# Need help with DTFT transform (sampling and reconstruction)

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 $$X \left( \theta \right)$$.

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