Consider the sinusoidal signal \(\rm X(t)=\cos \left(\frac{ω_s}{2}t+\phi\right)\) Suppose this signal is sampled at w, frequency. If this impulsesampled signal is applied as the input to an ideal lowpass filter with cutoff frequency w ω_s/2, the resulting output is: 

Concept:

We are given a sinusoidal signal: \( X(t) = \cos\left(\frac{\omega_s}{2}t + \phi\right) \)

This signal is sampled at frequency \(\omega_s\). After sampling, the signal becomes an impulse train containing frequency components centered at multiples of \(\omega_s\).

If this sampled signal is passed through an ideal lowpass filter with cutoff frequency \(\omega_s/2\), only baseband components (centered around 0 Hz) are retained.

Calculation:

Using the trigonometric identity:

\(\cos\left(\frac{\omega_s}{2}t + \phi\right) = \cos\phi \cdot \cos\left(\frac{\omega_s}{2}t\right) - \sin\phi \cdot \sin\left(\frac{\omega_s}{2}t\right)\)

After lowpass filtering, the component \(\sin\left(\frac{\omega_s}{2}t\right)\) is removed because it lies at or above the cutoff frequency and is attenuated. Only the baseband term \(\cos\left(\frac{\omega_s}{2}t\right)\) remains.

Thus, the output is: \(x_r(t) = \cos\phi \cdot \cos\left(\frac{\omega_s}{2}t\right)\)

Final Answer:

✅ The resulting output after ideal lowpass filtering is:

Correct Option: (1) \(x_r(t) = (\cos \phi) \cos\left(\frac{\omega_s}{2}t\right)\)