# Forward Euler Discretization

I don't understand why the substitution $$s=\frac{z-1}{T}$$ allows us to discretize a transfer function from laplace to z-transform through Forward Euler Discretization. Can you explain to me ?

Multiplication with $$s$$ in the Laplace transform domain equals differentiation in the time domain. In the discrete-time domain we can approximate differentiation by the equation

$$y[n]=\frac{x[n+1]-x[n]}{T}\tag{1}$$

where $$T$$ is the sampling interval.

In the Z-transform domain, Eq. $$(1)$$ becomes

$$Y(z)=X(z)\frac{z-1}{T}\tag{2}$$

I.e., the transfer function

$$H(z)=\frac{z-1}{T}\tag{3}$$

approximates differentiation, and replacing $$s$$ in a continuous-time transfer function by $$H(z)$$ is thus a way (usually not the best one) to approximate a continuous-time system by a discrete-time system.

One of the problems with the forward Euler method is that transforming a stable continuous-time system could result in an unstable discrete-time system.

• When you talk about instability, what definition of stability are you considering ? Commented Jan 12, 2020 at 17:44
• @themagiciant95: The one that is relevant for linear time-invariant systems: BIBO-stability. A continuous-time system with all its poles in the left-half plane could transform to a discrete-time system with poles outside the unit circle of the z-plane. Commented Jan 12, 2020 at 17:48
• Thanks. Can you help me here -> dsp.stackexchange.com/questions/63187/… ? Commented Jan 12, 2020 at 17:48
• Another thing, i read that Backward Euler maintain the stability, are there any drawbacks in using the Backward Euler ? Commented Jan 12, 2020 at 18:02
• @themagiciant95: Otherwise similar to forward Euler, it's usually only good at frequencies that are small compared to the sampling frequency, because replacing the differential quotient by differences only works well if the step $T$ is sufficiently small, i.e., if the sampling frequency is high (compared to the frequencies we're interested in). Commented Jan 12, 2020 at 18:04