diff --git a/PS2/doc.aux b/PS2/doc.aux index fbb50a0..f67c204 100644 --- a/PS2/doc.aux +++ b/PS2/doc.aux @@ -17,4 +17,6 @@ \@writefile{lof}{\contentsline {figure}{\numberline {5}{\ignorespaces Sketch of degenerate energy levels.\relax }}{6}{}\protected@file@percent } \@writefile{lof}{\contentsline {figure}{\numberline {6}{\ignorespaces (a) 1s probability density. (b) 2s probability density.\relax }}{8}{}\protected@file@percent } \@writefile{lof}{\contentsline {figure}{\numberline {7}{\ignorespaces Numerical result (black line) and analytical solution scaled by \(a\) (orange circles).\relax }}{9}{}\protected@file@percent } -\gdef \@abspage@last{12} +\@writefile{lof}{\contentsline {figure}{\numberline {8}{\ignorespaces (a) momentum wavefunction versus normalized momentum. (b) Probability density versus normalized momentum.\relax }}{11}{}\protected@file@percent } +\@writefile{lof}{\contentsline {figure}{\numberline {9}{\ignorespaces (a),(b) Previous plots but with the classical momentum marked in red.\relax }}{11}{}\protected@file@percent } +\gdef \@abspage@last{13} diff --git a/PS2/doc.fdb_latexmk b/PS2/doc.fdb_latexmk index f81710b..4fd9ade 100644 --- a/PS2/doc.fdb_latexmk +++ b/PS2/doc.fdb_latexmk @@ -1,8 +1,10 @@ # Fdb version 3 -["pdflatex"] 1614741649 "d:/Users/Speedee/Documents/University/Year5/Winter/ECE456/ProblemSets/Git/ECE_456/PS2/doc.tex" "d:/Users/Speedee/Documents/University/Year5/Winter/ECE456/ProblemSets/Git/ECE_456/PS2/doc.pdf" "doc" 1614741653 +["pdflatex"] 1614982838 "d:/Users/Speedee/Documents/University/Year5/Winter/ECE456/ProblemSets/Git/ECE_456/PS2/doc.tex" "d:/Users/Speedee/Documents/University/Year5/Winter/ECE456/ProblemSets/Git/ECE_456/PS2/doc.pdf" "doc" 1614982843 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d:\Users\Speedee\Documents\University\Year5\Winter\ECE456\ProblemSets\Git\ECE_456\PS2\q3bvi_fig2.png +INPUT d:\Users\Speedee\Documents\University\Year5\Winter\ECE456\ProblemSets\Git\ECE_456\PS2\q3bvi_fig2.png +INPUT d:\Users\Speedee\Documents\University\Year5\Winter\ECE456\ProblemSets\Git\ECE_456\PS2\q3bvi_fig2.png +INPUT d:\Users\Speedee\Documents\University\Year5\Winter\ECE456\ProblemSets\Git\ECE_456\PS2\q3bvi_fig2.png +INPUT d:\Users\Speedee\Documents\University\Year5\Winter\ECE456\ProblemSets\Git\ECE_456\PS2\q3bvi_fig2.png +INPUT d:\Users\Speedee\Documents\University\Year5\Winter\ECE456\ProblemSets\Git\ECE_456\PS2\q3bvi_fig2.png INPUT d:\Users\Speedee\Documents\University\Year5\Winter\ECE456\ProblemSets\Git\ECE_456\PS2\doc.aux INPUT C:\Users\speed\AppData\Local\MiKTeX\fonts\pk\ljfour\jknappen\ec\dpi600\tcrm1000.pk INPUT C:\Users\speed\AppData\Local\MiKTeX\fonts\pk\ljfour\jknappen\ec\dpi600\tcrm1000.pk @@ -774,6 +804,8 @@ INPUT C:\Users\speed\AppData\Local\Programs\MiKTeX\fonts\type1\public\amsfonts\c INPUT C:\Users\speed\AppData\Local\Programs\MiKTeX\fonts\type1\public\amsfonts\cm\cmr9.pfb INPUT C:\Users\speed\AppData\Local\Programs\MiKTeX\fonts\type1\public\amsfonts\cm\cmsy10.pfb INPUT C:\Users\speed\AppData\Local\Programs\MiKTeX\fonts\type1\public\amsfonts\cm\cmsy10.pfb +INPUT C:\Users\speed\AppData\Local\Programs\MiKTeX\fonts\type1\public\amsfonts\cm\cmsy5.pfb +INPUT