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4 years ago

What is the difference between Laplace Transform and fourier Transform?

2 answers:

6 0

The difference between laplace transform and fourier transform is :

Laplace transform is used to get directly the final response of any system. Having transient behavior just by knowing the initial condition of the system

Fourier transform is used to breakup any varying signal into its sin and cosin components

hope this helps

Gre4nikov [31]4 years ago

3 0

<span>What is the difference between Laplace Transform and fourier Transform?<span>

</span>Answer: The difference between Laplace Transform and Fourier Transform is that Laplace is used to shift the system transfer function from time domain to frequency domain and in Fourier we get the frequency spectrum of the signal and their relative amplitude.

I hope it helps, Regards.</span>

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Advantages of solid expansion

bulgar [2K]

Answer:

One of the most useful use of expansion by heat is for swiches onmany tyes of machinery and home appliancesThemostats in the millions of cars are poerated by using the eaxpansion of metal Themometers arre another example,

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3 years ago

How can we search for evidence that other planets were once habitable?

Slav-nsk [51]

Answer:

Hey

Good question

How can we know if I planet was once inhabited by life? There diferent tequnecs for this. The first would be searching in the soil for fossilized microbes. in June of 19, 1976 Viking 1 was launched on its way to mars to began the search of life els wear. What this little lab did was drill a hole in the marsion terrain and take a sample of its soil to search for microbes or microbe fosles. Sadly it did not find anything. Other technics are to search for ruins on the planet or exoplanet. I do not have much to say about this method because it has never been used.

Hope it helped (i wrote all that).

Spiky Bob your answerer

Below is a picture of Carl Sagan next to the Viking Voyager 1. Obveosly this is just a modal. you could not breathe on mars.

4 0

3 years ago

A ball is shot from the ground straight up into the air with initial velocity of 42 ft/sec. Assuming that the air resistance can

Volgvan

Answer:

Maximum height of the ball, h(t) = 27.56 m

Explanation:

It is given that, a ball is shot from the ground straight up into the air with initial velocity of 42 ft/sec.

The height of the ball as a function of time t is given by :

$h(t)=h_o+v_ot-16t^2$

h₀ is initial height, h₀ = 0

So, $h(t)=42t-16t^2$ .........(1)

For maximum/minimum height, $\dfrac{dh(t)}{dt}=0$

$42-32t=0$ ...(2)

t = 1.31 s

Differentiating equation (2) wrt t

h''(t) = -32 < 0

So, at t = 1.31 seconds we will get the maximum height.

Put the value of t in equation (1)

$h(t)=42\times 1.31-16\times (1.31)^2$

h(t) = 27.56 m

Hence, this is the required solution.

7 0

4 years ago

Consider two gratings with the same slit separation. The only difference between the two gratings is that one grating has 5 slit

lora16 [44]

Answer:

b. Both the gratings will produce the same separation of peaks.

e. The grating with 10 slits will produce better-defined(narrower) peaks.

6 0

4 years ago

The wavelength of red helium-neon laser light in air is 632.8 nm.

e-lub [12.9K]

(a) $4.74\cdot 10^14 Hz$

The frequency of an electromagnetic wave is given by:

$f=\frac{c}{\lambda}$

where

$c=3.0\cdot 10^8 m/s$ is the speed of the wave in a vacuum (speed of light)

$\lambda$ is the wavelength

In this problem, we have laser light with wavelength

$\lambda=632.8 nm=6.33\cdot 10^{-7} m$ . Substituting into the formula, we find its frequency:

$f=\frac{3.0\cdot 10^8 m/s}{6.33\cdot 10^{-7} m}=4.74\cdot 10^14 Hz$

(b) 427.6 nm

The wavelength of an electromagnetic wave in a medium is given by:

$\lambda=\frac{\lambda_0}{n}$

where

$\lambda_0$ is the original wavelength in a vacuum (approximately equal to that in air)

$n$ is the index of refraction of the medium

In this problem, we have

$\lambda_0=632.8 nm$

n = 1.48 (index of refraction of glass)

Substituting into the formula,

$\lambda=\frac{632.8 nm}{1.48}=427.6 nm$

(c) $2.03\cdot 10^8 m/s$

The speed of an electromagnetic wave in a medium is

$v=\frac{c}{n}$

where c is the speed of light in a vacuum and n is the refractive index of the medium.

Since in this problem n=1.48, we find

$v=\frac{3\cdot 10^8 m/s}{1.48}=2.03\cdot 10^8 m/s$

3 0

4 years ago