Subject: Principles of Communication Engineering

Difficulty : Medium

Marks : 05

Block diagram:

Transmitter:

Explanation:

Figure shows transmitter for 4 bit QAM system. The input bit stream is applied to a serial to parallel converter. Four successive bits are applied to the digital to analog converter. These bits are applied after every Ts second. Ts is the symbol period & Ts=4Tb. Bits Bk and Bk+1 are applied to upper digital to analog converter and Bk+2, Bk+3 are applied to lower D to A converter. Depending upon the two input bits, the output of D to A converter takes four output levels. Thus Ae(t) and Ao(t) takes 4 levels depending upon the combination of two input bits. Ae(t) modulates the carrier $\sqrt {Ps} cos(2πf_0t)$ and Ao(t) modulates $\sqrt{Ps} sin (2πf_0t)$.

The adder combines two signals to give QAM signal. It is given as,

$ S(t) = Ae(t) \sqrt{Ps} cos (2πf_0t) + Ao(t) \sqrt{Ps} sin (2πf_0t)$.

Receiver:

• The quadrature carriers are recovered from the received QAM signal. The input QASK signal is first raised to the 4th power and then by using a BPF, with a center frequency $4f_c$, along with a frequency divider (divide by 4), the required quadrature carriers are recovered.
• Then, two balanced modulators are used together with two integrators to recover the signal Ae(t) and Ao(t). Both the integrators integrate over one symbol interval Ts or 4Tb. The symbol time synchronizer is used along with each integrator.
• Integrator output is $Ao(t)\sqrt{2Ps}2Tb$ and $Ae(t)\sqrt{2Ps}2Tb$
• Finally, the original bits are obtained from Ae(t) and Ao(t) by using two A/D converters. The outputs of the two A/D converters are then applied to the serial to parallel converter to obtain the sequence b(t).