EE 445S Real-Time Digital Signal Processing Lab Fall 2013 Lab 6 Fundamentals with Quadrature Amplitude Modulation Zeina Sinno Credits to Professor Steven Tretter In Lab 6 we will discuss QAM. Lab 6 extends over two weeks. In the first week we will write the code for the transmitter and for the second week we will write the code for the receiver.

Outline What is QAM? Block Diagram of the Transmitter. Zoom in into the Different Components. An alternative representation of the Modulator.

What is QAM? QAM is an extension of PAM for Bandpass Channels. Recall that for PAM, we had no upconversion; e.g: no modulation by the carrier. It is used in a variety of applications: high speed cable, microwave and satellite systems. Why? Because of its spectral efficiency.

Block Diagram of the Transmitter s(t)=a(t)cosωct- b(t)sinωct Start with a stream of bits. Use PN sequences to produce this stream of bits. Pass this stream of bits to the serial to parallel converter to group the input bits into J bit binary words. Similarly to PAM, we will have now to map the “J” bits into some constellation. For PAM we had to do a one dimensional mapping. In case, we will have to do the mapping to a 2D constellation map. So QAM is an extension of PAM from 1D to 2D. Recall that last time for PAM we did the mapping into {-3,-1,+1,+3}. Today we will have “two axis”. So if we started initially with a rate Rd bps, the symbol rate will be equal to Rd/J. We will obtain a complex number that has the real part: an and the imaginary part bn. We will obtain an+jbn. The real and imaginary streams will be modulated separately. The obtained signals a*(t) and b*(t) will be convoluted with two baseband transmit filers to obtain: An inphase component a(t) A quadrature component b(t) We next take the inphase component, a(t), and multiply it by cos(wct), and we take the Quadrature component, b(t), and multiply it by sin(wct). Next we will add the contribution of the two components to form a signal s(t) equal to a(t)cosωct- b(t)sinωct Why the Minus? We will answer this question in few slides But now, let’s zoom in into the MAP to 2D constellation Point

The 16-Point Rectangular QAM Constellation As we mentioned in the slide before, the levels for the PAM were {-3,-1,1,3} Here we still have the same levels however we have two dimensions. As you can see gray coding still applies as well. So the difference between two adjacent point is a bit only. (show an example).

Block Diagram of the Transmitter s(t)=a(t)cosωct- b(t)sinωct So we zoomed in into the Map to 2D onstellation point.Let’s zoom in now into the other parts.

Expressions at the output of Impulse Modulator

Expressions at the output of Baseband Shaping Filter

How to choose the cutoff frequency ωc ? We choose it equal to the cutoff of the pulse shaping filter. Otherwise we will have aliasing.

Expressions of the output at the end of the receiver As we mentioned before we should combine the inphase and quadrature components after modulation. We obtain: Why the Minus?

Pre-Envelope Form Take the inside, call it s+(t) and express s+(t) as

An Alternative Representation of the Modulator We can alternatively represent the QAM Modulator in terms of Complex Signals