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 \documentclass[11pt,a4paper,notitlepage,fleqn,final]{article} \input{preambles/preamble2016.tex} \title{Θεωρία Σημάτων και Γραμμικών Συστημάτων \\ { \normalsize Σημειώσεις από τις παραδόσεις }} \date{2016 \\ { \small Τελευταία ενημέρωση: \today } } \author{ Για τον κώδικα σε \LaTeX, ενημερώσεις και προτάσεις: \\ \url{https://github.com/kongr45gpen/ece-notes}} \setmainfont{Linux Libertine O} \setsansfont{Ubuntu} %\newfontfamily\greekfont[Script=Greek]{Linux Libertine O} %\newfontfamily\greekfontsf[Script=Greek]{Linux Libertine O} \usepackage{polyglossia} \newfontfamily\greekfont[Script=Greek,Scale=0.95]{GFS Artemisia} \begin{document} \maketitle \tableofcontents \include{signals/chap0} \include{signals/chap1} \include{signals/chap2} \include{signals/chap3} \section{Θεώρημα Δειγματοληψίας} \subsection{Εισαγωγή} \begin{gather*} X(f) = \int_{-\infty}^{\infty} x(t) e^{-j2\pi ft}\dif t\\ x(t) = \int_{-\infty}^{\infty} X(f) e^{j2\pi f t}\dif t\\ x_1(t) x_2(t) \xrightarrow{\mathscr F} X_1(f)*X_2(f)\\ X_1(f) X_2(f) \xrightarrow{\mathscr F^{-1}} x_1(t)*x_2(t)\\ \mathrm{sinc} (t) = \frac{\sin(\pi t)}{\pi t} \end{gather*} Συνάρτηση δειγματοληψίας: $$\displaystyle S_{T_s}(t) = \sum_{n=-\infty}^\infty \delta(t-nT_s) \xrightarrow{\mathscr F} F_p \sum_{n=-\infty}^\infty \delta(f-nF_p) \qquad F_p=\frac{1}{T_s}$$ \begin{tikzpicture} \draw[->] (-3,0) -- (3,0); \draw (0,0) -- (0,3); \draw[thick,blue,->] (-2,0) -- (-2,1.5); \draw[thick,blue,->] (-1,0) -- (-1,1.5); \draw[thick,blue,->] (0,0) -- (0,1.5); \draw[thick,blue,->] (1,0) -- (1,1.5); \draw[thick,blue,->] (2,0) -- (2,1.5); \draw[<->] (1,-0.2) --(2,-0.2) node[below,midway] {$T_s$}; \draw (4,1.5) node {$$\xrightarrow{\qquad \mathlarger{\mathlarger{\mathlarger{\mathscr F}}}\qquad }$$}; \draw (6,0) -- (11,0); \draw[->] (8.4,0) -- (8.4,3); \draw[thick,blue,->] (6.8,0) -- (6.8,1.5); \draw[thick,blue,->] (7.6,0) -- (7.6,1.5); \draw[thick,blue,->] (8.4,0) -- (8.4,1.5); \draw[thick,blue,->] (9.2,0) -- (9.2,1.5); \draw[thick,blue,->] (10,0) -- (10,1.5); \draw[<->] (9.2,-0.2) --(10,-0.2) node[below,midway] {$F_p = \frac{1}{T_s}$}; \end{tikzpicture} \begin{gather*} S_{F_s}(f) \xrightarrow{\mathscr F^{-1}} T_p S_{T_p} (t) \qquad T_p=\frac{1}{F_s} \\ S_{\Xi_s}(\xi) \end{gather*} Αν κάπου δειγματοληπτώ, στον άλλο χώρο είναι περιοδικότητα \subsubsection [Συνάρτηση ορθογωνικού παραθύρου μήκους T στον κόσμο t]{% Συνάρτηση ορθογωνικού παραθύρου μήκους $$T$$ στον κόσμο $$t$$} $W_T(t) = \begin{cases} 1 \quad & |t| < \sfrac{T}{2} \\ 0 \quad & \text{αλλού} \end{cases}$ \begin{tikzpicture} \draw (-2,0) -- (2,0); \draw[->] (0,-0.2) -- (0,2); \draw[very thick, blue] (-1,0) node[black,below] {$-\frac{T}{2}$} --(-1,1) -- (1,1) -- (1,0) node[black,below] {$\frac{T}{2}$}; \draw (3,1) node {$$\xrightarrow{\qquad \mathlarger{\mathlarger{\mathlarger{\mathscr F}}}\qquad }$$}; \draw (5,0) -- (9,0); \draw[->] (7,-0.2) -- (7,2); \draw[very thick, xshift=7cm,xscale=0.5,samples=50,domain=-4:4,smooth,variable=\x,blue] plot ({\x},{sinc(pi*\x)}); \draw[thick,green!60!black,<->] (6.5,-0.1) -- ++(1,0) node[below,midway] {$\frac{2}{T}$}; \draw[thick,green!60!black,<->] (8,-0.1) -- ++(0.5,0) node[below,midway] {$\frac{1}{T}$}; \end{tikzpicture} \begin{tikzpicture} \draw[->] (-2,0) -- (2,0) node[below] {$f$}; \draw[->] (0,-0.2) -- (0,2) node[below right] {$\mathrm W_F(f)$}; \draw[very thick, blue] (-1,0) node[black,below] {$-\frac{F}{2}$} --(-1,1) -- (1,1) -- (1,0) node[black,below] {$\frac{F}{2}$}; \draw (3,1) node {$$\xrightarrow{\qquad \mathlarger{\mathlarger{\mathlarger{\mathscr F^{-1}}}}\qquad }$$}; \draw (6,1) node[yshift=-5pt] {$$\mathlarger{\mathlarger{\mathlarger{F\;\mathrm{sinc}(Ft)}}}$$}; \end{tikzpicture} \subsubsection{Δειγματοληψία} \begin{tikzpicture} \draw[thick] (0,0) circle(0.15); \fill[thick] (0,0) circle (0.03); \draw[snake,->] (0,2) node[right] {γινόμενο} -- (0,0.15); \draw[thick,->] (-2,0) -- ++(1.85,0) node[above,midway] {$g(t)$}; \draw[thick,->] (0,-2) node[below] {$S_{T_S}$} -- ++(0,1.85); \draw[thick,->] (0.15,0) -- (2,0) node[below] {$g_s(T)$}; \draw[->] (3,0) -- ++(1.5,0) node[above,midway] {$\mathscr F$}; \begin{scope}[xshift=7cm] \draw[thick] (0,0) node {$*$} circle(0.15); \draw[snake,->] (0,2) node[right] {συνέλιξη} -- (0,0.15); \draw[thick,->] (-2,0) -- ++(1.85,0) node[above,midway] {$G(f)$}; \draw[thick,->] (0,-2) node[below] {$F_P S_{F_P}$} -- ++(0,1.85); \draw[thick,->] (0.15,0) -- (2,0) node[below] {$G^S(f)$}; \end{scope} \draw (0,-3) node {Δειγματοληψία}; \end{tikzpicture} \begin{align*} G^S(f) &= G(f) * F_p S_{F_p}(f) \\ &= F_p G(f) * \left( \sum_n \delta(f-nF_p) \right) \\ &= F_p \sum_n G(f-nF_P) \end{align*} Παρατηρούμε τις επικαλύψεις μεταξύ των διαδοχικών φασμάτων (aliasing). Για να περιοριστεί αυτό μπορούμε να αυξήσουμε το $$F_p$$ ($$\implies$$ να αυξήσουμε τη συχνότητα δειγματοληψίας) \begin{tikzpicture} \draw[very thick,blue] plot [variable=\x,domain=-2:2,smooth,samples=20] ({\x}, { 1.5*exp( -\x*\x) }); \draw[->] (0,-0.5) -- (0,2) node[right] {$G(f)$}; \draw[->] (-2,0) -- (2,0) node[below] {$f$}; \draw (0,1.5) node[above left] {$A$}; \draw[->,thick] (2,1) -- ++(2,0) node[above,midway] {After}; \begin{scope}[xshift=7cm] \draw[very thick,blue!15] plot [variable=\x,xshift=2cm,domain=-2:2,smooth,samples=20] ({\x}, { 1.5*exp( -\x*\x) }); \draw[very thick,blue!40] plot [variable=\x,xshift=-1cm,domain=-2:2,smooth,samples=20] ({\x}, { 1.5*exp( -\x*\x) }); \draw[very thick,blue!40] plot [variable=\x,xshift=1cm,domain=-2:2,smooth,samples=20] ({\x}, { 1.5*exp( -\x*\x) }); \draw[very thick,blue] plot [variable=\x,domain=-2:2,smooth,samples=20] ({\x}, { 1.5*exp( -\x*\x) }); \draw[->] (0,-0.5) -- (0,2) node[right] {$G^S(f)$}; \draw[->] (-2,0) -- (2.5,0) node[above right] {$f$}; \draw (0,1.5) node[above left] {$A_{F_p}$}; \draw (-1,0) node[below] {$-F_P$}; \draw (1,0) node[below] {$F_P$}; \draw (2,0) node[below] {$2F_P$}; \end{scope} \end{tikzpicture} \begin{tikzpicture}[scale=1.1] \draw[very thick,blue] plot[smooth,tension=2] coordinates {(-1,0) (0,1.5) (1,0)}; \draw (-1,0) node[below] {$-\sigma$}; \draw (1,0) node[below] {$\vphantom{-} \sigma$}; \draw (0,-1) node {$\sigma$-ζωνοπερατό}; \draw[->] (0,-0.5) -- (0,2) node[right] {$G(f)$}; \draw[->] (-2,0) -- (2,0) node[below] {$f$}; \draw (0,1.5) node[above left] {$A$}; \filldraw (0,1.5) circle (1pt); \draw[->,thick] (1.8,1) -- ++(2,0) node[above,midway] {After}; \begin{scope}[xshift=8cm] \foreach \x in {-3,-1.5,...,1.5} { \begin{scope} \clip plot[xshift=\x cm,smooth,tension=2] coordinates {(-1,0) (0,1.5) (1,0)}; \fill[purple!40,postaction={pattern=north east lines,opacity=.2}] plot[xshift={\x cm+1.5 cm},smooth,tension=2] coordinates {(-1,0) (0,1.5) (1,0)}; \end{scope} } \draw[xshift=3cm,very thick,blue!70,path fading=east] plot[smooth,tension=2] coordinates {(-1,0) (0,1.5) (1,0)}; \draw[xshift=1.5cm,very thick,blue!85] plot[smooth,tension=2] coordinates {(-1,0) (0,1.5) (1,0)}; \draw[xshift=-3cm,very thick,blue!70,path fading=west] plot[smooth,tension=2] coordinates {(-1,0) (0,1.5) (1,0)}; \draw[xshift=-1.5cm,very thick,blue!85] plot[smooth,tension=2] coordinates {(-1,0) (0,1.5) (1,0)}; \draw[very thick,blue] plot[smooth,tension=2] coordinates {(-1,0) (0,1.5) (1,0)}; \draw[->] (0,-0.5) -- (0,2); \draw[->] (-3,0) -- (3,0); \draw[dashed] (1.5,1.5) -- ++(0,-1.5) node[below] {$F_P$}; \draw[->] (0.5,-0.05) -- (0.5,-0.4) node[below] {$\mathsmaller{F_P-\sigma}$}; \begin{scope}[yshift=-4cm] \fill[green!30,postaction={pattern=north west lines,opacity=.1}] (-1.3,0.7) rectangle (1.3,0); \draw[xshift=5cm,very thick,blue!70,path fading=east] plot[smooth,tension=2] coordinates {(-1,0) (0,1.5) (1,0)}; \draw[xshift=2.5cm,very thick,blue!85] plot[smooth,tension=2] coordinates {(-1,0) (0,1.5) (1,0)}; \draw[xshift=-5cm,very thick,blue!70,path fading=west] plot[smooth,tension=2] coordinates {(-1,0) (0,1.5) (1,0)}; \draw[xshift=-2.5cm,very thick,blue!85] plot[smooth,tension=2] coordinates {(-1,0) (0,1.5) (1,0)}; \draw[very thick,blue] plot[smooth,tension=2] coordinates {(-1,0) (0,1.5) (1,0)}; \draw[->] (0,-0.5) -- (0,2); \draw[->] (-3,0) -- (3,0); \draw[dashed] (2.5,1.5) -- ++(0,-1.5) node[below] {$F_P$}; \draw[->] (1.5,-0.05) -- (1.5,-0.4) node[below] {$\mathsmaller{F_P-\sigma}$}; \draw[->] (3.5,-0.05) -- (3.5,-0.4) node[below] {$\mathsmaller{F_P+\sigma}$}; \draw[green!50!black,very thick] (-1.3,0) -- ++(0,0.7) node[above,rotate=45,xshift=15pt,yshift=-1mm] {$W_F(f)$} -- ++(2.6,0) -- ++(0,-0.7); \draw[gray!50!black,thick,->] (0,-1) -- ++(0,-1); \draw[blue!50!black] (-1,0) node[below] {$-\sigma$}; \draw[blue!50!black] (1,0) node[below] {$\vphantom{-} \sigma$}; \end{scope} \begin{scope}[yshift=-8.5cm] \draw[->] (0,-0.5) -- (0,2); \draw[->] (-3,0) -- (3,0); \draw[very thick,blue] plot[smooth,tension=2] coordinates {(-1,0) (0,1.5) (1,0)}; \draw[blue!50!black] (-1,0) node[below] {$-\sigma$}; \draw[blue!50!black] (1,0) node[below] {$\vphantom{-} \sigma$}; \end{scope} \end{scope} \end{tikzpicture} Για να μην έχουμε aliasing πρέπει: $F_p -\sigma > \sigma \implies F_p > 2\sigma$ $\text{\Large Nyquist's Criterion:} \quad \boxed{ \hspace{120pt} \mathlarger{\mathlarger{\mathlarger{\mathlarger{\mathlarger{ \underbrace{F_p}_{\mathclap{\text{συχνότητα δειγματοληψίας}}} > 2\overbrace{\sigma}^{\mathclap{\text{max frequency}}} }}}}} \hspace{100pt} }$ \begin{center} \begin{tikzpicture}[scale=1.5] \fill[green!10,postaction={pattern=north west lines,opacity=.1}] (-1.7,0.7) rectangle (1.7,0); \draw[very thick,blue] plot[smooth,tension=2] coordinates {(-1,0) (0,1.5) (1,0)}; \draw[->] (0,-0.5) -- (0,2) node[right] {$F_P-\sigma$}; \draw (-3,0) -- (3,0); \draw[green!50!black,very thick] (-1.7,0) -- ++(0,0.7) -- ++(3.4,0) -- ++(0,-0.7); \draw (-1,0) node[below] {$-\sigma$}; \draw (1,0) node[below] {$\vphantom{-} \sigma$}; \draw[green!50!black] (-1.7,0) node[below] {$-\frac{F}{2}$}; \draw[green!50!black] (1.7,0) node[below] {$\frac{F}{2}$}; \end{tikzpicture} \end{center} \begin{gather*} W_F(f): \sigma < \sfrac{F}{2} < F_p-\sigma \\ \underbrace{F_p = W_F(f)\cdot G^S(f)}_{\downarrow ΙFT} \\ F_p g(t) = \mathscr F^{-1} \left\lbrace W_F(f) \right\rbrace * \mathscr F^{-1} \left\lbrace G^S(f) \right\rbrace \end{gather*} \begin{align*} F_p g(t) &= F \sinc(Ft) * \left( \underbrace{g(t)\cdot S_{T_s}(t)}_{g_s(t)} \right) \\ \\ &= F\int_{-\infty}^{\infty} g(t)\sum_n \delta(t'-nT_s)\sinc\left(F(t-t')\right)\dif t' \\ &= F\sum_n \int_{-\infty}^{\infty} g(t)\delta(t'-nT_s)\sinc\left(F(t-t')\right)\dif t' \\ &= \frac{F}{F_p} \sum_n g(nT_s) \sinc\left( F\cdot(t-nT_s) \right) \end{align*} \begin{align*} g(kT_s) &= \frac{F}{F_P} \sum_n g(nT_s) \sinc\left(F(kT_s-nT_s)\right) \\ F=F_p \qquad g(kT_s) &= \sum_n g(nT_s)\sinc(k-n) \end{align*} Γενικά, αν $$F = F_p$$: $g(t) = \sum_n g(nT_s) \sinc\left( \frac{1}{T_s} (t-nT_s) \right)$ \paragraph[Δειγματοληψία όταν Fp=2σ]{Δειγματοληψία όταν $$F_p=2\sigma$$} \hspace{0pt} \begin{tikzpicture} \draw[very thick,blue] plot[variable=\x,domain=-2:3,smooth,samples=50] ({\x}, { sin(\x r*pi/2*2) }); \draw[->] (0,-0.5) -- (0,2); \draw (-2,0) -- (3,0); \filldraw[very thick,draw=orange!60!black,fill=orange,fill opacity=.5] (0,0) circle (4pt) + (1,0) circle(4pt); \begin{scope}[yshift=-4cm,xshift=15pt] \draw[->] (0,-1.5) -- (0,2); \draw[->] (-2,0) -- (2,0); \draw[orange!70!black,ultra thick,->] (-1,0) -- ++(0,-1); \draw[orange!70!black,ultra thick,->] (1,0) -- ++(0,1); \begin{scope}[xshift=5cm] \draw[->] (0,-1.5) -- (0,2); \draw[->] (-2,0) -- (2.5,0); \draw[orange!70!black,ultra thick,->] (-1,0) -- ++(0,-1); \draw[orange!70!black,ultra thick,->] (1,0) -- ++(0,1); \draw[orange!70!black,ultra thick,->] (1,0) -- ++(0,-1); \draw[orange!70!black,ultra thick,->] (2,0) -- ++(0,1); \draw[orange!40!black,opacity=.5] (1,0) ellipse (0.5 and 1.5); \draw (1.5,1.5) node {$0$}; \end{scope} \end{scope} \end{tikzpicture} Παρατηρούμε ότι οι $$\delta$$ αφαιρούνται μεταξύ τους, οπότε $$G^S(f) = 0 \not\leftarrow \sin$$, επομένως δεν μπορούμε να έχουμε $$F_p = 2\sigma$$. \paragraph{Άσκηση για το σπίτι} $\phi_n^{F,T_s}(t) = \sinc\left( F(t-nF_p) \right)$ Να βρεθεί το $$\left\langle \phi_n(t),\phi_k(t)\right\rangle$$. \subsection{Υποδειγματοληψία (undersampling)} \begin{tikzpicture}[scale=2] \draw[very thick,red!50] (0.5,0) -- (0.5,1) -- (1.5,0); \draw[very thick,red!20] (-0.5,0) -- (-0.5,1) -- (0.5,0); \draw[very thick,red!10] (-1.5,0) -- (-1.5,1) -- (-0.5,0); \draw[very thick,green!70] (1.47,0) -- (1.47,1) -- (0.47,0); \draw[very thick,green!70] (3.5,0) -- (3.5,1) -- (2.5,0); \draw[very