I have a basic doubt about discrete time circuits.

A discrete - time signal is a sequence of values which are taken at specific time instants (let's call them sampling instants), as we can see from the following picture:

enter image description here

So, from what I have understood, a discrete - time signal is in general like an analog signal which has been sampled, and it is not exactly equal to a digital signal because its values have not been quantized yet (so they may assume values inside a continuous set of numbers).

My question is about the physical implementations of discrete time circuits (for instance discrete - time filters).

For instance, this is a switched capacitor integrator:

enter image description here

If I have understood it correctly, it performs integration during the sampling instants (in which the capacitor C1 acts like a resistor).

But my question is: obviously the circuital elements (op - amp etc) are ON during all time, in general, and at the output we will have a signal for each time. What does it happen during the time between two sampling instants and how discrete - time filters manage it?

  • \$\begingroup\$ What does C2 do in conjunction with the opamp? \$\endgroup\$
    – user16324
    Apr 22, 2020 at 19:28

2 Answers 2


While phase 1 is active, C1 charges to the level of the input, and retains the last voltage it sensed when phase 1 ended. When phase 2 is active, the negative of that voltage is applied to the op amp input, forcing the output to change until the input is at ground. This discharges C1 and charges C2. The resulting voltage will stabilize at Vin times C1/C2. When phase 2 is over, C2 maintains its charge, so the next phase 2 will add the input cumulatively, hence the integration.

Another way to look at it (depending on which is easiest for analysis) is that during phase 1 C1 charges up, and during phase 2 the charge is transferred to C2, in addition to any charge already on C2.

  • \$\begingroup\$ Ok, now I have understood how this circuit works. My last doubt is about the fact that the mechanism you have just described happens during all time (and not in discrete instants). So, why is this circuit discrete time? Is it supposed to be followed by a stage which looks only at the output at specific discrete instants? \$\endgroup\$
    – Kinka-Byo
    Apr 22, 2020 at 19:48
  • 1
    \$\begingroup\$ It's discrete time because the output is dependent on the value of the input when phase 1 ends. The value of the input at other times has (theoretically) no influence at the output, so it's regarded as sampling the input at the end of phase 1. \$\endgroup\$ Apr 23, 2020 at 14:13

Discrete-time sampling assumes that the signal is at zero between samples, that is, a series of ideally zero-width pulses representing the amplitude of the sampled signal at each instance.

The integrator you show is a switched-capacitor circuit, which will have a stair-step waveform that roughly follows the continuous-time (integral of) the input. It's not really a discrete-time signal in the normal sense, and it wouldn't be thought of that way in signal processing.

Instead, an ADC front-end tries to minimize the sample window in order to make an accurate numeric stream of pulse data. The smaller the sample window, the better the discrete-time representation.

The integrator circuit would be modified to zero out between each sample, so instead it would be measuring the captured charge that represents a sample point without influence from adjacent samples.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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