a1 (t) and 12 (t) are two given signals. #1 (t) is band-limited to 500 Hz and a2 (t) is bandlimited to 2.5 kHz. X|| f, Hz 500 O 500 S, Hz -2500 2500 QUESTION 2 These signals are multiplexed by impulse-sampling them at a rate T and combining the samples as shown in the figure. The multiplexer output is given by y (t) = 1 (t) E 5 (t – nT') + *2 (t) E 5 (t – nT T-00 n--00 (a) demultiplexing? What is the minimum value of T which makes it possible for us to recover the signals after (b) Plot the Fourier transform of r1 (t) 8 (t – nT). -00 NOTE: F (a (t)) = / # (t) e-jul dt (t) e-jut dt 8 (t – nT') | = E (w - kwo) %3D T km-00 T -00

Introductory Circuit Analysis (13th Edition)
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Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
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T1 (t) and r2 (t) are two given signals. r1 (t) is band-limited to 500 Hz and r2 (t) is bandlimited
to 2.5 kHz.
1.
f. Hz
500
O 500
1
f, Hz
-2500
2500
QUESTION 2
These signals are multiplexed by impulse-sampling them at a rate T and combining the samples as shown
in the figure. The multiplexer output is given by
y (t) = 1 (t) E 5(t – nT') + #2 (t) E 6 (t- nT –5)
n=-00
n=-00
(a)
demultiplexing?
What is the minimum value of T which makes it possible for us to recover the signals after
(b)
Plot the Fourier transform of r1 (t) E 8 (t – nT').
n=-00
NOTE:
F (# (t)] = / (t) e-jut dt
2n
FE 8 (t – nT)
) = E 6 (w - kuo)
k=-00
Transcribed Image Text:T1 (t) and r2 (t) are two given signals. r1 (t) is band-limited to 500 Hz and r2 (t) is bandlimited to 2.5 kHz. 1. f. Hz 500 O 500 1 f, Hz -2500 2500 QUESTION 2 These signals are multiplexed by impulse-sampling them at a rate T and combining the samples as shown in the figure. The multiplexer output is given by y (t) = 1 (t) E 5(t – nT') + #2 (t) E 6 (t- nT –5) n=-00 n=-00 (a) demultiplexing? What is the minimum value of T which makes it possible for us to recover the signals after (b) Plot the Fourier transform of r1 (t) E 8 (t – nT'). n=-00 NOTE: F (# (t)] = / (t) e-jut dt 2n FE 8 (t – nT) ) = E 6 (w - kuo) k=-00
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