The minimum sampling frequency of signal $[10 sin c^{3} (400 t) * 90 sin c^{4} (500 t)] 2 sin c (300 t)$ to avoid aliasing should be __________Hz.
1500

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Solution

The correct option is A 1500
Maximum frequency component in
10 $s i n c^{3} (400 t) i s (1200 π) r a d / s e c$

Maximum frequency component in
90 $s i n c^{4} (500 t) i s (2000 π) r a d / s e c$

$∴ M a x i m u m f r e q u e n c y c o m p o n e n t i n 10 s i n c^{3} (400 t) * 90 s i n c^{4} (500 t) i s (1200 π) r a d / s e c$

Maximum frequency component in $2 s i n c (300 t) i s (300 π) r a d / s e c$

Hence for overall signal, maximum frequency component

$= 1200 π + 300 π = 1500 π r a d / s e c .$

Thus for avoiding aliasing

$ω_{N} \geq 2 ω_{m a x} \geq 2 \times 1500 π \geq 3000 π r a d / s e c$

$f_{N} = \frac{3000 π}{2 π} = 1500 H z$

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The minimum sampling frequency of signal [10sinc3(400 t)∗90sinc4(500 t)]2sinc(300 t) to avoid aliasing should be __________Hz.1500

The minimum sampling frequency of signal $[10 sin c^{3} (400 t) * 90 sin c^{4} (500 t)] 2 sin c (300 t)$ to avoid aliasing should be __________Hz.
1500