MY-ASSIGNMENTEXPERT™可以为您提供iona.edu MTH432 Sampling theory抽样调查课程的代写代考和辅导服务！
这是愛納學院抽样调查课程的代写成功案例。
MTH432课程简介
Syllabus : Principles of sample surveys; Simple, stratified and unequal probability sampling with and without replacement; ratio, product and regression method of estimation, Varying Probability Scheme
An overview of probability and statistics. Experiments; sample spaces; random variables; probability measures and distributions; combinatorics; expectation; data collection and analysis; confidence intervals; selected hypothesis tests.
Lecture
Credits: 3
Prerequisite: MTH 232
Offered in Fall Semester
Prerequisites
Books: You can choose any one of the following book for your reference. Books at serial numbers 1 and 2 are easily available, so I will base my lectures on them. Other books are available in the library.
Sampling Techniques : W.G. Cochran, Wiley (Low price edition available)
Theory and Methods of Survey Sampling : Parimal Mukhopadhyay, Prentice Hall of India
Theory of Sample surveys with applications : P.V. Sukhatme, B.V Sukhatme, S. Sukhatme and C. Asok, IASRI, Delhi
Sampling Methodologies and Applications : P.S.R.S. Rao, Chapman and Hall/ CRC
Sampling Theory and Methods : M.N. Murthy, Statistical Publishing Society, Calcutta (Out of print)
Elements of sampling theory and methods : Z. Govindrajalu, Prentice Hall
Sampling Methods- Exercises and Solutions : Pascal Ardilly and Yves Tille’ (Download here through IITK Library link)
MTH432 Sampling theory HELP（EXAM HELP， ONLINE TUTOR）
In the text, we showed that if $h[n]$ is absolutely summable, i.e., if
$$
\sum_{k=-\infty}^{\infty}|h[k]|<\infty
$$
then the LTI system with impulse response $h[n]$ is stable. This means that absolute summability is a sufficient condition for stability. In this problem, we shall show that it is also a necessary condition. Consider an LTI system with impulse response $h[n]$ that is not absolutely summable; that is,
$$
\sum_{k=-\infty}^{\infty}|h[k]|=\infty
$$
(a) Suppose that the input to this system is
$$
x[n]= \begin{cases}0, & \text { if } h[-n]=9 \ \frac{h[-n]}{\mid h[-n],}, & \text { if } h[-n] \neq 9\end{cases}
$$
Does this input signal represent a bounded input? If so, what is the smallest number $B$ such that
$$
|x[n]| \leq B \text { for all } n ?
$$
(b) Calculate the output at $n=0$ for this particular choice of input. Does the result prove the statement that absolute summability is a necessary condition for stability?
(c) In a similar fashion, show that a continuous-time LTI system is stable if any only if its impulse response is absolutely integrable.
(a) It is a bounded input. $|x[n]| \leq 1=B_x$ for all $n$.
(b) Consider
$$
\begin{aligned}
y[0] & =\sum_{k=-\infty}^{\infty} x[-k] h[k] \
& =\sum_{k=-\infty}^{\infty} \frac{h^2[k]}{|h[k]|} \
& =\sum_{k=-\infty}^{\infty}|h[k]| \rightarrow \infty
\end{aligned}
$$
Therefore, the output is not bounded. Thus, the system is not stable and absolute summability is necessary.
(c) Let
$$
x[n]= \begin{cases}0, & \text { if } h(-t)=0 \ \frac{h(-t)}{|h(-t)|}, & \text { if } h(-t) \neq 0\end{cases}
$$
Now, $|x(t)| \leq 1$ for all $t$. Therefore, $x(t)$ is a bounded input. Now,
$$
\begin{aligned}
y(0) & =\int_{-\infty}^{\infty} x(\tau) h(\tau) d \tau \
& =\int_{-\infty}^{\infty} \frac{h^2(\tau)}{|h(\tau)|} d \tau \
& =\int_{-\infty}^{\infty}|h(t)| d t=\infty
\end{aligned}
$$
Therefore, the system is unstable if the impulse response is not absolutely integrable.
Consider a causal LTI system $S$ whose input $x[n]$ and output $y[n]$ are related by the difference equation
$$
2 y[n]-y[n-1]+y[n-3]=x[n]-5 x[n-4]
$$
(a) Verify that $S$ may be considered a cascade connection of two causal LTI systems $S_1$ and $S_2$ with the following input-output relationship:
$$
\begin{aligned}
& S_1: 2 y_1[n]=x_1[n]-5 x_1[n-4] \
& S_2: y_2[n]=\frac{1}{2} y_2[n-1]-\frac{1}{2} y_2[n-3]+x_2[n]
\end{aligned}
$$
(b) Draw a block diagram representation of $S_1$.
(c) Draw a block diagram representation of $S_2$.
(d) Draw a block diagram representation of $S$ as a cascade connection of the block diagram representation of $S_1$ followed by the block diagram representation of $S_2$.
(e) Draw a block diagram representation of $S$ as a cascade connection of the block diagram representation of $S_2$ followed by the block diagram representation of $S_1$.
(f) Show that the four delay elements in the block diagram representation of $S$ obtained in part (e) may be collapsed to three. The resulting block diagram is referred to as a Direct Form II realization of $S$, while the block diagrams obtained in parts (d) and (e) are referred to as Direct Form I realizations of $S$.
(a) Realizing that $x_2[n]=y_1[n]$, we may eliminate these from the two given difference equations. This would give us
$$
2 y_2[n]-y_2[n-1]+y_2[n-3]=x_1[n]-5 x_1[n-4] .
$$
This is the same as the overall difference equation.
(b) The figures corresponding to the remaining parts of this problem are shown in Figure 1.
Determine the Fourier series representations for the following signals. You shall not only give the Fourier series coefficients, but also give the Fourier series expression of the signals.
