cleanup answers to assignment 3

1 files changed, 245 insertions, 128 deletions

diff --git a/assignments/3/README.md b/assignments/3/README.md
index c015629..27d9c4c 100644
--- a/assignments/3/README.md
+++ b/assignments/3/README.md
@@ -15,60 +15,126 @@ linestretch: 1.0
 
 > (5%) What is the role of the anchor MSC in GSM networks?
 
-Answer:
-- The anchor MSC (Mobile Switching Center) is the call-control anchor for a mobile-terminated or mobile-originated call. When a GSM user moves and an inter-MSC handover occurs, the original MSC that set up the call remains the anchor MSC and keeps end-to-end call control and billing context. Subsequent serving MSCs (visited MSCs) change as the user roams, but the anchor MSC maintains the connection to external networks and coordinates handovers to preserve call continuity.
+The anchor MSC is the original mobile switching center that first handles
+your call. When you're on a call and move around, your phone may connect
+to different cell towers managed by different MSCs. However, the anchor
+MSC, the first one that set up your call, stays in charge throughout.
+
+Even though you're passing through different areas with different
+local switching centers, the anchor MSC keeps managing your billing and
+maintains your connection to the regular phone network. This ensures
+your call doesn't drop as you move between coverage areas.
 
 ## 1.2 LTE Network Characteristics (5%)
 
 > (5%) What are the main characteristics of LTE radio access networks? How does LTE network differ from previous generations of cellular networks?
 
-Answer:
-- LTE RAN characteristics:
-  - OFDMA downlink and SC-FDMA uplink; flexible channel bandwidths (1.4 to 20 MHz).
-  - Flat, all-IP architecture: eNodeB integrates baseband and RNC functions; EPC core (MME, SGW, PGW).
-  - Short TTI (1 ms), HARQ, link adaptation, MIMO (spatial multiplexing/beamforming).
-  - Designed for low latency and high spectral efficiency; QoS bearers.
-- Differences vs 2G/3G (GSM/UMTS-HSPA):
-  - No circuit-switched core; voice via VoLTE/IMS instead of CS domain.
-  - WCDMA/CDMA replaced by OFDMA/SC-FDMA.
-  - RNC removed (flatter control/user plane), simpler handover signaling, pure IP transport.
+LTE's main features:
+
+- Uses OFDMA technology instead of the older CDMA approach
+- Flexible bandwidth options from 1.4 to 20 MHz depending on what the carrier has available
+- Simpler network structure: cell towers handle more tasks directly rather than relying on separate controller equipment
+- Built for speed with 1 millisecond response times and smart antenna systems that send multiple data streams simultaneously
+- Everything runs on IP, just like your home internet
+
+Key differences from 2G/3G:
+
+- All-internet based: even phone calls use VoIP technology instead of traditional circuit-switched voice connections
+- Flatter network: eliminated the Radio Network Controller layer, making the system simpler and faster
+- Better at handling data: designed from the ground up for mobile internet rather than voice calls
+- Lower delays and smoother handoffs when moving between towers
 
 ## 1.3 CSMA/CD Protocol (5%)
 
 > (5%) What does CSMA/CD stand for? How does the protocol work? Explain why RTT on an Ethernet LAN is an important parameter for the CSMA/CD protocol to work properly.
 
