A signal $x (t) = 2 cos 6 π \times 10^{3} t + 3 cos 12 π \times 10^{3} t$ is sampled with a sampling interval of 0.125 msec. The sampled signal is passed through a low pass filter with a cut-off frequency of 9 kHz. The number of positive frequency components present in the output of low pass filter are____.
4

Open in App

Solution

The correct option is A 4
$Frequency component present in the input signal are f_{m 1} = 3 k H z a n d f_{m 2} = 6 k H z$
$Sampling Rate = \frac{1}{Sampling Interval} = \frac{1}{0.125 \times 10^{- 3}} = 8 k H z$
The speciman or sampled signal wound have $n t, x f_{m}$ ccomponents
$∴ f o r f_{m 1} = 3 k H z \Rightarrow 3, 8, \pm 3, 16 \pm 3.... = 3 k H z, 5 k H z, 11 k H z, . . . .$
$f o r f_{m 2} = 6 k H z \Rightarrow 6, 8, \pm 6, 16 \pm 6.... = 6 k H z, 2 k H z, 14 k H z, . . . .$
The cut-off frequency of low pass filter is 9 kHz
$∴$ The filter output will have 2 kHz, 3 kHz,5 kHz, 6 kHz
$\Rightarrow$ 4 frequency components will be present in the output

Suggest Corrections

Similar questions

ω_{c} = 1000 π

x (t) = 2 + cos (50 π t)

T_{s} = 0.025 s e c

(50 π t)

t_{s} = 0.025

A signal x (t) = e^{- t} u (t) is passed through an ideal low pass filter with cut-off frequency of ω_{c} rad/sec, then the value of ω_{c} for the output energy of low pass filter is 90% of the input energy will be ________rad/sec

R_{f}

k Ω

Join BYJU'S Learning Program