Prove Differentiator is Linear and Time-Invariant

Question

The differentiator gives an output equal to the derivative of its input. Show that the differentiator is a linear time invariant system. Consider the input $f(t)=\sin(t^2).$

Attempt

For time-invariant:

I'm not sure how to prove it is time invariant. We need to show that for a delayed input $f(t-t_0)$ the output will be $g(t-t_0).$ So in this case let $f=\sin (t^2 -a),$ then the output will be:

$$\frac{d}{dt} \sin (t^2 -a) = 2t \cos (t^2 -a)$$

How can I proceed after this step? Is this the correct approach?

For linearity:

For constants $c_1, c_2$, and $f_1 = \sin(t^2_1), f_2 = \sin(t^2_2)$:

$$\frac{d}{dt} c_1 \sin(t^2_1) + c_2 \sin(t^2_2) = c_1 2 t_1 \cos (t_1^2) + 2 t_2 c_2 \cos (t_2^2) = c_1 g_1(t) + c_2 g_2 (t).$$

Is this correct or is there a more rigorous way of proving this?

Answer 1

To prove that the differentiator $D=\frac{\mathrm d}{\mathrm d\,t}$ is LTI you must verify that $$ D\{f(t-t_0)\}=g(t-t_0) $$ and $$D\{c_1f_1(t)+c_2f_2(t)\}=c_1D\{f_1(t)\}+c_2D\{f_2(t)\} $$

so you have for $f(t)=\sin(b t^2)$ $$ D\left\{\sin\left(b[t-a]^2\right)\right\}=2b(t-a)\cos\left(b[t-a]^2\right)=g(t-a). $$ For linearity you must choose two different functions of the same variable $t$; for example $f_{1}(t)=\sin(b_{1} t^2)$ anf $f_{2}(t)=\sin(b_{2} t^2)$ different parameters $b_{n}$ $$\begin{align} D\left\{c_1\sin\left(b_1t^2\right)+c_2\sin\left(b_2t^2\right)\right\}&=c_12b_1 t\cos\left(b_1t^2\right)+c_22b_2 t\cos\left(b_2t^2\right)\\ &=c_1D\{\sin\left(b_1t^2\right)\}+c_2D\{\sin\left(b_2t^2\right)\} \end{align} $$

Answer 2

For sufficiently smooth functions $$f:\quad {\mathbb R}\to {\mathbb C},\qquad t\mapsto f(t)$$ ("time signals") we have the operations $D:\>f\mapsto f'$ and $T_a:\> f\mapsto f_a$, whereby $f_a$ is defined by $f_a(t):=f(t-a)$. Both these operations are obviously linear: $D(\lambda f+\mu g)=\lambda\> Df+\mu\> Dg$, and similarly for $T_a$.

It is claimed that $D$ is translation invariant, which is the same thing as stating that $T_a$ and $D$ commute, whatever $a\in{\mathbb R}$: $$D\circ T_a=T_a\circ D\ .$$

Proof. Let $g:=Df$. Then $g(t)=f'(t)$ for all $t$, hence $$g_a(t)=f'(t-a)=\lim_{h\to0}{f(t-a+h)-f(t-a)\over h}=\lim_{h\to0}{f_a(t+h)-f_a(t)\over h}=f_a'(t)$$ for all $t\in{\mathbb R}$. But this is saying that $$T_aD\>f=T_a\>g= D\>f_a=DT_a\>f\ .$$ Since this is valid for all $f$ the claim follows.