CTS Skills 2 — Questions and Answers
Question 1: What is the correct procedure for calculating amplifier power requirements for a loudspeaker system that must achieve 100 dB SPL at 10 meters, using a loudspeaker rated at 97 dB sensitivity (1W/1m)?
- 100 dB − 97 dB = 3 dB difference; 3 dB requires 2W; answer: 2W
- Sensitivity at 10m = 97 − 20 = 77 dB; need 100 − 77 = 23 dB above 1W; 23 dB = 200W; answer: approx. 200W (Correct answer)
- 100 dB at 1m requires 97W per meter; multiply by 10 = 970W
- 97 dB sensitivity × 10m = 970W
Correct answer: Sensitivity at 10m = 97 − 20 = 77 dB; need 100 − 77 = 23 dB above 1W; 23 dB = 200W; answer: approx. 200W
First, find SPL at 10m at 1W: 97 dB − 20×log10(10) = 97 − 20 = 77 dB. Then find power needed to reach 100 dB: 100 − 77 = 23 dB. Power = 10^(23/10) = 10^2.3 ≈ 200W.
Step 1: Calculate SPL at 1W at 10m using inverse square law: SPL = Sensitivity − 20×log10(distance) = 97 − 20×log10(10) = 97 − 20 = 77 dB. Step 2: Calculate required additional dB: 100 − 77 = 23 dB. Step 3: Convert dB to power ratio: P = 10^(dB/10) = 10^(23/10) = 10^2.3 ≈ 200 watts. Always calculate headroom and include a safety margin (typically 3–6 dB, requiring 400–800W) in real designs.
Question 2: An AV technician is configuring an EDID (Extended Display Identification Data) emulator in a matrix switcher installation. What problem does EDID management solve?
- EDID emulation increases the resolution of the source signal
- EDID emulation provides a stable handshake signal to the source device, preventing it from defaulting to a lower resolution or losing sync when displays are switched (Correct answer)
- EDID emulation converts HDMI signals to DisplayPort
- EDID emulation bypasses HDCP content protection
Correct answer: EDID emulation provides a stable handshake signal to the source device, preventing it from defaulting to a lower resolution or losing sync when displays are switched
When a display is switched away from a source, the EDID handshake is broken, causing the source to renegotiate and potentially output a lower-common-denominator resolution. An EDID emulator maintains a stable EDID to the source at all times, preventing resolution drops during switching.
HDMI and DisplayPort use EDID (a data structure stored in the display) to communicate display capabilities to the source. When a matrix switcher routes a source to a different display, or when the original display is powered off, the EDID connection is interrupted. The source then re-reads EDID from the new display (which may have different capabilities) and may change its output resolution. EDID emulators (also called EDID managers or ghost display emulators) present a fixed, consistent EDID to the source regardless of what display is connected downstream, ensuring stable output.
Question 3: What is the function of a delay line in a distributed audio system, and when must it be applied?
- A delay line adds reverb effect to improve speech intelligibility
- A delay line introduces a time offset to loudspeakers further from the source, aligning them with the primary speaker so listeners hear a single, coherent sound source (Correct answer)
- A delay line reduces amplifier power consumption during quiet passages
- A delay line converts analog audio to digital for network distribution
Correct answer: A delay line introduces a time offset to loudspeakers further from the source, aligning them with the primary speaker so listeners hear a single, coherent sound source
In distributed audio, delayed fill speakers must be time-aligned to the primary speaker. Without delay, a listener near a fill speaker hears it before the main system, creating an echo and destroying intelligibility. Haas effect alignment requires the fill speaker to arrive 0–25ms after the primary source.
Sound travels at approximately 344 m/s (1,130 ft/s) at 20°C. In a distributed system, a fill speaker 17.2 meters (56 ft) from the main speaker produces sound 50ms earlier than the main system for a listener near the fill speaker. The Haas (precedence) effect means listeners perceive the source as coming from the loudspeaker that arrives first. Adding a 50ms delay to the fill speaker aligns its output with the main system, so the listener perceives a single coherent source and speech intelligibility is preserved.
Question 4: A CTS technician is setting up a multi-camera videoconferencing system and needs to configure correct white balance on all cameras. What does white balance achieve, and what is the standard procedure?
