The Nyquist interval corresponding to the continuous time signal
$x (t) = [S a^{3} (150 π t) * S a^{2} (200 π t)] S a^{3} (200 π t)$ is ________msec .

1

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Solution

The correct option is A 1
$S a^{3} (150 π t) \to ω_{m_{1}} = 3 \times 150 π = 450 π r a d / s e c$

$S a^{2} (200 π t) \to ω_{m_{2}} = 2 \times 200 π = 400 π r a d / s e c$

for $(f_{1} (t) * f_{2} (t)) \to ω_{m f} = min {ω_{m_{1}}, ω_{m_{2}}} = 400 π r a d / s e c$

$S a^{3} (200 π t) \to 600 π$

Total band width = $400 π t + 600 π t = 1000 π t$

Nyquist frequency = $1000 π r a d / s e c$

Nyquist interval , $T_{N} = \frac{2 π}{2000 π}$ = 1 m sec

Nyquist interval , $T_{N} = \frac{π}{ω_{m}} = \frac{π}{1000 π} = 1 m s e c$

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