Fourier Series (cont.)

1. For a periodic signal s(t) of period P = 1/f, Fourier's showed that

   s(t) = (a_o)/(2) + _n=1 ( a_n cos( 2 n f t) + b_n sin(2 n f t ) )

   where

   a_n = (2)/(P) ₀^P s(t) cos( 2 n f t ) dt

   and

   b_n = (2)/(P) ₀^P s(t) sin( 2 n f t ) dt

2. Last time, we determined the Fourier coefficients for one component of the encoding of a digital signal using frequency modulation. Namely the d(t) sin( 2 f₁ t) term of the frequency modulated signal described by

   d(t) sin( 2 f₁ t) + (1 - d(t))sin( 2 f₂ t)

   In particular, we concluded that

   d(t) sin( 2 f₁ t) = (sin( 2 f₁ t))/(2) + _{n = 1 and n odd} (1)/(n)( cos( 2 (f₁ - n f) t ) + cos( 2 (f₁ + n f) t )

   • This still may not look like a sine wave to you, but recall that a cosine wave is just a sine wave with a phase shift. So, we really have expressed s₁(t) as the sum of sine waves with frequencies of the form (f₁ ±n f).

   • Unfortunately, we again have an infinite set of sine waves.

   • Fortunately, the higher frequency waves have smaller coefficients AND the frequencies are centered around the carrier frequency.

   • As we expected, the higher the data rate, the wider the spread of frequencies will be.

   • Fortunately, in "real" systems, replacing the square wave with squares with rounded corners can further reduce the range of frquencies that are significant.

   • If we took the time to account for the other component of the transmitted signal, (1 - d(t))sin( 2 f₂ t), we would obtain a similar series of sine waves with frequencies of the form (f₂ ±n f)

   • Thus, the spectrum of the actual signal will consist of two sequences of sine waves each centered around one or the frequencies used as a carrier wave.

3. If nothing else, these results should let you understand what is so wonderful about fiber. (The following figures are meant more to be suggestive of the scale of things than precisely grounded in reality.)

   • The spectrum of visible light falls roughly between 2.5 x 10¹⁵ and 5 x 10¹⁵Hertz. Thus, the range of frequencies available for multiplexed signals in this range is about 2.5 x 10¹⁵ hertz.

   • When transmitting at one Gigahertz, the span of the first five terms of a frequency modulated signal would be about 10¹⁰ Hertz.

   • The spectrum of visible light could be divided into roughly 2.5 x 10⁵ channels each capable of carrying a Gigahertz.

4. Now, think about what would happen if instead of using two frequencies, we used four frequencies to encode 2 bits per baud.

   • If we choose the new frequencies so that they fall between the two original frequencies, the range of the spectrum required for the signal will not increase!

   • Thus, in this case, the bandwidth available limits the baud rate, but not necessarily the bit rate.

   • What limits the bit rate is noise on the line (and in the receiver) which limits how closely together the frequencies used can fall before we can no longer distinguish them.