Major distinction: analog vs digital.

Analog information is continuous and has infinite number of values (traditional clock with moving hands, electrical current into your house)

Digital information is discrete and has limited number of values (clock with digital display, 0 or 1 for binary)

• Signal may be periodic (continuously repeated pattern -- like sine wave) or aperiodic.
• Periodic signal characterized by:

• Basic period unit is second.
• Basic frequency unit is Hertz.
• Basic phase unit is degrees or radians.
• Arbitrary periodic wave can be represented by composition of sine waves.

• Also characterized by Amplitude, Period and Phase.
• Represented by square waves.
• Can be approximated using analog waves.

• Time domain graphs Time on X-AXIS, Amplitude on Y-AXIS
• Frequency domain graphs Frequency on X-AXIS, Amplitude on Y-AXIS
• Frequency expressed in harmonics (defined below)
• Translate between these domains using Fourier analysis (details below) 

• Frequency graphed as Harmonic : multiple of "fundamental frequency"
• Harmonic is discrete.
• Frequency domain shows harmonic components of complex analog signal
• Complex signal is composition of sine waves, each having different harmonic and amplitude. This are discovered using Fourier Analysis.
• As number of harmonics increases, the approximation of original signal improves.
• As number of harmonics decreases, it becomes more difficult to accurately represent and recognize the signal.

• Bandwidth is width of frequency range (spectrum) available to coding signals.
• For binary (2 level) analog signals, the following (approximate) relationship holds.

• If bandwidth is fixed, then increase in xmit rate must be offset by decrease in # harmonics transmitted.
• Bandwidth of medium generally partitioned into fixed size subbands (channels), so fixed bandwidth is rule rather than exception.
• Usually, frequencies at channel boundaries cannot be used (guard).

Will use voice grade telephone as example: typical usable bandwidth is 3000 Hz.

Nyquist theorem (1924) provides theoretical upper bound on bits per second (BPS), assuming noiseless wire.

where:

D = max BPS rate

B = Baud rate (# signal changes per second - equivalent to bandwidth)

K = number of signal levels

As noted above, baud rate is limited by wire bandwidth. In the phone example, upper bound is about 3000 Hz.

So how can modems achieve high throughput (e.g. 9600, 14.4K, 28.8K BPS)? By increasing the value of K!

If the number of signal levels were two, you are limited in a practical (not theoretical) sense to 2400 BPS. Each doubling in the value of K adds another multiple of B. Notice then that BPS and Baud are not the same thing, although Baud is used almost universally to mean BPS.

Basically, a two-level signal allows you to encode one bit per signal change (or baud). A four-level signal allows you to encode two bits per baud (e.g. +30v is 00, +15v is 01, -15v is 10, -30v is 11). An eight-level signal allows to you encode three bits per baud, etc.

Nyquist applies only to noise-free. Extended by Shannon in 1948 to include noise.

where:

D = max BPS rate

H = available bandwidth

S/N = signal-to-noise ratio (thermal noise) on medium.

Regardless of number of signal levels. What is required S/N to double the D with fixed H?

Last reviewed: 20 January 1998