# Laplace Transform, Transfer Functions, Calculate the new output when input changed

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If the initial conditions are zero, we can write the differential equation as $\frac{1}{\hat{h}(s)}\hat{y}(s) = \hat{f}(s)$, or equivalently, $\hat{y}(s) = \hat{h}(s) \hat{f}(s)$.

You have worked out that $\hat{h}(s) = \frac{2-s^2}{3(s+1)(s+2)^2}$, and for Part (c), you have $\hat{f}(s) = \frac{1}{s}$. Hence we have $\hat{y}(s) = \frac{2-s^2}{3s(s+1)(s+2)^2}$.

Inversion is easier if we expand $\hat{y}$ in terms of partial fractions, that is $\hat{y}(s) = \frac{1}{6 ( s+2 ) }-\frac{1}{3 ( s+2 ) ^{2}}-\frac{1}{3( s+1) }+\frac{1}{6 s}$, from which we get $y(t) = (\frac{1}{6}e^{-2t} - \frac{1}{3} t e^{-2t}-\frac{1}{3}e^{-t}+\frac{1}{6})u(t)$.

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### user59036

Updated on October 30, 2020

Consider the initial value problem for $0<t<∞$:

$ay′′+by′+cy=f(t)$,$y(0)=0$,$y′(0)=0$,

where $a$,$b$,$c$ are constants and $f(t)$ is a known function. We can view this problem as defining a linear system, where $f(t)$ is a known input and the corresponding solution $y(t)$ is the output. Laplace transforms of the input and output functions satisfy the multiplicative relation $Y(s)=H(s)F(s)$, where $H(s)$ is the system transfer function.

Suppose an input $f(t)=6t$, when applied to the linear system above, produces the output $y(t)=2(e^{−t}−1)+t(e^{−t}+1)$, $t≥0$.

$(a)$Find $Y(s)=\mathcal L\left\{y(t)\right\}$ and $F(s)=\mathcal L\left\{f(t)\right\}$.

$(b)$ Use your answer to part $(a)$ to find the system transfer function, $H(s)$.

$(c)$ Suppose the input $f(t)$ is changed. What will be the output if a Heaviside unit step input $f(t)=u(t)$ is applied to the system? Use the result from part $(b)$ to find the answer.

I have solved $(a)$ and $(b)$

$Y(s)=\dfrac{2}{s+1}-\dfrac{2}{s}+\dfrac{1}{(s+1)^2}+\dfrac{1}{s^2}$

$F(s) = \dfrac{6}{s^2}$

$H(s)=\dfrac{\dfrac{2}{s+1}-\dfrac{2}{s}+\dfrac{1}{(s+1)^2}+\dfrac{1}{s^2}}{\dfrac{6}{s^2}}$

But how should I use the value of $H(s)$ above to calculate the new $y(t)$ when the $f(t)$ changed from $6t$ to $u(t)$?

Shouldn't that be $\dfrac{2}{s-1}$?
Are you saying that the new $y(t) = \dfrac{1}{s-1}$?
No, in your $Y(s)$, the first term. I should have put a $2$ there to be clearer.
This is an online assignment, and it told me that the values I calculated for $Y(s)$ and $F(s)$ are correct. So it is$\dfrac{2}{s+1}$