C:\Users\speed\AppData\Local\Programs\MiKTeX\fonts\type1\public\amsfonts\cm\cmsy5.pfb INPUT C:\Users\speed\AppData\Local\Programs\MiKTeX\fonts\type1\public\amsfonts\cm\cmsy7.pfb INPUT C:\Users\speed\AppData\Local\Programs\MiKTeX\fonts\type1\public\amsfonts\cm\cmsy7.pfb INPUT C:\Users\speed\AppData\Local\Programs\MiKTeX\fonts\type1\public\amsfonts\cm\cmti10.pfb diff --git a/PS2/doc.log b/PS2/doc.log index 9b108eb..9b867cb 100644 --- a/PS2/doc.log +++ b/PS2/doc.log @@ -1,4 +1,4 @@ -This is pdfTeX, Version 3.14159265-2.6-1.40.21 (MiKTeX 21.1) 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[5 ] File: q1di_fig1.png Graphic file (type png) -Package pdftex.def Info: q1di_fig1.png used on input line 359. +Package pdftex.def Info: q1di_fig1.png used on input line 363. (pdftex.def) Requested size: 227.98666pt x 170.99103pt. File: q1di_fig2.png Graphic file (type png) -Package pdftex.def Info: q1di_fig2.png used on input line 364. +Package pdftex.def Info: q1di_fig2.png used on input line 368. (pdftex.def) Requested size: 227.98666pt x 170.99103pt. File: q1div.jpg Graphic file (type jpg) -Package pdftex.def Info: q1div.jpg used on input line 382. +Package pdftex.def Info: q1div.jpg used on input line 386. (pdftex.def) Requested size: 227.98666pt x 230.517pt. [6 ] [7] File: q2c_fig1_1s.png Graphic file (type png) -Package pdftex.def Info: q2c_fig1_1s.png used on input line 485. +Package pdftex.def Info: q2c_fig1_1s.png used on input line 489. (pdftex.def) Requested size: 238.68115pt x 179.01422pt. 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PDF statistics: - 130 PDF objects out of 1000 (max. 8388607) + 143 PDF objects out of 1000 (max. 8388607) 0 named destinations out of 1000 (max. 500000) - 56 words of extra memory for PDF output out of 10000 (max. 10000000) + 76 words of extra memory for PDF output out of 10000 (max. 10000000) diff --git a/PS2/doc.pdf b/PS2/doc.pdf index c209bdb..3997640 100644 Binary files a/PS2/doc.pdf and b/PS2/doc.pdf differ diff --git a/PS2/doc.synctex.gz b/PS2/doc.synctex.gz index 1f0f1ca..5562864 100644 Binary files a/PS2/doc.synctex.gz and b/PS2/doc.synctex.gz differ diff --git a/PS2/doc.tex b/PS2/doc.tex index 47d8dcb..64faed6 100644 --- a/PS2/doc.tex +++ b/PS2/doc.tex @@ -25,6 +25,8 @@ \captionsetup[figure]{width=0.75\textwidth,labelfont=normalfont,font=it,labelsep=period} +\DeclareMathOperator*{\sinc}{sinc} +\DeclareMathOperator*{\rect}{rect} \renewcommand{\thesubsection}{\indent(\alph{subsection})} \definecolor{dkgreen}{rgb}{0,0.6,0} @@ -62,6 +64,8 @@ % } +\newcommand{\angstrom}{\textup{\AA}} + \title{ECE 456 - Problem Set 2} \date{2021-03-01} \author{David Lenfesty \\ lenfesty@ualberta.ca @@ -547,24 +551,126 @@ axis([0 1e-9 0 0.04]); \begin{enumerate}[align=left,leftmargin=*,labelsep=0em,label=(\roman*)] \item %3bi - \lipsum[1] + + Mapping the provided Fourier identities from \( t \) and \( \omega \) onto + \(x'\) and \(k'\), we can evaluate the Fourier transform \(A(k')\): + + \begin{align*} + \mathcal{F}\left[rect(\frac{x'}{L}) \right] &= \frac{L}{\sqrt{2 \pi}} sinc \left( \frac{k' L}{2 \pi} \right) \\ + \mathcal{F}\left[f(x')cos(\frac{\pi}{L}x')\right] &= \frac{1}{2} \left[ F(k' + k_1) + F(k' - k_1) \right] \\ + A(k') &= \frac{1}{2} \sqrt{\frac{L}{\pi}} \left\{ \sinc \left( \frac{L}{2\pi} (k' + k_1) \right) + \sinc \left( \frac{L}{2\pi} (k' - k_1) \right) \right\}. + \end{align*} + \item %3bii - \lipsum[1] + Beginning with the result for \(A(k')\) above, and writing + \begin{equation*} + \Phi(p') \equiv \frac{1}{\sqrt{\hbar}} A \left(\frac{p'}{\hbar}\right), + \end{equation*} + we can obtain + \begin{equation*} + \Phi(p') = \frac{1}{2} \sqrt{\frac{L}{\pi\hbar}} \left\{ \sinc \left( \frac{L}{2\pi\hbar} (p' + p_1) \right) + \sinc \left( \frac{L}{2\pi\hbar} (p' - p_1) \right) \right\}. + \end{equation*} + + \item %3biii - \lipsum[1] + \(\left|\Phi(p')\right|^2\) has units of [\si{\s.kg^{-1}.m^{-1}}], which are those of inverse momentum. Thus, multiplication + (or integration) by a differential of momentum results in a unitless probability, as we should expect. This holds in the 1D case + and can easily be generalized to higher dimensions. + + \item %3biv - \lipsum[1] + + \( sinc \) is a purely real function, so we can ignore the normalizing portion of the integral. + + % TODO just replace these in the formulas themselves + As well, to simplify the interim equations we will assign the constants \( A = \frac{1}{2} \sqrt{\frac{L}{\pi \hbar}} \) + and \( B = \frac{L}{2 \pi \hbar} \). + + \begin{equation*} + \int _{-\infty} ^{\infty} \Phi (p')^2 dp = \int _{-\infty} ^{\infty} A^2 \left[ sinc\left( B(p'+p_1) \right)^2 + 2 sinc \left( B(p'+p_1) \right) sinc \left( B(p'-p_1) \right) + sinc \left( B(p'-p_1) \right) \right] dp + \end{equation*} + + Given property (26) of the \( sinc \) function, we can simplify the left and right elements. % TODO better word than elements + Using a change of variable \( p'' = p_1 - p' \), and properties (27) and (28), we can further evaluate the central element. + + \begin{align} + \int _{-\infty} ^{\infty} \Phi (p')^2 dp = \int _{-\infty} ^{\infty} A^2 \left[2 sinc \left( B(2p_1 - p'') \right) sinc \left( B(-p'') \right) \right] dp + A^2 \frac{2}{B} \\ + &= \int _{-\infty} ^{\infty} A^2 \left[ 2 sinc \left(B(2p_1 - p'') \right) sinc(B p'') \right] dp + A^2 \frac{2}{B} \\ + &= \frac{A^2}{B} \left[ sinc(2p_1) + 2 \right] \\ + \end{align} + \item %3bv - \lipsum[1] + + + \begin{minipage}[t]{\linewidth} + + \begin{figure}[H] + \centering + \begin{subfigure}{0.5\textwidth} + \centering + \includegraphics[width=\textwidth]{q3bv_fig1.png} + \caption{} + \end{subfigure}% + \begin{subfigure}{0.5\textwidth} + \centering + \includegraphics[width=\textwidth]{q3bv_fig2.png} + \caption{} + \end{subfigure} + \caption{(a) momentum wavefunction versus normalized momentum. (b) Probability density versus normalized momentum.} + \end{figure} + \end{minipage} \item %3bvi - \lipsum[1] + The points of classical momentum are given by \( p_1 = \pm \sqrt{2mE}\). On the normalized plots, these occur at \(\pm 1\) on the \(p/p_1\) axis. + Given that \(L = \SI{101}{\angstrom}\), we can find the velocity of the electron by taking \(v = \frac{p_1}{m_e}\). We find that + \(v = \pm \SI{3.6e4}{m/s}\). + + \begin{minipage}[t]{\linewidth} + + \begin{figure}[H] + \centering + \begin{subfigure}{0.5\textwidth} + \centering + \includegraphics[width=\textwidth]{q3bvi_fig1.png} + \caption{} + \end{subfigure}% + \begin{subfigure}{0.5\textwidth} + \centering + \includegraphics[width=\textwidth]{q3bvi_fig2.png} + \caption{} + \end{subfigure} + \caption{(a),(b) Previous plots but with the classical momentum marked in red.