Free CWISE sample questions

Correct answer: Chirp Spread Spectrum (CSS)

Chirp Spread Spectrum (CSS) uses wideband linear frequency modulated chirp pulses to encode information. This technique is inherently resistant to multipath fading and Doppler effect, making it ideal for harsh industrial environments with metallic obstructions. ASK is highly susceptible to noise and fading. OFDM is robust but typically requires higher power and processing overhead than CSS for simple sensor networks. FSK is robust but less resilient to severe multipath than CSS in this context.

Correct answer: Connectivity Standards Alliance (CSA)

The Connectivity Standards Alliance (CSA), formerly the Zigbee Alliance, is the organization responsible for developing and maintaining the Matter specification. IEEE develops physical and MAC layer standards (like 802.15.4). IETF develops internet protocols (like 6LoWPAN). The Wi-Fi Alliance focuses on WLAN interoperability.

Correct answer: False

False. The noise floor is dynamic and changes based on environmental factors, new electronic device deployments, and other RF interference sources. In a smart city, the RF environment is particularly volatile due to the constant addition of new infrastructure and consumer devices. Continuous monitoring is required.

Correct answer: Full Mesh Topology

A Full Mesh topology allows every node to communicate directly with every other node (assuming they are within range), facilitating peer-to-peer communication without reliance on a central coordinator for local logic. A Star topology requires all traffic to pass through a central hub. A Cluster Tree is hierarchical and may not support arbitrary peer-to-peer without routing up the tree.

Correct answer: Time Slotted Channel Hopping (TSCH)

Time Slotted Channel Hopping (TSCH) is the MAC mechanism used by WirelessHART (and ISA100.11a). It combines Time Division Multiple Access (TDMA) with channel hopping. For every timeslot, a different frequency channel is used based on a hopping sequence, effectively mitigating frequency-selective fading and persistent interference on any single channel.

Correct answer: Class C

Class C devices have continuously open receive windows, closing only when transmitting. This allows for low-latency downlink communication initiated by the server at any time. Class A only receives after a transmission. Class B opens receive windows at scheduled intervals (beacons), which has higher latency than Class C. Class C consumes the most power but meets the low-latency downlink requirement. sequenceDiagram participant Server participant Device_ClassC Note over Device_ClassC: Always Listening Server->>Device_ClassC: Downlink (Immediate) Device_ClassC-->>Server: Ack (if confirmed)

Correct answers: BLE uses 40 channels with 2 MHz spacing, while Classic uses 79 channels with 1 MHz spacing; BLE supports sleep modes that allow for years of operation on a coin-cell battery

BLE utilizes 40 channels (3 advertising, 37 data) spaced 2 MHz apart in the 2.4 GHz ISM band. Bluetooth Classic uses 79 channels spaced 1 MHz apart.

BLE is optimized for short bursts of data transmission followed by long sleep periods, enabling extremely low power consumption suitable for coin-cell operation. Bluetooth Classic is connection-oriented and designed for continuous data streams like audio.

Correct answer: QoS 0 (At most once)

QoS 0 is 'fire and forget'. The message is delivered at most once, meaning it might be lost, but it will never be duplicated. This matches the requirement where missing a message is acceptable but duplicates are not. QoS 1 guarantees delivery but may result in duplicates. QoS 2 guarantees exactly once delivery but has higher overhead.

Correct answer: Header Compression and Fragmentation

The 6LoWPAN adaptation layer performs two critical functions: Header Compression (HC) to reduce the IPv6 header size, and Fragmentation/Reassembly to split IPv6 packets (MTU 1280 bytes) into 802.15.4 frames (max payload ~127 bytes). NAT64 is for translation between IPv6 and IPv4. DHCPv6 handles addressing, not sizing.

Correct answer: -62 dBm

RSL = Tx Power - Cable Loss + Tx Antenna Gain - FSPL + Rx Antenna Gain - Cable Loss (assuming split or total). Using total cable loss: RSL = 20 dBm - 2 dB + 10 dBi - 100 dB + 10 dBi = 38 - 100 = -62 dBm. Calculation: Tx Power: +20 Total Cable Loss: -2 Total Antenna Gain: +20 (10+10) FSPL: -100 Total: 20 - 2 + 20 - 100 = -62 dBm. graph LR Tx[Tx Power: 20 dBm] --> L1[Loss: -1 dB] L1 --> Ant1[Gain: +10 dBi] Ant1 -- FSPL: -100 dB --> Ant2[Gain: +10 dBi] Ant2 --> L2[Loss: -1 dB] L2 --> Rx[Rx Signal: -62 dBm]