what is the power after a signal is sampled?

say there's a continuous signal A(t), now we sample it with a series of impulse signal, then we get A = A(n*Ts)* delta(t - nTs).

If we look at the spectra, the fourier transform of A(t) is a bandlimited waveform. the FT( Not discrete Fourier transform) of A is a infinite replica of A(t)'s spectrum.

does this mean now the sampled version A is with infinite power? by looking at time domain of A it's absurd. But in the frequency domain there're infinite replica.

Where did I get wrong? please help...thanks.

Hi Bravoegg,
Welcome to Edaboard!

From my point of view, unfortunately there is a common misconception: to think that sampling a signal means to multiply it by an infinite train of Dirac deltas. It is unrealistic and easily leads to confusions.

To sample a signal means to keep only its values at some specific instants. When you take lectures of the voltmeter on your notepad once a minute you are sampling a signal. When your analog-to-digital converter writes numbers in a memory there is a samplig process. No dirac deltas, no infinite powers.

The essential is that from the samples taken at instants nT you can reconstruct the whole signal:
A = sum{A(nT)*sinc[(t-nT)/T]}

If you imagine a sequence of deltas whose "areas" are the values of the samples... well, you are right: you would have a signal with infinite power (think that a single delta has infinite energy) and infinite bandwidth.

For some aspects, it is a useful tool to imagine such a train of deltas as an intermediate stage (for example, for calculating the spectrum of a reconstructed signal and to know the amount of distortion). But When you think about a "sampled signal" it is better to think in a sequence of numbers rather than in a train of deltas.

Regards

Z

zorro, thank you for the detailed answer.

Can I say the practical sampling process is sampling at one instant, then hold it until next sampling point.
Thus it equals sum{x(nT) * square(Ts)} in time domain. The square is the hold circuitry with time length the same as sampling interval Ts.

Put this in frequency domain, it equals the original X(w) convolve with a series of sinc functions. Even if we use a ideal low pass filter to extract the baseband, due to the mainlobe width(if mainlobe width is 0 then it's dirac delta again) and sidelobe disturbances of sinc functions, it's impossible to get back the original frequency response.

regards
Sy

Hi Sy,

This is a case in wich it is a useful tool to think in the train of deltas as an intermediate step.
Your stepwise function can be thougth as if it were constructed as follows:

a) multiply the original signal by a train of deltas at intervals T
b) pass the result by a filter whose impulse function is a unit rectangle of duration T

Then:
- The spectrum after a) is a series of replicas of the original spectrum, centered at the multiples if 1/T. (They don't overlap if the signal is band limited to 1/(2T).)
- The filter in b) has frequency response of the type sinc.

You can recover the original signal with a lowpass filter that:
1) Removes all the replicas, i.e. its transfer function is (ideally) null for f>=1/(2T)
2) Corrects (equalizes) the sinc() decay in the main lobe.

Such a filter is commonly referred as "inverse sinc" or "x/sin(x)" filter.

Regards

Z

Since it's impossible to generate an ideal train of deltas function, so we have to do with a rather steep square(let's suppose it's an ideal square).

1. is it true that, in frequency domain, the original baseband spectrum, let's name it A(w), after it's convovled by a train of equally spaced sinc functons, would overlap with the other convolved and shifted versions of itself, because of the sidelobes introduced by SINC functions.

2. if 1) is true, then is it true that we can never get the original signal back in real practice? just a approximation?

regards...
Sy.

bravoegg said:

Since it's impossible to generate an ideal train of deltas function, so we have to do with a rather steep square(let's suppose it's an ideal square).

Click to expand...

0) I guess that you mean an infinite train of finite width pulses. Right?

1. is it true that, in frequency domain, the original baseband spectrum, let's name it A(w), after it's convovled by a train of equally spaced sinc functons, would overlap with the other convolved and shifted versions of itself, because of the sidelobes introduced by SINC functions.

Click to expand...

If 0) ir true, then the spectrum of that train is not a train of sinc funtions, but a train of deltas (recall that such a train is a periodic function) whose amplitudes are not all the same, but shaped by a sinc envelope.

Regards

Z

I got fourier transform of periodic signal completely wrong...
it is the fourier series that is being shifted in the frequency domain, not the frequency response as a whole.

thank you so much!