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Chapter 8: Problem 4

In electrical engineering, a continuous function like \(f(t)=\sin t,\) where \(t\) is in seconds, is referred to as an analog signal. To digitize the signal, we sample \(f(t)\) every \(\Delta t\) seconds to form the sequence \(s_{n}=f(n \Delta t) .\) For example, sampling \(f\) every \(1 / 10\) second produces the sequence \(\sin (1 / 10), \sin (2 / 10), \sin (3 / 10), \ldots\) Suppose that the analog signal is given by $$ f(t)=(t-0.5)^{2} $$ Give the first 6 terms of a sampling of the signal every \(\Delta t=0.25\) seconds: (Enter your answer as a comma-separated list.)

Short Answer

Expert verified
0.0625, 0, 0.0625, 0.25, 0.5625, 1

Step by step solution

01

Understand the function and sampling interval

The given analog signal function is \[ f(t) = (t - 0.5)^2 \]and we need to sample it every \[ \Delta t = 0.25 \] seconds.
02

Calculate the first sample

To find the first term, substitute \( n = 1 \) into \[ s_n = f(n \Delta t) \]Thus, we have:\[ s_1 = f(1 \times 0.25) = f(0.25) = (0.25 - 0.5)^2 \]Simplifying:\[ s_1 = (-0.25)^2 = 0.0625 \]
03

Calculate the second sample

To find the second term, substitute \( n = 2 \):\[ s_2 = f(2 \times 0.25) = f(0.5) = (0.5 - 0.5)^2 \]Simplifying:\[ s_2 = 0^2 = 0 \]
04

Calculate the third sample

To find the third term, substitute \( n = 3 \):\[ s_3 = f(3 \times 0.25) = f(0.75) = (0.75 - 0.5)^2 \]Simplifying:\[ s_3 = 0.25^2 = 0.0625 \]
05

Calculate the fourth sample

To find the fourth term, substitute \( n = 4 \):\[ s_4 = f(4 \times 0.25) = f(1) = (1 - 0.5)^2 \]Simplifying:\[ s_4 = 0.5^2 = 0.25 \]
06

Calculate the fifth sample

To find the fifth term, substitute \( n = 5 \):\[ s_5 = f(5 \times 0.25) = f(1.25) = (1.25 - 0.5)^2 \]Simplifying:\[ s_5 = 0.75^2 = 0.5625 \]
07

Calculate the sixth sample

To find the sixth term, substitute \( n = 6 \):\[ s_6 = f(6 \times 0.25) = f(1.5) = (1.5 - 0.5)^2 \]Simplifying:\[ s_6 = 1^2 = 1 \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Continuous Function
In electrical engineering, a continuous function is often used to describe an analog signal. An example of a continuous function is \(f(t) = \sin t\), where \(t\) represents time in seconds. This function represents an analog signal that varies smoothly over time. Unlike digital signals, a continuous function does not have interruptions or discrete steps. Every point in time has a corresponding value on the function's curve.
Digital Signal Processing
Digital Signal Processing (DSP) involves converting an analog signal into a digital form so it can be processed by digital systems. The first step in DSP is sampling, where the continuous analog signal is measured at regular intervals. These readings are then transformed into a sequence of numbers that represent the original signal. This digitized version can then be manipulated using algorithms to enhance, filter, or analyze the data. The process of transforming these signals is crucial for many applications, including audio processing, communications, and image manipulation.
Sequence Sampling
Sequence sampling means taking regular measurements of a continuous function to create a sequence of values. For example, if the function \(f(t) = (t - 0.5)^2\) is sampled every \(\Delta t = 0.25\) seconds, we derive values from the function at specific intervals. These values form a sequence like \(s_n = f(n \Delta t)\). This transformation allows us to analyze and process the signal more efficiently using digital tools. Each sampling point is crucial for preserving the signal's characteristics in its digital form.
Sampling Interval
The sampling interval, denoted by \(\Delta t\), refers to the time gap between each sample taken from the continuous function. For instance, in the given function \(f(t) = (t - 0.5)^2\), if \(\Delta t = 0.25\) seconds, the function is sampled every 0.25 seconds. The choice of this interval is critical; if it is too large, important information may be lost, but if it is too small, it may result in excessive data. The right balance ensures the digital representation of the signal closely resembles the original analog signal. Choosing the correct sampling interval is essential for accurate digital signal processing.

