# Homework Help: Laplace Transforms: Transfer Functions, and IVT/FVT Problems

1. Sep 26, 2015

### ConnorM

1. The problem statement, all variables and given/known data
I uploaded the problem statements as a picture as well. I have completed these and was wondering if someone could check my work, and let me know if it is correct.

Problem 1.3:
Find the expression for the transfer function of this linear time-invariant causal system with input $x(t)$ and output $y(t)$.

$H(s) = Y(s) / X(s)$

$y^{'''} + 5y^{''} + 3y^{'} + y = x^{''} + 4x^{'} + 7x$

Problem 1.4:
Consider the following Laplace Transforms:

1. $F_1 (s) = {\frac{2(s+2)}{s(s+1)(s+3)}}$

2. $F_2 (s) = {\frac{s}{s^2 + 2s +10}}$

(a) Determine the inverse laplace transform for each function.

(b) Determine the initial value $f(0^+)$ (if it exists) for each function using the initial value theorem (IVT).

(c) Determine the final value $f(\infty)$ (if it exists) for each function using the final value theorem (FVT).

(d) Show that the results obtained from using the IVT and FVT are the same as those obtained by directly
evaluating the inverse Laplace transforms.

2. Relevant equations

IVT - $x(0^+) = \displaystyle\lim_{t\rightarrow 0^+} x(t) = \displaystyle\lim_{s\rightarrow \infty} sX(s)$

FVT - $x(\infty) = \displaystyle\lim_{t\rightarrow \infty} x(t) = \displaystyle\lim_{s\rightarrow 0} sX(s)$

3. The attempt at a solution

Problem 1.3:

$s^{3}Y(s) + 5s^{2}Y(s) + 3sY(s) + Y(s) = s^{2}X(s) + 4sX(s) + 7X(s)$

$H(s) = Y(s) / X(s) = {\frac{s^2 + 4s + 7}{s^3 + 5s^2 + 3s + 1}}$

Problem 1.4:

Here are my inverse Laplace transforms,

1. $f_1 (t) = 4/3 - e^{-t} - e^{-3t}/3$

2. $f_2 (t) = e^{-t}cos(3t) - e^{-t}sin(3t)/3$

IVT for 1.

$F_1 (\infty) = 0$, $f_1 (0^+) = 0$

FVT for 1.

$F_1 (0) = 4/3$, $f_1 (\infty) = 4/3$.

IVT for 2.

$F_2 (\infty) = 1$, $f_2 (0) = 1$

FVT for 2.

$F_2 (0) = 0$, $f_2 (\infty) = 0$

Overall I found that the initial values and final value was the same regardless of the domain of the function.

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2. Sep 26, 2015

### axmls

Looks right to me!