CMPS 376 - Lab 2 Solution

Lab 2 Solution

A modem constellation diagram has data points at the coordinates: (1,1), (1,-1), (-1,1) and (-1,-1) (eg QPSK encoding). How many bps can a modem with this encoding achieve at 1200 baud?
QPSK encodes 2 bits/symbol. rate = baud * bits/symbol = 1200 * 2 = 2400bps
A modem constellation diagram has data points at the coordinates (0,1) and (0,2). Does the modem use phase modulation or amplitude modulation?
This is amplitude modulation because both points are the same angle from the positive x axis but are different distances away from origin (0,0).
Now suppose all the points lie in a circle centered on the origin. What kind of modulation is being used?
This is phase modulation. All points are the same distance from origin but they are at different angles to the positive x axis.
An ADSL system using DMT allocates 3/4 of the available data channels to the download stream. It uses QAM-64 modulation on each channel. What is the maximum capacity of the downstream link?
With DMT, there are 256 channels in total, but 8 channels are for voice data, control or are reserved. This leaves 248 available data channels. If 75% are used for the download stream, then the download stream has 186 channels. QAM-64 encoding has 6 bits/symbol and DSL is 4000 baud. The calculation is:
```
4000 baud * 6 bits/symbol = 24,000 bps per channel
186 channels * 24,000 bps per channel = 4.464 Mbps 
```
Why has PCM sampling time been set at 125 usec?
Nyquist Theorem states that 8000 samples/sec is sufficient for a 4000Hz line. 8000 samples/sec is equivalent to 125usec/sample. Higher rates gain no additional data.
What is the percent overhead (percentage of bps used for control) on a T1 carrier for:
1. A voice T1
  A voice T1 has 24 channels with 8 bits per channel. Of those bits, 7 are for data and 1 is for control. The 24 channels are grouped into a frame which contains one additional framing bit.
```
Total bits = 24 * 8 + 1 = 193 bits
Control bits = 24 * 1 + 1 = 25 bits
25/193 = 12.95% overhead
```
2. A data T1
  A data T1 also has 24 channels, but one channel is used exclussively for control purposes.
```
Total bits = 193 bits (same as voice T1)
Control bits = 23 * 1 + 8 + 1 = 32 bits
32/193 = 16.58% overhead
```
The user data rate for OC-3 is stated on page 146 to be 148.608 Mbps. Show how this number is derived from the SONET OC-3 parameters.
An OC-3 frame is in essence 3 OC-1 frames. So first we must look at the overhead of a basic OC-1 frame. An OC-1 frame is 810 bytes arranges in 90 columns x 9 rows. A total of 8000 frames are sent per second. The first 3 columns are section overhead. There is also an additional 1 column of SPE path overhead. So there are 4 columns of overhead overall.
```
4 columns * 9 rows = 36 bytes overhead per frame
36 bytes/frame * 8 bits/byte = 288 bits/frame * 8000 frames/second = 2.304Mbps overhead
OC-1 total frame size = 51.84Mbps
OC-1 user data size = 51.84Mbps - 2.304Mbps = 49.536Mbps
OC-3 user data size = 3 * OC-1 user data size = 3 * 49.536Mbps = 148.608Mbps
```
What is the difference between packet switching and message switching?
In message switching, the whole message is sent in one block. It is a store and forward network. It can take up a lot of memory and disk space to store the whole message. Additionally, users can monopolize the channel with long messages. Packet switching places an upper bound on the size of each transmitted block. This size is usually small enough to buffer the packet in memory rather than requiring hard disks. No one user can monopolize the system with a long message because the message is broken down into smaller packets and other users packets can be transmitted inbetween the packets of the large message. In other words, not all packets of the message have to be transmitted in sequence. Packets can be propogated along the network individually instead of waiting for the whole message.
D-AMPS has appreciably worse speech quality than GSM. Is this due to the requirement that D-AMPS be backwards compatible with AMPS, whereas GSM had no such constraint? Discuss your answer.
Yes, it is due in part to the backwards compatability with AMPS. In particular, D-AMPS has the same channel size (30kHz) as AMPS. GSM has a 200kHz channel size. Additionally, GSM has 8 users sharing one channel while D-AMPS has 3 users sharing one channel. These two factors combined means that GSM has a higher data rate per user, which tends to result in better voice quality.
A CDMA receiver is within the range of four stations with the following chip sequences:
- A: (-1 -1 -1 +1 +1 -1 +1 +1)
- B: (-1 -1 +1 -1 +1 +1 +1 -1)
- C: (-1 +1 -1 +1 +1 +1 -1 -1)
- D: (-1 +1 -1 -1 -1 -1 +1 -1)
The reciever receives the following chips: (-1 +1 -3 +1 -1 -3 +1 +1). Which stations transmitted and which bits did the transmitting stations send? (Hint: look at page 163, Fig 2-45 for six examples for these four stations)
There are two ways to solve this problem. Either you could try various combinations of sending bits from the 4 stations (brute force) or you could take the normalized inner product (S*T) of each station and the transmitted message. The answer will show the later. Let S equal the received chips. Then:
```
S*A = (+1 -1 +3 +1 -1 +3 +1 +1)/8 = 1
S*B = (+1 -1 -3 -1 -1 -3 +1 -1)/8 = -1
S*C = (+1 +1 +3 +1 -1 -3 -1 -1)/8 = 0
S*D = (+1 +1 +3 -1 +1 +3 +1 -1)/8 = 1 
```
So the transmitting stations were A, B and D. A sent 1. B sent 0. D sent 1.