On Wikipedia, I came across the following equation $$h[n] = Th_c(nT)$$ This equation relates the impulse response of the discrete time filter which is the impulse invariant of a continuous time filter. $h_c(t)$ denotes the impulse response of the continuous time filter, $h[n]$ denotes the discrete time filter.

I fail to understand why $T$ is multiplied on the right hand side.

I have seen a similar equation that relates the value of a discrete time signal and samples of a continuous time signal: $$x[n] = x_c(nT)$$ Which does make sense to me as we are sampling our continuous time signal $x_c(t)$ and formulating a discrete time signal with these individual samples.


1 Answer 1


Sampling the continuous-time impulse response will result in a scaled output signal. This can be seen as follows:

The output signal of the continuous-time system is given by the continuous-time convolution:


The integral in $(1)$ is the limit of a Riemann sum:

$$y(t)=\lim_{\Delta\tau\to 0}\sum_{k=-\infty}^{\infty}x(k\Delta\tau)h_c(t-k\Delta\tau)\Delta\tau\tag{2}$$

If we replace $\Delta\tau$ by a small but finite value $T$ we obtain the following approximation:

$$y(t)\approx \sum_{k=-\infty}^{\infty}x(kT)h_c(t-kT)\cdot T\tag{3}$$

Sampling the output at $t=nT$ results in

$$y(nT)\approx \sum_{k=-\infty}^{\infty}x(kT)h_c((n-k)T)\cdot T\tag{4}$$

The discrete-time sequence on the right-hand side of $(4)$ is our desired discrete-time output signal:


with $x[k]=x(kT)$ and $h[k]=Th_c(kT)$. Hence, the scaling factor $T$ is necessary for $y[n]$ to approximate a sampled version of the original continuous-time output signal.

  • 1
    $\begingroup$ And to note, similarly, if we differentiate in time, we divide by T when creating a matched discrete time process. I just remember d/dt is dividing by dt and integral...dt is multiplying by dt to keep that straight as this usually throws me off when trying to maintain actual time in both continuous and discrete time. So it's specifically the differential or integral processing. $\endgroup$ Commented Feb 15, 2023 at 14:16

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