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@ -25,6 +25,8 @@
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\captionsetup[figure]{width=0.75\textwidth,labelfont=normalfont,font=it,labelsep=period}
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\DeclareMathOperator*{\sinc}{sinc}
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\DeclareMathOperator*{\rect}{rect}
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% }
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\newcommand{\angstrom}{\textup{\AA}}
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\title{ECE 456 - Problem Set 2}
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\date{2021-03-01}
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\author{David Lenfesty \\ lenfesty@ualberta.ca
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@ -547,24 +551,126 @@ axis([0 1e-9 0 0.04]);
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\begin{enumerate}[align=left,leftmargin=*,labelsep=0em,label=(\roman*)]
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\item %3bi
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\lipsum[1]
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Mapping the provided Fourier identities from \( t \) and \( \omega \) onto
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\(x'\) and \(k'\), we can evaluate the Fourier transform \(A(k')\):
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\begin{align*}
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\mathcal{F}\left[rect(\frac{x'}{L}) \right] &= \frac{L}{\sqrt{2 \pi}} sinc \left( \frac{k' L}{2 \pi} \right) \\
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\mathcal{F}\left[f(x')cos(\frac{\pi}{L}x')\right] &= \frac{1}{2} \left[ F(k' + k_1) + F(k' - k_1) \right] \\
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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\}.
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\end{align*}
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\item %3bii
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\lipsum[1]
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Beginning with the result for \(A(k')\) above, and writing
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\begin{equation*}
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\Phi(p') \equiv \frac{1}{\sqrt{\hbar}} A \left(\frac{p'}{\hbar}\right),
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\end{equation*}
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we can obtain
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\begin{equation*}
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\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\}.
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\end{equation*}
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\item %3biii
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\lipsum[1]
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\(\left|\Phi(p')\right|^2\) has units of [\si{\s.kg^{-1}.m^{-1}}], which are those of inverse momentum. Thus, multiplication
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(or integration) by a differential of momentum results in a unitless probability, as we should expect. This holds in the 1D case
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and can easily be generalized to higher dimensions.
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\item %3biv
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\lipsum[1]
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\( sinc \) is a purely real function, so we can ignore the normalizing portion of the integral.
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% TODO just replace these in the formulas themselves
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As well, to simplify the interim equations we will assign the constants \( A = \frac{1}{2} \sqrt{\frac{L}{\pi \hbar}} \)
|
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and \( B = \frac{L}{2 \pi \hbar} \).
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\begin{equation*}
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\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
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\end{equation*}
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Given property (26) of the \( sinc \) function, we can simplify the left and right elements. % TODO better word than elements
|
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Using a change of variable \( p'' = p_1 - p' \), and properties (27) and (28), we can further evaluate the central element.
|
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\begin{align}
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\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} \\
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&= \int _{-\infty} ^{\infty} A^2 \left[ 2 sinc \left(B(2p_1 - p'') \right) sinc(B p'') \right] dp + A^2 \frac{2}{B} \\
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&= \frac{A^2}{B} \left[ sinc(2p_1) + 2 \right] \\
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\end{align}
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\item %3bv
|
||||
\lipsum[1]
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||||
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\begin{minipage}[t]{\linewidth}
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||||
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||||
\begin{figure}[H]
|
||||
\centering
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\begin{subfigure}{0.5\textwidth}
|
||||
\centering
|
||||
\includegraphics[width=\textwidth]{q3bv_fig1.png}
|
||||
\caption{}
|
||||
\end{subfigure}%
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\begin{subfigure}{0.5\textwidth}
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\centering
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||||
\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
|
||||
\(v = \pm \SI{3.6e4}{m/s}\).
|
||||
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||||
\begin{minipage}[t]{\linewidth}
|
||||
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||||
\begin{figure}[H]
|
||||
\centering
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\begin{subfigure}{0.5\textwidth}
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||||
\centering
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||||
\includegraphics[width=\textwidth]{q3bvi_fig1.png}
|
||||
\caption{}
|
||||
\end{subfigure}%
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\begin{subfigure}{0.5\textwidth}
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||||
\centering
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||||
\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]
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||||
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}
|
||||
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||||
\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}
|
||||
|
||||
49
PS2/problem_3_part_b_v.m
Normal file
49
PS2/problem_3_part_b_v.m
Normal file
@ -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]');
|
||||
BIN
PS2/q3bv_fig1.png
Normal file
BIN
PS2/q3bv_fig1.png
Normal file
Binary file not shown.
|
After Width: | Height: | Size: 30 KiB |
BIN
PS2/q3bv_fig2.png
Normal file
BIN
PS2/q3bv_fig2.png
Normal file
Binary file not shown.
|
After Width: | Height: | Size: 21 KiB |
BIN
PS2/q3bvi_fig1.png
Normal file
BIN
PS2/q3bvi_fig1.png
Normal file
Binary file not shown.
|
After Width: | Height: | Size: 30 KiB |
BIN
PS2/q3bvi_fig2.png
Normal file
BIN
PS2/q3bvi_fig2.png
Normal file
Binary file not shown.
|
After Width: | Height: | Size: 21 KiB |
@ -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);
|
||||
|
||||
145
PS2/q4b_main.m
Normal file
145
PS2/q4b_main.m
Normal file
@ -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]');
|
||||
113
PS2/q4b_subplots.m
Normal file
113
PS2/q4b_subplots.m
Normal file
@ -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
|
||||
5
PS2/sinc.m
Normal file
5
PS2/sinc.m
Normal file
@ -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
|
||||
Loading…
Reference in New Issue
Block a user