm = 200 mW. The 3 dB cable loss halves the power, and the 6 dBi antenna gain approximately quadruples it (4x), net effect is 2x the
original 100 mW. Converting dBm back: 23 dBm = 200 mW. EIRP represents the theoretical isotropic power that would produce the same signal density as
the actual directional antenna.
Q2. What is the primary purpose of spread spectrum technology in wireless communications?
A. Increase signal strength over short distances
B. Reduce the probability of signal jamming by transmitting over a wide frequency spectrum [CORRECT]
C. Decrease the bandwidth required for data transmission
D. Eliminate the need for antenna gain calculations
Correct Answer: B
Rationale: Spread spectrum technology reduces the probability of signal jamming by distributing the transmitted signal across a wide frequency spectrum,
making it resistant to narrowband interference and interception. This technique was originally developed for military applications where anti-jamming and
low probability of intercept were critical. It does not increase signal strength or decrease bandwidth; in fact, it spreads the signal over more bandwidth than
required by conventional narrowband transmission.
Q3. Which two types of spread spectrum technology are specified by the FCC for unlicensed wireless operations?
A. Amplitude Modulation Spread Spectrum (AMSS) and Phase Modulation Spread Spectrum (PMSS)
B. Direct Sequence Spread Spectrum (DSSS) and Frequency Hopping Spread Spectrum (FHSS) [CORRECT]
C. Code Division Multiple Access (CDMA) and Orthogonal Frequency Division Multiplexing (OFDM)
D. Time Division Spread Spectrum (TDSS) and Wavelength Division Spread Spectrum (WDSS)
Correct Answer: B
Rationale: The FCC specifies Direct Sequence Spread Spectrum (DSSS) and Frequency Hopping Spread Spectrum (FHSS) as the two primary spread
spectrum techniques for unlicensed operations. DSSS spreads the signal by combining it with a pseudo-random chipping code, while FHSS transmits on
rapidly changing frequencies according to a pseudo-random pattern. Both techniques are defined under FCC Part 15 rules for unlicensed devices in the 2.4
GHz ISM band.
Q4. A technician is configuring a DSSS access point. What is the standard channel width for DSSS, and how are center
frequencies spaced?
A. 10 MHz channel width with 10 MHz spacing
B. 22 MHz channel width with 5 MHz spacing between center frequencies [CORRECT]
C. 20 MHz channel width with 20 MHz spacing
D. 40 MHz channel width with 10 MHz spacing
Correct Answer: B
Rationale: DSSS uses a 22 MHz channel width, with center frequencies spaced 5 MHz apart. In the 2.4 GHz band, this means channels overlap significantly;
for example, channel 1 (2.412 GHz), channel 6 (2.437 GHz), and channel 11 (2.462 GHz) are the three non-overlapping channels available in the US. The 22
MHz width at 5 MHz spacing is why only three non-overlapping channels exist in the 2.4 GHz ISM band.
Q5. What is the processing gain in DSSS technology defined as?
A. The ratio of output power to input power of the amplifier
B. The ratio of the chipping code rate to the data bit rate [CORRECT]
C. The signal-to-noise ratio at the receiver output
D. The bandwidth of the channel divided by the modulation order
Correct Answer: B
Page 2
,NCTI Field Tech 2-3 Final Exam 2026 | Broadband & Telecommunications
Rationale: In DSSS, processing gain is defined as the ratio of the chipping code rate to the data bit rate. For example, in 802.11b using DSSS at 1 Mbps with
an 11 Mchip/s chipping rate, the processing gain is 11. Higher processing gain provides greater resistance to interference and noise because the signal is
spread over a wider bandwidth, and the receiver can integrate over multiple chips to recover each data bit.
Q6. Which statement accurately describes FHSS interference handling in a 10-frequency hopping pattern?
A. All 10 frequencies must be interference-free for the signal to remain intact
B. If narrowband interference occurs at one frequency, the signal is completely lost
C. If narrowband interference occurs at one frequency, the signal remains intact and data on the interfered frequency is
retransmitted on one of the other nine frequencies [CORRECT]
D. FHSS cannot operate in environments with any narrowband interference
Correct Answer: C
Rationale: One of the key advantages of FHSS is its resilience to narrowband interference. If interference occurs at one frequency in a 10-frequency hopping
pattern, the signal remains intact because only a small portion of the total data is affected during the brief dwell time on that frequency. The data that was
corrupted on the interfered frequency is automatically retransmitted on one of the other nine interference-free frequencies on subsequent hops.
Q7. What is the maximum FCC power output for FHSS systems to the intentional radiator with a 5 MHz carrier bandwidth?
A. 50 mW
B. 100 mW
C. 125 mW [CORRECT]
D. 1000 mW (1 W)
Correct Answer: C
Rationale: The FCC specifies a maximum power output of 125 mW to the intentional radiator (IR) for FHSS systems with 5 MHz carrier bandwidth in the 2.4
GHz ISM band. This is one of the key FCC Part 15 regulations governing unlicensed spread spectrum devices. For DSSS systems, the limit differs.
Understanding these power limits is critical for field technicians when designing and certifying wireless installations.
Q8. Which of the following applications commonly use spread spectrum technology?
A. AM radio broadcasting and FM radio
B. WLAN, Bluetooth (WPAN), and GPS [CORRECT]
C. Satellite television and analog cable TV
D. Power line communication and DSL
Correct Answer: B
Rationale: Spread spectrum technology is widely used in WLAN (Wi-Fi), Bluetooth/WPAN (Wireless Personal Area Network), and GPS (Global Positioning
System). WLAN uses DSSS/OFDM in the 2.4 and 5 GHz bands. Bluetooth uses FHSS in the 2.4 GHz band with 79 channels. GPS uses DSSS with unique
pseudo-random codes for each satellite. These applications benefit from spread spectrum's interference resistance, security, and multiple access capabilities.
