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2 years ago

How would you achieve the linear convolution of a 100 sample time series and a 20 tap filter in the frequency domain?

Engineering

1 answer:

Sindrei [870]2 years ago

4 0

Answer:

divide then add XD my guy this is easy

Explanation:

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An engineer is testing the shear strength of spot welds used on a construction site. The engineer's null hypothesis at a 5% leve

lilavasa [31]

Answer:

b) The null hypothesis should be rejected.

Explanation:

The null hypothesis is that the mean shear strength of spot welds is at least

3.1 MPa

H0: u ≥3.1 MPa against the claim Ha: u< 3.1 MPa

The alternate hypothesis is that the mean shear strength of spot welds is less than 3.1 MPa.

This is one tailed test

The critical region Z(0.05) < ± 1.645

The Sample mean= x`= 3.07

The number of welds= n= 15

Standard Deviation= s= 0.069

Applying z test

z= x`-u/s/√n

z= 3.07-3.1/0.069/√15

z= -0.03/0.0178

z= -1.68

As the calculated z= -1.68 falls in the critical region Z(0.05) < ± 1.645 the null hypothesis is rejected and the alternate hypothesis is accepted that the mean shear strength of spot welds is less than 3.1 MPa

8 0

2 years ago

A steel pipe of 400-mm outer diameter is fabricated from 10-mm-thick plate by welding along a helix that forms an angle of 20° w

Verdich [7]

Explanation:

Outer di ameter $d_{0}=400 \mathrm{mm}[tex] Thickness of the cylinder [tex]t=10 \mathrm{mm}$

$\therefore[tex] Inner diam eter [tex]d_{i}=d_{0}-2 t=400-2 \times 10$

$d_{1}=380 \mathrm{mm}$

Given loading on the cylinder $P=300 \mathrm{kN}$ Helix an gle of the weld form $\theta=20^{\circ}$

(i) Normal stress on the plane at angle $\theta=20^{\circ}$ is

$\sigma=\frac{P \cos ^{2} \theta}{A_{0}}$

$\text { Where } A_{0}=\frac{\pi}{4}\left(d_{0}^{2}-d_{1}^{2}\right)$

$\quad=\frac{\pi}{4}\left(400^{2}-380^{2}\right)$

$=12252.21 \mathrm{mm}^{2}$

$=12.25221 \times 10^{-9} \mathrm{m}^{2}$

$\sigma=\frac{-300 \times 10^{2} \times \cos ^{2} 20}{12.25221 \times 10^{-1}}$

$=-21.6 \mathrm{MPa}$

(ii) Shear stress along an angle of $\theta=20^{\circ}$ is $\tau=\frac{P}{A_{0}} \cos \theta \sin \theta$

$=\frac{-300 \times 10^{-1} \times \cos 20 \times \sin 20}{12.25221 \times 10^{-3}}$

$=-7.86 \mathrm{MPa}$

3 0

2 years ago

Someone claims that in fully developed turbulent flow in a tube, the shear stress is a maximum at the tube surface. Is this clai

Alika [10]

Answer:

Yes this claim is correct.

Explanation:

The shear stress at any point is proportional to the velocity gradient at any that point. Since the fluid that is in contact with the pipe wall shall have zero velocity due to no flow boundary condition and if we move small distance away from the wall the velocity will have a non zero value thus a maximum gradient will exist at the surface of the pipe hence correspondingly the shear stresses will also be maximum.

5 0

3 years ago

I’m building a pc, but I only have 2 sata cables. What are the other required cables to connect to or parts, such as the gpu?

Usimov [2.4K]

Answer:

You would also need power cables.

3 0

2 years ago

Read 2 more answers

Suppose that all the dislocations in 3000 mm3 of crystal were somehow removed and linked end to end. Given 1 m =0.0006214 mile,

zalisa [80]

Answer: