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Alexxandr [17]
3 years ago
8

Consider the cascade of the three LTI systems having impulse responses: h-1(t) = e^-tu(t + 3) h_2(t) = rect((1 -1)/2) h_3(t) = d

elta(t - 3) What is the equivalent impulse response of the resulting system?

Engineering
1 answer:
omeli [17]3 years ago
8 0

Explanation:

There are two ways to find out the equivalent impulse response of the system.

1. Convolution in time domain

2. Simple multiplication in Laplace domain

2nd method is efficient, easy and is less time consuming.

Step 1: Take the Laplace transform of the given three impulse response functions to convert time domain signals into s-domain

Step 2: Once we get signals in s-domain, multiply them algebraically to get the equivalent s-domain response.

Step 3: Take inverse Laplace transform of the equivalent impulse response to convert from s-domain into time domain.

Solution using Matlab:

Step 1: Take Laplace Transform

Ys1 =  1/(s + 1)

Ys2 =  1/s - exp(-s/2)/s

Ys3 =  exp(-3*s)

Step 2: Multiplication in s-domain

Y =  (exp(-(7*s)/2)*(exp(s/2) - 1))/(s*(s + 1))

Step 3: Inverse Laplace Transform (Final Solution in Time Domain)

h =  heaviside(t - 7/2)*(exp(7/2 - t) - 1) - heaviside(t - 3)*(exp(3 - t) - 1)

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Answer:

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2 years ago
A specimen of a 4340-steel alloy with a plane strain fracture toughness of 54.8 MPa sqrt(m) (50 ksi sqrt(in.)) is exposed to a s
Lunna [17]

Answer:

critical stress \sigma _c = 1382.67 MPa

Explanation:

given data

plane strain fracture toughness = 54.8 MP

length of surface creak = 0.5 mm

we take here

parameter Y = 1.0

solution

we apply critical stress formula that is

critical stress \sigma _c = \frac{K}{Y\sqrt{\pi \times a} }   .............................1

here K is design stress plane strain fracture toughness and a is length of surface creak so put all these value in equation 1

critical stress \sigma _c  =  \frac{54.8 \times 10^6}{1 \sqrt{\pi \times 5 \times 106{-4}}}    

solve it we get

critical stress  = 1382.67 MPa

As exposed stress 1030 MPa is less than critical stress 1382 MPa

so that fracture will not be occur here

7 0
3 years ago
Tech A says that a transistor has a single P–N junction. Tech B says that a transistor is a semiconductor device used as a switc
Aloiza [94]

Answer:

Both Technician A and technician B are correct.

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A transistor also can be described as a semiconductor which acts as a switch and can be used to amplify currents. Transistors are very key and vital to electronic devices especially the mobile phones in recent times, it helps to ensure that electronic systems perform optimally.

The charges in the P-N junction is controlled by the availability of Positive and negative electrons.

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2 years ago
BRAINLIAST IF ANSWERED DETAILED AND PRECISE
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4 0
3 years ago
Link BD consists of a single bar 36 mm wide and 18 mm thick. Knowing that each pin has a 12-mm diameter, determine the maximum v
MAXImum [283]

Answer:

hello the diagram attached to your question is missing attached below is the missing diagram

answer :

a) 48.11 MPa

b) - 55.55 MPa

Explanation:

First we consider the equilibrium moments about point A

∑ Ma = 0

( Fbd * 300cos30° ) + ( 24sin∅ * 450cos30° ) - ( 24cos∅ * 450sin30° ) = 0

therefore ;<em> Fbd = 36 ( cos ∅tan30° - sin∅ ) kN  ----- ( 1 )</em>

A ) when ∅ = 0

Fbd = 20.7846 kN

link BD will be under tension when ∅ = 0, hence we will calculate the loading area using this equation

A = ( b - d ) t

b = 12 mm

d = 36 mm

t = 18

therefore loading area ( A ) = 432 mm^2

determine the maximum value of average normal stress in link BD  using the relation below

бbd = \frac{Fbd}{A}  = 20.7846 kN / 432 mm^2  =  48.11 MPa

b) when ∅ = 90°

Fbd = -36 kN

the negativity indicate that the loading direction is in contrast to the assumed direction of loading

There is compression in link BD

next we have to calculate the loading area using this equation ;

A = b * t

b = 36mm

t = 18mm

hence loading area = 36 * 18 = 648 mm^2

determine the maximum value of average normal stress in link BD  using the relation below

бbd = \frac{Fbd}{A} = -36 kN / 648mm^2 = -55.55 MPa

4 0
2 years ago
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