# Derivative Notation as a factor

In the figure below, from the book "Principles of Linear Systems and Signals - Lathi", the author uses the notation D to replace the Leibniz's notation $d/dt$, and after that he rewrites the equation, disconnecting the symbol of the derivative, $D$, of its function, considering each as a factor. I don't understand how he did that. Is that mathematically possible? • Can you add a link to the figure to your question? – Matt L. Mar 31 '15 at 19:24
• I already upload the figure. – user15266 Mar 31 '15 at 21:10
• it's a sloppy abuse of notation that is equivalent to Laplace transformation of the linear differential equations (with the assumption of 0 initial conditions). essentially "$D$" means "$s$" of the L.T. and Eq. (2.1b) should be $$(s^N + a_1 s^{N-1} + ... + a_{N-1}s + a_N) Y(s) = (b_{N-M} s^M + b_{N-M+1} s^{M-1} + ... + b_{N-1} s + b_N) X(s)$$ and you would have polynomials $Q(s)$ and $P(s)$ and Eq. (2.1c) would be $$Q(s)Y(s) = P(s)X(s)$$. – robert bristow-johnson Mar 31 '15 at 21:41
• @robertbristow-johnson, you're very mistaken. This is neither sloppy nor abusive. The theory of linear operators on vector spaces uses this notation very rigorously and consistently, and there are many upsides to it. It is also fully compatible with the usual notion of an operator or a matrix product in linear algebra. – Jazzmaniac May 1 '15 at 13:15
• You always have the last word @robertbristow-johnson, don't you? I studied mathematics, not engineering, so I doubt that you share my perspective. Mathematics is very comfortable with polynomials of linear operators, and those products are very well defined and as "real" as products between numbers. There's nothing fishy, misleading, sloppy, abusive or confusing about it. It's perfectly rigorous mathematics. But I know you enjoy leaning far out and uttering your opinion, so please keep going. – Jazzmaniac May 1 '15 at 20:41

As stated in the text, it's called operational notation, i.e. $D$ is an operator being applied to a continuous-time signal:

$$D^mx(t)=\frac{d^mx(t)}{dt^m}$$

In discrete time you have the operator $T$, defined by

$$T^mx[n]=x[n+m]$$

(see e.g. here).

What is often done by people is that they use the complex Laplace transform variable $s$ instead of $D$, or the $\mathcal{Z}$-transform variable $z$ instead of $T$. This is what Robert Bristow-Johnson referred to in his comment as 'sloppy abuse of notation'. I would even go further and would call it simply wrong. The use of different letters for $D$ and $T$ may seem an unimportant triviality, but the real problem here is that you mix different domains (Laplace/$\mathcal{Z}$ transform domain and time domain), which can be a great source of misunderstandings and errors.

I personally do not see a great advantage in the use of this type of operational notation for engineering applications. It rather makes things more confusing for beginners because they have a tendency to mix domains anyway, and operational notation seems to suggest that this is legitimate.