Given the continuous time signal

$$ x \left( t \right)= 2 \cos \left( 100 \pi t \right) + 3 \sin \left( 250 \pi t \right) \tag{1} $$

The signal is sampled in point sampling with sampling interval $T_{1} = 0.006 \textrm{s}$ and reconstructed (assuming that sampled above Nyquist sampling rate) by an ideal low-pass filter which sampling interval is $T_{2} = 0.002 \textrm{s}$.

  • A Find the reconstructed signal $x \left( t \right)$ by using the DTFT transform.

My Solution:

$$ X \left[ n \right] = 2 \cos \left( 100 \pi n T_{1} \right) + 3 \sin \left( 250 \pi n T_{1} \right) \tag{2} $$

$$ X^{f} \!\left( \theta \right) = 2 \sum_{n = -\infty}^{\infty} \cos \left( 100 \pi n T_{1} \right) \cdot e^{-j n \theta} + 3 \sum_{n = -\infty}^{\infty} \sin \left( 250 \pi n T_{1} \right) \cdot e^{-j n \theta} \tag{3} $$

From lectures, I know that:

$$ X^{f} \!\left( \theta \right) = \frac{1}{T_{s}} \sum_{k = -\infty}^{\infty} X^{F} \left( \frac{\theta - 2 \pi k}{T_{s}} \right) \tag{4} $$

So I calculated $X \!\left( \omega \right)$ for the equation above and I got:

$$ X^{F} \!\left( \omega \right) = 2 \pi \left[ \delta \!\left( \omega - 100 \pi \right) + \delta \!\left( \omega + 100 \pi \right) \right] + j 3 \pi \left[ \delta \!\left( \omega + 250 \pi \right) - \delta \!\left( \omega - 250 \pi \right) \right] \tag{5} $$

At this point I'm stuck. I am unsure how to transition from $X \left( \omega \right)$ to $X \left( \theta \right)$.

Thank you in advance for your help!

  • 2
    $\begingroup$ Welcome. Please learn to use $\LaTeX$. $\endgroup$ Feb 16 at 16:30


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