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）
Determine the values of $P_{\infty}$ and $E_{\infty}$ for each of the following signals:
(a) $x_1(t)=e^{-2 t} u(t)$
(b) $x_2(t)=e^{j(2 t+\pi / 4)}$
(c) $x_3(t)=\cos (t)$
(d) $x_1[n]=\left(\frac{1}{2}\right)^n u[n]$
(e) $x_2[n]=e^{j(\pi / 2 n+\pi / 8)}$
(f) $x_3[n]=\cos \left(\frac{\pi}{4} n\right)$
Solution
(a) $E_{\infty}=\int_0^{\infty} e^{-2 t} d t=\frac{1}{4} . P_{\infty}=0$, because $E_{\infty}<\infty$.
(b) $x_2(t)=e^{j(2 t+\pi / 4)},\left|x_2(t)\right|=1$. Therefore,
$$
\begin{gathered}
E_{\infty}=\int_{-\infty}^{\infty}\left|x_2(t)\right|^2 d t=\int_{-\infty}^{\infty} d t=\infty . \
P_{\infty}=\lim {T \rightarrow \infty} \frac{1}{2 T} \int{-T}^T\left|x_2(t)\right|^2 d t=\lim {T \rightarrow \infty} \frac{1}{2 T} \int{-T}^T d t=\lim {T \rightarrow \infty} 1=1 . \end{gathered} $$ (c) $x_3(t)=\cos (t)$. Therefore, $$ \begin{gathered} E{\infty}=\int_{-\infty}^{\infty}\left|x_3(t)\right|^2 d t=\int_{-\infty}^{\infty} \cos ^2(t) d t=\infty \
P_{\infty}=\lim {T \rightarrow \infty} \frac{1}{2 T} \int{-T}^T\left|x_3(t)\right|^2 d t=\lim {T \rightarrow \infty} \frac{1}{2 T} \int{-T}^T \cos ^2(t) d t=\lim {T \rightarrow \infty} \frac{1}{2 T} \int{-T}^T\left(\frac{1+\cos (2 t)}{2}\right) d t=\frac{1}{2} .
\end{gathered}
$$
(d) $x_1[n]=\left(\frac{1}{2}\right)^n u[n],\left|x_1[n]\right|^2=\left(\frac{1}{4}\right)^n u[n]$. Therefore,
$$
E_{\infty}=\sum_{n=-\infty}^{\infty}\left|x_1[n]\right|^2=\sum_{n=0}^{\infty}\left(\frac{1}{4}\right)^n=\frac{4}{3} .
$$
$P_{\infty}=0$, because $E_{\infty}<\infty$.
(e) $x_2[n]=e^{j(\pi / 2 n+\pi / 8)},\left|x_2[n]\right|^2=1$. Therefore,
$$
\begin{gathered}
E_{\infty}=\sum_{n=-\infty}^{\infty}\left|x_2[n]\right|^2=\sum_{n=-\infty}^{\infty} 1=\infty \
P_{\infty}=\lim {N \rightarrow \infty} \frac{1}{2 N+1} \sum{n=-N}^N\left|x_2[n]\right|^2=\lim {N \rightarrow \infty} \frac{1}{2 N+1} \sum{n=-N}^N 1=1 .
\end{gathered}
$$
(f) $x_3[n]=\cos \left(\frac{\pi}{4} n\right)$. Therefore,
$$
\begin{gathered}
E_{\infty}=\sum_{n=-\infty}^{\infty}\left|x_3[n]\right|^2=\sum_{n=-\infty}^{\infty} \cos ^2\left(\frac{\pi}{4} n\right)=\infty \
P_{\infty}=\lim {N \rightarrow \infty} \frac{1}{2 N+1} \sum{n=-N}^N\left|x_3[n]\right|^2=\lim {N \rightarrow \infty} \frac{1}{2 N+1} \sum{n=-N}^N \cos ^2\left(\frac{\pi}{4} n\right)=\lim {N \rightarrow \infty} \frac{1}{2 N+1} \sum{n=-N}^N\left(\frac{1+\cos \left(\frac{\pi}{2} n\right)}{2}\right)=\frac{1}{2} .
