-We will now have a look at QAM modulations, Quadrature Amplitude Modulation. We will introduce these new modulations in order to decrease the bandwidth used by the signal for a given bit rate. The I-Q modulator is the same as the one we saw for the QPSK modulation. But the bits-symbols mapping functions are different. We will consider the specific case of "square" QAM modulations. That is to say modulations for which the I- and Q-channel constellations are identical. So we have equal bit rates on the in-phase and in-quadrature channels. The total bit rate equals the sum of the elementary bit rates on each channel. We have said that compared to the QPSK, the bits-symbols mapping function was different. Indeed, we no longer consider a binary modulation. In the general case, we will consider that we have n bits per symbol. So after the bits-symbols mapping, on the in-phase channel, the RsI symbol rate equals the RbI bit rate divided by n, n the number of bits per symbol. Let us take a practical example. n = 2. A symbol is made of two bits. So there are four possible symbols. Here is the correspondence we can make between bits and symbols. Symbols are -3, -1, 1 and 3. We get the following constellation which includes the four symbols, -3, -1, 1 and 3. Let us look at the in-quadrature channel. We will get the same results as for the in-phase channel and a constellation also made of four symbols in the studied example. -3, -1, 1 and 3. We have defined the constellations of the in-phase and in-quadrature channels. We will now define the constellation of the global modulation. It is a bi-dimensional modulation just like the QPSK. We will thus draw it in a plane. In abscissa, we draw constellation I. In ordinate, constellation Q. The total number of points of the constellation will be the number of points of constellation I multiplied by the number of points of constellation Q, thus 2 to the power of m with m = 2n. Let us illustrate this with an example, the 16-QAM modulation. So we have two bits per symbol on the real and imaginary channels. On the in-phase channel, we get constellation I. The correspondence between bits and symbols is also indicated. For the in-quadrature channel, we get the same constellation and the same bits and symbols correspondence. The 16-QAM constellation will be obtained by putting constellation I in abscissa and constellation Q in ordinate. We get this constellation made of 16 symbols. We add here the bits associated to each symbol of the modulation. Let us consider the symbol circled in blue for example. This symbol corresponds to abscissa -1 and ordinate -3. So it is the (-1,-3) symbol corresponding to bits 0100. Let us get back to the constellation symbols. These symbols are made of two symbols. A symbol on channel I and a symbol on channel Q. We can deduce that the symbol rates are equal as shown on the slide. We have seen that the symbols on each channel are made of n bits. We can thus deduce the symbol rate relation written on the slide with RbI = Rb/2. We can deduce that the total symbol rate equals the total bit rate divided by m. m is such that the number of symbols in the alphabet equals 2 to the power of m. Let us now have a look at the shaping function. It is identical to what we have seen in the previous episodes about BPSK and QPSK modulations. Let us take the example of the NRZ-type shaping function. Let us take two bits per symbol on the in-phase channel. We get the signal shown on the screen that you already saw during week 3. We also note that the symbol duration in this case equals two times the bit duration. What is happening in the spectral domain? In the case of the NRZ-type shaping, still on the in-phase channel, power is mainly concentrated in a bandwidth that is no longer two times the bit rate, but two times the symbol rate. This is due to the fact that the symbols change at a speed of Rs. Power is thus mainly concentrated in a 2RsI-large bandwidth since we consider channel I here. Remember that RsI equals Rs, the total symbol rate. So this power is concentrated in 2Rs. 2Rs equals 2Rb/m. Let us look at the in-quadrature channel. We get the same result. Power is concentrated in a 2Rb/m-large bandwidth. The frequency transposition is identical to what we have seen for the QPSK modulation. Here is a reminder of the scheme of this transposition which will allow us to separate the in-phase and in-quadrature channels at receiver side. Let us take the example of a NRZ signal with a transposition around the f0 frequency. We note that after the shift, the bandwidth used has not changed and thus equals 2Rb/m. Remember that we have suggested using the QAM modulation in order to reduce the used bandwidth. Let us see this by comparing QAM to the BPSK modulation. We will take the example of the NRZ modulation to compare the modulations. Power, in the case of BPSK, is mainly concentrated in a 2Rb-large bandwidth. In the case of a QAM modulation, this bandwidth equals 2Rb/m as we just showed it. To conclude, for a given bit rate, the bandwidth used was divided by m compared to the BPSK modulation, with m being the total number of bits per symbol. It is now time to conclude on these square QAM modulations. For these square modulations, there are two baseband channels, I and Q, that use the same symbol alphabet and the same bit rate. The carrier frequency transposition allows channels I and Q to be separated upon reception. The bandwidth used is divided by m compared to BPSK, m being the total number of bits per symbol.