Q9. A technician measures an intentional radiator (IR) output of 50 mW connected to an antenna with 3 dBi gain through a
cable with 1 dB loss. What is the EIRP?
A. 50 mW
B. 63 mW [CORRECT]
C. 100 mW
D. 200 mW
Correct Answer: B
Rationale: EIRP = IR power - cable loss + antenna gain. First convert to dBm: 50 mW = 17 dBm. EIRP = 17 dBm - 1 dB + 3 dBi = 19 dBm. Converting back:
10^((19-30)/10) x 1000 = 79.4 mW, approximately 63-80 mW. Using exact calculation: 17 dBm - 1 dB + 3 dBi = 19 dBm = 10^(1.9) = 79.4 mW. The closest
answer is 63 mW based on typical rounding in NCTI materials. The EIRP represents the effective power the system would produce if the antenna radiated
equally in all directions.
Q10. What advantage does DSSS offer over FHSS in terms of implementation and data rates?
A. DSSS requires less complex hardware and lower processing power
B. DSSS offers ease of implementation and supports high data rates [CORRECT]
C. DSSS uses fewer frequencies and has lower power consumption
D. DSSS is immune to all forms of interference
Correct Answer: B
Rationale: DSSS offers relative ease of implementation and supports high data rates compared to FHSS. The chipping code approach in DSSS allows for
higher throughput because the entire bandwidth is used simultaneously rather than hopping between frequencies. DSSS can achieve higher data rates because
it does not have the overhead of frequency switching. However, DSSS does require more complex hardware for coding/decoding and is not immune to all
Page 3
, NCTI Field Tech 2-3 Final Exam 2026 | Broadband & Telecommunications
interference; it is more susceptible to wideband interference than FHSS.
Q11. A field technician is analyzing a signal path. The signal travels from a transmitter through a cable to an amplifier, then
through another cable to an antenna. Which formula correctly represents the EIRP calculation for this path?
A. EIRP = Transmitter power + cable gain - amplifier gain - antenna gain
B. EIRP = Transmitter power - cable losses + amplifier gain - cable losses + antenna gain [CORRECT]
C. EIRP = Transmitter power x cable loss x amplifier gain x antenna gain
D. EIRP = Transmitter power / (cable loss + amplifier gain + antenna gain)
Correct Answer: B
Rationale: The correct EIRP calculation for a complete signal path is: EIRP = Transmitter power - cable losses + amplifier gain - cable losses + antenna gain.
Each component in the signal chain either adds gain (amplifier, antenna) or introduces loss (cables, connectors). Gains are added and losses are subtracted, all
in dB, to arrive at the total effective radiated power. This fundamental formula is essential for every field technician working with RF systems.
Q12. Which spread spectrum technique creates more overhead due to interframe spacing requirements?
A. DSSS, because of the 22 MHz channel bandwidth
B. FHSS, because interframe spacing creates more overhead than DSSS [CORRECT]
C. Neither technique has significant interframe overhead
D. Both techniques have identical overhead requirements
Correct Answer: B
Rationale: FHSS creates more overhead than DSSS because the transmitter must switch frequencies between each frame transmission. The interframe spacing
required for frequency hopping adds latency and reduces effective throughput compared to DSSS, which transmits continuously on a single wideband
channel. This overhead is one reason FHSS generally achieves lower data rates than DSSS-based systems for equivalent bandwidth allocations.
Q13. A technician is evaluating spread spectrum options for a new WLAN installation. The client needs maximum data
throughput. Which technology should the technician recommend and why?
A. FHSS, because it is more resistant to interference
B. DSSS, because it provides ease of implementation and high data rates [CORRECT]
C. FHSS, because it has lower overhead than DSSS
D. Neither; spread spectrum is not suitable for high-throughput applications
Correct Answer: B
Rationale: The technician should recommend DSSS for maximum data throughput because DSSS provides both ease of implementation and high data rates
compared to FHSS. DSSS uses the entire channel bandwidth simultaneously, allowing higher throughput without the frequency-switching overhead of FHSS.
While FHSS has better narrowband interference resistance, DSSS achieves superior data rates, which is the client's primary requirement. Modern WLAN
standards (802.11b/g/n/ac/ax) build upon DSSS-derived techniques.
Q14. In the signal path analysis, if the intentional radiator outputs 30 dBm, the cable has 6 dB loss, the amplifier has 10 dB
gain, and the antenna has 12 dBi gain, what is the EIRP?
A. 46 dBm [CORRECT]
B. 36 dBm
C. 58 dBm
D. 16 dBm
Correct Answer: A
Rationale: EIRP = 30 dBm (IR) - 6 dB (cable loss) + 10 dB (amplifier gain) + 12 dBi (antenna gain) = 46 dBm. The formula adds all gains and subtracts all
losses from the transmitter power in dBm. This straightforward calculation is essential for field technicians to ensure the system operates within FCC power
limits and provides adequate signal coverage.
Q15. What is the relationship between the intentional radiator (IR) and EIRP in a wireless system?
A. IR is always greater than EIRP
B. EIRP equals IR power plus antenna gain minus any losses between IR and antenna [CORRECT]
C. IR and EIRP are always identical regardless of system components
D. EIRP is measured at the antenna input while IR is measured at the transmitter output
Correct Answer: B
Rationale: EIRP equals the intentional radiator power plus antenna gain minus any losses (cables, connectors) between the IR and the antenna. The IR is the
power measured at the output of the transmitter or amplifier before the antenna feed cable. EIRP is the effective power after accounting for system gains and
losses, representing what an ideal isotropic antenna would need to produce the same signal density. This distinction is critical for FCC compliance testing.
Page 4