\end{gathered}
$$
Let $x(t)$ be the continuous-time complex exponential signal
$$
x(t)=e^{j w_0 t}
$$
with fundamental frequency $\omega_0$ and fundamental period $T_0=2 \pi / \omega_0$. Consider the discrete-time signal obtained by taking equally spaced samples of $x(t)$ – that is,
$$
x[n]=x(n T)=e^{j \omega_0 n T}
$$
(a) Show that $x[n]$ is periodic if and only if $T / T_0$ is a rational number – that is, if and only if some multiple of the sampling interval exactly equals a multiple of the period of $x(t)$.
(b) Suppose that $x[n]$ is periodic – that is, that
$$
\frac{T}{T_0}=\frac{p}{q}
$$
where $p$ and $q$ are integers. What are the fundamental period and fundamental frequency of $x[n]$ ? Express the fundamental frequency as a fraction of $\omega_0 T$.
(c) Again assuming that $\frac{T}{T_0}$ satisfies equation (1), determine precisely how many periods of $x(t)$ are needed to obtain the samples that form a single period of $x[n]$.
Solution
(a) If $x[n]$ is periodic, then $e^{j \omega_0(n+N) T}=e^{j \omega_0 n T}$, where $\omega_0=2 \pi / T_0$. This implies that
$$
\frac{2 \pi}{T_0} N T=2 \pi k \Rightarrow \frac{T}{T_0}=\frac{k}{N}=\text { a rational number. }
$$
If $\frac{T}{T_0}=\frac{k}{N}=$ a rational number, then we have
$$
\frac{T}{T_0}=\frac{k}{N} \Rightarrow \frac{2 \pi}{T_0} N T=2 \pi k
$$
This implies that $e^{j \omega_0(n+N) T}=e^{j \omega_0 n T}$, where $\omega_0=2 \pi / T_0 . x[n]$ is periodic.
Combining the above two conditions, we can conclude that $x[n]$ is periodic if and only if $T / T_0$ is a rational number.
(b) If $\frac{T}{T_0}=\frac{p}{q}$ then $x[n]=e^{j 2 \pi n\left(\frac{p}{q}\right)}$.The fundamental period is $N=q / \operatorname{gcd}(p, q)$ (gcd refer to the greatest common divisor). The fundamental frequency is
$$
\frac{2 \pi}{q} g c d(p, q)=\frac{2 \pi}{p} \frac{p}{q} g c d(p, q)=\frac{\omega_0 T}{p} \operatorname{gcd}(p, q)
$$
(c) We know that the fundamental period of (b) is $N=q / \operatorname{gcd}(p, q)$, so overall $\frac{N T}{T_0}=p / \operatorname{gcd}(p, q)$ periods of $x(t)$ is needed to obtain the samples that form a single period of $x[n]$.
Show that causality for a continuous-time linear system is equivalent to the following statement:
For any time $t_0$ and any input $x(t)$ such that $x(t)=0$ for $t<t_0$, the corresponding output $y(t)$ must also be zero for $t<t_0$.
Assumption: If $x(t)=0$ for $tt_0, x_2(t) \neq x_1(t)$. Since the system is linear,
$$
x_1(t)-x_2(t) \rightarrow y_1(t)-y_2(t)
$$
Since $x_1(t)-x_2(t)=0$ for $t<t_0$, by our assumption $y_1(t)-y_2(t)=0$ for $t<t_0$. This implies that $y_1(t)=y_2(t)$ for $t<t_0$. In other words, the output is not affected by input values for $t \geq t_0$. Therefore, the system is causal.
Assumption: The system is causal. To prove that: If $x(t)=0$ for $t<t_0$, then $y(t)=0$ for $t<t_0$.
Let us assume that the signal $x(t)=0$ for $t<t_0$. Then we may express $x(t)$ as $x(t)=x_1(t)-x_2(t)$, where $x_1(t)=x_2(t)$ for $t<t_0$. Since the system is linear, the output to $x(t)$ will be $y(t)=y_1(t)-y_2(t)$. Now, since the system is causal, $x_1(t)=x_2(t)$ for $t<t_0$ implies that $y_1(t)=y_2(t)$ for $t<t_0$. Therefore, $y(t)=0$ for $t<t_0$.
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！。
