Answer all parts in MATLAB code.

Laplace transform and transfer function

A system's step response is $y(t)=[1-e^{-4t}-0\mathrm{.}2cos(4t)]u(t).$

Part $1$: Step response and transfer function

$1$a$)$ Use the symbolic math toolbox to express the step response and assign to variable $y.$

$1$b$)$ Plot the step response. The plot command cannot be used since $y$ is a symbolic variable, but fplot can plot a function or expression.

$1$c$)$ Use the symbolic laplace function to find the Laplace transform of $y(t)$ and assign the result to variable $Y.$

$1$d$)$ Use the symbolic simplifyFraction function to merge the partial fractions into a single $Y(s)=\frac{-s^3+16s^2+320}{5s(s^2+16)(s+4)}$ saved in variable Ys$.$

$1$e$)$ Find the system transfer function $H(s)=\frac{Y(s)}{x(s)}=\frac{Y(s)}{\frac{1}{s}}=ss(s).$ Assign the result to variable $H.$

Part $2$: Finding the system's response to a cosine input, using the transfer function

$2$a$)$ Find the system's response to the input $xz(t)=2cos(3t)u(t).$ The Laplace transform of $x2(t)$ should be saved in variable $x2.$

The Laplace transformed output is Y$2.$

$Y_2(s)=x_2(s)H(s)$ because $H(s)=\frac{Y(s)}{x(s)}=\frac{Y_2(s)}{x_2(s)}.$

$2$b$)$ Find the output in time domain using the ilaplace function, saved as $y2.$

$2$c$)$ Plot the output $y2$ using fplot

$2$d$)$ What is the maximum value of $y2?$ Save the value to the variable y$2$Max.

Note: the max $()$ function cannot be used on $22$ directly, without converting the symbolic $y2$ to remove the variable t$.$ But

double $(max(subs(y2,t,T)))$

does the conversion, subs $()$ substitutes the numbers in vector $T$ for the symbolic $t$ and evaluates $y2$ to a vector of symbolic expressions. Then max $()$ finds the symbolic max value, and double$()$ evaluates the expression to a number. $T$ can be $0$ to $10$ seconds with a time step of $0\mathrm{.}05$ seconds.

$2$e$)$ What is the period of $y2?$ The period can be found either by calculation or by inspecting the plot. Call the value y$2$Period.