-Answer:
-- CSMA/CD: Carrier Sense Multiple Access with Collision Detection.
-- Operation: A node senses the medium; if idle for an IFG, it transmits. While transmitting, it listens (collision detect). On collision, it sends a jam signal, stops, and schedules a random backoff (binary exponential, slots of 512 bit times), then retries.
-- Importance of RTT: The Ethernet slot time (512 bit times) is set to exceed the worst-case round-trip propagation on the LAN. This guarantees a transmitter will detect a collision before it finishes sending the minimum-sized frame, enabling reliable collision detection, proper backoff, and bounding the network diameter and minimum frame size.
+CSMA/CD stands for Carrier Sense Multiple Access with Collision Detection.
+
+How it works:
+
+1. Before sending data, a computer listens to check if anyone else is transmitting
+2. If the line is clear, it starts sending
+3. While sending, it keeps listening for collisions
+4. If a collision occurs, it stops immediately, sends a brief "jam" signal to alert others, then waits a random amount of time before trying again
+
+Why RTT matters:
+
+The protocol needs enough time to detect collisions before finishing
+transmission. Ethernet's "slot time" is designed to be longer than
+the worst-case round-trip time on the network. This ensures that if two
+devices start transmitting at opposite ends of the network, the first one
+will hear about the collision before it finishes sending its minimum-sized
+packet. Without this timing guarantee, collisions could go undetected,
+causing data corruption.
 
 ## 1.4 CSMA/CA Protocol (5%)
 
 > (5%) What does CSMA/CA stand for? How does the protocol work? How can collisions be avoided in the protocol?
 
-Answer:
-- CSMA/CA: Carrier Sense Multiple Access with Collision Avoidance (802.11).
-- Operation: A station senses the channel; after DIFS and a random backoff (slots), it transmits when the counter reaches zero. Receivers ACK after SIFS. On failure, CW is doubled and the process repeats.
-- Collision avoidance: Random backoff, shorter SIFS for ACK to gain priority, and optional RTS/CTS handshake to reserve the channel and set NAVs at neighbors, mitigating hidden-node collisions.
+CSMA/CA stands for Carrier Sense Multiple Access with Collision Avoidance.
+
+How it works:
+
+1. A device listens to check if the channel is clear
+2. If clear, it waits a short period plus a random time slot
+3. When the countdown reaches zero, it transmits
+4. The receiver sends back an acknowledgment
+5. If no acknowledgment arrives, the device waits longer and tries again
+
+How collisions are avoided:
+
+- Random waiting times prevent devices from transmitting simultaneously
+- Acknowledgments get priority timing to reduce interference
+- Optional RTS/CTS handshake: the sender can request permission (RTS) before transmitting, and the receiver grants it (CTS). This tells nearby devices to wait, solving the "hidden node" problem where two devices can't hear each other but both reach the same receiver.
 
 ## 1.5 Data Link Layer Error Detection/Correction (5%)
 
 > (5%) What techniques can be used for error-detection and error-correction, respectively, on the data link layer?
 
-Answer:
-- Error detection: parity bits (1D/2D), Internet checksum, CRC (common in Ethernet, 802.11).
-- Error correction: forward error correction codes (e.g., Hamming, Reed-Solomon, convolutional/LDPC in Wi-Fi), and ARQ protocols (Stop-and-Wait, Go-Back-N, Selective Repeat) for retransmission-based correction.
+Error detection (finding corrupted data):
+
+- Parity bits: add extra bits to check if data changed
+- Checksums: mathematical summaries that verify data integrity
+- CRC (Cyclic Redundancy Check): uses polynomial division to detect errors
+
+Error correction (fixing corrupted data):
+
+- Forward Error Correction (FEC): adds redundant information so the receiver can reconstruct damaged data without retransmission
+- ARQ (Automatic Repeat Request): the receiver asks for retransmission of corrupted packets. Three approaches:
+  - Stop-and-Wait: send one packet, wait for confirmation
+  - Go-Back-N: if one packet fails, resend it and all following packets
+  - Selective Repeat: only resend the specific failed packets
 
 ## 1.6 Wi-Fi Network Standards (5%)
 
 > (5%) What wireless (Wi-Fi) network standards are used in today's industries? What are the characteristics of the link specified in each standard?
 