} + \end{figure} + \end{minipage} \item %3bvii - \lipsum[1] + From the plot of the probability density, we can clearly see that the particle can take a continuum of momentum values. + Thus the statement is false. \end{enumerate} + \item %3c - \lipsum[1] + Because the probability density is even about \(p' = 0\), we can surmise that \(\left\langle p'\right\rangle = 0\). + \newline + To verify this, we find \(\left\langle p'\right\rangle\) + from \(\phi(x') = \sqrt{\frac{2}{L}}\sin{\left(\frac{\pi}{L}x'\right)}\times\rect{\left(\frac{x'}{L}\right)}\) according to + \begin{align*} + \left\langle p'\right\rangle\ &= \int_{-\infty}^\infty \phi^*(x')\:\hat{p}\:\phi(x')\:dx'\\ + \left\langle p'\right\rangle\ &= -i\hbar\int_{-L/2}^{L/2} \sqrt{\frac{2}{L}}\sin{\left(\frac{\pi}{L}x'\right)}\:\frac{d}{dx'}\left[\sqrt{\frac{2}{L}}\sin{\left(\frac{\pi}{L}x'\right)}\right]\:dx'\\ + \left\langle p'\right\rangle\ &= \frac{-2i\pi\hbar}{L^2}\int_{-L/2}^{L/2} \sin{\left(\frac{\pi}{L}x'\right)}\:\cos{\left(\frac{\pi}{L}x'\right)}\:dx'. + \end{align*} + Using Equation (31) in the assignment we can write + + \begin{align*} + \left\langle p'\right\rangle\ &= \left.\frac{-i\hbar}{L}\sin^2{\left(\frac{\pi}{L}x'\right)}\right|_{-L/2}^{L/2}\\ + \left\langle p'\right\rangle\ &= \frac{-i\hbar}{L}\left(1 - 1\right) = 0, + \end{align*} + which verifies our above inference. \item %3d - \lipsum[1] + + The momentum associated with the wavefunction \( \theta (x') = e^{ik'x'} \) is sharp, + and the corresponding value is \( p' = \hbar k' \). + + \begin{equation*} + \hat{p} = -i \hbar \frac{d}{dx'} e ^{i k' x'} = -i^2 \hbar k' e ^{i k' x'} = \hbar k' e^{i k' x'} + \end{equation*} \end{enumerate} \newpage @@ -573,15 +679,27 @@ axis([0 1e-9 0 0.04]); \begin{enumerate}[align=left,leftmargin=*,labelsep=1em,label=\bfseries(\alph*)] \item %4a - \lipsum[1] + + % TODO I really don't like this explanation... + An \( a \) value of \( 0.62 \angstrom \) was chosen, in order to provide an adequately + shaped graph without sacrificing too much computation time. + + % maybe talk about normalization? + \item %4b \begin{enumerate}[align=left,leftmargin=*,labelsep=0em,label=(\roman*)] \item %4bi \lipsum[1] \item %4bii - \lipsum[1] + + + \item %4biii - \lipsum[1] + + Negative Drain Resistance (NDR) is present in this design from a \(V_D\) of + approximately \(0.27V\) to \(0.45V\), with a mostly linear region from + aroung \(0.28V\) to \(0.3V\). + \end{enumerate} \end{enumerate} diff --git a/PS2/problem_3_part_b_v.m b/PS2/problem_3_part_b_v.m new file mode 100644 index 0000000..90d2e05 --- /dev/null +++ b/PS2/problem_3_part_b_v.m @@ -0,0 +1,49 @@ +% A simple MATLAB code to plot the momentum +% wave function for a particle in an escape-proof "box" + +clear all + +hbar = 1.054e-34; +L = 101e-10; + +% Set up k' and p' axes + +k_prime = linspace(-2*pi/L,2*pi/L,100); +dk_prime = k_prime(2) - k_prime(1); +p_prime = hbar * k_prime; +dp_prime = p_prime(2) - p_prime(1); + +% k1 and p1 + +k1 = (pi/L); p1 = hbar * k1; + +% Fourier transform A(k') + +A1 = 0.5*sqrt(L/pi)*( sinc((k_prime+k1)*L/(2*pi)) ); +A2 = 0.5*sqrt(L/pi)*( sinc((k_prime-k1)*L/(2*pi)) ); +A = A1 + A2; +A_sum = sum(abs(A).