thick,green!40] (3.5,0) -- (3.5,1) -- (4.5,0); \draw[->] (-3,0) -- (5,0) node[below] {$f$}; \draw[->] (0,-0.2) -- (0,2) node[right] {$x(f)$}; \draw[very thick,blue] (1.5,0) node[black,below] {$f_L$} -- (1.5,1) -- (2.5,0) node[black,below] {$f_H$}; \draw[very thick,blue] (-1.5,0) node[black,below] {$-f_L$} -- (-1.5,1) -- (-2.5,0) node[black,below] {$-f_H$}; \end{tikzpicture} Για να μην πέφτουν τα "πλακάκια" το ένα πάνω στο άλλο: \begin{align*} \left. \begin{array}{ll} \kappa f_s-f_L < f_L \\ (\kappa+1)f_s - f_H > f_H \end{array} \right\rbrace &\implies \begin{array}{l} f_s < \frac{2f_L}{\kappa} \\ f_s > \frac{2f_H}{\kappa + 1} \end{array} \\ &\frac{2f_H}{\kappa+1} < f_s < \frac{2f_L}{\kappa} \intertext{Ψάχνω το μέγιστο $\kappa$, έτσι ώστε να βρω το ελάχιστο $f_s$} \\& \frac{2f_H}{\kappa+1} < \frac{2f_L}{\kappa} \\ & k \leq \frac{f_L}{f_H-f_L} \\ \underset{\min}{f_s} \leftarrow & \kappa_{\text{best}} = \left\lfloor \frac{f_L}{f_H - f_L} \right\rfloor \end{align*} Θα ψάξω το ελάχιστο $$f_s$$: \begin{gather*} \frac{2f_H}{\left\lfloor\frac{f_L}{f_H-f_L}\right\rfloor+1} < f_s < \frac{2f_L}{\left\lfloor\frac{f_L}{f_H-f_L}\right\rfloor} \\ \frac{2f_H}{\left\lfloor\frac{f_L}{f_H-f_L}+1\right\rfloor} < f_s < \frac{2f_L}{\left\lfloor\frac{f_L}{f_H-f_L}+1\right\rfloor} \\ \frac{2f_H}{\left\lfloor\frac{f_H}{f_H-f_L}\right\rfloor} < f_s \\ \frac{2f_H}{\left\lfloor\frac{f_H}{f_H-f_L}\right\rfloor} = \frac{2f_H}{\frac{f_H}{f_H-f_L}-\underset{\mathclap{0<\epsilon<1}}{\epsilon}} =\frac{2f_H}{\frac{f_H-\epsilon f_H + \epsilon f_L}{f_H-f_L}} = \frac{2f_H (f_H-f_L)}{f_H(1-\epsilon)+\epsilon f_L} = \frac{2(f_H-f_L)}{(1-\epsilon)+\epsilon\frac{f_L}{f_H}} \\ = \frac{2(f_H-f_L)}{1-\epsilon\left(1-\frac{f_L}{f_H}\right)} > 2(f_H - f_L) \quad \text{ αλλά όχι πολύ κοντά εκεί!} \end{gather*} $2(f_H-f_L) < \boxed{ \frac{2f_H}{\left\lfloor\frac{f_L}{f_H-f_L}\right\rfloor + 1} < f_s }$ \begin{tikzpicture}[scale=1.5] \draw[thick,blue!50] (2,0) -- (2,1) --(1,0); \draw[thick,blue!50] (0,0) -- (0,1) -- (1,0); \draw[thick,blue!50] (-2,0) -- (-2,1) -- (-1,0); \draw[thick,blue!50] (0,0) -- (0,1) -- (-1,0); \filldraw[very thick,blue,fill=green!30] (2,0) node[black,below] {$f_L$} -- (2,1) -- (3,0) node[black,below] {$f_H$}; \filldraw[very thick,blue,fill=green!30] (-2,0) node[black,below] {$-f_L$} -- (-2,1) -- (-3,0) node[black,below] {$-f_H$}; \draw[->] (-4,0) -- (4,0) node[below] {$f$}; \draw[->] (0,-0.2) -- (0,2) node[right] {$X^S(f)$}; \draw[thick,green!60!black] (-3.05,0) -- ++(0,1.5) -- ++(1.1,0) -- ++(0,-1.5); \draw[thick,green!60!black] (3.05,0) -- ++(0,1.5) -- ++(-1.1,0) -- ++(0,-1.5); \draw[thick,orange!70] (-3.1,0) -- ++(0,3.5) -- ++(6.2,0) -- ++(0,-3.5); \draw[thick,red!70] (-1.9,0) -- ++(0,3) node[below left] {-} -- ++(3.8,0) -- ++(0,-3); \end{tikzpicture} \subsection{Gibbs' Phenomenon} \begin{tikzpicture} \draw (2.5,0) node[below] {$\vphantom{-}\sigma$} -- ++(0,{1.2*exp(-2.5^2/7)}); \draw (-2.5,0) node[below] {$-\sigma$} -- ++(0,{1.2*exp(-2.5^2/7)}); \fill[green!20] plot[variable=\x,domain=-2.5:2.5,smooth,samples=20] ({\x}, { 1.2*exp(-\x*\x/7) }) -- (2.5,0) -- (-2.5,0) -- cycle; \draw[very thick,blue] plot[variable=\x,domain=-3:3,smooth,samples=20] ({\x}, { 1.2*exp(-\x*\x/7) }); \draw[->] (0,-0.5) -- (0,2) node[right] {$X(\omega)$}; \draw[->] (-3,0) -- (3,0) node[right] {$\omega$}; \end{tikzpicture} \begin{align*} x(t) &\xrightarrow{FT} X(\omega ) \\ x_\sigma(t) &= \frac{1}{2\pi} \int_{-\sigma}^{\sigma} X(\omega ) e^{j\omega t} \dif\omega = \frac{1}{2\pi} \int_{-\sigma}^{\sigma} \boxed{\int_{-\infty}^{\infty} x(\tau) e^{j\omega\tau}\dif\tau } e^{j\omega t}\dif \omega \\ &= \frac{1}{2\pi} \int_{-\infty}^{\infty} x(\tau) \int_{-\sigma}^{\sigma} e^{-j\omega \tau +j\omega t}\dif\omega\dif t = \int_{-\sigma}^{\sigma} e^{j\omega (t-\tau)}\dif\omega \\ &= \left.\frac{1}{j(t-\tau)}e^{j\omega (t-\tau)} \right|_{-\sigma}^\sigma \\ &= \frac{1}{j(t-\tau)}\left[ e^{j\sigma(t-\tau)}-e^{-j\sigma (t-\tau)} \right] \\ &= \frac{2}{t-\tau} \sin \left( \sigma(t-\tau) \right) \intertext{δηλαδή} X_\sigma(f) &= \int_{-\infty}^{\infty} x(\tau) \frac{\sin\left(\sigma(t-\tau)\right)}% {\pi(t-\tau)}\dif\tau \end{align*} \begin{tikzpicture} \draw[very thick,draw=blue] (-1,0) node[below] {$-\sigma$} -- (-1,1) -- (1,1) -- (1,0) node[below] {$\sigma$}; \draw (0,1) node[above right] {$1$}; \draw[->] (0,-0.5) -- (0,2) node[right] {$X(\omega)$}; \draw[->] (-2,0) -- (2,0) node[right] {$\omega$}; \draw[thick,->] (2,1) -- (3,1) node[midway,above] {IFT}; \draw[thick,->] (0,-1) -- (0,-2) node[midway,right] {$\sigma\to\infty$}; \draw[thick,->] (4,-1) -- (4,-2) node[midway,right] {$\sigma\to\infty$}; \draw (4,1) node {$\mathlarger{\frac{\sin(\sigma t)}{\pi t}}$}; \begin{scope}[yshift=-4.5cm] \draw[very thick,draw=blue] (-1.9,1) -- (1.9,1); \draw[->] (0,-0.5) -- (0,2) node[right] {$X(\omega)$}; \draw[->] (-2,0) -- (2,0) node[right] {$\omega$}; \draw[thick,->] (2.2,1) -- (3.2,1) node[midway,above] {IFT}; \draw (4,1) node {$\mathlarger{\delta(t)}$}; \end{scope} \end{tikzpicture} \begin{align*} \lim_{\sigma\to \infty } x_\sigma(t) &= \int_{-\infty}^{\infty} x(\tau) \lim_{\sigma\to \infty} \frac{\sin\left(\sigma(t-\tau)\right)}{\pi(t-\tau)}\dif\tau \\ \lim_{\sigma\to \infty} x_\sigma(t) &= \int_{-\infty}^{\infty} x(\tau)\delta(t-\tau) \dif\tau \\ \lim_{\sigma \to \infty} x_\sigma(t) = x(t) \end{align*} \begin{tikzpicture}[scale=0.8] \draw[very thick,blue] plot[variable=\x,domain=-3:0,smooth,samples=40] ({\x}, { 0.4*sin(5*\x r)-\x }); \draw[very thick,blue] plot[variable=\x,domain=0:3,smooth,samples=40] ({\x}, { 1.5+0.3*sin(3*\x r)-\x/1.5 }); \draw (0,0) node[below right] {$x(0^-)$}; \draw (0,1.5) node[left,xshift=3pt] {$x(0^+)$}; \draw[blue] (0.4,0.3) node[right] {$x(t)$}; \draw[->] (0,-0.5) -- (0,2); \draw[->] (-3,0) -- (3,0) node[right] {$t$}; \draw (4,1) node {$=$}; \begin{scope}[xshift=7cm] \draw[very thick,blue] plot[variable=\x,domain=-3:0,smooth,samples=40] ({\x}, { 0.4*sin(5*\x r)-\x }); \draw[very thick,blue] plot[variable=\x,domain=0:3,smooth,samples=40] ({\x}, { 0.3*sin(3*\x r)-\x/1.5 }); \draw (-2,1) node {$x(0^-)$}; \draw (1.7,1) node {$x(0^-)$}; \draw[->] (0,-0.5) -- (0,2) node[right] {$X_C(t)$}; \draw[->] (-3,0) -- (3,0) node[right] {$t$}; \draw (4,1) node {$+$}; \end{scope} \begin{scope}[xshift=14cm] \draw[very