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A signal is sampled at the rate of 15 Hz and applied to ideal rectangular LPF with cut-off frequency of 10Hz, then the output of filter contains

  • a)
      rad/sec and components

  • b)
    Only rad/sec components

  • c)
     Only rad/sec components

  • d)
    rad/sec and  components

     

Correct answer is option 'C'. Can you explain this answer?
Verified Answer
A signal is sampled at the rate of 15 Hz and applied to ideal rectang...
The signal given is:


x(t)=6cos⁡(10πt)


 The goal is to determine what frequency components will pass through the filter.


Step 1: Angular Frequency of the Signal


The signal has a frequency component of cos⁡(10πt). The angular frequency ω is given by:


ω=2πf 


From the given cosine term, we can directly see that:


ω=10π rad/sec 


 So, the signal has a frequency of 5Hz (since 10π/2π=5 Hz).


Step 2: Sampling at 15 Hz


The signal is sampled at a frequency of 15 Hz. This will create replicas of the spectrum at intervals of the sampling frequency ωs= 2π × 15 rad/sec = 30π rad/sec.


Step 3: Ideal LPF with 10 Hz Cutoff


The filter has a cutoff frequency of 10 Hz, or 20π rad/sec.


The filter will pass frequency components only within this range (±20πrad/sec).


Step 4: Frequency Components at the Output


The original frequency component of ω=10π rad/sec (5 Hz) is within the filter’s passband, so it will pass through.


There will be no additional components from aliasing (since the filter only allows up to 20 Hz components).


Conclusion:


The output of the filter will contain only the ±10π rad/sec components.


Thus, the correct answer is (c) Only ±10π rad/sec components.


 
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A signal is sampled at the rate of 15 Hz and applied to ideal rectangular LPF with cut-off frequency of 10Hz, then the output of filter containsa) rad/sec and componentsb)Only rad/sec componentsc)Only rad/sec componentsd)rad/sec and componentsCorrect answer is option 'C'. Can you explain this answer?
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