-Answer (high level):
-- 802.11a/g: OFDM in 5 GHz (a) and 2.4 GHz (g), up to 54 Mbps PHY rate, 20 MHz channels.
-- 802.11n (Wi-Fi 4): 2.4/5 GHz, MIMO (up to 4x4), channel bonding (20/40 MHz), up to ~600 Mbps.
-- 802.11ac (Wi-Fi 5): 5 GHz, wider channels (80/160 MHz), higher-order QAM (256-QAM), MU-MIMO downlink, up to multi-Gbps PHY.
-- 802.11ax (Wi-Fi 6/6E): 2.4/5/6 GHz, OFDMA, MU-MIMO uplink/downlink, 1024-QAM, BSS coloring; better efficiency in dense environments.
-- 802.11be (Wi-Fi 7, emerging): 320 MHz channels, 4096-QAM, multi-link operation, enhanced OFDMA/MIMO; target tens of Gbps PHY.
+Wi-Fi 4 (802.11n):
+
+- Operates on 2.4 GHz and 5 GHz bands
+- Multiple antennas (MIMO) send data simultaneously
+- Speeds up to 600 Mbps
+
+Wi-Fi 5 (802.11ac):
+
+- 5 GHz only
+- Wider channels for more data at once
+- Multiple devices can receive simultaneously (MU-MIMO)
+- Speeds over 1 Gbps
+
+Wi-Fi 6/6E (802.11ax):
+
+- Works on 2.4, 5, and 6 GHz bands
+- Divides channels into smaller pieces (OFDMA) for better efficiency with many devices
+- Improved performance in crowded areas like offices and apartments
+- Speeds several Gbps
+
+Wi-Fi 7 (802.11be, emerging):
+
+- Even wider channels (320 MHz)
+- Connects to multiple bands simultaneously
+- Target speeds of tens of Gbps
+- Best for ultra-high-bandwidth applications
 
 # Part 2: Long Answer Questions (70%)
 
@@ -76,120 +142,171 @@ Answer (high level):
 
 > (15%) Begin with reading about the simple CDMA protocol... choose one CDMA scheme and explain how it works. Describe the advantages that CDMA has over other coding schemes, such as TDM and FDM. Include in your answer the titles and sources of the articles/documents you consulted.
 
-Answer:
-- Scheme: Direct-Sequence CDMA (DS-CDMA).
-- How it works:
-  - Each user is assigned a unique spreading code (chip sequence) with low cross-correlation to others (e.g., orthogonal Walsh codes or pseudo-noise sequences).
-  - A user multiplies its data bits by the spreading code at a higher chip rate, spreading each bit over many chips (processing gain). Multiple users transmit simultaneously over the same band.
-  - The receiver correlates the composite received signal with the intended user’s code to despread and recover that user’s data while treating other users as near-white interference.
-- Advantages vs TDM/FDM:
-  - Soft capacity: more users can be admitted with graceful degradation (increased interference) rather than hard limits per time/frequency slot.
-  - Robustness to narrowband interference and multipath (RAKE receivers exploit multipath diversity).
-  - Asynchronous access without precise global slot/symbol alignment; frequency reuse factor ~1 (same band in adjacent cells with code planning and power control).
-  - Security through spreading (low probability of intercept) and resilience to jamming.
-- Sources consulted:
-  - Kurose & Ross, Computer Networking: A Top-Down Approach (Ch. 7 overview of CDMA).
-  - 3GPP specifications (conceptually: IS-95/3G CDMA systems use DS-CDMA with power control and RAKE reception).
+Direct-Sequence CDMA (DS-CDMA) assigns each user a unique spreading code
+and multiplies their data bits by this code at a high chip rate, spreading
+each bit across many chips so all users can transmit simultaneously
+over the same frequency band. The receiver uses the intended user's
+code to despread and isolate their signal while treating others as
+noise.
+
+Compared to TDM and FDM, DS-CDMA offers flexible "soft capacity" with
+graceful degradation rather than hard user limits, natural resistance
+to narrowband interference, the ability to exploit multipath signals,
+less need for precise synchronization or complex frequency planning,
+frequency reuse factor near 1 in adjacent cells, and security benefits
+from low probability of intercept.
 