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Most popular questions from this chapter

Let $$ a_{n}=\frac{2 n}{10 n+7} $$ For the following answer blanks, decide whether the given sequence or series is convergent or divergent. If convergent, enter the limit (for a sequence) or the sum (for a series). If divergent, enter 'infinity' if it diverges to \(\infty\), '-infinity' if it diverges to \(-\infty\) or 'DNE' otherwise. (a) The series \(\sum_{n=1}^{\infty} \frac{2 n}{10 n+7}\). (b) The sequence \(\left\\{\frac{2 n}{10 n+7}\right\\}\).

Compute the value of the following improper integral. If it converges, enter its value. Enter infinity if it diverges to \(\infty\), and -infinity if it diverges to \(-\infty\). Otherwise, enter diverges. $$ \int_{1}^{\infty} \frac{3 d x}{x^{2}+1}= $$ Does the series \(\sum_{n=1}^{\infty} \frac{3}{n^{2}+1}\) converge or diverge? [Choose: converges | diverges to +infinity | diverges to -infinity | diverges]

In the Limit Comparison Test we compared the behavior of a series to one whose behavior we know. In that test we use the limit of the ratio of corresponding terms of the series to determine if the comparison is valid. In this exercise we see how we can compare two series directly, term by term, without using a limit of sequence. First we consider an example. a. Consider the series \(\sum \frac{1}{k^{2}}\) and \(\sum \frac{1}{k^{2}+k}\) We know that the series \(\sum \frac{1}{k^{2}}\) is a \(p\) -series with \(p=2>1\) and so \(\sum \frac{1}{k^{2}}\) converges. In this part of the exercise we will see how to use information about \(\sum \frac{1}{k^{2}}\) to determine information about \(\sum \frac{1}{k^{2}+k} .\) Let \(a_{k}=\frac{1}{k^{2}}\) and \(b_{k}=\frac{1}{k^{2}+k} .\) i) Let \(S_{n}\) be the \(n\) th partial sum of \(\sum \frac{1}{k^{2}}\) and \(T_{n}\) the \(n\) th partial sum of \(\sum \frac{1}{k^{2}+k}\). Which is larger, \(S_{1}\) or \(T_{1}\) ? Why? ii) Recall that $$ S_{2}=S_{1}+a_{2} \text { and } T_{2}=T_{1}+b_{2} $$ Which is larger, \(a_{2}\) or \(b_{2}\) ? Based on that answer, which is larger, \(S_{2}\) or \(T_{2}\) ? iii) Recall that $$ S_{3}=S_{2}+a_{3} \text { and } T_{3}=T_{2}+b_{3} $$ Which is larger, \(a_{3}\) or \(b_{3}\) ? Based on that answer, which is larger, \(S_{3}\) or \(T_{3}\) ? iv) Which is larger, \(a_{n}\) or \(b_{n}\) ? Explain. Based on that answer, which is larger, \(S_{n}\) or \(T_{n}\) ? v) Based on your response to the previous part of this exercise, what relationship do you expect there to be between \(\sum \frac{1}{k^{2}}\) and \(\sum \frac{1}{k^{2}+k} ?\) Do you expect \(\sum \frac{1}{k^{2}+k}\) to converge or diverge? Why? b. The example in the previous part of this exercise illustrates a more general result. Explain why the Direct Comparison Test, stated here, works.

