Chap4 Sampling Lecture
Transcript of Chap4 Sampling Lecture
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April 16, 2013 Digital Signal Processing 1
EEE & ECE DepartmentBITS-Pilani, Hyderabad campus
Sampling &
Reconstruction
Digital Signal Processing
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nT)-(t(nT)g(t)p(t)g(t)gn
-n
aap
Since impulse is a periodic signal of period T, it
can be expressed as trigonometric Fourier series.
............2cos32cos22cos1T1)()( ooo tttnTttp
n
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............(t)cos32g
(t)cos22g(t)cos2g(t)g
T
1)()()(
oa
oaoaa
t
tttptgtg ap
The FT of gp(t) is Gp(j)
............)cos3(j2G)cos2(j2G)cos(j2G)(jG
T1)(
oa
oaoaa
t
ttjGp
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Recovery of The Signal
The discrete time signal must pass through
an analog lowpass filter.
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Recovery of The Signal
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Recovery of The Signal
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ADC
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DAC
Recovery of The Signal
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Recovery of The Signal
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Recovery of The Signal
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Recovery of The Signal
Ideal DACoutput
Practical DACoutput
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Recovery of The Signal
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Recovery of The Signal
FT
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Recovery of The Signal
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Aliasing
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Critical Sampling
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Under Sampling
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Over Sampling
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Problem
(1) A continuous time signal xa(t) is composedof a linear combination of sinusoidalsignals of frequencies 300 Hz, 500 Hz, 1.2kHz, 2.15 kHz and 3.5 kHz. The signal xa(t)
is sampled at a 2.0 kHz rate and thesampled sequence is passed through anideal low pass filter with a cut-offfrequency of 900 Hz, genearting acontinuos time signal of ya(t)
What are the frequency componentspresent in the output signal ?
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Problem
The continuous-time signal
xa (t) = 4 sin (20t) 5 cos (24t) + 3 sin (120 t)
+ 2 cos (176
t)
is sampled at a 50 Hz rate, generating the
sequence x[n].
Determine the exact expression of x[n].
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Filtering Using FDA tool
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Filtering Using FDA tool
M4.2 (SK. Mitra)
Determine the lowest order of a lowpass
Chebyshev Type I filter with a 0.25 dB passbandfrequency at 1.5 kHz and minimum attenuation
0f 25 dB at 6.0 kHz.