- White balance adjusts the camera's zoom to fill the frame — set by focusing on the far wall
- White balance calibrates the camera's color channels so that a white object under the room's lighting appears neutral white on screen — set by pointing the camera at a white reference card under actual room lighting (Correct answer)
- White balance sets the camera's frame rate to match the room's AC frequency
- White balance adjusts audio gain to match the conference codec settings
Correct answer: White balance calibrates the camera's color channels so that a white object under the room's lighting appears neutral white on screen — set by pointing the camera at a white reference card under actual room lighting
White balance ensures the camera renders the room's color temperature correctly. Under tungsten lighting (2,700–3,200K) without white balance, an image appears orange; under fluorescent lighting (4,000–5,000K), it may appear green. Manual white balance using a reference card under actual room lighting gives the most accurate result.
Camera sensors have fixed spectral sensitivity. White balance adjusts the red, green, and blue channel gains so a calibration reference (white card or gray card) produces equal R, G, B values — appearing neutral on screen. Auto white balance can drift during a videoconference (as lighting changes or people move). For professional videoconferencing, manual white balance set under stable room lighting conditions produces consistent, accurate color reproduction. Color temperature of room lighting should ideally be consistent (all sources within 200K of each other).
Question 5: In AV control system programming, what is the difference between a 'set' command and a 'get' (query) command sent to a controlled device?
- 'Set' commands are sent via RS-232; 'get' commands are sent via TCP/IP
- 'Set' commands instruct a device to change state (e.g., power on, volume to 50%); 'get' or query commands request the device to report its current state (Correct answer)
- 'Set' and 'get' are identical — they both change the device state
- 'Get' commands are only available on IP-controlled devices, not RS-232
Correct answer: 'Set' commands instruct a device to change state (e.g., power on, volume to 50%); 'get' or query commands request the device to report its current state
Control system programming uses 'set' commands to send configuration changes to devices (output state, input selection, volume) and 'get' or query commands to retrieve the current device state for feedback to the control interface, enabling accurate status displays.
Robust AV control system design relies on feedback from controlled devices to confirm that commands were received and executed. 'Set' (or 'command') strings change device parameters. 'Get' (or 'query') strings request device status, and the device responds with its current state. For example, a DSP might respond to a volume query with the current gain level in dB. This feedback loop allows the control system to display accurate status on touch panels and handle failed commands gracefully. Both RS-232 and TCP/IP control protocols support bidirectional communication.
Question 6: What is throw ratio in projection, and how do you calculate the required projector throw distance for a 120-inch wide screen using a lens with a throw ratio of 1.5:1?
- Throw ratio = screen height ÷ lens focal length; distance = 120 ÷ 1.5 = 80 inches
- Throw ratio = throw distance ÷ screen width; distance = throw ratio × screen width = 1.5 × 120 = 180 inches (Correct answer)
- Throw ratio = projector lumens ÷ screen size; distance = 1.5 × 120 lm = 180 lm
- Throw ratio = screen area ÷ projector area; distance = 1.5 × 120 = 180 inches
Correct answer: Throw ratio = throw distance ÷ screen width; distance = throw ratio × screen width = 1.5 × 120 = 180 inches
Throw ratio = throw distance ÷ screen width (or image width). Rearranging: throw distance = throw ratio × screen width = 1.5 × 120 inches = 180 inches (15 feet). This tells you how far from the screen to mount the projector.
Throw ratio (TR) is defined as: TR = D / W, where D = throw distance (lens to screen) and W = projected image width. Rearranging: D = TR × W. For TR = 1.5 and W = 120 inches: D = 1.5 × 120 = 180 inches = 15 feet. Short-throw lenses have ratios below 0.8; standard zoom lenses range from 1.3–2.5; long-throw lenses exceed 3.0. Variable zoom lenses have a range expressed as (e.g., 1.5–2.0:1). This calculation is a common CTS exam question type.
CTS Skills 2
What is the correct procedure for calculating amplifier power requirements for a loudspeaker system that must achieve 100 dB SPL at 10 meters, using a loudspeaker rated at 97 dB sensitivity (1W/1m)?