(a) Each $x[n]$ or $x(t)$ illustrated in Figure 2.
(b) $x(t)$ periodic with period 2 and
$$
x(t)=e^{-t} \text { for }-1 \leq t \leq 1
$$
(c) $x(t)$ periodic with period 4 and
$$
x(t)= \begin{cases}\sin (\pi t), & 0 \leq t \leq 2, \ 0, & 2<t \leq 4 .\end{cases}
$$
Solution
(a) (i) It is known that $N=2$. So that
$$
\begin{aligned}
a_0 & =\frac{1}{N} \sum_N x[n] e^{-j k(2 \pi / N) n} \
& =\frac{1}{2} \sum_{n=0}^1 x[n] e^{-j \cdot 0 \cdot \pi n} \
& =\frac{1}{2} \sum_{n=0}^1 x[n] \
& =-\frac{1}{2}
\end{aligned}
$$
Meanwhile,
$$
\begin{aligned}
a_k & =\frac{1}{N} \sum_N x[n] e^{-j k(2 \pi / N) n} \
& =\frac{1}{2} \sum_{n=0}^1 x[n] e^{-j k \pi n} \
& =\frac{1}{2}-(-1)^k
\end{aligned}
$$
The Fourier series representation is
$$
x[n]=\sum_{k=-\infty}^{\infty} a_k e^{j k(2 \pi / N) n}=\sum_{k=-\infty}^{\infty}\left(\frac{1}{2}-(-1)^k\right) e^{j k \pi n}
$$
(ii) It is known that $T=6$. So that
$$
\begin{aligned}
a_0 & =\frac{1}{N} \int_T x(t) e^{-j k(2 \pi / T) t} d t \
& =\frac{1}{6} \int_0^6 x(t) e^{-j \cdot 0 \cdot(\pi / 3) t} d t \
& =\frac{1}{6} \int_0^6 x(t) d t \
& =0
\end{aligned}
$$
Meanwhile,
$$
\begin{aligned}
a_k & =\frac{1}{T} \int_T x(t) e^{-j k(2 \pi / T) t} d t \
& =\frac{1}{6} \int_0^6 x(t) e^{-j k(\pi / 3) t} d t \
& =\frac{1}{6}\left{-\left(\int_1^2 e^{-j k(\pi / 3) t} d t\right)+\left(\int_4^5 e^{-j k(\pi / 3) t} d t\right)\right} \
& =\frac{1}{j k \pi}\left(\cos \left(\frac{2 k \pi}{3}\right)-\cos \left(\frac{k \pi}{3}\right)\right)
\end{aligned}
$$
Note that not only $a_0=0$, but also $a_k$ even $=0$.
The Fourier series representation is
$$
x[n]=\sum_{k=-\infty}^{\infty} a_k e^{j k(2 \pi / T) t}=\sum_{k=-\infty}^{\infty}\left(\frac{1}{j k \pi}\left(\cos \left(\frac{2 k \pi}{3}\right)-\cos \left(\frac{k \pi}{3}\right)\right)\right) e^{j k(\pi / 3) t}(k \neq 0)
$$
(b) It is known that $T=2$. So that
$$
\begin{aligned}
a_k & =\frac{1}{T} \int_T x(t) e^{-j k(2 \pi / T) t} d t \
& =\frac{1}{2} \int_{-1}^1 e^{-t} e^{-j k \pi t} d t \
& =\frac{1}{2} \int_{-1}^1 e^{(-j k \pi-1) t} d t \
& =\frac{1}{2(1+j k \pi)}\left[e^{1+j \pi k}-e^{-1-j \pi k}\right]
\end{aligned}
$$
for all $k$.
The Fourier series representation is
$$
x[n]=\sum_{k=-\infty}^{\infty} a_k e^{j k(2 \pi / T) t}=\sum_{k=-\infty}^{\infty}\left(\frac{1}{2(1+j k \pi)}\left[e^{1+j \pi k}-e^{-1-j \pi k}\right]\right) e^{j k \pi t}
$$
(c) It is known that $T=4$. So that
$$
\begin{aligned}
a_0 & =\frac{1}{T} \int_T x(t) e^{-j k 2 \pi / T t} d t \
& =\frac{1}{4} \int_0^2 \sin (\pi t) e^{-j \cdot 0 \cdot(2 \pi / 4) t} d t \
& =\frac{1}{4} \int_0^2 \sin (\pi t) d t \
& =0 \
a_k= & \frac{1}{T} \int_T^2 x(t) e^{-j k 2 \pi / T t} d t \
= & \frac{1}{4} \int_0^2 \sin (\pi t) e^{-j k(\pi / 2) t} d t \
= & \frac{1}{8 j} \int_0^2\left(e^{j \pi t}-e^{-j \pi t}\right) e^{-j k(\pi / 2) t} d t \
= & \frac{1}{8 j}\left[\frac{e^{j \pi(2-k)-1}}{j \pi(1-k / 2)}+\frac{e^{j \pi(2+k)-1}}{j \pi(1+k / 2)}\right] \
= & \frac{e^{-j \pi k}}{\pi\left(k^2-4\right)}=\frac{(-1)^k}{\pi\left(k^2-4\right)}
\end{aligned}
$$
The Fourier series representation is
$$
x[n]=\sum_{k=-\infty}^{\infty} a_k e^{j k(2 \pi / T) t}=\sum_{k=-\infty}^{\infty}\left(\frac{(-1)^k}{\pi\left(k^2-4\right)}\right) e^{j k(\pi / 2) t}(k \neq 0)
$$
MY-ASSIGNMENTEXPERT™可以为您提供IONA.EDU MTH432 SAMPLING THEORY抽样调查课程的代写代考和辅导服务！。
