Flat Top Sampling

During transmission, noise is introduced at top of the transmission pulse which can be easily removed if the pulse is in the form of flat top. Here, the top of the samples are flat i.e. they have constant amplitude. Hence, it is called as flat top sampling or practical sampling. Flat top sampling makes use of sample and hold circuit.

Theoretically, the sampled signal can be obtained by convolution of rectangular pulse $p(t)$ with ideally sampled signal say $y_\delta(t)$, i.e.

$$y(t)=p(t)\star y_\delta(t)\tag{1}$$

To get the sampled spectrum, consider Fourier transform on both sides for equation $(1)$

$$Y\left[\omega\right]=\mathcal F\left\{p(t)\star y_\delta(t)\right\}$$

By the knowledge of convolution property,

$$Y\left[\omega\right]=P(\omega)Y_\delta(\omega)$$

Here $P(\omega)=TS_a(\omega T_2)=2\sin \omega T/\omega$

My question

If the continue signal $x(t)$ is like: It is a step shape wave. I sampled it by flat top sampling mentioned above. The rectangular pulse width is the same as the step width of $x(t)$. It is easy to find that the sampled signal is the same as the original signal. So the Fourier transform will not change before and after sampling. However, according the above equation $Y[\omega]=P(\omega)Y_\delta(\omega)$, the Fourier transform of the sampled signal is different from the original. What is going on?

• You just claim that the two functions are different, but have you actually tried to show it? – Matt L. Apr 28 '16 at 6:58
• Hm, check your equations. There are two things that might be problematic: 1) Your sampled signal $y(t)$ is not fully given by each sample point multiplied by a convolution of $\delta(t)$ and the rectangular function. You have to take into account that an (infinite) sum of $\delta$-functions behaves as a Dirac-comb. Its FT will scale the distance of the deltas. 2) Shouldn't $Y[ω]=P(ω)Yδ(ω)$ be $Y[ω]=P(ω)Y \ast δ(ω)$? – M529 Apr 28 '16 at 7:53
• I guess you remember the fact that $x_c(t)$ is not a bandlimited signal, which should not have been so, and hence is improperly sampled. Nevertheless your question is interesting. – Fat32 Apr 28 '16 at 10:10
• I have corrected the equation $Y\left[\omega\right]=P(\omega)Y_\delta(\omega)$, $Y_\delta(\omega)$ is the Fourier transform of ideally sampled signal. The derivation above comes from tutorialspoint . It should suitable for any signal, no matter it is bandlimited or not. – zeleven Apr 28 '16 at 13:06
• As the answer of Matt L., maybe the two functions are essentially the same. Thanks for all the answers. – zeleven Apr 28 '16 at 13:46