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scZoUnD [109]
3 years ago
8

Use the convolutional integral to find the response of an LTI system with impulse response ℎ(????) and input x(????). Sketch the

output signal. a) x(????) = ????(????) and ℎ(????) = ????????(????) b) x(????) = ????????c???? ( ???? 2???? ) and ℎ(????) = exp(−????????) ????(????) (You may write ????????c???? ( ???? 2???? ) as ????(???? + ????) − ????(???? − ????).)
Engineering
1 answer:
Thepotemich [5.8K]3 years ago
8 0

Answer:

Use the convolution integral to determine the output of an LTI system with impulse response h(t) to input x(t) specified below.

h(t) = 4e^-6t-1 u(t+1) x(t)=2u(t – 2)

Explanation:

y(t)=∫0th(t−λ)f(λ)dλ

Start with a linear time-invariant (LTI) system (in box).  There is an input to the system, x(t), and an output from the system, y(t).

Now consider the same system with the input as an impulse, δ(t).  The output is then, by definition, the impulse response, h(t).

If we use a delayed impulse into the system, the output is just the impulse response with the same delay.

Because of linearity, if we scale the input by any factor, the output will be scaled by the same factor.  In particular, we can scale the input (and hence the output) by the factor x(λ)dλ.

If we integrate the input, the output is also integrated.  In this case we'll take the upper limit of the integration to be t+; but it could just as well have been positive infinity.  We could also take the lower lower limit to be negative infinity.

By the sifting property of the impulse function we know that (if λ>0, t>0):

     

∫0tδ(t−λ)x(λ)dλ=x(t)

Therefore the input to the system is x(t)

If the input is x(t), then the output must be y(t) (by definition, from the first diagram).  

Equating the outputs of the last two diagrams leads directly to the convolution theorem:

y(t)=∫0th(t−λ)f(λ)dλ

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Explanation:

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The yield strength of mild steel is 150 MPa for an average grain diameter of 0.038 mm ; yield strength is 250 MPa for average gr
djyliett [7]

Answer:

Explanation:

Hall-Petch equation provides direct relations between the strength of the material and the grain size:

σ=σ0+k/√d , where d- grain size, σ- strength for the given gran size, σ0 and k are the equation constants.

As in this problem, we don't know the constants of the equation, but we know two properties of the material, we are able to find them from the system of equations:

σ1=σ0+k/√d1

σ2=σ0+k/√d2 , where 1 and 2 represent 150MPa and 250MPa strength of the steel.

Note, that for the given problem, there is no need to convert units to SI, as constants can have any units, which are convenient for us.

From the system of equations calculations, we can find constant: σ0=55.196 MPa, k=18.48 MPa*mm^(0.5)

Now we are able to calculate strength for the grain diameter of 0.004 mm:

σ=55.196+18.48/(√0.004)=347.39 MPa

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Wet steam at 15 bar is throttled adiabatically in a steady-flow process to 2 bar. The resulting stream has a temperature of 130°
cricket20 [7]

Answer:

\Delta s = 0.8708\,\frac{kJ}{kg\cdot K}

Explanation:

The adiabatic throttling process is modelled after the First Law of Thermodynamics:

m\cdot (h_{in} - h_{out}) = 0

h_{in} = h_{out}

Properties of water at inlet and outlet are obtained from steam tables:

State 1 - Inlet (Liquid-Vapor Mixture)

P = 1500\,kPa

T = 198.29\,^{\textdegree}C

h = 2726.9\,\frac{kJ}{kg}

s = 6.3068\,\frac{kJ}{kg\cdot K}

x = 0.967

State 2 - Outlet (Superheated Vapor)

P = 200\,kPa

T = 130\,^{\textdegree}C

h = 2726.9\,\frac{kJ}{kg}

s = 7.1776\,\frac{kJ}{kg\cdot K}

The change of entropy of the steam is derived of the Second Law of Thermodynamics:

\Delta s = 7.1776\,\frac{kJ}{kg\cdot K} - 6.3068\, \frac{kJ}{kg\cdot K}

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