^2)*dk_prime; + +% Momentum wave function Phi(p') + +Phi1 = 0.5*sqrt(L/(pi*hbar))*( sinc((p_prime+p1)*L/(2*pi*hbar)) ); +Phi2 = 0.5*sqrt(L/(pi*hbar))*( sinc((p_prime-p1)*L/(2*pi*hbar)) ); +Phi = Phi1 + Phi2; +Phi_sum = sum(abs(Phi).^2)*dp_prime; + +% Normalized p' axis for plotting purposes; the points of classical +% momenta occur when the variable pp_N is plus or minus unity + +pp_N = p_prime / p1; + +figure(1); clf; +h = plot(pp_N,Phi1,'kx',pp_N,Phi2,'ko',pp_N,Phi,'k--'); +grid on; set(h,'linewidth',[2.0]); set(gca,'Fontsize',[18]); +xlabel('NORMALIZED MOMENTUM [p^{\prime}/p_1]'); +ylabel('WAVE FUNCTION [sqrt(s / kg m)]'); +legend('Phi1','Phi2','Phi'); + +figure(2); clf; h = plot(pp_N,abs(Phi).^2,'k-'); +grid on; set(h,'linewidth',[2.0]); set(gca,'Fontsize',[18]); +xlabel('NORMALIZED MOMENTUM [p^{\prime}/p_1]'); +ylabel('PROBABILITY DENSITY [s / kg m]'); diff --git a/PS2/q3bv_fig1.png b/PS2/q3bv_fig1.png new file mode 100644 index 0000000..e3a93ce Binary files /dev/null and b/PS2/q3bv_fig1.png differ diff --git a/PS2/q3bv_fig2.png b/PS2/q3bv_fig2.png new file mode 100644 index 0000000..e868488 Binary files /dev/null and b/PS2/q3bv_fig2.png differ diff --git a/PS2/q3bvi_fig1.png b/PS2/q3bvi_fig1.png new file mode 100644 index 0000000..a4a7614 Binary files /dev/null and b/PS2/q3bvi_fig1.png differ diff --git a/PS2/q3bvi_fig2.png b/PS2/q3bvi_fig2.png new file mode 100644 index 0000000..551b314 Binary files /dev/null and b/PS2/q3bvi_fig2.png differ diff --git a/PS2/q4a.m b/PS2/q4a.m index 7917caf..3909d57 100644 --- a/PS2/q4a.m +++ b/PS2/q4a.m @@ -18,7 +18,7 @@ w0 = (0.2879*q/hbar); % YOU MUST ENTER AN APPROPRIATE VALUE OF "a" -a = ; +a = 12 / sqrt(m*w0/hbar) / 100; beta = sqrt(m*w0/hbar); x = [-6/beta:a:6/beta]; N = length(x); diff --git a/PS2/q4b_main.m b/PS2/q4b_main.m new file mode 100644 index 0000000..1557ba3 --- /dev/null +++ b/PS2/q4b_main.m @@ -0,0 +1,145 @@ +clear all; + +% Physical constants in MKS units + +hbar = 1.054e-34; +q = 1.602e-19; + +% Energy parameters in eV; included are the single-electron charging +% energy U0; the kBT product; the equilibrium Fermi level mu; and the +% energy levels cal_E1 and cal_E2, where "cal_E" is short-form for "calligraphic E" + +% Please especially note that ALL ENERGY VARIABLES IN THIS CODE +% ARE IN eV (NOT joules); the equations from class must be adjusted +% accordingly, multiplying or dividing appropriate terms by a factor of q + +% YOU MUST ENTER THE APPROPRIATE VALUES OF cal_E1 and cal_E2 + +U0 = 0.025; +kBT = 0.025; +mu = 0; +cal_E1 = 0.2879 / 2; +cal_E2 = 0.2879 * 3 / 2; + +% Capacitance parameters + +alpha_G = 0.5; +alpha_D = 0.5; +alpha_S = 1 - alpha_G - alpha_D; + +% Energy grid in eV, from -1 eV to 1 eV + +NE = 501; +E = linspace(-1,1,NE); +dE = E(2) - E(1); + +% Coupling coefficients, which are now grids over E, with +% gamma_1 equal to 0.005 eV for E > 0, and 0 for E <= 0, and +% with gamma_2 equal to 0.005 eV for all E; the 1e-6 +% term is included to avoid divide-by-zero errors + +gamma_1 = 0.005*(E + abs(E)) ./ (E + E + 1e-6); +gamma_2 = 0.005*ones(1,NE); +gamma = gamma_1 + gamma_2; + +% Reference number of electrons in the channel, assumed to be zero in +% this code + +N0 = 0; + +% Voltage values to consider for the final plots + +NV = 121; +VV = linspace(-0.4,0.8,NV); +dV = VV(2) - VV(1); + +% Loop over voltage values and compute number of electrons and current +% for each voltage value in a self-consistent manner + +for count = 1:NV + + % Set terminal voltages + + VG = 0; + VD = VV(count); + VS = 0; + + % Values of mu1 and mu2; notice that the usual factor of q multiplying + % the voltages is omitted, because in this code, energy is in eV + + mu1 = mu - VS; + mu2 = mu - VD; + + % Compute source and drain Fermi functions + + f1 = 1./(1+exp((E - mu1)./kBT)); + f2 = 1./(1+exp((E - mu2)./kBT)); + + % Value of Laplace potential in eV + + UL = - (alpha_G*VG) - (alpha_D*VD) - (alpha_S*VS); + + % Initial value of Poisson part in eV + + UP = 0; + + % Iterate until self-consistent potential is achieved by monitoring + % the Poisson part (the Laplace part does not change) + + dUP = 1; + while dUP > 1e-6 + + % Lorentzian SHIFTED density of states for levels 1 and 2, + % each normalized so that its integral is unity + + D1 = (0.01/(2*pi))./((E - (UL + UP) - cal_E1).^2+(0.01/2)^2); + D1 = D1./(dE*sum(D1)); + + D2 = (0.01/(2*pi))./((E - (UL + UP) - cal_E2).^2+(0.01/2)^2); + D2 = D2./(dE*sum(D2)); + + % Total density of states + + D = D1 + D2; + + % Compute number of channel electrons + + N(count) = dE*sum( ((gamma_1./gamma).*f1 + (gamma_2./gamma).*f2).*D ); + + % Newly calculated Poisson part of self-consistent potential + + UPnew = U0*( N(count) - N0 ); + + % Change in Poisson part between iterations + + dUP = abs(UP - UPnew); + + % New guess for next iteration, found by adding a fraction of the + % difference between iterations to the old guess + + UP = UP + 0.1*(UPnew - UP); + + end + + % Compute the current in A after the self-consistent potential + % has been achieved; notice the extra factor of q preceding the + % equation, which is needed since the gammas are in eV + + I(count) = q*(q/hbar)... + *dE*sum((f1-f2).*D.*gamma_1.*gamma_2./gamma); + + run q4b_subplots; + +end + +% Plotting commands, including lines to modify the linewidth +% and Fontsize, just to make the plots look nicer; you don't +% need to worry about how these work + +figure(1); h = plot(VV,N,'k'); grid on; +set(h,'linewidth',[2.0]); set(gca,'Fontsize',[18]); +xlabel('DRAIN VOLTAGE [V]'); ylabel('NUMBER OF ELECTRONS'); + +figure(2); h = plot(VV,I,'k'); grid on; +set(h,'linewidth',[2.0]); set(gca,'Fontsize',[18]); +xlabel('DRAIN VOLTAGE [V]'); ylabel('CURRENT [A]'); diff --git a/PS2/q4b_subplots.m b/PS2/q4b_subplots.m new file mode 100644 index 0000000..bf9d930 --- /dev/null +++ b/PS2/q4b_subplots.m @@ -0,0 +1,113 @@ +% Code to create the required subplots for Problem Set 2, Problem 5(b) + +% If you've used the same variable names as in the sample code, then you +% should be able to simply insert this into the appropriate spot in your +% own code; otherwise, you'll have to modify this accordingly, which +% should be easy to do---if disaster strikes and it doesn't