I'm trying to calculate the Fourier transform of an impulse-train sampled signal in two differnt ways but I end up with different results.
Impulse-train sampling of a continous signal $x(t)$ with sampling period $T_{s}$ is defined as
$$x_{p}(t) = x(t)p(t) \textrm{ where } p(t) = \sum_{n=-\infty}^{\infty} \delta(t - nT_{s})$$Now consider the input signal $ x(t) = e^{-t}u(t)$ where the Fourier transform is $X(\omega) = \frac{1}{1 + j \omega}$. According to common procedure I calculate the Fourier transform of $x_{p}(t)$, using convolution in the frequency domain, as:$$X_{p}(\omega) = \frac{1}{T_{s}}\sum_{k=-\infty}^{\infty}X_{}(\omega- \frac{2\pi k}{T_{s}})$$$$ \textrm{Inserting $X(\omega) $ I get } X_{p}(\omega)= \frac{1}{T_{s}}\sum_{k=-\infty}^{\infty}\frac{1}{1 + j (\omega - \frac{2\pi k}{T_{s}})} \tag{1}$$The other way is to calculate $X_{p}$ directly from $x_{p}(t) = x(t)p(t) = \sum_{n=-\infty}^{\infty} x(nT_{s}) \delta(t - nT_{s})$.Now I get the result$$X_{p}(\omega) = \int_{-\infty}^{\infty} e^{-nT_{s}}\sum_{n=0}^{\infty} \delta(t - nT_{s}) e^{-j \omega t }dt = \sum_{n=0}^{\infty} e^{-nT_{s}(1+j \omega)} \tag{2}$$
I can't see that 1 and 2 are equal? Or are they? If not, what misstake did I do?
