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EE C128 / ME C134 Spring 2014 HW1 - Solutions

UC Berkeley

Homework 1 - Solutions 1. Deriving Laplace Transforms Derive the Laplace transforms of the following time functions using the definition of the Laplace transform. Do not simply just look them up in a table. (a) δ(t) = impulse Solution: Z

∞

L(δ(t)) =

δ(t)e−st dt = 1

0−

(b) sin(ωt)u(t) Solution: Z

∞

L(sin(ωt)u(t)) =

sin(ωt)u(t)e−st dt

0−

∞ e−st (s sin(ωt) − ω cos(ωt)) 2 2 s +ω 0 ω = 2 s + ω2 =

2. Using Laplace Transform Pairs Using Laplace transform pairs in Table 2.1 and theorems in Table 2.2 in the book of Nise, derive the Laplace transforms for the following time function: (a) e−at cos(ωt)u(t) Solution: Using the Frequency Shift Theorem and the Laplace Transform of cos(ωt), we get L(e−at cos(ωt)u(t)) =

s+a (s + a)2 + ω 2

3. Solving Differential Equations Using Laplace Transforms Solve the following differential equation using Laplace transforms. Assume all forcing functions are zero prior to t = 0− . (Hint: you will need to use partial fraction decomposition) d2 x dx (a) +6 + 8x = 5 sin 3t 2 dt dt x(0) = 4, x0 (0) = 1 Solution: Taking the Laplace Transform with the given initial conditions, we get s2 X(s) − 4s − 1 + 6(sX(s) − 4) + 8X(s) = 5

s2

3 +9

Solving for X(s), we get X(s) =

Rev. 1.0, 02/01/2014

4s3 + 25s2 + 36s + 240 (s2 + 9)(s2 + 6s + 8) 1 of 3

EE C128 / ME C134 Spring 2014 HW1 - Solutions

UC Berkeley

Using Partial Fraction expansion, X(s) =

−3 118 24 −18s + + − 2 2 65(s + 9) 65(s + 9) 13(s + 2) 5(s + 4)

L−1 (X(s)) = x(t) =

−18 1 118 −2t 24 −4t cos(3t) − sin(3t) + e − e 65 65 13 5

4. Differential Equation To Transfer Function in Laplace Domain A system is described by the following differential equation (see below). Find the expression for the transfer function of the system, Y (s)/X(s), assuming zero initial conditions. d3 y d2 y dy d3 x d2 x dx (a) + 3 + 5 + y = + 4 +6 + 8x dt3 dt2 dt dt3 dt2 dt Solution: The Laplace Trasform assuming zero initial conditions is given by s3 Y (s) + 3s2 Y (s) + 5sY (s) + Y (s) = s3 X(s) + 4s2 X(s) + 6sX(s) + 8X(s) Y (s)[s3 + 3s2 + 5s + 1] = X(s)[s3 + 4s2 + 6s + 8] s3 + 4s2 + 6s + 8 Y (s) = 3 X(s) s + 3s2 + 5s + 1

5. Transfer Function Review Write the corresponding differential equation for the following transfer function: s+3 X(s) = 3 (a) F (s) s + 11s2 + 12s + 18 Solution: X(s)[s3 + 11s2 + 12s + 18] = F (s)[s + 3] s3 X(s) + 11s2 X(s) + 12sX(s) + 18X(s) = sF (s) + 3F (s) d3 x d2 x dx df + 11 + 12 + 18x = + 3f 3 2 dt dt dt dt 6. Electrical networks Find the transfer function, G(s) = VL (s)/V (s) for the following network: Solution:

Rev. 1.0, 02/01/2014

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EE C128 / ME C134 Spring 2014 HW1 - Solutions

UC Berkeley

Writing mesh equations (2s + 2)I1 (s) − 2I2 (s) = Vi (s)

(1)

−2I1 (s) + (2s + 4)I2 (s) = 0

(2)

From Eq(2), I1 (s) = (s + 2)I2 (s). Substituting this in Eq(1) yields (2s + 2)(s + 2)I2 (s) − 2I2 (s) = Vi (s) ⇒

I2 (s) 1 = 2 Vi (s) 2s + 6s + 2

But VL (s) = 2sI2 (s) ∴

Rev. 1.0, 02/01/2014

VL (s) s = 2 Vi (s) s + 3s + 1

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