3 Sampling and aliasing The aim of this part is to demonstrate the effects of aliasing arising from improper sampling. A given analog signal x(t) is sampled at a rate fs = 1/T, the resulting samples (nT) are then reconstructed by an ideal reconstructor into the analog signal a(t). Improper choice of f, will result in different signals ra(t) + x(t), even though they agree at their sample values, that is, xa (nT) = x(nT). The procedure is illustrated by the following block diagram: x(t) x(nT) = x(nT) Xa(1) ideal sampler ideal reconstructor analog input sampled signal analog output rate fs [-fs/2, fs/2] Fig. 4: Block diagram showing the sampling and reconstruction operations. a) (10 points). Consider an analog signal x(t) consisting of three sinusoids of frequencies of 1, 4, and 6 kHz: x(t) = cos(2πt) + cos(8πt) + cos(12πt) where t is in milliseconds. Show that if this signal is sampled at a rate of fs = 5 kHz, it will be aliased with the following signal, in the sense that their sample values will be the same (5 points): xa(t) = 3 cos(2πt). On the same graph, plot (2.5 points) the two signals (t) and x(t) versus t in the range 0 <t <2 msec. To this plot, add the time samples (nT) (2.5 points) and verify that x(t) and xa(t) intersect precisely at these samples (see Fig. 5 for example). f = 5 kHz x(t), x(t) 2 7 2 -3 original aliased samples 0 0.2 0.4 0.6 0.8 1 1.2 1.4 t (msec) 1.6 1.8 2 b) => (10 points). Repeat part (a) with f, = 10 kHz. In this case, determine the signal x(t) with which x(t) is aliased (5 points). Plot both (t) and (t) on the same graph (2.5 points) over the same range 0 <t<2 msec, and add the samples (2.5 points) (nT). Verify again that the two signals intersect at the sampling instants (see Fig. 6 for example). f= = 10 kHz (4)" x x(t), 0 0.2 0.4 0.6 0.8 original aliased samples 1 1.2 1.4 1.6 1.8 t (msec) Fig. 6: Plot for Problem 3.2. 2