chapter 2

 

1: Compute the Fourier coefficients for the function f(t) = t (0 ≤ t ≤ 1)

Solution:

To compute the Fourier series of the function $f(t) = t$ on the interval $0 \leq t \leq 1$, we first need to express it as a periodic function of period 1. The Fourier series of a function $f(t)$ on the interval $[0,1]$ is given by:

$$f(t) = a_0 + \sum_{n=1}^{\infty} \left( a_n \cos(2\pi nt) + b_n \sin(2\pi nt) \right)$$

Where:

$$a_0 = \int_0^1 f(t) \, dt = \int_0^1 t \, dt = \frac{1}{2}$$ $$a_n = 2 \int_0^1 t \cos(2\pi nt) \, dt = 0 \quad (\text{since this is an odd function about } t = 0.5)$$ $$b_n = 2 \int_0^1 t \sin(2\pi nt) \, dt$$

Using integration by parts for $b_n$:

$$b_n = \frac{-1}{\pi n}$$

So the Fourier series becomes:

$$f(t) = \frac{1}{2} - \sum_{n=1}^{\infty} \frac{1}{\pi n} \sin(2\pi nt)$$

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3: Television channels are 6 MHz wide. How many bits/sec can be sent if four-level digital signals are used? Assume a noiseless channel.

Solution:

In a noiseless channel, the maximum data rate is given by the Nyquist formula:

$$\text{Maximum Data Rate} = 2 \times \text{Bandwidth} \times \log_2(L)$$

Where:

• Bandwidth = 6 MHz

• $L = 4$ (4 levels = 2 bits per symbol)

$$\text{Maximum Data Rate} = 2 \times 6 \times 10^6 \times \log_2(4) = 2 \times 6 \times 10^6 \times 2 = 24 \, \text{Mbps}$$

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5: What signal-to-noise ratio is needed to put a T1 carrier on a 50-kHz line?

Solution:

A T1 carrier has a data rate of 1.544 Mbps. The Shannon capacity formula is:

$$C = B \log_2(1 + \text{SNR})$$

Given:

• $C = 1.544 \times 10^6$

• $B = 50 \times 10^3$

$$1.544 \times 10^6 = 50 \times 10^3 \log_2(1 + \text{SNR}) \Rightarrow \log_2(1 + \text{SNR}) = \frac{1.544 \times 10^6}{50 \times 10^3} = 30.88 \Rightarrow 1 + \text{SNR} = 2^{30.88} \Rightarrow \text{SNR} = 2^{30.88} - 1 \approx 1.17 \times 10^9$$

So, the SNR must be around 1.17 billion, which is extremely high.

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7: How much bandwidth is there in 0.1 microns of spectrum at a wavelength of 1 micron?

Solution:

Bandwidth in frequency is related to the change in wavelength by:

$$\Delta f = \frac{c}{\lambda^2} \Delta \lambda$$

Where:

• $c = 3 \times 10^8 \, \text{m/s}$

• $\lambda = 1 \, \mu m = 10^{-6} \, \text{m}$

• $\Delta \lambda = 0.1 \, \mu m = 10^{-7} \, \text{m}$

$$\Delta f = \frac{3 \times 10^8}{(10^{-6})^2} \times 10^{-7} = 3 \times 10^8 \times 10^5 = 3 \times 10^{13} \, \text{Hz} = 30 \, \text{THz}$$

So the bandwidth is 30 terahertz (THz).

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9: Is the Nyquist theorem true for high-quality single-mode optical fiber or only for copper wire?

Answer:

The Nyquist theorem is a theoretical limit that applies to all noiseless channels, including both copper and optical fiber. However, in practice:

• For copper wire, the Nyquist theorem is often limited by noise and attenuation over distance.

• For single-mode optical fiber, the channel is more ideal (less noise, high bandwidth), so the Nyquist theorem holds more effectively and allows higher data rates.

Therefore, the Nyquist theorem is theoretically true for both, but is more applicable and practical in optical fibers due to their better transmission properties.

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11: A laser beam 1 mm wide is aimed at a detector 1 mm wide 100 m away. How much of an angular diversion (in degrees) does the laser have to have before it misses the detector?

Solution:

Use the small angle approximation:

$$\theta = \frac{w}{d} = \frac{1 \, \text{mm}}{100 \, \text{m}} = \frac{10^{-3}}{100} = 10^{-5} \, \text{radians}$$

Convert to degrees:

$$\theta = 10^{-5} \times \frac{180}{\pi} \approx 0.00057^\circ$$

So an angular deviation of just 0.00057 degrees will cause the laser to miss the detector.

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13: Calculate the end-to-end transit time for a packet for both GEO (35,800 km), MEO (18,000 km), and LEO (750 km) satellites.

Solution:

Transit time is:

$$t = \frac{2 \times h}{c}$$

Where $h$ is the altitude and $c = 3 \times 10^8 \, \text{m/s}$

GEO:

$$t = \frac{2 \times 35,800 \times 10^3}{3 \times 10^8} \approx 0.238 \, \text{seconds}$$

MEO:

$$t = \frac{2 \times 18,000 \times 10^3}{3 \times 10^8} \approx 0.12 \, \text{seconds}$$

LEO:

$$t = \frac{2 \times 750 \times 10^3}{3 \times 10^8} \approx 0.005 \, \text{seconds}$$