我在使用环境时遇到了问题\wrapimage。
当我插入一张图像并换到右边时，它显示得很好，但是当我想添加第二张图像时，它却不出现。
我究竟做错了什么？
以下是代码：
\documentclass{article}
\usepackage[utf8]{inputenc}
\usepackage{geometry}
\usepackage{amsmath}
\usepackage{amsfonts}
\usepackage{amssymb}
\usepackage{array}
\usepackage{siunitx}
\usepackage{tablefootnote}
\usepackage{graphicx}
\usepackage{wrapfig}
\usepackage{float}
\newcommand\tab[1][.5cm]{\hspace*{#1}}
\geometry{
a4paper,
total={170mm,257mm},
left=20mm,
top=20mm,
}
\begin{document}
\title{Characterization of the amplifier using the protocol}
\author{Group 11C - Handy Audio}
\maketitle
In order to verify the performance of our amplifier we applied the protocol described in \textit{Amplifier characteristics measurement protocol} obtaining the following results.
\section{Output power. $P_o$}
\begin{table}[!h]
\begin{tabular}{||m{0.3\textwidth} | m{0.7\textwidth} ||}
\hline \hline
Instruments and measure set up &
\begin{itemize}
\item Power supply
\item Waveform generator ($f=1\mathrm{kHz}, V_{pp}=0.8943\mathrm{V}$)
\item Oscilloscope
\end{itemize}
\\
\hline
Measured data &
\begin{itemize}
\item Output voltage: $V_o$
\end{itemize}
\\
\hline
Involved variables, measured values and procedure &
Using \textit{Math Mode}, measure the voltage drop amplitude across the $8\mathrm\Omega$ resistor taking into account the $\frac{1}{\sqrt{2}}$ factor of the RMS.\newline
Applying the following formula:
$$ P_o = I\cdot V_{o_{RMS}} = \frac{V_{o_{RMS}}^{2}}{R_L}
= \frac{V_o^{2}}{2\cdot8\mathrm{\Omega}}$$
we obtain the power delivered by the amplifier to the speaker.
\\
\hline \hline
\end{tabular}
\end{table}
\begin{wrapfigure}{R}{0.6\textwidth}
\includegraphics[width=0.6\textwidth,keepaspectratio=true]{../Fotos/20161221/mesures_amplitud}
\caption{Output signal at $f=1$kHz}
\end{wrapfigure}
\noindent
Specification: $1\mathrm{W} \le P_o \le 3\mathrm{W}$\\ \\
Measurements:
\begin{itemize}
\item $V_o=\frac{V_{pp}}{2}=\frac{7.68\mathrm V}{2}=3.84$V
\end{itemize}
\noindent
\\
{\Large Result: $P_o=0.92$W }
\\ \\
As we can see on Figure 1, the output signal amplitude does not reach the 4V required to achieve 1W of power, we could only reach 3.89V of amplitude at the output which gives us 0.92W of output power.
\pagebreak
\section{Energetic efficiency. $\eta$}
\begin{table}[H]
\begin{tabular}{||m{0.3\textwidth} | m{0.7\textwidth} ||}
\hline \hline
Instruments and measure set up &
\begin{itemize}
\item Power supply
\item Waveform generator ($f=1\mathrm{kHz}, V_{pp}=0.8943\mathrm{V}$)
\item Oscilloscope
\end{itemize}
\\
\hline
Measured data &
\begin{itemize}
\item Current consumption (power supply readout): $I_g$
\item Power supply voltage: $V_{cc}$
\end{itemize}
\\
\hline
Involved variables, measured values and procedure &
Using the readout provided by the power supply ($I_g$) and the previous measurement, using the following formula:
$$ \eta = \frac{P_o}{I_g \cdot V_{cc}} $$
we obtain the power efficiency of the amplifier.
\\
\hline \hline
\end{tabular}
\end{table}
\begin{wrapfigure}{R}{0.6\textwidth}
\includegraphics[width=0.6\textwidth,keepaspectratio=true]{../Fotos/20161221/mesures_consum}
\caption{Current consumption at $f=1$kHz}
\end{wrapfigure}
\noindent
Specification: $\eta \ge 80\%$ \\ \\
Measurements:
\begin{itemize}
\item $I_g=0.21$A
\item $V_{cc}=5.0$V
\end{itemize}
\noindent
\\
{\Large Result: $\eta=87.62\%$ }
\\
\pagebreak
\section{Frequency response}
\begin{table}[H]
\begin{tabular}{||m{0.3\textwidth} | m{0.7\textwidth} ||}
\hline\hline
Instruments and measure set up &
\begin{itemize}
\item Power supply
\item Oscilloscope
\item Waveform generator ($f\in(20, 20\mathrm k)\mathrm{Hz}, V_{pp}=0.8943\mathrm{V}$)
\end{itemize}
\\
\hline
Measured data &
\begin{itemize}
\item Output voltage: $V_o$
\end{itemize}
\\
\hline
Involved variables, measured values and procedure&
First we have to set up a fixed amplitude input signal and measure the output (voltage drop across the resistor: $V_R$). Using the cursors, we set up two thresholds:
\begin{itemize}
\item $-1\mathrm{dB}$: $0.89 V_R$
\item $-3\mathrm{dB}$: $0.70 V_R$
\end{itemize}
Using the knob on the waveform generator we have to first sweep the frequency between 100Hz and 10kHz and check that the output signal does not go below the first threshold.\linebreak Then we have to repeat the frequency sweep but this time between 20Hz and 20kHz and check that the output signal does not go below the second threshold.\\
\hline \hline
\end{tabular}
\end{table}
\vspace{5mm}
\noindent
{\Large Result:\,} {\large OK} if it satisfies both conditions
\pagebreak
\section{Total harmonic distortion + noise. $THD+N$}
\begin{table}[H]
\begin{tabular}{||m{0.3\textwidth} | m{0.7\textwidth} ||}
\hline \hline
Instruments and measure set up &
\begin{itemize}
\item Power supply
\item Floating ground differential amplifier
\item Waveform generator ($f=1\mathrm{kHz}, V_{pp}=0.8943\mathrm{V}$)
\item Spectrum analyzer (or an oscilloscope with FFT function)
\end{itemize}
\\
\hline
Measured data &
\begin{itemize}
\item FFT of the output signal
\end{itemize}
\\
\hline
Involved variables, measured values and procedure&
Visualizing the spectrum of the output signal, we have to measure the amplitude of the harmonics and using the following formula\tablefootnote{$V_0$ is the amplitude of the fundamental harmonic\\\tab$V_n$ is the amplitude of the $n$th harmonic}:
$$THD=\frac{\sqrt{V_1^2+V_2^2+\dots+V_N^2}}{V_0}$$
we obtain the total harmonic distortion introduced by the amplifier.\\
\hline \hline
\end{tabular}
\end{table}
% \begin{wrapfigure}{r}{0.6\textwidth}
% \includegraphics[width=0.6\textwidth,keepaspectratio=true]{../Fotos/20161221/mesures_espectre}
% \caption{Output signal spectrum at $f=1$kHz}
% \end{wrapfigure}
\noindent
Specification: $THD \le 1\%$ \\ \\
{\Large Result: $THD|_{N=5} =1.845\%$ }
\end{document}
下面是我得到的输出：
提前致谢！