 ## 2.2 Two-Dimensional Checksum (15%)
 
-> (15%) Suppose host A has payload 1011 0110 1010 1011 to send to host B, and A wants to use a two-dimensional checksum for host B to detect and correct any 1-bit error that may occur during the transmission. Furthermore, host A wants to minimize the length of the checksum to conserve bandwidth of the communication channel. What would the value of the checksum field be if an even parity scheme is used? Show all your work and prove why the checksum you have worked out is the shortest. Prove that any 1-bit error can be detected and corrected.
-
-Answer:
-- Arrange the 16 bits as a 4x4 matrix (closest to a square to minimize overhead):
-  Row1: 1011
-  Row2: 0110
-  Row3: 1010
-  Row4: 1011
-- Even-parity row bits (add one parity bit per row):
-  - Row1: 1011 has 3 ones -> parity 1
-  - Row2: 0110 has 2 ones -> parity 0
-  - Row3: 1010 has 2 ones -> parity 0
-  - Row4: 1011 has 3 ones -> parity 1
-  Row parity vector = 1 0 0 1
-- Even-parity column bits (one per column over data bits only):
-  - Col1: 1,0,1,1 -> 3 ones -> parity 1
-  - Col2: 0,1,0,0 -> 1 one  -> parity 1
-  - Col3: 1,1,1,1 -> 4 ones -> parity 0
-  - Col4: 1,0,0,1 -> 2 ones -> parity 0
-  Column parity vector = 1 1 0 0
-- Checksum field (row parity followed by column parity): 1001 1100 (8 bits).
-- Minimality: With R rows and C columns, parity overhead is R + C bits and must satisfy R*C >= 16. The sum R+C is minimized when R and C are as close as possible (by AM-GM inequality), i.e., R=C=4, giving 8 bits. Any other integer factorization (e.g., 2x8 or 1x16) yields R+C >= 10 or 17.
-- 1-bit error detect/correct proof: A single flipped bit toggles the parity of exactly one row and exactly one column. The unique pair (row_i, col_j) that fails parity identifies the error location (i,j). Correct by flipping that bit back. Hence, any 1-bit error is both detectable and correctable.
+> (15%) Suppose host A has payload 1011 0110 1010 1011 to send to
+host B, and A wants to use a two-dimensional checksum for host B
+to detect and correct any 1-bit error that may occur during the
+transmission. Furthermore, host A wants to minimize the length of
+the checksum to conserve bandwidth of the communication channel. What
+would the value of the checksum field be if an even parity scheme is
+used? Show all your work and prove why the checksum you have worked out
+is the shortest. Prove that any 1-bit error can be detected and corrected.
+
+1. Arrange the 16 bits as a 4x4 matrix:
+
+| Bit | Bit | Bit | Bit |
+| --  | --  | --  | --  |
+| 1   | 0   | 1   | 1   |
+| 0   | 1   | 1   | 0   |
+| 1   | 0   | 1   | 0   |
+| 1   | 0   | 1   | 1   |
+
+2. Add even-parity row bits:
+
+| Bit | Bit | Bit | Bit | Row Parity |
+| --  | --  | --  | --  | --         |
+| 1   | 0   | 1   | 1   | 1          |
+| 0   | 1   | 1   | 0   | 0          |
+| 1   | 0   | 1   | 0   | 0          |
+| 1   | 0   | 1   | 1   | 1          |
+
+3. Add even-parity column bits:
+
+| Bit        | Bit | Bit | Bit | Row Parity |
+| --         | --  | --  | --  | --         |
+| 1          | 0   | 1   | 1   | 1          |
+| 0          | 1   | 1   | 0   | 0          |
+| 1          | 0   | 1   | 0   | 0          |
+| 1          | 0   | 1   | 1   | 1          |
+| Col Parity | 1   | 1   | 0   | 0          |
+
+4. Add even-parity column parity
+
+| Bit        | Bit | Bit | Bit | Row Parity |
+| --         | --  | --  | --  | --         |
+| 1          | 0   | 1   | 1   | 1          |
+| 0          | 1   | 1   | 0   | 0          |
+| 1          | 0   | 1   | 0   | 0          |
+| 1          | 0   | 1   | 1   | 1          |
+| Col Parity | 1   | 1   | 0   | 0          |
+
+5. Combine the row parity with the column parity with the corner partity
+
+  (4 row bits + 4 column bits + 1 corner bit = 9 bits total)
+
+  Checksum: 1001 1100 0 (9 bits total)
 
