The signal $x (t) = c o s (5 π t) + 0.5 c o s (10 π t)$ is instaneously sampled. The interval between the sample is $T_{s}$ . The maximum allowable value for $T_{s}$ is ____ sec.

0.1

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Solution

The correct option is A 0.1
The maximum frequency component present in the signal x(t) is $10 π$

$f_{m a x} = \frac{ω_{m a x}}{2 π} = 5 H z$

$∴$ Nyquist rate = $f_{s} = 2 f_{m a x}$

= 10 samples/sec

Hence maximum allowable value for $T_{s}$ is

$T_{s} = \frac{1}{f_{s}} = \frac{1}{10} = 0.1 s e c$

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