The system $y(n) = c\cdot x(n) +d$ is given. Is the system linear, time-invariant, stable and causal?

Question

The system $y(n) = c\cdot x(n) +d $ is given. (Unfortunately, nothing more is given.) Is the system linear, time-invariant, stable and/or causal?

I assumed, that $c,d\in\mathbb{R}$

Linearity

We need to show, that $T\{k_1x_1(n)+k_2x_2(n)\}=k_1\cdot T(x_1(n)\}+k_2\cdot T\{x_2(n)\}$ (I assume for $k_i\in\mathbb{R}$ and T might be the shift-function)

$T\{cx(n)+d\}=T\{cx(n)\}+T\{d\}=c\cdot T\{x(n)\}+d$

Time-invariant

We need to show, that $T\{x(n)\}=y(n)\implies T\{x(n-n_0)\}=y(n-n_0)$

$T\{c\cdot x(n-n_0)+d\}=c\cdot T\{x(n-n_0)\}+d=c\cdot T\{x(n)-x(n_0)\}+d=T\{c\cdot (x(n)-x(n_0))+d\}=T\{c\cdot x(n)-c(n_0)+d\}=y(n)-y(n_0)=y(n-n_0)$

Stability

We need to show, that $\mid y(n)\mid <N<\infty \text{ with } \mid x(n)\mid < M,\; \text{ I assume for some fixed } M,N\in \mathbb{R}$

$\sum\limits^{\infty}_{-\infty}\mid T\{cx(n)+d\}\mid<N<\infty=\sum\limits^{\infty}_{-\infty}\mid c\cdot T\{x(n)\}+d\mid<N<\infty=\sum\limits^{\infty}_{-\infty}\mid T\{x(n)\}\mid<N<\infty=\sum\limits^{\infty}_{-\infty} M<N<\infty\implies M<N<\infty$

Causality

We need to show, that $x(n) = 0$ for $n<n_0 \implies y(n) = 0$ for $n<n_0$

$y(n) = cx(n)+d=c\cdot 0 +d = d \implies y(n) = d$ The system is causal for $d=0$

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Is this ok or complete nonsense? Furthermore, I apologize for wrong/non-complete definitions.

Answer 1

I feel you may be over thinking it, in that there are simpler proofs.

Linear: Not unless $d=0$. Let $x[n] = x_1[n]+x_2[n]$. Then:

$$y[n] = cx[n] + d = cx_1[n] + cx_2[n] + d$$

But now let $y_1[n] = cx_1[n] + d$ and $y_2[n] = cx_2[n] + d$. Then $y_1 + y_2 = cx_1 + cx_2 + 2d$, so if $d \ne 0$ this is nonlinear.

Time invariant: Yes as long as $c,d$ are constant. Let $n = n_0 + K$. Then $y[n_0 + K] = cx[n_0 + K] + d$

Stable: The coarsest stability criterion we have is BIBO. Presuming again that $c,d$ are constant, then let $x$ be bounded such that $|x[n]| \le X \forall n$. Then $y[n] \le c|X| + d \lt \infty$

Causal: by definition of causality, the output can depend only on current and previous inputs. It does, in that $y[n] = y(x[n])$.