A signal x(t) = 5 cos(150 $π t - 60$ ) is sampled at 200 Hz. The fundamental period of the sampled sequence x[n] is

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\frac{2}{200}

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\frac{1}{200}

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Solution

The correct option is B 8
The given signal is

$x (t) = 5 c o s (150 π t - 60)$

$f_{s} = 200 H z$

$\Rightarrow T_{s} = \frac{1}{f_{s}} = \frac{1}{200} s e c$

$x (n T s) = 5 c o s (150 π n T_{s} - 60)$

= $5 c o s (150 π \times \frac{1}{200} n - 60)$

= $5 c o s (\frac{3 π}{4} n - 60)$

$H e r e, ω_{0} = \frac{3 π}{4}$

$\frac{ω_{0}}{2 π} = \frac{m}{N}$

$\Rightarrow \frac{m}{N} = \frac{3 π}{\frac{4}{2 π}} = \frac{3 π}{8 π} = \frac{3}{8}$

$Hence, N = 8$

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A signal x(t) = 5 cos(150 πt−60) is sampled at 200 Hz. The fundamental period of the sampled sequence x[n] is

The correct option is B 8The given signal is x(t)=5cos(150πt−60) fs=200Hz ⇒Ts=1fs=1200sec x(nTs)=5cos(150πnTs−60) = 5cos(150π×1200n−60) = 5cos(3π4n−60) Here, ω0=3π4 ω02π=mN ⇒mN=3π42π=3π8π=38 Hence, N = 8

A signal x(t) = 5 cos(150 $π t - 60$ ) is sampled at 200 Hz. The fundamental period of the sampled sequence x[n] is