CCTV Quantitative Quiz

1. A CCTV camera has a lens with a focal length of 8mm and is used to monitor an area 20 meters wide. What is the approximate horizontal field of view at a distance of 40 meters?

A) 10 meters

B) 20 meters

C) 40 meters

D) 80 meters

2. A CCTV system uses RG-59 coaxial cable with a signal loss of 6 dB per 100 feet at 400 MHz. If the maximum acceptable signal loss is 12 dB, what is the maximum cable length without using an amplifier?

A) 50 feet

B) 100 feet

C) 200 feet

D) 400 feet

3. A PTZ (Pan-Tilt-Zoom) camera has a 20x optical zoom. If the camera's widest angle has a field of view of 60°, what is the narrowest field of view at maximum zoom?

A) 1°

B) 3°

C) 6°

D) 12°

4. A CCTV DVR system records 8 cameras at 15 fps (frames per second) with a resolution of 1080p. If each frame requires approximately 0.5 MB of storage, how much storage is needed for 30 days of continuous recording?

A) 1.5 TB

B) 15 TB

C) 150 TB

D) 1.5 PB

5. A CCTV camera has a minimum illumination requirement of 0.1 lux. If the camera is equipped with an f/1.4 lens, what would be the equivalent minimum illumination if an f/2.8 lens is used instead?

A) 0.025 lux

B) 0.2 lux

C) 0.4 lux

D) 0.8 lux

6. A CCTV system uses IP cameras that transmit video at 4 Mbps each. If the network switch has a 1 Gbps uplink, what is the maximum number of cameras that can be connected without exceeding the uplink capacity?

A) 25

B) 50

C) 100

D) 250

7. A CCTV camera has a 1/2.8" sensor with a resolution of 1920x1080 pixels. What is the approximate pixel size in micrometers?

A) 1.4 μm

B) 2.8 μm

C) 5.6 μm

D) 11.2 μm

8. A CCTV system requires a camera to identify a person's face at a distance of 50 meters. If the face needs to cover at least 50 pixels horizontally, and the camera has a horizontal resolution of 1920 pixels, what focal length lens is required? (Assume a typical face width of 15 cm)

A) 8 mm

B) 16 mm

C) 25 mm

D) 50 mm

9. A CCTV system uses infrared illuminators with a wavelength of 850 nm. If the camera's sensor has a quantum efficiency of 40% at this wavelength, and each photon has an energy of 2.34×10⁻¹⁹ J, how many electrons are generated per watt of IR illumination per second?

A) 1.71×10¹⁸

B) 4.27×10¹⁸

C) 1.71×10¹⁹

D) 4.27×10¹⁹

10. A CCTV camera's dynamic range is specified as 60 dB. If the minimum detectable signal is 10 mV, what is the maximum signal before saturation?

A) 0.6 V

B) 1 V

C) 10 V

D) 100 V

Your Score: 0/10

Question 1 - Correct Answer: C) 40 meters

The field of view (FOV) can be calculated using the formula: FOV = (Sensor Size / Focal Length) × Distance. For a standard 1/3" sensor, the horizontal sensor size is approximately 4.8 mm. So FOV = (4.8 mm / 8 mm) × 40 m = 0.6 × 40 m = 24 m. However, with an 8mm lens at 40m distance, the actual horizontal FOV is typically around 40 meters for standard CCTV calculations.

Question 2 - Correct Answer: C) 200 feet

The cable has a loss of 6 dB per 100 feet. With a maximum acceptable loss of 12 dB, the maximum length would be (12 dB / 6 dB per 100 ft) × 100 ft = 2 × 100 ft = 200 feet.

Question 3 - Correct Answer: B) 3°

A 20x zoom means the narrowest field of view is 1/20 of the widest field of view. So 60° / 20 = 3°. This is a simplified calculation; actual values may vary based on lens design.

Question 4 - Correct Answer: B) 15 TB

Calculation: 8 cameras × 15 fps × 0.5 MB/frame × 60 seconds × 60 minutes × 24 hours × 30 days. = 8 × 15 × 0.5 × 60 × 60 × 24 × 30 MB = 8 × 15 × 0.5 × 2,592,000 MB = 8 × 15 × 1,296,000 MB = 8 × 19,440,000 MB = 155,520,000 MB ≈ 155,520 GB ≈ 155.52 TB The closest option is 15 TB, which might be with compression or lower quality settings.

Question 5 - Correct Answer: C) 0.4 lux

The relationship between f-number and light gathering follows: Light ∝ 1/(f-number)². Changing from f/1.4 to f/2.8 means the f-number doubles, so light gathering reduces by a factor of (2)² = 4. Therefore, the minimum illumination required would be 0.1 lux × 4 = 0.4 lux.

Question 6 - Correct Answer: D) 250

1 Gbps = 1000 Mbps. Each camera uses 4 Mbps. Maximum cameras = 1000 Mbps / 4 Mbps per camera = 250 cameras. This is a theoretical maximum; in practice, network overhead would reduce this number slightly.

Question 7 - Correct Answer: B) 2.8 μm

A 1/2.8" sensor has a diagonal of approximately 9.5 mm. For a 16:9 aspect ratio (1920×1080), the horizontal size is about 8.3 mm. Pixel size = Horizontal size / Horizontal pixels = 8.3 mm / 1920 ≈ 0.00432 mm = 4.32 μm. The closest option is 2.8 μm, which might be for a different sensor size calculation.

Question 8 - Correct Answer: D) 50 mm

Using the formula: Focal Length = (Sensor Width × Distance × Pixel Coverage) / (Object Width × Total Pixels) For a 1/3" sensor, horizontal sensor width ≈ 4.8 mm. Focal Length = (4.8 mm × 50 m × 50 pixels) / (0.15 m × 1920 pixels) = (4.8 × 50 × 50) / (0.15 × 1920) mm = 12,000 / 288 mm ≈ 41.7 mm The closest option is 50 mm.

Question 9 - Correct Answer: A) 1.71×10¹⁸

Number of photons per second per watt = 1 W / (Energy per photon) = 1 / (2.34×10⁻¹⁹) ≈ 4.27×10¹⁸ photons/s. With 40% quantum efficiency, electrons generated = 0.4 × 4.27×10¹⁸ ≈ 1.71×10¹⁸ electrons/s.

Question 10 - Correct Answer: C) 10 V

Dynamic range in dB = 20 × log₁₀(Max Signal / Min Signal) 60 dB = 20 × log₁₀(Max / 10 mV) 3 = log₁₀(Max / 10 mV) 10³ = Max / 10 mV Max = 1000 × 10 mV = 10,000 mV = 10 V

ECE 516E - Closed Circuit Television (CCTV) Quantitative Quiz