work, then +% please ask for help + +% The "if" statement is used to choose VD values closest to the required +% values of 0.0, 0.2, 0.3, ..., 0.8 V, and you don't need to +% worry about how this works + +if (abs(VD-0.0) <= dV/2) + figure(3); title('VD = 0.0 V'); + subplot(2,3,1); plot(f1,E,'k-'); axis([-0.1 1.1 -1 1]); + xlabel('f1(E)'); ylabel('ENERGY [eV]'); title('VD = 0.0 V'); + subplot(2,3,2); plot(D/100,E,'k-'); axis([-0.1 1.1 -1 1]); + xlabel('D(E-U)/100'); ylabel('ENERGY [eV]'); title('VD = 0.0 V'); + subplot(2,3,3); plot(f2,E,'k-'); axis([-0.1 1.1 -1 1]); + xlabel('f2(E)'); ylabel('ENERGY [eV]'); title('VD = 0.0 V'); + subplot(2,3,5); plot(f1-f2,E,'--',D/100,E,'k-'); axis([-0.1 1.1 -1 1]); + xlabel('f1(E)-f2(E), D(E-U)/100'); ylabel('ENERGY [eV]'); title('VD = 0.0 V'); +elseif (abs(VD-0.2) <= dV/2) + figure(4); title('VD = 0.2 V'); + subplot(2,3,1); plot(f1,E,'k-'); axis([-0.1 1.1 -1 1]); + xlabel('f1(E)'); ylabel('ENERGY [eV]'); title('VD = 0.2 V'); + subplot(2,3,2); plot(D/100,E,'k-'); axis([-0.1 1.1 -1 1]); + xlabel('D(E-U)/100'); ylabel('ENERGY [eV]'); title('VD = 0.2 V'); + subplot(2,3,3); plot(f2,E,'k-'); axis([-0.1 1.1 -1 1]); + xlabel('f2(E)'); ylabel('ENERGY [eV]'); title('VD = 0.2 V'); + subplot(2,3,5); plot(f1-f2,E,'--',D/100,E,'k-'); axis([-0.1 1.1 -1 1]); + xlabel('f1(E)-f2(E), D(E-U)/100'); ylabel('ENERGY [eV]'); title('VD = 0.2 V'); +elseif (abs(VD-0.25) <= dV/2) + figure(5); title('VD = 0.25 V'); + subplot(2,3,1); plot(f1,E,'k-'); axis([-0.1 1.1 -1 1]); + xlabel('f1(E)'); ylabel('ENERGY [eV]'); title('VD = 0.25 V'); + subplot(2,3,2); plot(D/100,E,'k-'); axis([-0.1 1.1 -1 1]); + xlabel('D(E-U)/100'); ylabel('ENERGY [eV]'); title('VD = 0.25 V'); + subplot(2,3,3); plot(f2,E,'k-'); axis([-0.1 1.1 -1 1]); + xlabel('f2(E)'); ylabel('ENERGY [eV]'); title('VD = 0.25 V'); + subplot(2,3,5); plot(f1-f2,E,'--',D/100,E,'k-'); axis([-0.1 1.1 -1 1]); + xlabel('f1(E)-f2(E), D(E-U)/100'); ylabel('ENERGY [eV]'); title('VD = 0.25 V'); +elseif (abs(VD-0.3) <= dV/2) + figure(6); title('VD = 0.3 V'); + subplot(2,3,1); plot(f1,E,'k-'); axis([-0.1 1.1 -1 1]); + xlabel('f1(E)'); ylabel('ENERGY [eV]'); title('VD = 0.3 V'); + subplot(2,3,2); plot(D/100,E,'k-'); axis([-0.1 1.1 -1 1]); + xlabel('D(E-U)/100'); ylabel('ENERGY [eV]'); title('VD = 0.3 V'); + subplot(2,3,3); plot(f2,E,'k-'); axis([-0.1 1.1 -1 1]); + xlabel('f2(E)'); ylabel('ENERGY [eV]'); title('VD = 0.3 V'); + subplot(2,3,5); plot(f1-f2,E,'--',D/100,E,'k-'); axis([-0.1 1.1 -1 1]); + xlabel('f1(E)-f2(E), D(E-U)/100'); ylabel('ENERGY [eV]'); title('VD = 0.3 V'); +elseif (abs(VD-0.4) <= dV/2) + figure(7); title('VD = 0.4 V'); + subplot(2,3,1); plot(f1,E,'k-'); axis([-0.1 1.1 -1 1]); + xlabel('f1(E)'); ylabel('ENERGY [eV]'); title('VD = 0.4 V'); + subplot(2,3,2); plot(D/100,E,'k-'); axis([-0.1 1.1 -1 1]); + xlabel('D(E-U)/100'); ylabel('ENERGY [eV]'); title('VD = 0.4 V'); + subplot(2,3,3); plot(f2,E,'k-'); axis([-0.1 1.1 -1 1]); + xlabel('f2(E)'); ylabel('ENERGY [eV]'); title('VD = 0.4 V'); + subplot(2,3,5); plot(f1-f2,E,'--',D/100,E,'k-'); axis([-0.1 1.1 -1 1]); + xlabel('f1(E)-f2(E), D(E-U)/100'); ylabel('ENERGY [eV]'); title('VD = 0.4 V'); +elseif (abs(VD-0.5) <= dV/2) + figure(8); title('VD = 0.5 V'); + subplot(2,3,1); plot(f1,E,'k-'); axis([-0.1 1.1 -1 1]); + xlabel('f1(E)'); ylabel('ENERGY [eV]'); title('VD = 0.5 V'); + subplot(2,3,2); plot(D/100,E,'k-'); axis([-0.1 1.1 -1 1]); + xlabel('D(E-U)/100'); ylabel('ENERGY [eV]'); title('VD = 0.5 V'); + subplot(2,3,3); plot(f2,E,'k-'); axis([-0.1 1.1 -1 1]); + xlabel('f2(E)'); ylabel('ENERGY [eV]'); title('VD = 0.5 V'); + subplot(2,3,5); plot(f1-f2,E,'--',D/100,E,'k-'); axis([-0.1 1.1 -1 1]); + xlabel('f1(E)-f2(E), D(E-U)/100'); ylabel('ENERGY [eV]'); title('VD = 0.5 V'); +elseif (abs(VD-0.6) <= dV/2) + figure(9); title('VD = 0.6 V'); + subplot(2,3,1); plot(f1,E,'k-'); axis([-0.1 1.1 -1 1]); + xlabel('f1(E)'); ylabel('ENERGY [eV]'); title('VD = 0.6 V'); + subplot(2,3,2); plot(D/100,E,'k-'); axis([-0.1 1.1 -1 1]); + xlabel('D(E-U)/100'); ylabel('ENERGY [eV]'); title('VD = 0.6 V'); + subplot(2,3,3); plot(f2,E,'k-'); axis([-0.1 1.1 -1 1]); + xlabel('f2(E)'); ylabel('ENERGY [eV]'); title('VD = 0.6 V'); + subplot(2,3,5); plot(f1-f2,E,'--',D/100,E,'k-'); axis([-0.1 1.1 -1 1]); + xlabel('f1(E)-f2(E), D(E-U)/100'); ylabel('ENERGY [eV]'); title('VD = 0.6 V'); +elseif (abs(VD-0.65) <= dV/2) + figure(10); title('VD = 0.65 V'); + subplot(2,3,1); plot(f1,E,'k-'); axis([-0.1 1.1 -1 1]); + xlabel('f1(E)'); ylabel('ENERGY [eV]'); title('VD = 0.65 V'); + subplot(2,3,2); plot(D/100,E,'k-'); axis([-0.1 1.1 -1 1]); + xlabel('D(E-U)/100'); ylabel('ENERGY [eV]'); title('VD = 0.65 V'); + subplot(2,3,3); plot(f2,E,'k-'); axis([-0.1 1.1 -1 1]); + xlabel('f2(E)'); ylabel('ENERGY [eV]'); title('VD = 0.65 V'); + subplot(2,3,5); plot(f1-f2,E,'--',D/100,E,'k-'); axis([-0.1 1.1 -1 1]); + xlabel('f1(E)-f2(E), D(E-U)/100'); ylabel('ENERGY [eV]'); title('VD = 0.65 V'); +elseif (abs(VD-0.7) <= dV/2) + figure(11); title('VD = 0.7 V'); + subplot(2,3,1); plot(f1,E,'k-'); axis([-0.1 1.1 -1 1]); + xlabel('f1(E)'); ylabel('ENERGY [eV]'); title('VD = 0.7 V'); + subplot(2,3,2); plot(D/100,E,'k-'); axis([-0.1 1.1 -1 1]); + xlabel('D(E-U)/100'); ylabel('ENERGY [eV]'); title('VD = 0.7 V'); + subplot(2,3,3); plot(f2,E,'k-'); axis([-0.1 1.1 -1 1]); + xlabel('f2(E)'); ylabel('ENERGY [eV]'); title('VD = 0.7 V'); + subplot(2,3,5); plot(f1-f2,E,'--',D/100,E,'k-'); axis([-0.1 1.1 -1 1]); + xlabel('f1(E)-f2(E), D(E-U)/100'); ylabel('ENERGY [eV]'); title('VD = 0.7 V'); +elseif (abs(VD-0.8) <= dV/2) + figure(12); title('VD = 0.8 V'); + subplot(2,3,1); plot(f1,E,'k-'); axis([-0.1 1.1 -1 1]); + xlabel('f1(E)'); ylabel('ENERGY [eV]'); title('VD = 0.8 V'); + subplot(2,3,2); plot(D/100,E,'k-'); axis([-0.1 1.1 -1 1]); + xlabel('D(E-U)/100'); ylabel('ENERGY [eV]'); title('VD = 0.8 V'); + subplot(2,3,3); plot(f2,E,'k-'); axis([-0.1 1.1 -1 1]); + xlabel('f2(E)'); ylabel('ENERGY [eV]'); title('VD = 0.8 V'); + subplot(2,3,5); plot(f1-f2,E,'--',D/100,E,'k-'); axis([-0.1 1.1 -1 1]); + xlabel('f1(E)-f2(E), D(E-U)/100'); ylabel('ENERGY [eV]'); title('VD = 0.8 V'); +end diff --git a/PS2/sinc.m b/PS2/sinc.m new file mode 100644 index 0000000..8758064 --- /dev/null +++ b/PS2/sinc.m @@ -0,0 +1,5 @@ +function out = sinc(x) +% Deal with the removable singularity at 0 explicitly. +out = sin(pi*x)./(pi*x); +out(x == 0) = 1; +end \ No newline at end of file