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.