0
$\begingroup$

I've found a material explaining the sample and hold operation pretty well, but there's one thing I can't understand.

http://www.springer.com/cda/content/document/cda_downloaddocument/9781848001183-c1.pdf?SGWID=0-0-45-484012-p173780368

Next, we would like to take this ideally sampled signal and hold the values between samples. As we have a train of impulses with the correct areas, we need a “block” that takes an impulse with area A, transforms it into a rectangular pulse of height A that starts at the time at which the delta function is input to the block, and persists for exactly Ts seconds. A little bit of thought shows that what we need is a linear, time-invariant (LTI) filter whose impulse response, h(t), is 1 between t = 0 and t = Ts and is zero elsewhere.

How do we know this is the impulse response that will give us a pulse of HEIGHT A, not area A? Could anyone explain this "a little bit of thought" part? Thanks in advance.

$\endgroup$
  • $\begingroup$ It's sad to see, as late as 2008, that some author is still confused about the role and functions of the Sample-and-Hold (a device that sometimes precedes an A/D converter in the signal path) and the Zero-order Hold which is a hypothetical device (which is LTI) that is needed to model the hold state of a conventional D/A converter between sample instances. The author, S. Engelberg, is mistaken and repeats a mistake that was in early textbooks on DSP. Sample-and-Hold (S/H) is not the same as Zero-order Hold (ZOH). $\endgroup$ – robert bristow-johnson Mar 20 '14 at 3:07
2
$\begingroup$

An impulse of "area" $A$ means $A\delta(t)$. The nomenclature arises because $\displaystyle \int_{-\infty}^\infty A\delta(t) \,\mathrm dt = A,$ that is, the "area under the curve" is $A$.

Since the system is linear and time-invariant, and the response to input $\delta(t)$ is a pulse of height $1$ and duration from $0$ to $T_s$, the response to $A\delta(t)$ is a pulse of height $A$ and duration from $0$ to $T_s$. Its "area" is $AT_s$, that is, $A$ multiplied by the area $T_s$ of the response to the unit impulse.

$\endgroup$
0
$\begingroup$

Some little bits of thought:

  • the system behaves constantly at any time, turning an impulse into a rectangle => it is time invariant. One can also say that the system needs to transform an impulse into a rectangle but does not now when it will arrive;
  • linear, because if two impulses are superimposed then the heights will be added, not mixed or masked in any other way;
  • the time range between t and Ts should be obvious;
  • the desired output is the height of the impulse (which carries the information), but this value needs to be kept during a certain amount of time Ts to be able to be observed. Thus, what the sample-and-hold should output is a rectangle height A and width Ts. The sample part needs to detect the amplitude A of the signal, while the hold part need to maintain it during Ts seconds.
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.