Impulse invariance: why the factor $T$ in $h[n] = Th_c(nT)$?

Question

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.

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:

$$y(t)=\int_{-\infty}^{\infty}x(\tau)h_c(t-\tau)d\tau\tag{1}$$

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:

$$y[n]=\sum_{k=-\infty}^{\infty}x[k]h[n-k]\tag{5}$$

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.