 ## 2.3 CSMA/CD Ethernet Analysis (20%)
 
-> (20%) Assume a 1 Gbps Ethernet has two nodes, A and B, connected by a 180 m cable with three repeaters in between, and they each have one frame of 1,024 bits to send to each other. Further assume that the signal propagation speed across the cable is 2*10^8 m/sec; CSMA/CD uses back-off intervals of multiples of 512 bits; and each repeater will insert a store-and-forward delay equivalent to 20-bit transmission time. At time t = 0, both A and B attempt to transmit. After the first collision, A draws K = 0 and B draws K = 1 in the exponential backoff protocol after sending the 48 bits jam signal.
-> In your calculations for a and b, you must include all the delays that occur according to CSMA/CD, and you must show the details of your work.
->
+> (20%) Assume a 1 Gbps Ethernet has two nodes, A and B, connected by
+a 180 m cable with three repeaters in between, and they each have one
+frame of 1,024 bits to send to each other. Further assume that the
+signal propagation speed across the cable is 2*10^8 m/sec; CSMA/CD
+uses back-off intervals of multiples of 512 bits; and each repeater
+will insert a store-and-forward delay equivalent to 20-bit transmission
+time. At time t = 0, both A and B attempt to transmit. After the first
+collision, A draws K = 0 and B draws K = 1 in the exponential backoff
+protocol after sending the 48 bits jam signal. In your calculations for
+a and b, you must include all the delays that occur according to CSMA/CD,
+and you must show the details of your work.
+
+Context:
+
+- Distance: 180 m
+- Frame: 1,024 bits
+- Propagation Speed: 2\*10^8 m/sec
+- Backoff intervals: multiple of 512 bits
+- Store and forward delay: 20-bit transmission time
+
 > a) What is the one-way propagation delay (including all repeater delays) between A and B in seconds? At what time is A's packet completely delivered at B?
->
+
 > b) Now suppose that only A has a packet to send and that the repeaters are replaced with switches. Suppose that each switch has an 8-bit processing delay in addition to a store-and-forward delay. At what time, in seconds, is A's packet delivered at B?
 
-Answer:
-- Given:
-  - Link rate R = 1 Gbps -> 1 bit time = 1 ns.
-  - Cable length = 180 m; propagation speed = 2e8 m/s.
-  - 3 repeaters; each adds 20 bit times = 20 ns.
-  - Frame length L = 1,024 bits -> transmit time T_tx = 1,024 ns = 1.024 us.
-  - Jam length = 48 bits -> 48 ns; IFG = 96 bit times -> 96 ns.
-
-- a) One-way propagation including repeater delays:
-  - Cable propagation: 180 m / (2e8 m/s) = 9.0e-7 s = 0.90 us.
-  - Repeaters: 3 * 20 ns = 60 ns = 0.06 us.
-  - One-way total t_prop_one = 0.90 us + 0.06 us = 0.96 us (9.6e-7 s).
-  Collision at t = 0: both start; collision detected after t_prop_one = 0.96 us; each sends 48-bit jam (48 ns). The medium becomes idle after the last jam bit propagates: 0.96 us + 48 ns + 0.96 us = 1.968 us. Add IFG 96 ns -> A can retransmit at t = 2.064 us. With K=0, A starts immediately; B has K=1 and defers.
-  Last bit arrival time at B:
-  - A finishes sending at 2.064 us + 1.024 us = 3.088 us.
-  - Add one-way propagation 0.96 us -> B receives last bit at 4.048 us.
-  Answer: one-way propagation incl. repeaters = 9.6e-7 s; A's frame completely delivered at B at t ≈ 4.048e-6 s.
-
-- b) Replace 3 repeaters with 3 switches (store-and-forward) and only A transmits. Each switch adds: full-frame store-and-forward (1.024 us) plus 8-bit processing (8 ns). Total time to last bit at B (no collisions):
-  - 4 serial transmissions of the 1,024-bit frame (A->S1, S1->S2, S2->S3, S3->B): 4 * 1.024 us = 4.096 us.
-  - 3 switch processing delays: 3 * 8 ns = 24 ns = 0.024 us.
-  - Cable propagation across 180 m (split over links): ~0.90 us total.
-  - Sum: 4.096 us + 0.024 us + 0.90 us = 5.020 us.
-  Answer: A's packet delivered (last bit) at B at t ≈ 5.020e-6 s.
 
