Analog Transmission

Analog Transmission

Binary amplitude shift keying
Binary phase shift keying
QPSK

Example


Find the bandwidth for a signal transmitting at 12 Mbps for QPSK. The value of d = 0.

Solution
For QPSK, 2 bits is carried by one signal element.
This means that r = 2.
So the signal rate (baud rate) is S = N × (1/r) = 6 Mbaud.
With a value of d = 0, we have B = S = 6 MHz.


Concept of a constellation diagram
Constellation diagrams for ASK (OOK), BPSK, and QPSK
Constellation diagrams for some QAMs

Modem Standards

Modem = Modulator/Demodulator
Telephone modem:

Telephone Line Bandwidth

V.32 Modem

Use 32-QAM
Data is divided into 4-bit sections, each adding a redundant bit to form a 5-bit => reduce value density => reduce noise interference (how?)
baud rate = 2400 (why?)


V.32bis Modem

1st modem standard to support 14,400bps
Can adjust upstream or downstream speed depending on line or signal quality
Use 128-QAM => 6-bit data => Bit rate = 14,400 bps

V.34bis Modem

960-point constellation => bit rate = 28,800 bps
1664-point constellation => bit rate = 33,600 bps

33.6Kbps: max bit rate of traditional modems


Traditional Modems

56K Modem: V.90
56K Modem: V.92

Similar to V.90
Modem can adjust speed
If noise allows => upload max 48 Kbps, download still 56 Kbps

V.92: can interrupt the Internet connection when there is an incoming call (if call-waiting service is installed)