Consider the system 
$$x \to \fbox{F} \to y $$

where
$$y = F(x) \mbox{, or } y(t) = (F(x))(t)$$

Let $\widehat{x}=x(t-T)$ be a time-shifted version of $x(t)$. Then the system $F$ is \emph{Time Invariant} if $\widehat{y} = y(t-T) \forall x\in X, \forall T \in \R$. There is an assumption that all shifts of $x$ along time are signals that are in $X$. In other words, $X$ is closed under time shifts and so is $Y$.

\begin{problem}
Is the system $y(t) = \cos (\omega_0t)x(t)$ time invariant?

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\rput(-3.2,0){$x(t)$}
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\rput(3.2,0){$y(t)$}
\psline[linewidth=1.25 pt, arrowscale=1.1]{->}(0.25,0)(2.7,0)
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\rput(0,-2){$\cos(\omega_0 t)$}
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\end{problem}

\begin{solution}
$y(t) = \cos (\omega_0t)x(t)$ is time varying. 
 
Let $\widehat{x}(t) = x(t-T)$. Then $\widehat{y}(t) = \cos( \omega_0 t)\widehat{x}(t) = \cos(\omega_0t)x(t-T) \neq y(t-T)$

\end{solution}

\begin{problem}
Is this system time invariant if the system involving a capacitor, where the voltage $y(t)$ is defined by:
\begin{enumerate}
\item $y(t) = \frac{1}{C}\int^t_{-\infty}x(\tau)d\tau$
\item $y(t) = \frac{1}{C}\int^t_0x(\tau)d\tau$
\end{enumerate}

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%\includegraphics{images/circuits/capacitor.ps}
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\end{problem}

\begin{solution}
\begin{enumerate}
\item $y(t) = \frac{1}{C}\int^t_{-\infty}x(\tau)d\tau$ is time invariant.

Let $\widehat{x}(t) = x(t-T)$. Then $$\widehat{y}(t) = \frac{1}{C} \int^t_{-\infty}\widehat{x}(\tau)d\tau  = \frac{1}{C} \int^t_{-\infty} x(\tau-T) d\tau .$$ Let $\lambda = \tau -T \Rightarrow d\lambda = d\tau$, and the limits of the integral change from $-\infty$ to $t-T$ since $\tau = -\infty \Rightarrow \lambda = -\infty$, and $\tau = t \Rightarrow \lambda = t - T$. So now we have$\frac{1}{C} \int ^{t-T}_{-\infty}x(\lambda)d\lambda = y(t-T)$

\item $y(t) = \frac{1}{C}\int^t_0x(\tau)d\tau$ is time varying.

When we do a variable change, the integral becomes $\frac{1}{C} \int^{t-T}_{-T}x(\lambda)d\lambda \neq y(t-T)$

\end{enumerate}
\end{solution}

\begin{problem}

Given a system involving a resistor with voltage $y(t)$, is the system LTI when the voltage is given by:
\begin{enumerate}
\item $y(t) = Rx(t)$
\item $y(t) = R(t)x(t)$
\end{enumerate}

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%\includegraphics{images/circuits/resistor.ps}
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\end{problem}

\begin{solution}
\begin{enumerate}
\item $y(t) = Rx(t)$ is time invariant.

Let $\widehat{x}(t) = x(t-T)$. Then $\widehat{y}(t) = R\widehat{x} = R x(t-T) = y(t-T)$

\item $y(t) = R(t)x(t)$ is time varying.

Let $\widehat{x}(t) = x(t-T)$. Then $\widehat{y}(t) = R(t)\widehat{x}(t) = R(t) x(t-T) \neq y(t-T)$ 
\end{enumerate}
\end{solution}

\begin{problem}
Determine if the following signals are time-invariant.
\begin{enumerate}
\item $y(t) = x^2(t)$
\item $y(n) = \frac{x(n) + x(n-1)}{2}$
\end{enumerate}
\end{enumerate}

\begin{solution}
Here are the solutions:
\begin{enumerate}
\item $y(t) = x^2(t)$ is time invariant.

Let $\widehat{x}(t) = x(t-T)$. Then $$\widehat{y}(t) = \widehat{x}^2(t) = x^2(t-T) = y(t-T).$$ Therefore it is time invariant $\forall T \in \R, \forall x \in X$

\item $y(n) = \frac{x(n) + x(n-1)}{2}$ is time invariant.

Let $\widehat{x}(n) = x(n-N)$. Then $\widehat{y}(n) = \frac{\widehat{x}(n) + \widehat{x}(n-1)}{2} = \frac{x(n-N) + x(n-N-1)}{2} = y(t-T)$

\end{enumerate}

\end{solution}


Time Invariance