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Interview Question :: Laplace Transform Question: Consider an LCR circuit arranged in series


I'm trying to solve the question below:

Consider an LCR circuit arranged in series such that L, C, and R are the values of inductance, capacitance and resistance respectively. The voltage across the capacitor satisfies the ordinary differential equation:

(LC)d²Vc/dt² + (RC)dVc/dt + Vc = V(t)

where V(t) is the supply voltage.

for LC = 1/2 and RC = 1, use the method of Laplace transforms to solve the equation above when the supply voltage is given by:

V(t) = Vo(d(t-b) + 2H(t-b))

where b > 0 and the circuit is initiated according to Vc(0) = 1, Vc'(0) = -1. What are the values of Vc(t) at t = b/2 and t = 2b?

I've been trying to do this question for some time now and haven't got anywhere! Any help would be greatly appreciated!

Thanks :)

Malcom
Answers to "Laplace Transform Question: Consider an LCR circuit arranged in series"
RE: Laplace Transform Question: Consider an LCR circuit arranged in series?

(1/2) · d²Vc/dt² + dVc/dt + Vc = V0( d(t-b) + 2H(t-b) )

transform the whole equation

<=>

L{ (1/2) · d²Vc/dt² + dVc/dt + Vc } = L{ V0( d(t-b) + 2H(t-b) ) }

<=>

(1/2)·L{d²Vc/dt²} + L{dVc/dt} + L{Vc} = V0·L{d(t-b)} + 2·V0·L{H(t-b)}



For LHS use differentiation rule:

L{df/dt} = s·L{f} - f(0)

L{d²f/dt²} = s·L{df/dt} - f'(0) = s²·L{f} - s·f(0) - f'(0)

supposed d on RHS denotes Dirac delta function

L{d(t-b)} = exp{-b·s}

supposed H on RHS denotes Heaviside step function

L{H(t-b)} = exp{-b·s}/s



Hence:

(s²/2)·L{Vc} - s·Vc(0) - Vc'(0) + s·L{Vc} - Vc(0) + L{Vc} = V0·exp{-b·s} + 2·V0·exp{-b·s}/s

<=>

(s²/2)·L{Vc} - s + 1 + s·L{Vc} - 1 + L{Vc} = V0·exp{-b·s}·(1 + 2/s)

<=>

(s² + 2s + 2)·L{Vc} = 2·V0·exp{-b·s}·(s + 2)/s + 2·s

<=>

L{Vc} = 2·V0·exp{-b·s}·(s + 2)/[s·(s² + 2s + 2)] + 2·s/(s² + 2s + 2)

=>

Vc = L?¹{2·V0·exp{-b·s}·(s + 2)/[s·(s² + 2s + 2)] + 2·s/(s² + 2s + 2)}

= 2·V0·L?¹{exp{-b·s}·(2s + 4)/[s·(s² + 2s + 2)]} + 2·L?¹{s/(s² + 2s + 2)}



expand non-exponential terms in first transform to partial fractions:

(s + 2)/[s·(s² + 2s + 2)] = A/s + (Bs + C)/(s² + 2s + 2)

<=>

(s + 2) = A·(s² + 2s + 2) + (Bs + C)·s

<=>

s + 2 = s²·(A+B) + s·(2A + C) + 2A

compare coefficients of same power of s

A + B = 0

2A + C = 1

2A = 2

=>

A = 1

B = -1

C = -1

=>

(2s + 4)/[s·(s² + 2s + 2)] = 2/s - (2s + 2)/(s² + 2s + 2)

hence:

Vc = 2·V0·L?¹{exp{-b·s}/s} - 2·V0·L?¹{exp{-b·s}·(s + 1)/(s² + 2s + 2)} + 2·L?¹{s/(s² + 2s + 2)}





To keep it comprehensible continue piecewise:

(i) 2·V0·L?¹{exp{-b·s}/s}

This term is simple. A shown above exp{-b·s}/s is the transform of a heaviside step function

2·V0·L?¹{exp{-b·s}/s} = 2·V0·H(t-b)



(ii) 2·L?¹{s/(s² + 2s + 2)}

You can convert such a transform to that of a frequency shifted sine and/or cosine, which are:

L{sin(?·t)} = ?/(s² + ?²)

=>L{exp{-a·t)·sin(?·t)} = ?/((s+a)² + ?²)

and

L{cos(?·t)} = s/(s² + ?²)

=> L{exp{-a·t)·cos(?·t)} = (s+a)/((s+a)² + ?²)

use quadratic supplement to convert denominator

L?¹{s/(s² + 2s + 2)}

= L?¹{s/((s + 1)² + 1)}

= L?¹{(s+1 - 1)/((s + 1)² + 1)}

= L?¹{(s+1)/((s + 1)² + 1)} - L?¹{ 1/((s + 1)² + 1)}

= exp{-t)·cos(t) - exp{-)·sin(t)

= exp{-t)·[cos(t) - sin(t)]



(iii) - 2·V0·L?¹{exp{-b·s}·(s + 1)/(s² + 2s + 2)}

For a moment ignore the exponential term, because it just represents a time shift. The inverse transform of the rest can by found by same method as in (ii)

L?¹{(s + 1)/(s² + 2s + 2)}

= L?¹{(s + 1)/((s+1) + 1)}

= exp{-t)·cos(t)

To take exponential term into account use time shift property:

L{f(t-b)·H(t-b)} = exp(-b·t)·L{f(t)}



So all you have to do is to replace t in the calculated inverse transform by t-b and multiply the result by Heavside step function.

-2·V0·L?¹{exp{-b·s}·(s + 1)/(s² + 2s + 2)}

=- 2·V0·exp{-(t-b))·cos(t-b)·H(t-b)





So the solution of the IVP is:

Vc = 2·V0·H(t-b) - 2·V0·exp{-(t-b))·cos(t-b)·H(t-b) + exp{-t)·[ cos(t) - sin(t) ]

= 2·V0·H(t-b)·[1 - exp{-(t-b))·cos(t-b)] + exp{-t)·[cos(t) - sin(t)]





H(t-b) = 0 for t <b

so

Vc(b/2) = exp{-b/2)·[cos(b/2) - sin(b/2)]



H(t-b) = 1 for t>b

so

Vc(2b) = 2·V0·[1 - exp{-b))·cos(b)] + exp{-2b)·[cos(2b) - sin(2b)]
 
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