NWCA Study Guide ACTUAL EXAM
2026/2027 | NWCA Coaching
Certification | Verified Q&A | Pass
Guaranteed - A+ Graded
Section 1: Radio Frequency (RF) Fundamentals
Q1: A transmitter outputs 0.5 watts of power. What is this power in dBm? (Round to nearest whole
number)
A. 17 dBm
B. 20 dBm
C. 27 dBm [CORRECT]
D. 30 dBm
Correct Answer: C
Rationale: [CORRECT] Formula:
dBm=10×log
10
(power in mW)
.
0.5 W=500 mW
.
10×log
10
,(500)≈27 dBm
. A – 17 dBm is 50 mW. B – 20 dBm is 100 mW. D – 30 dBm is 1000 mW (1 W).
Q2: If the frequency of a signal is 2.4 GHz, what is the approximate wavelength? (Round to nearest cm)
A. 6.25 cm
B. 12.5 cm [CORRECT]
C. 24.0 cm
D. 30.0 cm
Correct Answer: B
Rationale: [CORRECT] Formula:
λ(meters)=c/f
.
c≈3×10
8
m/s
.
λ=(3×10
8
)/(2.4×10
9
)=0.125 meters=12.5 cm
. A – Incorrect calculation using
4.8 GHz
. C – Incorrect calculation using
,1.25 GHz
. D – Wavelength for 1 GHz.
Q3: Which modulation technique varies the instantaneous phase of the carrier signal to represent data?
A. Amplitude Modulation (AM)
B. Frequency Modulation (FM)
C. Phase Shift Keying (PSK) [CORRECT]
D. Quadrature Amplitude Modulation (QAM)
Correct Answer: C
Rationale: [CORRECT] PSK (Phase Shift Keying) encodes data by changing the phase of the carrier wave.
A – Varies amplitude/signal strength. B – Varies frequency. D – Varies both amplitude and phase.
Q4: A system has a transmitter power of 40 dBm, a cable loss of 3 dB, and an antenna gain of 10 dBi.
What is the Effective Isotropic Radiated Power (EIRP)?
A. 47 dBm [CORRECT]
B. 37 dBm
C. 53 dBm
D. 43 dBm
Correct Answer: A
Rationale: [CORRECT] EIRP = Tx Power - Losses + Gain.
40 dBm−3 dB+10 dBi=47 dBm
. B – Subtracts gain instead of adding. C – Adds loss instead of subtracting. D – Incorrect math.
Q5: What is the primary cause of "multipath fading" in an urban environment?
A. Atmospheric absorption.
B. Reflections and scattering of the RF signal off buildings and obstacles. [CORRECT]
, C. Interference from neighboring cells.
D. The curvature of the earth.
Correct Answer: B
Rationale: [CORRECT] Multipath occurs when signals reach the receiver via multiple paths (reflections),
causing constructive and destructive interference. A – Causes signal attenuation over distance. C –
Causes co-channel interference. D – Relates to Earth bulge/line of sight.
Q6: Convert 30 dBm to Watts.
A. 0.5 Watts
B. 1 Watt [CORRECT]
C. 2 Watts
D. 10 Watts
Correct Answer: B
Rationale: [CORRECT] 30 dBm is the reference for 1 Watt (
10log
10
(1000 mW)=30 dBm
). A – 27 dBm. C – 33 dBm. D – 40 dBm.
Q7: Which parameter defines the range of frequencies occupied by a modulated signal?
A. Duty cycle
B. Bandwidth [CORRECT]
C. Latency
D. Skew
Correct Answer: B
2026/2027 | NWCA Coaching
Certification | Verified Q&A | Pass
Guaranteed - A+ Graded
Section 1: Radio Frequency (RF) Fundamentals
Q1: A transmitter outputs 0.5 watts of power. What is this power in dBm? (Round to nearest whole
number)
A. 17 dBm
B. 20 dBm
C. 27 dBm [CORRECT]
D. 30 dBm
Correct Answer: C
Rationale: [CORRECT] Formula:
dBm=10×log
10
(power in mW)
.
0.5 W=500 mW
.
10×log
10
,(500)≈27 dBm
. A – 17 dBm is 50 mW. B – 20 dBm is 100 mW. D – 30 dBm is 1000 mW (1 W).
Q2: If the frequency of a signal is 2.4 GHz, what is the approximate wavelength? (Round to nearest cm)
A. 6.25 cm
B. 12.5 cm [CORRECT]
C. 24.0 cm
D. 30.0 cm
Correct Answer: B
Rationale: [CORRECT] Formula:
λ(meters)=c/f
.
c≈3×10
8
m/s
.
λ=(3×10
8
)/(2.4×10
9
)=0.125 meters=12.5 cm
. A – Incorrect calculation using
4.8 GHz
. C – Incorrect calculation using
,1.25 GHz
. D – Wavelength for 1 GHz.
Q3: Which modulation technique varies the instantaneous phase of the carrier signal to represent data?
A. Amplitude Modulation (AM)
B. Frequency Modulation (FM)
C. Phase Shift Keying (PSK) [CORRECT]
D. Quadrature Amplitude Modulation (QAM)
Correct Answer: C
Rationale: [CORRECT] PSK (Phase Shift Keying) encodes data by changing the phase of the carrier wave.
A – Varies amplitude/signal strength. B – Varies frequency. D – Varies both amplitude and phase.
Q4: A system has a transmitter power of 40 dBm, a cable loss of 3 dB, and an antenna gain of 10 dBi.
What is the Effective Isotropic Radiated Power (EIRP)?
A. 47 dBm [CORRECT]
B. 37 dBm
C. 53 dBm
D. 43 dBm
Correct Answer: A
Rationale: [CORRECT] EIRP = Tx Power - Losses + Gain.
40 dBm−3 dB+10 dBi=47 dBm
. B – Subtracts gain instead of adding. C – Adds loss instead of subtracting. D – Incorrect math.
Q5: What is the primary cause of "multipath fading" in an urban environment?
A. Atmospheric absorption.
B. Reflections and scattering of the RF signal off buildings and obstacles. [CORRECT]
, C. Interference from neighboring cells.
D. The curvature of the earth.
Correct Answer: B
Rationale: [CORRECT] Multipath occurs when signals reach the receiver via multiple paths (reflections),
causing constructive and destructive interference. A – Causes signal attenuation over distance. C –
Causes co-channel interference. D – Relates to Earth bulge/line of sight.
Q6: Convert 30 dBm to Watts.
A. 0.5 Watts
B. 1 Watt [CORRECT]
C. 2 Watts
D. 10 Watts
Correct Answer: B
Rationale: [CORRECT] 30 dBm is the reference for 1 Watt (
10log
10
(1000 mW)=30 dBm
). A – 27 dBm. C – 33 dBm. D – 40 dBm.
Q7: Which parameter defines the range of frequencies occupied by a modulated signal?
A. Duty cycle
B. Bandwidth [CORRECT]
C. Latency
D. Skew
Correct Answer: B
