Week 3

A system is \emph{linear} if a system defined by $y = F(x)$ such that
$$x \to \fbox{F} \to y $$
satisfies two properties. The first property is \emph{homogeneity}, where $\alpha y = F(\alpha x)$ such that
$$\alpha x \to \fbox{F} \to \alpha y $$
where $\alpha \in \C$. The second property is \emph{additivity}, where $y_1 = F(x_1)$ and $y_2 = F(x_2)$ such that
$$x_1 + x_2 \to \fbox{F} \to y_1 + y_2. $$
As a result, these two properties together form \emph{superposition}, where we have
$$\alpha_1 x_1 + \alpha_2 x_2 \to \fbox{F} \to \alpha_1 y_1 + \alpha_2 y_2 $$


Lets look at some examples:

\begin{problem}

Consider a circuit where the voltage across the circuit is given by Ohm's Law, $V=IR$, where $I$ is the current, and $R$ is the resistance.

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

\begin{enumerate}
\item If the resistance is constant, is $y(t)=Rx(t)$ linear?
\item What if $R$ varies with time, $y(t)=R(t)x(t)$?
\end{enumerate}
\end{problem}

\begin{solution}
Let $\hat{x} = \alpha x$., then $\hat{y} = R\hat{x} = \alpha R x = \alpha y$. Let $\tilde{x} = x_1 +  x_2$, then $\tilde{y}=R\tilde{x}=Rx_1+Rx_2=y_1+y_2$. Therefore the system is linear. Since we never used the constancy of $R$, we can observe that in both cases we have linearity.
\end{solution}



\begin{problem}
Are the following signals linear?
\begin{enumerate}
\item $y(t) = x^2(t)$
\item $y(t) = \Re(x(t))$
\item $y(t)=\log _i e$
\end{enumerate}
\end{problem}

\begin{solution}

\begin{enumerate}
None of the systems were linear.
\item Not linear. Let $\hat{x} = \alpha x(t)$, then $\hat{y} = \left( \alpha x(t) \right)^2= \alpha^2 x^2(t) \neq \alpha y(t)$
\item Not linear. Let $\hat{x} = \alpha x(t)$, then $\hat{y} = \Re(\alpha x(t))$ where $\alpha \in \C$. When $\alpha = i$, then we have $\hat{y} = \Re(i a(t) - b(t))=-b(t)$, and if $\alpha = 1$, then $\hat{y} = \Re( a(t) +i b(t))=a(t)$. In both cases, $\hat{y} \neq \alpha y(t)$.
\item Not linear, but a fun problem to solve for. Let $\log _i e=y$, then $i^y = e \Rightarrow e^{i \frac{\pi}{2}y} = e \Rightarrow  i \frac{\pi}{2}y = 1 \Rightarrow y = \frac{2}{\pi i}$
\end{enumerate}
\end{solution}


Linearity

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

If we multiply the input signal by $0$, we should have a $0$ output:
$$ 0 \cdot x \to \fbox{F} \to 0 \cdot y $$

This property known as \emph{ZIZO}, can be stated as follows:

$$  \mbox{system is linear} \Rightarrow \mbox{zizo}$$

ZIZO simply means \emph{zero input zero output}. Consider the system defined by: 

%\begin{center}
%\includegraphics{images/modulators/nonzizo.ps}
%\end{center}

\begin{center}
\begin{pspicture}(-3,-3)(4,1.2)

% in from x
\rput(-3.2,0){$x(t)$}
\psline[linewidth=1.25 pt, arrowscale=1.25]{->}(-2.7,0)(-0.25,0)


% out to y
\rput(3.9,0){$y(t)$}

\psline[linewidth=1.25 pt, arrowscale=1.25]{->}(0.25,0)(1.25,0)
% extended
\psline[linewidth=1.25 pt, arrowscale=1.25]{->}(1.75,0)(3.5,0)

% up arrow
\rput(1.5,-2){$g(t)$}
\psline[linewidth=1.25 pt, arrowscale=1.25]{->}(1.5,-1.65)(1.5,-0.25)

% up arrow
\rput(0,-2){$f(t)$}
\psline[linewidth=1.25 pt, arrowscale=1.25]{->}(0,-1.65)(0,-0.25)

% multiplier
\psline{-}(-0.1315,-0.2125)(0.1315,0.2125)
\psline{-}(-0.1315,0.2125)(0.1315,-0.2125)
\pscircle(0,0){0.25}

% adder
\psline{-}(1.25,0)(1.75,0)
\psline{-}(1.5,0.25)(1.5,-0.25)
\pscircle(1.5,0){0.25}


% box
\pspolygon(-1.25,-2.65)(-1.25,1)(2.65,1)(2.65,-2.65)

\end{pspicture}
\end{center}

$y(t) = x(t)f(t) + g(t)$, since zero input still gives us $g(t)$, the system is not linear, since the contrapositive is also true:
$$ \lnot \mbox{zizo} \Rightarrow \lnot \mbox{system is linear} $$

Although some systems may not be linear, they may be \emph{incrementally linear}, for example, consider the following two signals given by
$y_1(t)=f(t)x_1(t)+g(t)$, and $y_2(t)=f(t)x_2(t)+g(t)$. Then $y_1(t) - y_2(t)=f(t)\left[x_1(t)-x_2(t)\right]$ is incrementally linear, or \emph{affine}, meaning that the difference between the two functions is linear.


ZIZO

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?

%\begin{center}
%\includegraphics{images/modulators/amcos.ps}
%\end{center}

\begin{center}
\begin{pspicture}(-3,-3)(3,1.2)

% in from x
\rput(-3.2,0){$x(t)$}
\psline[linewidth=1.25 pt, arrowscale=1.1]{->}(-2.7,0)(-0.25,0)
% out to y
\rput(3.2,0){$y(t)$}
\psline[linewidth=1.25 pt, arrowscale=1.1]{->}(0.25,0)(2.7,0)
% up arrow
\rput(0,-2){$\cos(\omega_0 t)$}
\psline[linewidth=1.25 pt, arrowscale=1.1]{->}(0,-1.65)(0,-0.25)

% multiplier
\pscircle(0,0){0.25}
\psline(-0.175,0.175)(0.175,-0.175)
\psline(0.175,0.175)(-0.175,-0.175)

% box
\pspolygon(-1.5,-2.65)(-1.5,1)(1.5,1)(1.5,-2.65)

\end{pspicture}
\end{center}

\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}

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

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

\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}


3.1 Properties of Systems