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Alexxandr [17]
4 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]4 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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An immersion heater has a resistance of 50Ω and carries a current of 2.5A current. What will be the final temperature of 500 g o
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The diffusion coefficients for species A in metal B are given at two temperatures:
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Answer:

a) 149 kJ/mol, b) 6.11*10^-11 m^2/s ,c) 2.76*10^-16 m^2/s

Explanation:

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I will be using R in the equation instead of k_b as the problem asks for molar activation energy

I will be using

R = 8.314\ J/mol*K

and

°C + 273 = K

here, adjust your precision as neccessary

Since we got 2 difusion coefficients at 2 temperatures alredy, we can simply turn these into 2 linear equations to solve for a) and b) simply by taking logarithm

So:

ln(6.69*10^{-17})=ln(D_0) -\frac{Q_d}{R*(1030+273)}

and

ln(6.56*10^{-16}) = ln(D_0) -\frac{Q_d}{R*(1290+273)}

You might notice that these equations have the form of  

d=y-ax

You can solve this equation system easily using calculator, and you will eventually get

D_0 =6.11*10^{-11}\ m^2/s\\ Q_d=1.49 *10^3\ J/mol

After you got those 2 parameters, the rest is easy, you can just plug them all   including the given temperature of 1180°C into the Arrhenius equation

6.11*10^{-11}e^{\frac{149\ 000}{8.143*(1180+273)}

And you should get D = 2.76*10^-16 m^/s as an answer for c)

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