Communication Principles

• Frequency Spectrum
• Modulation and Demodulation
• Phase-locked loop
• Nyquist's Sampling Theorem

Notes:

Frequency Transform

• given a complex function x(t), -oo < t < +oo$.
• j^2 = -1$
• e^{j2\pi ft} = \cos(2 \pi f t) + j\sin(2 \pi f t)$
• Fourier (frequency) transform:

x(t) = \int[-oo]^{+oo}X(f)e^{j2\pi ft}\,df$

• for some X(f), the frequency spectrum of x(t)$
• in every case we will look at, X(f)$ is symmetric about zero

Notes:

Frequency Transform for
Periodic Functions

• if x(t) is periodic with period T (f = 1/T$)
• transform of x(t)$ is a sum instead of an integral:

x(t) = \sum[n=-oo]^{+oo}c[n] e^{j(2\pi n/T)t}$

cn = (1 / T) \int[-T/2]^{+T/2} x(t) e^{-j(2\pi n / T)t}\,dt$

Example of Periodic Functions:
Cosine Wave

• x(t) = \cos(2\pi f1 t)$
• since e^{j \theta} = \cos(\theta) + j \sin(\theta)$
• and \cos(-x) = \cos(x) whereas \sin(-x) = -\sin(x)$
• therefore \cos(\theta) = (1 / 2) (e^{j\theta} + e^{-j\theta})$
• therefore \cos(2\pi f1 t) = (1 / 2) (e^{j2\pi f1 t} + e^{-j2\pi f1 t})$
• which is a form of:

x(t) = \sum[n=-oo]^{+oo}c[n] e^{j(2\pi n/T)t}$ with c[(-1)] = c1 = (1 / 2) , and cn = 0 for all other n$.

Square Wave

• x(t) = 1 for (4n-1)T / 4 <= t < (4n + 1) T / 4$
• x(t) = 0$ otherwise
• decomposing into sine waves:

x(t) = \sum[n=-oo]^{+oo}c[n] e^{j(2\pi n/T)t}$

Notes:

Dirac Impulse

x(s) = \int[-oo]^{+oo} x(t) delta(t - s) \, dt$ $\sum[n=-oo]^{oo} delta(t - nT) = (1 / T) \sum[n=-oo]^{oo} e^{j(2\pi n/T)t}$

• dirac impulse samples the signal
• integral of dirac impulse is always one

Notes:

Amplitude Modulation

• x(t) = Ae^{j2\pi f1 t}$ is the signal to be modulated
• assume f0 > f1$ is the modulation frequency
• multiply x(t) by a sine wave \cos(2\pi f0 t)$ to give:
• y(t) = A/2 (e^{j2\pi(f1 + f0)t} + e^{j2\pi(f1 - f0)t}$
• Radio transmission: use more of the spectrum, use shorter antennas

Notes:

• Draw x(t) (sine wave).
• Draw frequency spectrum of x(t), single arrow of scale A, f1$ to the right of the zero.
• Draw frequency spectrum of y(t), two arrow of scale A/2, displaced f1$ to the right of the f0 and -f0$.
• Draw modulated signal.

Amplitude Demodulation

• y(t) = x(t) \cos(2\pi f0 t)$ is the input signal.
• multiply y(t) by a sine wave \cos(2\pi f0 t) to give z(t)$
• result is z(t) = x(t)/2 + x(t)\cos(4\pi f0 t)$
• only retain frequencies in the range [-f1, f1]$
• if f0 > f1, result is x(t)/2$

Phase-Locked Loop

• voltage-Controlled oscillator: f(V) = f'0 + alpha V$
• transmission at frequency f0 =~ f'0$ (frequency of oscillator)
• multiply output of oscillator with incoming signal, average over time
• if phase is the same, average product will be positive, f'0$ will increase
• if phase is opposite, average product will be negative, f'0$ will decrease
• since f0 =~ f'0$, over the long term f'0 will exactly track f0$

Notes:

Nyquist's sampling theorem

• frequency-limited signal x(t)$ with all energy between -f[max] < f < f[max] (e.g. f[max] = 20KHz$)
• Take period T < (1/2)f[max]$ (e.g. T = 20 u s$, corresponding to 50KHz)
• Sample x(t) at time nT for all -oo < n < oo$
• Given only the samples, I can reconstruct x(t)$ exactly.

Sampling in the Frequency Domain

Notes:

• Explain that the reason we need to limit the frequencies of x$ is to avoid overlap between adjacent shifts in the frequency domain.

• Explain how limiting the frequency guarantees that we cannot "miss" a significant event.

Homework

• Due Monday, September 14
• Problem 2-1.
• Problem 2-9. Assume:
  • packet sizes of 1000 bits
  • 10Mb/s$ bandwidth of IP links
  • 100Mb/s$ bandwidth of ATM links
  • 10 ms = 0.01s$ delay for ATM connection setup
  • 10u s = 0.00001s$ delay through each ATM switch
  • 100u s = 0.0001s$ delay through each IP router
  • 50u s = 0.00005s$ delay through each link (ATM or IP)

Homework -- continued

Hints:

1. Compute the delay for a single packet going through each of the networks (you need to use all the above data except the connection setup time).
2. compute the difference in time between sending a single packet over ATM and over IP
3. compute the number of packets required to add up to the ATM connection setup time.