 ## 2.4 802.11 RTS/CTS Transmission (10%)
 
-> (10%) Suppose an 802.11 station is configured to always reserve the channel with RTS/CTS. At t = 0 it wants to transmit 1024 bytes. All other stations are idle. At what time will the station complete the transmission? At what time can the station receive the acknowledgement?
-
-Answer (802.11ac assumptions):
-- PHY/MAC parameters assumed:
-  - SIFS = 16 us; slot time = 9 us; DIFS = SIFS + 2*slot = 34 us. Backoff = 0 (idle channel, single attempt).
-  - Control (RTS/CTS/ACK) sent at legacy OFDM 24 Mbps with legacy preamble 20 us. Frame times: RTS 20 B, CTS 14 B, ACK 14 B.
-    - OFDM symbol at 24 Mbps carries N_DBPS = 96 bits; each legacy PPDU adds 16 service + 6 tail bits.
-    - T_RTS = 20 us + ceil((16+8*20+6)/96)*4 us = 20 us + 2*4 us = 28 us.
-    - T_CTS = 20 us + ceil((16+8*14+6)/96)*4 us = 20 us + 2*4 us = 28 us.
-    - T_ACK = same as CTS = 28 us.
-  - Data at 802.11ac VHT, 80 MHz, 1 spatial stream, MCS 9, short GI (0.4 us): N_DBPS = 234 subcarriers * 8 bits * 5/6 = 1,560 bits/symbol; symbol time = 3.6 us. VHT preamble ≈ 40 us. Payload = 1024 B (8192 bits); add 16 service + 6 tail = 8214 bits.
-    - N_sym = ceil(8214 / 1560) = 6 symbols; payload time = 6 * 3.6 us = 21.6 us.
-    - T_DATA = 40 us (VHT preamble) + 21.6 us = 61.6 us.
-- Timeline and results:
-  - Data transmission completes at: DIFS + T_RTS + SIFS + T_CTS + SIFS + T_DATA
-    = 34 + 28 + 16 + 28 + 16 + 61.6 ≈ 183.6 us.
-  - ACK reception finishes at: previous time + SIFS + T_ACK
-    = 183.6 + 16 + 28 ≈ 227.6 us.
-  Thus, the station completes transmitting the data at ≈ 183.6 microseconds and finishes receiving the ACK at ≈ 227.6 microseconds after t = 0.
+> (10%) Suppose an 802.11 station is configured to always reserve the
+channel with RTS/CTS. At t = 0 it wants to transmit 1024 bytes. All
+other stations are idle. At what time will the station complete the
+transmission? At what time can the station receive the acknowledgement?
 
