\begin{problem}
Determine if the following signals are time-invariant.
\begin{enumerate}
\item $y(t) = \Re(x(t))$
\item $y(t) = \cos ( x(t) )$
\item $y(t) = \sum \limits_{k=0}^{n}x(k)$
\item $y(t) = x(4n+1)$
\item $y(t) = \int^{2t}_{-\infty}x(\tau)d\tau$
\end{enumerate}
\end{problem}

\begin{solution}
Here are the solutions:
\begin{enumerate}
\item $y(t) = \Re(x(t))$ is time invariant.
Let $\hat{x}(t) = x(t-T)$, then $\hat{y}(t) = \Re(\hat{x}(t)) = \Re{x(t-T)} = \frac{x(t-T) + x^*(t-T)}{2} = y(t-T) $

\item $y(t) = \cos ( x(t) )$ is time invariant.
Let $\hat{x}(t) = x(t-T)$, then $\hat{y}(t) = \cos(\hat{x}(t)) = \cos(x(t-T)) = y(t-T)$

\item $y(n) = \sum \limits_{k=0}^{n} x(k)$ is not time invariant.
Let $\hat{x}(n) = x(n-N)$. Then we have:
\begin{align*}
\hat{y}(n) &= \sum \limits_{k=0}^{n} \hat{x}(k)  \quad \mbox{note that the sum uses }k \mbox{, } n \mbox{ is in the limit} \\
\hat{y}(n) &= \sum \limits_{k=0}^{n} x(k-N)\\
\hat{y}(n) &= \sum \limits_{\omega=-N}^{n-N} x(\omega) \quad \mbox{change of variable } (\omega = k-N) \\
\hat{y}(n) &\neq y(n-N)
\end{align*}

\item $y(n) = x(4n+1)$ is not time invariant. This one can be slightly tricky:

Let $\hat{x}(n) = x(n-N)$, then $\hat{y}(n) = \hat{x}(4n+1) = x((4n + 1) - N) \neq x(4(n-N) + 1) = y(n-N)$

\item $y(t) = \int^{2t}_{-\infty}x(\tau)d\tau$ is not time invariant.
Let $\hat{x}(t) = x(t-T)$, then
\begin{align*}
\hat{y}(t) &= \int^{2t}_{-\infty}\hat{x}(\tau)d\tau \\
\hat{y}(t) &= \int^{2t}_{-\infty}x(\tau -T)d\tau \\
\hat{y}(t) &= \int^{2t-T}_{-\infty}x(\lambda)d\lambda \quad \mbox{variable change } (\lambda = \tau - T) \\
\hat{y}(t) &\neq \int^{2t-2T}_{-\infty}x(\lambda)d\lambda = y(t-T)
\end{align*}
\end{enumerate}
\end{solution}



Additional Problems