统计代写|STAT506 Sampling theory

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这是宾夕法尼亚州立大学 抽样调查课程的代写成功案例。

STAT506课程简介

The aim of this course is to cover sampling design and analysis methods that would be useful for research and management in many field. A well designed sampling procedure ensures that we can summarize and analyze data with a minimum of assumptions and complications.

Prerequisites 

Topics typically covered in STAT 506 are basic methods of sampling and estimation including:

• Simple random sampling with associated estimation and confidence interval methods,
• Selecting sample sizes,
• Estimating proportions,
• Unequal probability samping,
• Ratio and regression estimation,
• Stratified sampling,
• Cluster and systematic sampling,
• Multistage designs,
• Double or Two-stage sampling,
• Capture – recapture sampling, and
• Random response model

STAT506 Sampling theory HELP（EXAM HELP， ONLINE TUTOR）

Solution
(a)
$$
x(t)=\cos (4 \pi t)=\frac{1}{2} e^{4 \pi t}+\frac{1}{2} e^{-4 \pi t}
$$
So that the nonzero FS coefficients of $x(t)$ are $a_1=a_{-1}=1 / 2$.
(b)
$$
y(t)=\sin (4 \pi t)=\frac{1}{2 j} e^{4 \pi t}-\frac{1}{2 j} e^{-4 \pi t}
$$
So that the nonzero FS coefficients of $x(t)$ are $b_1=1 / 2 j, b_{-1}=-1 / 2 j$.
(c) Using the multiplication property, we know that
$$
z(t)=x(t) y(t) \stackrel{F S}{\leftrightarrow} c_k=\sum_{t=-\infty}^{\infty} a_l b_{k-l}
$$
Therefore,
$$
c_k=a_k * b_k=\sum_{t=-\infty}^{\infty} a_l b_{k-l}=\frac{1}{4 j} \delta[k-2]-\frac{1}{4 j} \delta[k+2]
$$
This implies that the nonzero Fourier series coefficients of $z(t)$ are $c_2=\frac{1}{4 j}, c_{-2}=-\frac{1}{4 j}$.
(d) We have
$$
z(t)=\sin (4 t) \cos (4 t)=\frac{1}{2} \sin (8 t)=\frac{1}{4 j} e^{8 \pi t}-\frac{1}{4 j} e^{-8 \pi t}
$$
Therefore, the nonzero Fourier series coefficients of $z(t)$ are $c_2=\frac{1}{4 j}, c_{-2}=-\frac{1}{4 j}$, the same with part (c).

Solution
(a) If $x(t)$ is real, then $x(t)=x^(t)$. This implies that for $x(t), a_k=a_{-k}^$. Since this is not true in this case, for example,
$$
a_1=j\left(\frac{1}{2}\right) \neq-j\left(\frac{1}{2}\right)=a_{-1}^* .
$$
Therefore, $x(t)$ is not real.
(b) If $x(t)$ is even, then $x(t)=x(-t)$. This implies that for $x(t), a_k=a_{-k}$. Since this is true in this case, because,
$$
a_k=j\left(\frac{1}{2}\right)^{|k|}=j\left(\frac{1}{2}\right)^{|k|}=a_{-k} .
$$
Therefore, $x(t)$ is even.
(c) We have
$$
g(t)=\frac{d x(t)}{d t} \stackrel{F S}{\leftrightarrow} b_k=j k \frac{2 \pi}{T} a_k,
$$
where $T$ refer to the period of signal $x(t)$. Therefore,
$$
b_k= \begin{cases}0, & k=0 \ -k(1 / 2)^{|k|}(2 \pi / T), & \text { otherwise }\end{cases}
$$
Since $b_k$ is not even, $g(t)$ is not even.

(a) For $0 \leq n \leq 9$, We have
$$
g[n]=x[n]-x[n-1]= \begin{cases}1, & n=0 \ 0, & 1 \leq n \leq 7 \ -1, & n=8 \ 0, & n=9\end{cases}
$$
This period begin to show again in the following 10 points, it’s clearly to draw the conclusion that $g[n]$ is periodic with period of 10 . The graph of signal $g[n]$ is shown in Figure 3 .

(b) It is known that $T=10$. So that the FS coefficients of $g[n]$ is $b_k$, which is
$$
\begin{aligned}
b_k & =\frac{1}{N} \sum_N g[n] e^{-j k(2 \pi / N) n} \
& =\frac{1}{10} \sum_{n=0}^9 g[n] e^{-j k(\pi / 5) n} \
& =\frac{1}{10}\left(1-e^{-j 8 k(\pi / 5)}\right)
\end{aligned}
$$
(c) Since $g[n]=x[n]-x[n-1]$, the FS coefficients $a_k$ and $b_k$ must be related as
$$
b_k=a_k-e^{-j k(\pi / 5)} a_k .
$$
Therefore,
$$
a_k=\frac{b_k}{1-e^{-j k(\pi / 5)}}=\frac{1}{10} \frac{1-e^{-j 8 k(\pi / 5)}}{1-e^{-j k(\pi / 5)}}
$$

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