text{Laplace transforms A powerful tool in solving problems in engineering and physics is the Laplace transform. Given a function f(t), the Laplace transform is a new function F(s) defined by } F(s)=int_0^infty e^{-st} f(t)dt text{where we assume s is a positive real number. For example, to find the Laplace transform of } f(t)=e^{-t} text{ , the following improper integral is evaluated using integration by parts:} F(s)=int_0^infty e^{-st}e^{-t}dt=int_0^infty e^{-(s+1)t}dt=frac{1}{s+1} text{ Verify the following Laplace transforms, where u is a real number. } f(t)=t rightarrow F(s)=frac{1}{s^2}

Question
Laplace transform
$$\text{Laplace transforms A powerful tool in solving problems in engineering and physics is the Laplace transform. Given a function f(t), the Laplace transform is a new function F(s) defined by }$$
$$F(s)=\int_0^\infty e^{-st} f(t)dt \(\text{where we assume s is a positive real number. For example, to find the Laplace transform of } f(t)=e^{-t} \text{ , the following improper integral is evaluated using integration by parts:} \(F(s)=\int_0^\infty e^{-st}e^{-t}dt=\int_0^\infty e^{-(s+1)t}dt=\frac{1}{s+1}$$
$$\text{ Verify the following Laplace transforms, where u is a real number. }$$
$$f(t)=t \rightarrow F(s)=\frac{1}{s^2}$$

2021-03-03
$$\text{Step 1}$$
$$\text{Here, the objective is to prove that }\ L(t)=\frac{1}{s^2}$$
$$\text{Step 2}$$
$$\text{Let } f(t) \text{ be t}$$
$$\text{Laplace transformation of f(t) is defined as:}$$
$$L(t)=F(s)=\int_0^\infty e^{-st}f(t)dt$$
$$\text{Substitute } f(t)=t$$
$$L(t)=F(s)=\int_0^\infty e^{-st} tdt$$
$$\text{Use integration by part } \int fg'=fg-\int f'g$$
$$\text{Here } f=t \Rightarrow f'=1 \text{ and } g'=e^{-st}\Rightarrow g=-\frac{e^{-st}}{s}$$
$$F(s)=\left[-\frac{e^{-st}t}{s}\right]_0^\infty -\int_0^\infty -\frac{e^{-st}}{s}dt \ (Where \int e^x dx=e^x+c)$$
$$F(s)=0+\int_0^\infty \frac{e^{-st}}{s}dt$$
$$F(s)=\left[-\frac{e^{-st}}{s^2}\right]_0^\infty$$
$$F(s)=\left[\(\lim_{t\rightarrow\infty}-\frac{e^{-st}}{s^2}-\bigg(-\frac{e^0}{s^2}\bigg)\right]$$
$$F(s)=\left[0-\bigg(-\frac{1}{s^2}\bigg)\right]$$
$$\text{Hence }$$
$$f(t)=t \rightarrow F(s)=\frac{1}{s^2}$$

Relevant Questions

Laplace transforms A powerful tool in solving problems in engineering and physics is the Laplace transform. Given a function f(t), the Laplace transform is a new function F(s) defined by
$$F(s)=\int_0^\infty e^{-st}f(t)dt$$
where we assume s is a positive real number. For example, to find the Laplace transform of f(t) = e^{-t}, the following improper integral is evaluated using integration by parts:
$$F(s)=\int_0^\infty e^{-st}e^{-t}dt=\int_0^\infty e^{-(s+1)t}dt=\frac{1}{(s+1)}$$
Verify the following Laplace transforms, where u is a real number.
$$f(t)=1 \rightarrow F(s)=\frac{1}{s}$$

The function
$$\begin{cases}t & 0\leq t<1\\ e^t & t\geq1 \end{cases}$$
has the following Laplace transform,
$$L(f(t))=\int_0^1te^{-st}dt+\int_1^\infty e^{-(s+1)t}dt$$
True or False

$$f(t)=3e^{2t}$$
Determine L[f]
Let f be a function defined on an interval $$[0,\infty)$$
The Laplace transform of f is the function F(s) defined by
$$F(s) =\int_0^\infty e^{-st}f(t)dt$$
provided that the improper integral converges. We will usually denote the Laplace transform of f by L[f].
a) Write the sigma notation formula for the right Riemann sum $$R_{n}$$ of the function $$f(x)=4-x^{2}$$ on the interval $$[0,\ 2]$$ using n subintervals of equal length, and calculate the definite integral $$\int_{0}^{2}f(x) dx$$ as the limit of $$R_{n}$$ at $$n\rightarrow\infty$$.
(Reminder: $$\sum_{k=1}^{n}k=n(n+1)/2,\ \sum_{k=1}^{n}k^{2}=n(n+1)(2n+1)/6)$$
b) Use the Fundamental Theorem of Calculus to calculate the derivative of $$F(x)=\int_{e^{-x}}^{x}\ln(t^{2}+1)dt$$
Explain why each of the following integrals is improper.
(a) $$\int_6^7 \frac{x}{x-6}dx$$
-Since the integral has an infinite interval of integration, it is a Type 1 improper integral.
-Since the integral has an infinite discontinuity, it is a Type 2 improper integral.
-The integral is a proper integral.
(b)$$\int_0^{\infty} \frac{1}{1+x^3}dx$$
Since the integral has an infinite interval of integration, it is a Type 1 improper integral.
Since the integral has an infinite discontinuity, it is a Type 2 improper integral.
The integral is a proper integral.
(c) $$\int_{-\infty}^{\infty}x^2 e^{-x^2}dx$$
-Since the integral has an infinite interval of integration, it is a Type 1 improper integral.
-Since the integral has an infinite discontinuity, it is a Type 2 improper integral.
-The integral is a proper integral.
d)$$\int_0^{\frac{\pi}{4}} \cot x dx$$
-Since the integral has an infinite interval of integration, it is a Type 1 improper integral.
-Since the integral has an infinite discontinuity, it is a Type 2 improper integral.
-The integral is a proper integral.

