The Beauty of Mathematics in Communications R. C. T. Lee Dept. of Information Management and Dept. of Computer Science National Chi Nan University

We often hear the term “modulation”. For instance, we are all familiar with the term “amplitude modulation”. Can we transmit our voice or music without going through the process of modulation?

A Music Signal Lasting 1 Second

What is inside the signal? Fourier transform. for t = 0, 1, 2, …, n-1.

A Discrete Fourier Transform Spectrum of the Signal

It can be seen that the frequencies of the signal are in the range of 100 to 3000 Hertz.

A Signal

The Discrete Fourier Transform Spectrum of the Signal

The above analysis shows that the signal is as follows:

Through Fourier transform, we can find out that the frequencies of our human voice are in the range of 300 to 3000 Hertz. Through Maxwell’s equations, we know that the length of an antenna to transmit a signal with frequency f has to be where v is the speed of light and f is the frequency of the signal. If f =3000 Hertz, the wavelength is 100km. No such an antenna is possible.

How can we lift the base band frequency to a higher one such that it can be transmitted through an antenna? Through the use of “modulation” which effectively lifts the frequencies of the base band.

Modulation: Multiplying the base band signal by a cosine signal. m(t): the base band signal. Carrier wave signal:

From Fourier transform theory, every signal component in m(t) with frequency f will have frequencies fc+f and fc-f after m(t) is multiplied by . This can be easily understood as follows: Consider any cosine function with frequency f :

Now, after multiplication with , we have Thus, every signal with frequency f, which is relatively low, becomes two signals with frequencies fc+f and fc-f which are relatively high because fc is a high frequency.

We shall show later that high frequency means small wavelength. Because the frequencies of the signals are high now, their wavelengths are now relatively small. They can now be transmitted by an antenna with reasonable length.

Illustrating the Amplitude Modulation Process. (a) Baseband Signal m(t). (b) AM Wave where for All t.

Spectrum of the Message Signal. (b) Spectrum of the DSB-SC Modulated Signal.

But the signal we receive with frequencies fc+f and fc-f are not comprehensible because they are of high frequencies. We must have a scheme to recover the base band signal thoroughly. How? Through demodulation.

The modulation is done by multiplication of . The demodulation is also done by multiplication. Consider any base band signal component with frequency f.

After modulation, the sent signal is

After demodulation, we have

Thus, after demodulation, we have two signals: and

Illustrating the Spectrum of the Demodulation Process.

The receiver receives two functions, and . Do we really need both of them to recover the sent signal ? Ans: No.

Let us assume that the receiver uses a filter so that only is passed. Then let us demodulate this signal by multiplying it by . The result is .

But , after demodulation, Thus the sent signal is recovered. This is the principle of Single Side Band.

Suppose we do not want to use a filter, what can we do? We can do the following trick: Let m(t) be . Let be .

Then we send the following signal:

This time, only one signal, namely is sent. But, in a signal such as human voice, there are a lot of cosine functions, how can we transform each cosine function into a sine function?

This is done by Fourier transform. Note that and

For the Fourier transform coefficient of f, we multiply it by 2j and for that of -f, we multiply it by –2j. We then perform an inverse Fourier transform, the result is: Each signal in the original signal undergoes a phase shift.

Digital Modulation Binary Phase Shift Keying

Signals of BPSK.

Binary Frequency Shift Keying for some fixed integer nc and i=1,2 If we want to send 1(0), we send s1(t)(s2(t)).

We can prove that and s1(t) and s2(t) are orthogonal.

To detect s1(t), we multiply the sent signal by s1(t). If 1 is transmitted, If 0 is transmitted,

Coherent Detector for FSK System.

Signals for an FSK System.

Can we mix two bits together and send them out at the same time? Yes, QPSK.

To detect si1, we multiply si(t) by .

In QPSK, we transmit 2 bits together. Can we further extend the idea into, say, sending n bits together? Yes, OFDM.

To compute s(t), we can use the fast Fourier transform. OFDM is used in our ADSL system.

Multiple Access Communications

FDMA

Spectrum of an FDMA System

TDMA

CDMA

m1=+1 or -1 m2=+1 or -1.

We can prove that s1(t) and s2(t) are orthogonal. Let us represent s1(t) by (1,1) and s2(t) by (1,-1). We can see that ( 1, 1)( 1, -1)=0

To detect m1, we multiply r(t) by s1(t) and integrate. To detect m2, we multiply r(t) by s2(t) and

Signature Signals Generated from H8.

Electromagnetic Theories Some basic notations: The cross product of two vectors. Let A and B be two vectors. Then

Then

E: electric field D: electric flux density B: magnetic flux density H: magnetic field Maxwell’s Equations

The plane EM wave:

Let In free space, and . Thus, This means that the EM wave travels in free space with the velocity of light.

The Electric and Magnetic Fields in a Plane Electromagnetic Wave

It can be proved that the wavelength of the wave is as follows: where f is the frequency of the signal. Let f=60 Hertz.

If the frequency is very high, the wavelength becomes smaller. The length of an antenna is usually . This is why we always have to elevate the signal frequency into a higher one so that some antenna can be used.

Transmission Lines If the wavelength of the signal on a line is small as compared with the length of the line, this line is considered a transmission line. Therefore, the power line in our houses is not a transmission line because the frequency of the signal is only 60 Hertz ( =5000 km).

A Co-Axial Cable Transmission Line Twin-Strip Parallel Plate Transmission Line

There are two voltage waves and two current waves on the transmission line.

The speed of the waves v is . Therefore the waves travel with the speed of light in a transmission line.

Standing Waves

Voltage and Current with Respect to Time along the Transmission Line A Lumped Circuit whose length Is Way Short as Compared with the Wavelength of the Signal

But, is it true that we are now sending digital signals (pulses) in the transmission lines?

If we want to send a large number of bits per second, the width of the pulses will become smaller. This in turn means that the corresponding frequency spectrum will become wider. This is why we call the advanced transmission lines which transmits pulses with high speed wide band transmission lines.

Antennas: Consider any transmission line From very far away, there is no current. No current, no radiation. This is not an antenna.

Simple Dipole Antenna

What happens if the length of the antenna is larger than ? No good at all.

Conclusion: We can not have modern communications without mathematics.