s(0) = +0.0460-0.0460j s(8)= +0.0460 -0.0460j
s(1) = -0.1320-0.0020js(9)= +0.0020 +0.1320j
s(2) = -0.0130 +0.0790j s(10) = -0.0790 +0.0130j
s(3) = +0.1430 +0.0130j s(11) = -0.0130 -0.1430j
s(4) = +0.0920 + 0.0000j s(12) = +0.0000 - 0.0920j
s(5) = +0.1430+ 0.0130j s(13)= -0.0130 - 0.1430j
s(6) = -0.0130 +0.0790j s(14)= -0.0790+ 0.0130j
s(7) = -0.1320-0.0020j s(15)= +0.0020 +0.1320j.
The signal is then
15
s(t) = s(n)pr (t-nT).
#0
This signal is used in the preamble of the IEEE 802.11 system. The signal is filtered with a filter with impulse response
h(t) s*(16T-1).
=
(a) Find and plot the magnitude of the output of the filter.
(b) If the signal is repeated eight times, plot the real part, imaginary part, and magnitude of the output of the same filter.
Fig: 1