A random sample of $$n_1 = 14$$ winter days in Denver gave a sample mean pollution index $$x_1 = 43$$.
Previous studies show that $$\sigma_1 = 19$$.
For Englewood (a suburb of Denver), a random sample of $$n_2 = 12$$ winter days gave a sample mean pollution index of $$x_2 = 37$$.
Previous studies show that $$\sigma_2 = 13$$.
Assume the pollution index is normally distributed in both Englewood and Denver.
(a) State the null and alternate hypotheses.
$$H_0:\mu_1=\mu_2.\mu_1>\mu_2$$
$$H_0:\mu_1<\mu_2.\mu_1=\mu_2$$
$$H_0:\mu_1=\mu_2.\mu_1<\mu_2$$
$$H_0:\mu_1=\mu_2.\mu_1\neq\mu_2$$
(b) What sampling distribution will you use? What assumptions are you making? NKS The Student's t. We assume that both population distributions are approximately normal with known standard deviations.
The standard normal. We assume that both population distributions are approximately normal with unknown standard deviations.
The standard normal. We assume that both population distributions are approximately normal with known standard deviations.
The Student's t. We assume that both population distributions are approximately normal with unknown standard deviations.
(c) What is the value of the sample test statistic? Compute the corresponding z or t value as appropriate.
(Test the difference $$\mu_1 - \mu_2$$. Round your answer to two decimal places.) NKS (d) Find (or estimate) the P-value. (Round your answer to four decimal places.)
(e) Based on your answers in parts (i)−(iii), will you reject or fail to reject the null hypothesis? Are the data statistically significant at level \alpha?
At the $$\alpha = 0.01$$ level, we fail to reject the null hypothesis and conclude the data are not statistically significant.
At the $$\alpha = 0.01$$ level, we reject the null hypothesis and conclude the data are statistically significant.
At the $$\alpha = 0.01$$ level, we fail to reject the null hypothesis and conclude the data are statistically significant.
At the $$\alpha = 0.01$$ level, we reject the null hypothesis and conclude the data are not statistically significant.
(f) Interpret your conclusion in the context of the application.
Reject the null hypothesis, there is insufficient evidence that there is a difference in mean pollution index for Englewood and Denver.
Reject the null hypothesis, there is sufficient evidence that there is a difference in mean pollution index for Englewood and Denver.
Fail to reject the null hypothesis, there is insufficient evidence that there is a difference in mean pollution index for Englewood and Denver.
Fail to reject the null hypothesis, there is sufficient evidence that there is a difference in mean pollution index for Englewood and Denver. (g) Find a 99% confidence interval for
$$\mu_1 - \mu_2$$.
lower limit
upper limit
(h) Explain the meaning of the confidence interval in the context of the problem.
Because the interval contains only positive numbers, this indicates that at the 99% confidence level, the mean population pollution index for Englewood is greater than that of Denver.
Because the interval contains both positive and negative numbers, this indicates that at the 99% confidence level, we can not say that the mean population pollution index for Englewood is different than that of Denver.
Because the interval contains both positive and negative numbers, this indicates that at the 99% confidence level, the mean population pollution index for Englewood is greater than that of Denver.
Because the interval contains only negative numbers, this indicates that at the 99% confidence level, the mean population pollution index for Englewood is less than that of Denver.
Use the table of Laplace transform and properties to obtain the Laplace transform of the following functions. Specify which transform pair or property is used and write in the simplest form.
a) $$x(t)=\cos(3t)$$
b)$$y(t)=t \cos(3t)$$
c) $$z(t)=e^{-2t}\left[t \cos (3t)\right]$$
d) $$x(t)=3 \cos(2t)+5 \sin(8t)$$
e) $$y(t)=t^3+3t^2$$
f) $$z(t)=t^4e^{-2t}$$
We will now add support for register-memory ALU operations to the classic five-stage RISC pipeline. To offset this increase in complexity, all memory addressing will be restricted to register indirect (i.e., all addresses are simply a value held in a register; no offset or displacement may be added to the register value). For example, the register-memory instruction add x4, x5, (x1) means add the contents of register x5 to the contents of the memory location with address equal to the value in register x1 and put the sum in register x4. Register-register ALU operations are unchanged. The following items apply to the integer RISC pipeline:
a. List a rearranged order of the five traditional stages of the RISC pipeline that will support register-memory operations implemented exclusively by register indirect addressing.
b. Describe what new forwarding paths are needed for the rearranged pipeline by stating the source, destination, and information transferred on each needed new path.
c. For the reordered stages of the RISC pipeline, what new data hazards are created by this addressing mode? Give an instruction sequence illustrating each new hazard.
d. List all of the ways that the RISC pipeline with register-memory ALU operations can have a different instruction count for a given program than the original RISC pipeline. Give a pair of specific instruction sequences, one for the original pipeline and one for the rearranged pipeline, to illustrate each way.
Hint for (d): Give a pair of instruction sequences where the RISC pipeline has “more” instructions than the reg-mem architecture. Also give a pair of instruction sequences where the RISC pipeline has “fewer” instructions than the reg-mem architecture.
$$x^{n}-a^{n}=1x-a21x^{n-1}+ax^{n-2}+a^{2}x^{n-3}+...+a^{n-2}x+a^{n-1}2$$, where n is a positive integer a is a real number.
$$\lim_{x\rightarrow 2}\frac{x^{5}-32}{x-2}$$