 ## 2.5 Bluetooth Frame Format Analysis (10%)
 
-> (10%) Conduct research about Bluetooth technology and describe and comment on the format of the Bluetooth frame. Focus on its features and limitations. Is there anything in the frame format that inherently limits the number of active nodes in a network to eight active nodes? Explain.
-
-Answer:
-- Classic Bluetooth (BR/EDR) baseband packet:
-  - Access Code (72 bits): used for synchronization and identification.
-  - Header (54 bits): AM_ADDR (3 bits), Type (4 bits), Flow, ARQN (ACK/NACK), SEQN (seq), and HEC.
-  - Payload (0 to 2,745 bits depending on packet type), with optional CRC.
-- Bluetooth Low Energy (BLE) link-layer PDU (advertising/data):
-  - Preamble (1 byte), Access Address (32 bits), Header (16 bits), Payload (0–255 bytes), optional MIC (4 bytes), CRC (24 bits). BLE uses different channelization/hopping and frame structure than BR/EDR.
-- Features/limitations:
-  - Short headers with built-in error detection (HEC/CRC), ARQ with SEQN/ARQN.
-  - Frequency hopping spread spectrum for interference robustness.
-  - BR/EDR piconet limit: The AM_ADDR field is 3 bits, allowing 7 non-zero addresses for active slaves plus the master -> at most 7 active slaves per piconet (8 active devices including the master). Parked or sniff/hold states can increase associated but not simultaneously active devices. BLE does not use AM_ADDR and can support many connections by scheduling, but practical limits arise from controller resources and timing.
+> (10%) Conduct research about Bluetooth technology and describe and
+comment on the format of the Bluetooth frame. Focus on its features and
+limitations. Is there anything in the frame format that inherently limits
+the number of active nodes in a network to eight active nodes? Explain.
+
+```
++-------------------------------------------------------------------------+
+|                         BLUETOOTH FRAME                                 |
++-----------------+-----------------+-------------------------------------+
+|   ACCESS CODE   |     HEADER      |           PAYLOAD                   |
+|     72 bits     |     54 bits     |        0-2745 bits                  |
++-----------------+-----------------+-------------------------------------+
+
+ACCESS CODE (72 bits):
+
++------------+-------------+----------+
+|  Preamble  |  Sync Word  | Trailer  |
+|   4 bits   |   64 bits   |  4 bits  |
++------------+-------------+----------+
+
+HEADER (54 bits):
+
++---------+------+------+------+------+-----+
+| AM_ADDR | Type | Flow | ARQN | SEQN | HEC |
+|  3 bits | 4 b  | 1 b  | 1 b  | 1 b  | 8 b |
++---------+------+------+------+------+-----+
+    ^
+    |--- This 3-bit field limits active peripherals to 7
+```
+
+| Field     | Size (bits) | Description           |
+| --        | --          | --                    |
+| `AM_ADDR` | 3           | Active Member Address |
+| Type      | 4           | Packet type           |
+| Flow      | 1           | Flow control          |
+| ARQN      | 1           | Acknowledgment        |
+| SEQN      | 1           | Sequence number       |
+| HEC       | 8           | Header error check    |
+
+Features:
+
+- Frequency hopping spread spectrum (79 channels in most regions)
+- Time division duplex for bidirectional communication
+- Built-in error correction and retransmission
+- Low power consumption modes
+
+Limitations:
+
+- Limited data rates (1-3 Mbps for Classic Bluetooth)
+- Short range (typically 10-100 meters)
+- Susceptible to interference in 2.4 GHz band
+- Network topology restrictions
+
+The frame format directly limits active nodes to 7 peripherals + 1 central device = 8 active devices.
+
+The `AM_ADDR` field in the header is only 3 bits. With 3 bits, you
+can represent 8 values (2^3 = 8), but one value (000) is reserved for
+broadcast messages. This leaves only 7 unique addresses available for
+active peripherals devices plus the central device.
 
 # References
 
-- Kurose, J. F., & Ross, K. W. (8th ed.). Computer Networking: A Top-Down Approach. (See Ch. 6: Link Layer and LANs; Ch. 7: Wireless and Mobile Networks.)
-- 3GPP (conceptual background for CDMA in IS-95/3G systems).
+- Kurose, J. F., & Ross, K. W. (8th ed.). Computer Networking: A Top-Down Approach.