thick,draw=blue,->] (0,0) -- (0,1.2) node[midway,above,xshift=10pt,sloped] {$\mathsmaller{\mathsmaller{\left[x(0^+)-x(0^-)\right]}}$} -- ++(3.2,0) node[midway,above] {$\left[x(0^+)-x(0^-)\right]\mathrm u(t)$}; \draw (0,-0.5) -- (0,2); \draw[->] (-3,0) -- (3,0) node[right] {$t$}; \end{scope} \end{tikzpicture} \begin{align*} x(t) &= x_c(t) + \left[x(0^+)-x(0^-)\right]\mathrm u(t) \\ x_\sigma(t) &= \int_{-\infty}^{\infty} x_c(t) \frac{\sin\left(\sigma(t-\tau)\right)}{\pi(t-\tau)}\dif\tau -\frac{\left[x(0^+)-x(0^-)\right]}{\pi} \int_{0}^{\infty} \frac{\sin\left(\sigma(t-\tau)\right)}{(t-\tau)}\dif \tau \\ \int_0^\infty \frac{\sin\left(\sigma(t-\tau)\right)}{t-\tau}\dif\tau & \\ \text{Θέτουμε }&\begin{array}{l} \sigma(t-\tau)=x \\ \dif\tau = -\frac{1}{\sigma}\dif x \\ t-\tau = \frac{x}{\sigma} \end{array} \\ &= \int_{\sigma t}^{\infty} \frac{\sin x}{\frac{x}{\sigma}} \left( \frac{\sin x}{x}\dif x \right) = \int_{-\infty}^{0} \frac{\sin x}{x}\dif x + \int_0^{\sigma t}\frac{\sin x}{x} \dif x \\ &= \frac{\pi}{2} + \int_{0}^{\sigma t} \frac{\sin x}{x}\dif x = \frac{\pi}{2} + \underbrace{\mathrm{Si}(\sigma t)}_{\mathclap{\text{Sine Integral}}} \end{align*} Άρα \begin{align*} x_\sigma(t) &= \int_{-\infty}^{\infty} x_c(\tau) \frac{\sin\left(\sigma(t-\tau)\right)}{\pi(t-\tau)}\dif\tau + \frac{\left[x(0^+)-x(0^-)\right]}{2} + \frac{\left[x(0^+)-x(0^-)\right]}{\pi} \cdot \mathrm{Si}\; (\sigma t) \end{align*} \paragraph{} $\mathrm{Si}(t) = \int_0^t \frac{\sin x}{x}\dif x$ \begin{tikzpicture} \draw[dashed,draw=brown!40!black] (0,pi/2) node[left] {$\frac{\pi}{2}$} -- ++(5,0); \draw[dashed,draw=brown!40!black] (0,-pi/2) node[right] {$-\frac{\pi}{2}$} -- ++(-4,0); \draw[dashed,brown!80!black] (0,{pi*(1/2+0.0895)}) -- ++(5,0); \draw[thick,<->] (4.5,pi/2) -- ++(0,0.0895*pi) node[midway,right,yshift=-1pt] {$\pi \cdot 0,0895\dots$}; \draw[very thick,blue] plot[xscale={1/pi},smooth] file{data/sine_integral.data}; \draw[->] (0,-2) -- (0,2); \draw[->] (-4,0) -- (4,0) node[right] {$t$}; \draw (1,0.1) -- ++(0,-0.2) node[below] {$\frac{\pi}{\sigma}$}; \draw (2,0.1) -- ++(0,-0.2) node[below] {$\frac{2\pi}{\sigma}$}; \draw (3,0.1) -- ++(0,-0.2) node[below] {$\frac{3\pi}{\sigma}$}; \draw (-1,0.1) -- ++(0,-0.2) node[below] {$\frac{-\pi}{\sigma}$}; \draw (-2,0.1) -- ++(0,-0.2) node[below] {$\frac{-2\pi}{\sigma}$}; \draw (-3,0.1) -- ++(0,-0.2) node[below] {$\frac{-3\pi}{\sigma}$}; \draw (5,0) node {$\mathlarger{\mathrm{Si}\, (\sigma t)}$}; \end{tikzpicture} Χρησιμοποιώντας τον Leibniz Rule (παραγωγίζοντας το ολοκλήρωμα) μπορούμε να αποδείξουμε την θέση των μεγίστων της $$\mathrm{Si}$$. \begin{align*} x_\sigma(t) &= \int_{-\infty}^{\infty} x_c(t) \frac{\sin\left[\sigma(t-\tau)\right]}{\pi(t-\tau)} \dif\tau + \frac{x(0^+)-x(0^-)}{2} + \frac{\left[x(0^+)-x(0^-)\right]}{\pi}\mathrm{Si}(\sigma t) \\ \lim_{\sigma\to \infty} x_\sigma(t) &= x_c(t) + \frac{x(0^+)-x(0^-)}{2} + \frac{\left[x(0^+)-x(0^-)\right]}{\pi} \lim_{\sigma\to \infty}\mathrm{Si}(\sigma t) \\ \lim_{\sigma\to \infty} x_\sigma(0) &= x(0^-) + \frac{x(0^+)-x(0^-)}{2} = \frac{x(0^+)+x(0^-)}{2} \end{align*} \begin{tikzpicture} \draw[ultra thick,red!90!blue] (-2,pi/2) -- (0,pi/2) -- (0,-pi/2) -- (2,-pi/2); \draw[thick,blue] plot[xscale={-0.5/pi},smooth] file{data/sine_integral.data}; \filldraw[very thick, draw=brown!50!black, fill=orange, fill opacity=.8] (0,0) circle (4pt); \end{tikzpicture} Παρατηρώ ότι τα ζιγκζακωτά παραμένουν και το ύψος του κυματισμού δεν αλλάζει. Το μόνο που μπορεί να μεταβληθεί είναι η θέση του μεγίστου, με αύξηση του $$\sigma$$ (ώστε να έρθει πιο κοντά στο σημείο ασυνέχειας). Αυτό είναι το φαινόμενο Gibbs. Εάν όμως το $$\sigma$$ δεν τείνει στο $$\infty$$, το αποτέλεσμα στο $$0$$ δεν είναι απαραίτητα το ημιάθροισμα των $$x(0^+)$$ και $$x(0^-)$$. \subsection{Ευστάθεια συστήματος} Ευσταθές ονομάζεται ένα σύστημα όταν πεπερασμένη (φραγμένη στο πλάτος) είσοδος δίνει πεπερασμένη έξοδο - ΠΕΠΕ (Πεπερασμένη Είσοδος - Πεπερασμένη Έξοδος) / BIBO (Bounded Input - Bounded Output) ευστάθεια Φραγμένη συνάρτηση $$f(t) \in BF$$ σημαίνει ότι $\exists M>0 : \forall t \quad \left|f(t)\right| < M$ όπου $$BF$$ ο κόσμος των φραγμένων συναρτήσεων. Ένα σύστημα είναι ευσταθές όταν $$\forall$$ είσοδο $$x(t) \in BF$$ η έξοδος του συστήματος είναι επίσης φραγμένη ($$y(t) \in BF$$). ( Αν το σύστημά μας είναι γραμμικό και ΑΚΜ (Αμετάβλητο Κατά τη Μετατόπιση) $\exists h(t) \text{ κρουστική απόκριση }\quad y(t) = h(t)*x(t) = \int_{-\infty}^{\infty} h(\tau)x(t-\tau)\dif\tau$ Έστω \begin{align*} \exists M : \forall t,\quad & \left|x(t)\right| < M \implies \\& \left|x(t-\tau)\right| < M \end{align*} Επίσης έστω \begin{align*} \exists N:& \left|y(t)\right| < N \quad \forall t \\ & \left|y(t)\right| = \left| \int_{-\infty}^{\infty} h(t) x(t-\tau)\dif \tau \right| < \int_{-\infty}^{\infty} \left|h(t)x(t-\tau)\right|\dif\tau < \int_{-\infty}^{\infty} \left|h(\tau)\right|\dif\tau < \frac{N}{M} \end{align*} Εάν η κρουστική απόκριση είναι απολύτως ολοκληρώσιμη, τότε το σύστημα είναι ευσταθές. \subsection{Ιδανικό φίλτρο} Έστω ένα ιδανικό χαμηλοπερατό φίλτρο: \begin{tikzpicture} \draw[->] (0,-0.2) -- (0,2.5) node[right] {$H(f)$}; \draw[->] (-3,0) -- (3,0) node[below,right] {$f$}; \draw[very thick,blue] (-1.5,0) node[below,black] {$-\sfrac{F}{2}$} -- ++(0,1.5) -- ++(3,0) -- ++(0,-1.5) node[below,black] {$\sfrac{F}{2}$}; \end{tikzpicture} $W_F(f) \xrightarrow{\text{IFT}} h(t) = F\sinc (Ft) = F \frac{\sin(\pi Ft)}{\pi Ft}$ \begin{tikzpicture}[scale=1.2] \filldraw[xscale=0.5,samples=200,domain=-5:5,smooth,variable=\x,blue, fill=blue!20] plot ({\x},{abs(sinc(pi*\x))}); \draw[->] (0,-0.5) -- (0,1.5) node[right] {$h(t)$}; \draw[->] (-2.5,0) -- (2.5,0) node[below,right] {$t$}; \draw[very thick,xscale=0.5,samples=200,domain=-5:5,smooth,variable=\x,blue] plot ({\x},{sinc(pi*\x)}); \draw (0,0) node[below right] {$0$}; \draw (0.5,0) node[below] {$\frac{1}{F}$}; \draw(1,0) node[below] {$\frac{2}{F}$}; \draw(1.5,0) node[below] {$\frac{3}{F}$}; \draw (-0.5,0) node[below] {$-\frac{1}{F}$}; \draw(2,0) node[below] {$\cdots$}; \end{tikzpicture} Παρατηρώ ότι η κρουστική απόκριση του ιδανικού αυτού φίλτρου δεν είναι αιτιατή, οπότε ένα τέτοιο φίλτρο δεν είναι υλοποιήσιμο. Άσκηση για το σπίτι: Η $$h(t)$$ είναι απολύτως ολοκληρώσιμη; \subparagraph{Απάντηση} \begin{tikzpicture}[baseline=(current bounding box.north)] \filldraw[xscale=0.8,samples=200,domain=-3:3,thick,variable=\x,blue, fill=blue!20] plot ({\x},{ abs(sinc(pi*\x)) }); \filldraw[xscale=0.8,samples=25,domain=-1:1,smooth,thick,variable=\x,blue, fill=blue!30] plot ({\x},{ abs(sinc(pi*\x)) }); \draw (0,-0.5) -- (0,1.5) ; \draw[->] (-2.5,0) -- (2.5,0) node[below,right] {$t$}; \draw (0.8,0) node[below] {$k$}; \draw (1.6,0) node[below] {$k+1$}; \draw (0,0.5) node[right] {$\Xi$}; \end{tikzpicture} \begin{gather*} \int_0^\pi \sin(t) \dif t = 2 k \leq t \leq k+1 \quad \left|h(t)\right| = \left|\frac{\sin(t)}{t}\right| > \frac{\left|\sin(t)\right|}{k+1} \\ \int_1^\infty \left|h(t)\right|\dif t > \sum_{k=1}^\infty \int_k^{k+1} \left|\frac{\sin(t)}{t}\right|\dif t > \sum_{n=1}^\infty \frac{1}{k+1} \cancelto{2}{\int \left|\sin(t)\right|} \\ \int_{-\infty}^{\infty} \left|h(t)\right|\dif t > \Xi + 2\sum_{k=1}^\infty \frac{2}{k+1} \end{gather*} Δηλαδή το σύστημα του ιδανικού χαμηλοπερατού φίλτρου δεν είναι ευσταθές. \subsection{Kramers - Kronig} Έστω $$h(t)$$ αιτιατή - πραγματική συνάρτηση με Μ$$F$$: $$H(\omega )= H_R(\omega ) + jH_I(\omega )$$ \begin{gather*} \text{Ξέρω } h(t) = \underbrace{h_e(t)}_{\mathclap{\text{even}}} + \underbrace{h_o(t)}_{\mathclap{\text{odd}}} \\ \text{Ξέρω } h_o(t) \xrightarrow{\text{FT}} H_o(\omega ) \in \mathbb I \quad \text{(επίσης περιττή συνάρτηση)}\\ \text{Ξέρω } h_e(t) \xrightarrow{\text{FT}} H_e(\omega ) \in \mathbb R \quad \text{(επίσης άρτια συνάρτηση)} \end{gather*} \begin{gather*} h_o(t) = -h_e(t) \quad t<0 \\ h_0(t) = h_e(t) \quad t \geq 0 \\ \text{Συνεπώς } h(t) = h_e(t) + \mathrm{sgn}(t) h_e(t) \\ \text{ } h(t) = h_o(t) + \mathrm{sgn}(t) h_o(t) \end{gather*} \begin{gather*} X(\omega ) = \int_{-\infty}^{\infty} x(t)e^{-j\omega t}\dif t \\ x_1(t) x_2(t) \xrightarrow{\text{FT}} \frac{1}{2\pi}X_1(\omega )*X_2(\omega ) \\ \mathrm{sgn}(t) \xrightarrow{\text{FT}} \frac{2}{j\omega } \\ h(t) = h_e(t) + \mathrm{sgn}(t) h_e(t) \xRightarrow{\text{FT}} H(\omega ) = H_e(\omega ) + \frac{1}{2\pi}\frac{2}{j\omega }*H_e(\omega ) = H_e(\omega ) - j\frac{1}{\pi} \int_{-\infty}^{\infty} \frac{1}{\omega -\omega'} H_e(\omega')\dif \omega' \\ \boxed{ H(\omega ) = \underbrace{H_e(\omega ) }_{H_R(\omega )} - \underbrace{j\frac{1}{\pi}\int_{-\infty}^{\infty} \frac{H_e(\omega' )}{\omega -\omega'}\dif \omega' }_{jH_I(\omega )} } \\ \boxed{ H_I(\omega ) = -\frac{1}{\pi} \int_{-\infty}^{\infty} \frac{H_R(\omega')}{\omega -\omega'}\dif \omega' } \end{gather*} ομοίως από πάνω υπολογίζουμε το $$H_R$$. \section{Συστήματα} \subsection{Ανακεφαλαίωση} \paragraph{Γραμμικά Αναλογικά Συστήματα} \begin{enumerate} \item \underline{Γραμμικότητα } \qquad $$T\left[ax_1(t)+bx_2(t)\right] = ay_1(t)+by_2(t)$$ όπου $$\begin{array}{l} y_1(t) = T\left[x_1(t)\right] \\ y_2(t) = T\left[x_2(t)\right] \end{array} \qquad \forall x_1(t),x_2(t),\ a,b$$ \item \underline{Χρονοαμετάβλητο} \qquad $$T\left[x(t-k)\right] = y(t-k)$$ όπου $$y(t)=T\left[x(t)\right] \qquad \forall k,x(t)$$ \item \underline{\textbf{Στιγμιαίο} ($$\neq$$ δυναμικό)} \qquad $$y(t) = f\left(x(t)\right)$$ Η έξοδος οποιαδήποτε χρονική στιγμή $$t$$ εξαρτάται μόνο από την είσοδο την ίδια στιγμή $$t$$. Η αντίσταση είναι ένα στιγμιαίο σύστημα, ενώ ο πυκνωτής και το πηνίο δεν είναι. \item \underline{Αιτιατό} \qquad Η έξοδος δεν εξαρτάται από μελλοντικές τιμές της εισόδου $\left[\begin{matrix} \text{\small ΞΕΡΩ} \\ \text{\small Γραμμικό} \\ \text{\small Χρον. Αμετάβλητο} \end{matrix}\right] \implies h(t) = 0 \quad \forall t < 0$ \item \underline{Ευσταθές} \qquad Για κάθε φραγμένη είσοδο, η έξοδος είναι φραγμένη. $\left[\begin{matrix} \text{\small ΞΕΡΩ} \\ \text{\small Γραμμικό} \\ \text{\small Χρον. Αμετάβλητο} \end{matrix}\right] \implies \int_{-\infty}^{\infty} \left|h(t)\right|\dif t < \infty$ \item \underline{Συγκεντρωμένο και Κατανεμημένο} \qquad Συγκεντρωμένο λέγεται όταν μοναδική ελεύθερη μεταβλητή είναι ο χρόνος $$t$$. \end{enumerate} \paragraph{Παράδειγμα} Το σύστημα $$y(t) = x^2(t)$$ είναι: \begin{enumerate}[itemsep=-1mm] \item Μη γραμμικό \item Χρονοαμετάβλητο \item Στατικό (στιγμιαίο) \item Αιτιατό \item Ευσταθές \item Συγκεντρωμένο \end{enumerate} \subsection{Περιγραφή συστήματος με διαφορική εξίσωση} $\sum_{i=0}^{n} a_i \od[i]{}{t} y(t) = \sum_{l=0}^m b_l \od[l]{}{t}x(t) \qquad t \geq 0$ \begin{enumerate}[itemsep=-.5mm] \item Γραμμικό \item Χρονοαμετάβλητο \item Δυναμικό, εκτός αν $$n=m=0$$ \item Αιτιατό \end{enumerate} \begin{align*} \text{Ολική λύση } &= \text{Ομογενής $+$ Μερική Λύση}\\ \text{Ολική λύση } &= \underbrace{\text{Εξαναγκασμένη λύση }}_{\text{% λύση θεωρώντας δεδομένη $x(t)$ και μηδενικές Α.Σ.}} + \underbrace{\text{Ελεύθερη λύση}}_{\text{Λύση θεωρώντας $x(t)$=0 και δεδομένες Α.Σ.}} \intertext{(Τα ζευγάρια ομογενής/ελεύθερη \& εξαναγκασμένη/μερική δεν ταυτίζονται!)} \text{Ολική λύση } &= \text{Λύση μόνιμης κατάστασης $+$ μεταβατική λύση} \end{align*} \paragraph{Μ. Laplace} $\int_{i=0}^n a_i\left( s^i Y(s) - \sum_{k=0}^{i-1} s^{i-k-1}y_0^{(k)} \right) = \int_{l=0}^m b_l\left( s^i X(s) - \sum_{k=0}^{l-1} s^{l-k-1}x_0^{(k)} \right)$ \begin{gather*} Y(s) = \underbrace{\dfrac{\sum_{l=0}^m b_ls^l}{\sum_{i=0}^n a_is^i} X(s) - \dfrac{\sum_{l=0}^m\sum_{k=0}^{l-1} b_l s^{l-k-1}x_0^{(k)}}{\sum_{i=0}^n a_is^i}}_% {\mathclap{\text{εξαναγκασμένη}}} + \underbrace{\dfrac{\sum_{i=0}^n \sum_{k=0}^{l-1} a_i s^{i-k-1}y_0^{(k)}}% {\sum_{i=0}^n a_is^i}}_{\mathclap{\text{ελεύθερη}}} \end{gather*} \subparagraph{} \begin{align*} y(t) &= x(t)*h(t) \\ Y(s) &= X(s)H(s) \end{align*} Συγκρίνοντας με τον παραπάνω τύπο, παρατηρούμε ότι η $$y(t)=x(t)*h(t)$$ μόνο\ όταν οι αρχικές συνθήκες είναι 0! \paragraph{Μ. Fourier} Για να λύσω το σύστημα $$\forall t > 0$$, θα πρέπει να μετασχηματίσω το αιτιατό κομμάτι του συστήματος μόνο: $$x_1(t) = x(t)\mathrm u(t)$$ $$y_1(t) = y(t)\mathrm u(t)$$ Άρα: \begin{gather*} \od{x_1}{t} = \od{x}{t}\mathrm u(t) + x(0)\delta(t) \\ \od{y_1}{t} = \od{y}{t}\mathrm u(t) + y(0)\delta(t) \\ \od[i]{y_1(t)}{t} = \od[i]{y}{t}\mathrm u(t) + \sum_{k=0}^{i-1} y_0(k) \delta(\tau)^{(\tau -1 -k)} \\ \od[i]{x_1(t)}{t} = \od[i]{x}{t}\mathrm u(t) + \sum_{k=0}^{l-1} x_0(k) \delta(\tau)^{(i -1 -l)} \\ \sum_{i=1}^N a_i\od[i]{y_1}{t} = \sum_{l=0}^{m} b_l\od[l]{x_1}{t} \end{gather*} Τελικά προκύπτει ότι η λύση της διαφορικής εξίσωσης $$y_1$$ που μας ενδιαφέρει είναι: $y_1(t) = \mathrm{IFT} \left\lbrace \frac{\sum_{l=0}^m \log(j\omega )^l}{\sum_{i=0}^{n} a_i(j\omega)^i}X_1(\omega ) -\frac{\sum_{l=0}^{m}\sum_{k=0}^{l-1} b_l(j\omega)^{l-k-1}x_0^{(k)}}{\sum_{i=0}^{n} a_i(j\omega)^i} +\frac{\sum_{i=0}^{n}\sum_{k=0}^{i-1} a_i(j\omega )^{i-1-k} y_0^{(k)} }{\sum_{i=0}^n a_i(j\omega )^i} \right\rbrace$ Για την έξοδο σε ένα αιτιατό σύστημα θα χρησιμοποιώ Μ/Σ Laplace. Για μελέτη στη μόνιμη κατάσταση ($$t \to \infty$$), όπου οι αρχικές συνθήκες δεν συνεισφέρουν, θα χρησιμοποιώ Μ/Σ Fourier, παίρνοντας τον μετασχηματισμό όλης της συνάρτησης και όχι μόνο του αιτιατού μέρους της. \paragraph{Ex.1} \mbox{} \\ \begin{circuitikz} \draw (0,0) to[battery1=$v$,i_=$i$] (0,2) to[switch] (2,2) to[R=$R$,i] (4,2) to[C=$C_Q$,v=$V_0$] (4,0) -- (0,0); \end{circuitikz} \begin{align*} v(t) &= Ri(t) + \frac{1}{c} \int_0^t i(\tau)\dif\tau + V_0\mathrm u(t) \\ V(s) &= RI(s) + \frac{1}{C}\frac{I(s)}{s}+\frac{V_0}{s} \\ \xrightarrow{V_0=0} H(s) &= \frac{I(s)}{V(s)} = \frac{1}{R+\frac{1}{Cs}} = \frac{Cs}{RCs+1}\cdot \frac{\sfrac{S}{R} }{\sfrac{S+1}{RC} } \\ &= \frac{1}{R} \left(1-\frac{\sfrac{1}{RC} }{s+\sfrac{1}{RC} }\right) \\ H(\omega ) &= \frac{1}{R} \left(1-\frac{\sfrac{1}{RC} }{j\omega+\sfrac{1}{RC} }\right) \\ h(t) &= \mathrm{ILT}\left\lbrace H(s) \right\rbrace = \frac{1}{R}\delta(t) - \frac{1}{R^2 C}e^{-\frac{1}{RC}t}\mathrm u(t) \end{align*} Έξοδος $$i(t) = h(t) * V(t) = \left[\frac{1}{R}\delta(t) -\frac{1}{R^2 C}e^{-\sfrac{1}{RC}\cdot t } \right]*\left[V\mathrm u(t)\right] = \cdots = \frac{V}{R}e^{-\sfrac{t}{RC}\mathrm u(t)}$$ Επειδή όμως οι αρχικές συνθήκες δεν είναι μηδενικές (ο πυκνωτής είναι φορτισμένος), το παραπάνω αποτέλεσμα είναι λάθος! Η σωστή λύση είναι: \begin{align*} v(t) &= Ri(t)+\frac{1}{C}\int_0^t i(\tau)\dif\tau + V_0 \mathrm u(t) \\ V(s) &= RI(s) + \frac{1}{C}\frac{I(s)}{s}+\frac{V_s}{s} \\ I(s) \left[R+\frac{1}{Cs}\right] &= V(s) - \frac{V_0}{s} \\ I(s) &= \frac{V(s) -\sfrac{V_0}{s} }{R+\frac{1}{Cs}} = \frac{(V-V_0)}{s}\cdot \underbrace{\frac{1}{R+\frac{1}{Cs}}}_{H(s)} \\ i(t) &= \frac{V-V_0}{R}e^{\sfrac{-t}{RC}}\mathrm u(t) \\ &= \underbrace{ \underbrace{\frac{V}{R}e^{\sfrac{-t}{RC}}}_{\mathclap{\text{εξαναγκασμένη λύση}}} - \frac{V_0}{R}e^{\sfrac{-t}{RC}}}_% {\mathclap{\text{μεταβατική λύση}}} \end{align*} \subsection{Decibels (dB)} Έστω \begin{align*} x_A(t) &= A\sin(\omega_1 t) \\ y_B(t) &= B\sin(\omega_1 t) \end{align*} \begin{gather*} x_Α \text{ πλάτος } A > 0 \\ x_B \text{ πλάτος } B > 0 \\ \boxed{\mathrm{dB} = 20\log_{10} \frac{A}{B}} \end{gather*} \paragraph{Για την ισχύ:} \begin{gather*} c\cdot x_Α^2 \text{ ισχύς } A \\ c\cdot x_B^2 \text{ ισχύς } B \\ \boxed{\mathrm{dB} = 10\log_{10} \frac{A^2}{B^2}} \end{gather*} Παρατηρούμε ότι τα dB πλάτους και dB ισχύος είναι το ίδιο! \begin{tikzpicture} \draw[->] (-3,0) -- (3,0) node[right] {$\omega$}; \draw[->] (0,-0.7) -- (0,2.5) node[right] {$H(\omega)$}; \draw[dashed] (0.79,0) node[below] {$\omega_1$} -- ++(0,1.83) -- ++(-0.79,0); \draw[very thick, blue] plot [smooth] coordinates { (-2,0) (-1.5,1.7) (1.5,1.7) (2,0)}; \begin{scope}[yshift=-4cm] \draw[->] (-3,0) -- (3,0) node[right] {$\omega$}; \draw[->] (0,-0.7) -- (0,2.5) node[right] {$\mathrm{dB}$}; \draw (0,0) node[below left] {$0$}; \draw[dashed] (.79,0) node[below] {$50$Hz} -- ++(0,1) -- (0,1) node[left] {$3$}; \draw[very thick,samples=100,yscale=1.8,domain=-3:2.96,smooth,variable=\x,blue] plot ({\x},{exp(-\x*\x)}); \end{scope} \end{tikzpicture} $$\pm 3\mathrm{dB}$$ σημαίνει (υπο)διπλασιασμό της ισχύος, ή πολλαπλασιασμό/διαίρεση του πλάτους με $$\sqrt{2}$$. \section{Ασκήσεις} \paragraph{Άσκηση} $$x(t) = 4000\sinc(4000t)$$ \begin{enumgreekparen} \item $$X(f) = ?$$ \item Nyquist Συχνότητα για τα $$x(t)$$ και $$x^2(t)$$ \end{enumgreekparen} \begin{enumgreekparen} \item \begin{tikzpicture}[baseline=(current bounding box.north)] \draw[->] (-3,0) -- (3,0) node[below right] {$f (\mathrm{Hz})$}; \draw[->] (0,-0.7) -- (0,2); \draw[very thick, blue] (-2,0) -- (-2,1.2) -- (2,1.2) -- (2,0); \draw (-2,0) node[below] {$\sfrac{-T}{2}$}; \draw (2,0) node[below] {$\sfrac{T}{2}$}; \draw (0,1.2) node[above right] {$1$}; \draw[->] (4,0.5) -- ++(2,0) node[above,midway] {FT} node[below,midway] {$f$}; \draw (7,0.5) node {$\mathlarger{T\sinc(Tf)}$}; \draw[->] (3.5,1) node[left] {$\mathrm W_T(t)$} to[bend left=45] (6.5,1); \end{tikzpicture} \begin{gather*} T\sinc(Tt) \xrightarrow{\text{FT}} \mathrm W_T (f) \\ x(t) = 4000\sinc (4000t) \xrightarrow[f]{\text{FT}} W_{4000}(f) = \begin{tikzpicture}[baseline=-0.65ex,scale=0.5] \draw[->] (-3,0) -- (3,0) node[below right] {$f (\mathrm{Hz})$}; \draw[->] (0,-0.7) -- (0,2); \draw[very thick, blue] (-2,0) -- (-2,1.2) -- (2,1.2) -- (2,0); \draw (-2,0) node[below] {$-2000$}; \draw (2,0) node[below] {$2000$}; \draw (0,1.2) node[above right] {$1$}; %\draw (4, .5) node %{$\mathlarger{\mathlarger{\mathlarger{\mathlarger{\mathlarger{*}}}}}$}; \end{tikzpicture} \end{gather*} \item Nyquist συχνότητα για το $$x(t)$$ είναι $$2\times 2000\ \mathrm{Hz} = 4000\ \mathrm{Hz}$$ $\mathrm{FT}\left\lbrace x^2(t) \right\rbrace = \mathrm{FT}\left\lbrace x(t)x(t) \right\rbrace = X(f) * X(f)$ \begin{tikzpicture}[scale=0.7] \draw (-3,0) -- (3,0); \draw (0,-1) -- (0,2); \draw[very thick, blue] (-2,0) -- (-2,1.3) -- (2,1.3) -- (2,0); \draw (-2,0) node[below] {$-2\mathrm{k}$}; \draw (2,0) node[below] {$2\mathrm{k}$}; \draw (4, .5) node {$\mathlarger{\mathlarger{\mathlarger{\mathlarger{\mathlarger{*}}}}}$}; \begin{scope}[xshift=8cm] \draw (-3,0) -- (3,0); \draw (0,-1) -- (0,2); \draw[very thick, blue] (-2,0) -- (-2,1.3) -- (2,1.3) -- (2,0); \draw (-2,0) node[below] {$-2\mathrm{k}$}; \draw (2,0) node[below] {$2\mathrm{k}$}; \end{scope} \draw (12, .5) node {$\mathlarger{\mathlarger{\mathlarger{\mathlarger{\mathlarger{=}}}}}$}; \begin{scope}[xshift=16cm] \draw (-3,0) -- (3,0); \draw (0,-1) -- (0,2); \draw[very thick, blue] (-2.5,0) -- (0,1.3) -- (2.5,0); \draw (-2.5,0) node[below] {$-4\mathrm{k}$}; \draw (2.5,0) node[below] {$4\mathrm{k}$}; \end{scope} \end{tikzpicture} Για το $$x^2(t)$$ η Nyquist είναι $$2\cdot 4\ \mathrm{kHz} = 8\ \mathrm{kHz}$$ \end{enumgreekparen} \paragraph{Άσκηση} \hspace{0pt} \begin{tikzpicture} \draw[->] (0.25,0) -- (2-.2,0) node[above,midway] {$x(t)$}; \draw[->] (2,0) -- ++(0,-3) -- ++(1.5,0) node[midway,above] {$a$} -- ++(0,0.75); \draw (3,-2.25) rectangle ++(1,1.5) node[midway] {$\mathlarger{\mathlarger{\mathlarger{\int}}}$}; \draw[->] (3.5,-0.75) -- (3.5,-0.25); \draw (3.5,-0.25) -- (3.5, 0); \draw[->] (5.5,-2.25) -- (5.5,-3); \draw[->] (5.5,-3) -- ++(1.5,0) node[midway,above] {$b$} -- ++(0,3-.2); \draw (5,-2.25) rectangle ++(1,1.5) node[midway] {$\mathlarger{\mathlarger{\mathlarger{\int}}}$}; \draw (5.5,-0.75) -- (5.5,-0.25); \draw[<-] (5.5,-0.35) -- (5.5, 0); \draw[<->] (2+.2,0) -- (7-.2,0) node[midway,above,green] {$\mathsmaller{r(t) \text{ (βοηθητική συνάρτηση)}}$}; \draw[->] (7+.2,0) -- ++(1.5-.2,0) node[above] {$y(t)$}; \filldraw[fill=white] (2,0) node {$+$} circle (.2); \filldraw[fill=white] (7,0) node {$+$} circle (.2); \end{tikzpicture} \begin{infobox}{Σημείωση} \begin{itemize} \item \begin{tikzpicture} \draw[->] (0,0) -- (1.5,0) node[midway, above] {$a$}; \end{tikzpicture}\hspace{.7em} δηλώνει πολλαπλασιασμό με τον αριθμό $$a$$ \item \begin{tikzpicture}[scale=0.7] \draw (0,0) rectangle (0.7,1) node[midway] {$\int$}; \end{tikzpicture}\hspace{.7em} σημαίνει ολοκλήρωση \end{itemize} \end{infobox} \begin{align} \int a(x+r) \dif t &= r \xRightarrow{\sfrac{\dif}{\dif t} } \od{r}{t} = ax+ar \label{eq:ex21} \\ r+b\int r\dif t &= y \implies \od{r}{t} + br = \od{y}{t} \label{eq:ex22} \end{align} \begin{align*} \eqref{eq:ex21} - \eqref{eq:ex22} \implies r(a+b) &= \od{y}{t} - ax \\ r &= \frac{1}{a+b} \left(\od{y}{t} - ax\right) \\ \frac{\od[2]{y}{t}-a\od{x}{t}}{a+b} + \frac{-a}{a+b} \left( \od{y}{t}-ax \right) = ax \\ \od[2]{y}{t} - a\od{y}{t} &= a\od{x}{t} + abx \implies \\ \implies s^2Y(s) - asY(s) &= asX(s) + abX(s) \\ \implies \frac{Y(s)}{X(s)} &= H(s) = \frac{a(s+b)}{s(s-a)} \end{align*} Αν $$\Re\left\lbrace a \right\rbrace < 0$$, τότε $$\displaystyle H(\omega ) = \frac{a(j\omega +b)}{j\omega (j\omega -a)}$$ \begin{infobox}{Ένα διάγραμμα όπως τα παρακάτω μπορεί να περιέχει:} \subparagraph{Τελεστή άθροισης} $$y(t) = x(t) + v(t) \pm w(t)$$ \begin{tikzpicture}[scale=1.5] \def\cr{4pt}; \draw (0,0) node{$+$} circle (\cr); \draw[->] (-1,0) -- (-\cr,0) node[midway,above] {$x(t)$}; \draw[->] (0,-1) node[right] {$w(t)$} -- (0,-\cr) node[midway,above left] {$\pm$}; \draw[->] (0,1) node[right] {$v(t)$} -- (0,\cr); \draw[->] (\cr,0) -- (1,0) node[midway,above] {$y(t)$}; \end{tikzpicture} \subparagraph{Γινόμενο συνάρτησης με αριθμό} $$y(t) = ax(t)$$ \begin{tikzpicture}[scale=1.5] \draw[->] (0,0) node[left] {$x(t)$} -- (1,0) node[right] {$y(t)$} node[midway,above] {$a$}; \end{tikzpicture} \subparagraph{Συνέλιξη} \hspace{0pt} \begin{tikzpicture}[scale=1.2] \draw[->] (0,0) -- (1,0) node[above,midway] {$x(t)$} ; \draw (1,0.25) rectangle (2.3,-0.25) node[midway] {$h_1(t)$}; \draw[->] (2.3,0) -- (3.3,0) node[above,midway] {$y(t)$} node[xshift=5mm,right] {$y(t) = h_1(t) * x(t)$}; \begin{scope}[yshift=-1cm] \draw[->] (0,0) -- (1,0) node[above,midway] {$x(t)$} ; \draw (1,0.25) rectangle (2.3,-0.25) node[midway] {$H_1(s)$}; \draw[->] (2.3,0) -- (3.3,0) node[above,midway] {$y(t)$} node[xshift=5mm,right] {$Y(s) = H_1(s) X(s)$}; \end{scope} \draw[thick,decorate,decoration={brace,amplitude=10pt}] (3.4,0.5) -- (3.4,-1.5); \draw (8,-1) node {$\mathsmaller{\mathsmaller{ H_1(s) = \mathscr L \left\lbrace h_1(t)\right\rbrace}}$}; \end{tikzpicture} \subparagraph{Συνδυασμός (σε σειρά)} $$y(t) = h_2(t) * v(t) = h_2(t) * \left(x(t)*h_1(t)\right) = \left[h_1(t) * h_2(t)\right] * x(t)$$ \begin{tikzpicture}[scale=1.3] \draw[->] (0,0) -- (1,0) node[above,midway] {$x(t)$} ; \draw (1,0.25) rectangle (2.3,-0.25) node[midway] {$h_1(t)$}; \draw[->] (2.3,0) -- (3.3,0) node[midway] (V) {}; \draw (3.3,0.25) rectangle (4.6,-0.25) node[midway] {$h_2(t)$}; \draw[->] (4.6,0) -- (5.6,0) node[right] {$y(t)$}; \filldraw[blue!50!green] (V) circle (1pt) node[above] {$v(t)$}; \draw ({(1+2.3)/2},-0.7) node {$\mathlarger{\mathlarger{\mathlarger{\mathlarger{\Updownarrow}}}}$}; \begin{scope}[yshift=-1.4cm] \draw[->] (0,0) -- (1,0) node[above,midway] {$x(t)$} ; \draw (1,0.25) rectangle (2.3,-0.25) node[midway] {$\mathsmaller{h_1(t)*h_2(t)}$}; \draw[->] (2.3,0) -- (3.3,0) node[above,midway] {$y(t)$}; \end{scope} \end{tikzpicture} ή \begin{tikzpicture}[scale=1.3] \draw[->] (0,0) -- (1,0) node[above,midway] {$x(t)$} ; \draw (1,0.25) rectangle (2.3,-0.25) node[midway] {$H_1(s)$}; \draw[->] (2.3,0) -- (3.3,0); \draw (3.3,0.25) rectangle (4.6,-0.25) node[midway] {$H_2(s)$}; \draw[->] (4.6,0) -- (5.6,0) node[right] {$y(t)$}; \draw ({(1+2.3)/2},-0.7) node {$\mathlarger{\mathlarger{\mathlarger{\mathlarger{\Updownarrow}}}}$}; \begin{scope}[yshift=-1.4cm] \draw[->] (0,0) -- (1,0) node[above,midway] {$x(t)$} ; \draw (1,0.25) rectangle (2.3,-0.25) node[midway] {$\mathsmaller{H_1(s)H_2(s)}$}; \draw[->] (2.3,0) -- (3.3,0) node[above,midway] {$y(t)$}; \end{scope} \end{tikzpicture} \subparagraph{Συνδυασμό (παράλληλα)} \hspace{0pt} \begin{tikzpicture}[scale=1.3] \def\cr{4pt}; \draw[->] (-1,0) -- (0,0) node[above,midway] {$x(t)$} ; \draw[->] (0,0) -- (0,0.75) -- ++(1,0); \draw (1,1) rectangle ++(1.3,-0.5) node[midway] {$h_1(t)$}; \draw[->] (2.3,0.75) -- ++(1,0) -- (3.3,\cr); \draw[->] (0,0) -- (0,-0.75) -- ++(1,0); \draw (1,-1) rectangle ++(1.3,0.5) node[midway] {$h_2(t)$}; \draw[->] (2.3,-0.75) -- ++(1,0) -- (3.3,-\cr); \draw (3.3,0) node {$+$} circle (\cr); \draw[->] (3.3cm+\cr,0) -- ++(1,0) node[above,midway] {$y(t)$}; \draw ({(1+2.3)/2},-1.4) node {$\mathlarger{\mathlarger{\mathlarger{\mathlarger{\Updownarrow}}}}$}; \begin{scope}[yshift=-2.1cm] \draw[->] (0,0) -- (1,0) node[above,midway] {$x(t)$} ; \draw (1,0.25) rectangle (2.3,-0.25) node[midway] {$\mathsmaller{h_1(t)+h_2(t)}$}; \draw[->] (2.3,0) -- (3.3,0) node[above,midway] {$y(t)$}; \end{scope} \end{tikzpicture} \subparagraph{Τελεστές} \hspace{0pt} \begin{tikzpicture}[scale=1,baseline] \draw[->] (0,0) -- (1,0) node[above,pos=.4] {$x(t)$}; \draw (1,0.35) rectangle (1.4,-0.35) node[midway] {$\int$}; \draw[->] (1.4,0) -- ++(1,0) node[above,pos=.5] {$y(t)$}; \end{tikzpicture} $$y(t) = \int_{-\infty}^{t} x(\tau) \dif\tau$$ \begin{tikzpicture}[scale=1,baseline] \draw[->] (0,0) -- (1,0) node[above,pos=.4] {$x(t)$}; \draw (1,0.35) rectangle (1.4,-0.35) node[midway] {$\od{}{t}$}; \draw[->] (1.4,0) -- ++(1,0) node[above,pos=.5] {$y(t)$}; \end{tikzpicture} $$y(t) = \od{x}{t}$$ \end{infobox} \newpage \paragraph{Άσκηση} Να βρεθεί το ισοδύναμο του block diagram: \begin{tikzpicture} \def\cr{7pt}; \draw[->] (0,0) -- (1cm-\cr,0) node[midway,above] {$x(t)$} node[midway,below] {$X(s)$}; \draw (1,0) node {$+$} circle (\cr); \draw[->] (1cm+\cr,0) -- (2.1,0) node [midway,below] (A) {}; \draw (2.1,0.25) rectangle ++(1.5,-0.5) node[midway] {$H_1(s)$}; \draw[->,green!50!black] (A) -- ++(0,1) node[above] {$\Xi(s)$}; \draw[->] (3.6,0) -- (5,0) node[right] {$y(t)$} node[below right,yshift=-2mm] {$Y(s)$}; \draw[->] (4.6,0) -- ++(0,-1) -- (3.6,-1); \draw (2.1,0.25-1) rectangle ++(1.5,-0.5) node[midway] {$H_2(s)$}; \draw[->] (2.1,-1) -- (1,-1) -- (1,-\cr); \draw (2.85,-2) node {$\mathlarger{\mathlarger{\mathlarger{\mathlarger{\Updownarrow}}}}$}; \begin{scope}[yshift=-3cm] \draw[->] (1,0) -- (2.1,0) node[midway,above] {$x(t)$}; \draw (2.1,0.25) rectangle ++(1.5,-0.5) node[midway] {$H(s)$}; \draw[->] (3.6,0) -- (5,0) node[midway,above] {$y(t)$}; \end{scope} \end{tikzpicture} \begin{align*} Y(s) &= H_1(s) \cdot \Xi(s) \\ &= H_1(s) \cdot \left[ X(s) + Y(s) \right] \implies \\ Y(s) &= H_1(s) \left[ X(s)+H_2(s)Y(s) \right] \implies \\ Y(s) \left[1-H_1(s)H_2(s)\right] &= H_1(s)X(s) \\[.3em] Y(s) &= H(s) X(s) \\ H(s) &= \frac{Y(s)}{X(s)} = \frac{H_1(s)}{1-H_1(s)H_2(s)} \end{align*} \paragraph{Άσκηση} \hspace{0pt} \begin{tikzpicture}[scale=1.2] \def\cr{7pt}; \draw[->] (-0.5,0) -- (1cm-\cr,0) node[midway,above] {$x(t)$}; \draw (1,0) node {$+$} circle (\cr); \draw[->] (1cm+\cr,0) -- (2.1,0) node [midway,below] (A) {}; \draw (2.1,0.3) rectangle ++(3,-0.6) node[midway] {$\ddot o + 5\dot o +6o = \dot\imath + \imath$}; \draw[->] (5.1,0) -- (6.5,0) node[right] {$y(t)$} node[below right,yshift=-2mm] {}; \draw[->] (6,0) -- ++(0,-1) -- (4.55,-1); \draw (2.55,0.25-1) rectangle ++(2,-0.5) node[midway] {$\dot o + o = \imath$}; \draw[->] (2.55,-1) -- (1,-1) -- (1,-\cr); \draw (3.6,0.25) node[above,green] {$\mathsmaller{H_1: \ s^2O(s)+5sO(s)+6O(s)=sI(s)+I(s)}$}; \draw (3.6,-1.25) node[below,green] {$\mathsmaller{H_2: \ sO(s)+O(s)=I(s)}$}; \draw[dashed,gray] (0.6, 0.8) rectangle (6.3,-1.8); \draw (2.85,-2) node {$\mathlarger{\mathlarger{\mathlarger{\mathlarger{\Updownarrow}}}}$}; \end{tikzpicture} \begin{align*} H_1(s) &= \frac{O(s)}{I(s)} = \frac{s+1}{s^2+5s+6} = \frac{s+1}{(s+3)(s+2)} \\ H_2(s) &= \frac{O(s)}{I(s)} = \frac{1}{s+1} \intertext{$$H_1,H_2$$ ευσταθή, αφού οι πόλοι τους βρίσκονται στο αριστερό ημιεπίπεδο} Y(s) &= H_1(s) \cdot \left[ X(s) - H_2(s)Y(s) \right] \implies \frac{Y(s)}{X(s)} = H(s) = \frac{H_1(s)}{1+H_1(s)H_2(s)} \end{align*} \paragraph{Άσκηση} \hspace{0pt} \begin{tikzpicture} \draw[->] (0,-2) -- (0,2) node[above] {$i\omega$}; \draw[->] (-2,0) -- (2,0) node[below] {$\sigma$}; \draw (1,1) node[above right] {$s$-plane}; \draw (-0.75,0) node[cross=3pt,thick,red] {} node[below,yshift=-1mm] {$-0.5$}; \draw (-1.5,0) node[cross=3pt,thick,red] {} node[below,yshift=-1mm] {$-1$}; \draw plot [smooth] coordinates {(-1.5,-0.7) ({(-0.75-1.5)/2},-1.1) (-0.75,-0.7)}; \draw[->] ({(-0.75-1.5)/2},-1.1) -- ++(0,-0.5) node[below,rectangle,align=center] {απλοί πόλοι,\\δεν έχει μηδενικά}; \end{tikzpicture} Αν η είσοδος $$x(t) = \mathrm u(t)$$, βρίσκω μετά από "πολλά χρόνια" ότι $$y(t)=1 \implies \lim_{t\to \infty} y(t) = 1$$. Να βρεθεί ολόκληρη η $$y(t)$$. \begin{align*} H(s) &= \frac{k}{(s+1)\left(s+\sfrac{1}{2} \right)} \\ Y(s) &= H(s)X(s) = \frac{k}{(s+1)\left(s+\sfrac{1}{2} \right)}\frac{1}{s} \\ \text{Όμως } \lim_{t\to \infty} y(t) &= \lim_{s \to 0} sY(s) \\ 1 &= \lim_{s\to 0} \left[\frac{k}{(s+1)\left(s+\sfrac{1}{2} \right)}\frac{1}{s}s\right] = 2k \implies \boxed{k = \sfrac{1}{2} } \\ \text{Άρα } Y(s) &= \frac{\sfrac{1}{2} }{s(s+1)\left(s+\sfrac{1}{2} \right)} = \frac{1}{s} + \frac{1}{s+1} + \frac{-2}{s+\sfrac{1}{2} } \\ y(t) &= \left[ 1+e^{-t} -2e^{\sfrac{-t}{2} } \right]\mathrm{u}(t) \end{align*} \paragraph{Άσκηση} \begin{tikzpicture}[baseline=(current bounding box.north)] \draw[->] (0,0) -- (2,0); \draw (2,-0.5) rectangle (4,0.5) node[midway] {$M$}; \draw[->] (4,0) -- (6,0) node[above right] {$y(t) = \left. 2\pi X(-\omega )\right|_{\omega = t}$}; \end{tikzpicture} όπου $$X(\omega ) = \mathrm{FT}\left\lbrace x(t) \right\rbrace$$ \begin{tikzpicture}[scale=1.2] \draw[->,yshift=-0.5mm] (0,0) -- (1,0) node[midway,above] {$x(t)$}; \draw (1,0.25) rectangle (2,-0.25) node[midway] {$M$}; \draw[->] (2,0) -- (3,0) node[blue,midway,below] {$\mathsmaller{(1)}$}; \draw (3,0.25) rectangle (4,-0.25) node[midway] {$M$}; \draw[->] (4,0) -- (5,0) node[blue,midway,below] {$\mathsmaller{(2)}$}; \draw (5,0.25) rectangle (6,-0.25) node[midway] {$M$}; \draw[->] (6,0) -- (7,0) node[blue,midway,below] {$\mathsmaller{(3)}$}; \draw (7.2,0) node {$\times$} circle (0.2); \draw[->] (7.4,0) -- (8,0) node[right] {$w(t)$}; \draw[<-] (7.2,-0.2) -- ++(0,-0.5) node[below] {$\left(\frac{1}{2\pi}\right)^4$}; \end{tikzpicture} Για να υπάρχει συνάρτηση μεταφοράς του $$M$$, πρέπει να είναι γραμμικό \& αμετάβλητο: \begin{align*} x(t) &\to 2\pi X(-t) \\ y(t) &\to 2\pi Y(-t) \\ ax+\beta y\ (t) &\to a2\pi X(-t) + \beta 2\pi Y(-t) \text{ good} \\[.3em] x(t) &\to \left. 2\pi X(-\omega ) \right|_{\omega = t} \\ y(t) &\to 2\pi e^{-j\tau} X(-\tau) \end{align*} Έχω: \begin{align*} y_1(t) &= \left. 2\pi X(-\omega )\right|_{\omega =t} = \left.2\pi \int_{-\infty}^{\infty} x(\tau_1) e^{-j\tau_1(-\omega )}\dif\tau_1 \right|_{\omega =t} =2\pi \int_{-\infty}^{\infty} x(\tau_1) e^{j\tau_1 t}\dif\tau_1 \\ y_2(t) &= \left. 2\pi Y_1(-\omega )\right|_{\omega =t} = \left.2\pi \int_{-\infty}^{\infty} y_1(\tau_2) e^{-j\tau_2(-\omega )}\dif\tau_2 \right|_{\omega =t} \\ &=\left. 2\pi \int_{-\infty}^{\infty} 2\pi \int_{-\infty}^{\infty} x(\tau_1) e^{j\tau_1\tau_2}\dif\tau_1 e^{-j\tau_2(-\omega)} \dif\tau_2\right|_{\omega = t} \\ &= (2\pi)^2 \int_{-\infty}^{\infty} x(\tau_1) \int_{-\infty}^{\infty} e^{j\tau_2(\tau_1+t)}\dif\tau_2\dif\tau_1 \\ &= (2\pi)^2 \int_{-\infty}^{\infty} x(\tau_1)2\pi \delta(\tau_1+t)\dif\tau_1 = (2\pi)^3 \int_{-\infty}^{\infty} x(\tau_1)\delta(\tau_1+t)\dif\tau_1 = (2\pi)^3 x(-t) \\ Y_2(\omega) &= (2\pi)^3 X(-\omega) \\ y_3(t) &= \left. 2\pi Y_2(-\omega ) \right|_{\omega =t} = \left. (2\pi)^3 X(\omega ) \right|_{\omega = t} \\ &= (2\pi)^4 X(t) \end{align*} \begin{attnbox}{} Στις εξετάσεις θα κουβαλήσουμε ένα δικό μας μαθηματικό τυπολόγιο (π.χ. Schaum's Tables \& Formulas) \end{attnbox} \end{document}