We have shown that if \(\sum(-1)^{k+1} a_{k}\) is a convergent alternating series, then the sum \(S\) of the series lies between any two consecutive partial sums \(S_{n}\). This suggests that the average \(\frac{S_{n}+S_{n+1}}{2}\) is a better approximation to \(S\) than is \(S_{n}\). a. Show that \(\frac{S_{n}+S_{n+1}}{2}=S_{n}+\frac{1}{2}(-1)^{n+2} a_{n+1}\). b. Use this revised approximation in (a) with \(n=20\) to approximate \(\ln (2)\) given that $$ \ln (2)=\sum_{k=1}^{\infty}(-1)^{k+1} \frac{1}{k} . $$ Compare this to the approximation using just \(S_{20} .\) For your convenience, \(S_{20}=\frac{155685007}{232792560}\).

Airy's equation \(^{2}\) $$ y^{\prime \prime}-x y=0 $$ can be used to model an undamped vibrating spring with spring constant \(x\) (note that \(y\) is an unknown function of \(x\) ). So the solution to this differential equation will tell us the behavior of a spring-mass system as the spring ages (like an automobile shock absorber). Assume that a solution \(y=f(x)\) has a Taylor series that can be written in the form $$ y=\sum_{k=0}^{\infty} a_{k} x^{k} $$ where the coefficients are undetermined. Our job is to find the coefficients. (a) Differentiate the series for \(y\) term by term to find the series for \(y^{\prime}\). Then repeat to find the series for \(y^{\prime \prime}\). (b) Substitute your results from part (a) into the Airy equation and show that we can write Equation \((8.6 .4)\) in the form$$ \sum_{k=2}^{\infty}(k-1) k a_{k} x^{k-2}-\sum_{k=0}^{\infty} a_{k} x^{k+1}=0 $$ (c) At this point, it would be convenient if we could combine the series on the left in \((8.6 .5)\), but one written with terms of the form \(x^{k-2}\) and the other with terms in the form \(x^{k+1}\). Explain why $$ \sum_{k=2}^{\infty}(k-1) k a_{k} x^{k-2}=\sum_{k=0}^{\infty}(k+1)(k+2) a_{k+2} x^{k} $$ (d) Now show that $$ \sum_{k=0}^{\infty} a_{k} x^{k+1}=\sum_{k=1}^{\infty} a_{k-1} x^{k} $$ (e) We can now substitute \((8.6 .6)\) and \((8.6 .7)\) into \((8.6 .5)\) to obtain $$ \sum_{n=0}^{\infty}(n+1)(n+2) a_{n+2} x^{n}-\sum_{n=1}^{\infty} a_{n-1} x^{n}=0 $$ Combine the like powers of \(x\) in the two series to show that our solution must satisfy $$ 2 a_{2}+\sum_{k=1}^{\infty}\left[(k+1)(k+2) a_{k+2}-a_{k-1}\right] x^{k}=0 $$ (f) Use equation (8.6.9) to show the following: i. \(a_{3 k+2}=0\) for every positive integer \(k\), ii. \(a_{3 k}=\frac{1}{(2)(3)(5)(6) \cdots(3 k-1)(3 k)} a_{0}\) for \(k \geq 1\), iii. \(a_{3 k+1}=\frac{1}{(3)(4)(6)(7) \cdots(3 k)(3 k+1)} a_{1}\) for \(k \geq 1\). (g) Use the previous part to conclude that the general solution to the Airy equation \((8.6 .4)\) is $$ \begin{aligned} y=& a_{0}\left(1+\sum_{k=1}^{\infty} \frac{x^{3 k}}{(2)(3)(5)(6) \cdots(3 k-1)(3 k)}\right) \\ &+a_{1}\left(x+\sum_{k=1}^{\infty} \frac{x^{3 k+1}}{(3)(4)(6)(7) \cdots(3 k)(3 k+1)}\right) \end{aligned} $$ Any values for \(a_{0}\) and \(a_{1}\) then determine a specific solution that we can approximate as closely as we like using this series solution.
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