I've the following matlab code written by myself, for the low pass filter and the input of x$($t$).$ How to get the output, in both time domain and frequency domain, in plot? If possible, how to get the frequency domain of the x$($t$)?$ How to get the function of the output in both time domain and frequency domain? All by Matlab. Will upvote if good answer, and vice versa, thank you.

$\%$ Parameters

f$0=20$e$3$; $\%$ Fundamental frequency $($Hz$)$

T$0=1/$ f$0$;

$\%$How to convert to fundamental angular frequency $($rad$/$s$)?$

omega$0=2*$ pi $*$ f$0$;

$\%$Define the x$-$axis,

t $=-2*$T$0$ : $1$e$-6$ : $2*$T$0$; $\%$ Time vector for $5$ periods

$\%$The dc component was obtained from manual solution

DC$\_$Component $=1/2$;

$\%$ Initialize the signal

$\%------------------------5$ harmonics$------------------------------------$

AC$\_$Component $=$ zeros$($size$($t$))$;

$\%$Since n $=1,3,5,7,...($Odd Harmonics$)$ until the number of harmonics wanted

$\%$Therefore, Loop from $1$ to X$,$ with increment of $2$ has syntax as shown

for n $=1$:$2$:$5$

$\%$The information below was copied directly from manual solution

bn $=2/($n $*$ pi$)$;

AC$\_$Component $=$ AC$\_$Component $+$ bn $*$ sin$($n $*$ omega$0*$ t$)$;

end

$\%$ Add the DC component here, because DC component is constant, and don't

$\%$ need to be looped! Only the AC component is looped

x$\_$t $=$ DC$\_$Component $+$ AC$\_$Component;

figure; $\%$This "figure" use only once, so that all subplots appear in $1$ fugure

subplot$(3,1,1)$

$\%$How to let the line to be blue, and line width to be $1\mathrm{.}5?$ so that it

$\%$is beautiful?

plot$($t$,$ x$\_$t$,\text{'}$r$\text{'},$ 'LineWidth', $1\mathrm{.}5)$;

title$(\text{'}$Fourier Series Approximation of x$($t$)\text{'})$;

xlabel$(\text{'}$Time $($s$)\text{'})$;

ylabel$(\text{'}$Amplitude$\text{'})$;

$\%$Since I want to show the y axis from $-1\mathrm{.}25$ to $1\mathrm{.}25,$ then following

$\%$syntax is used, because I want y$=0$ to be at center exactly like

$\%$the question given, it's put here not infront of "plot" or it

$\%$won't work

ylim$([-1\mathrm{.}251\mathrm{.}25])$;

$\%------------------------10$ harmonics$------------------------------------$

AC$\_$Component $=$ zeros$($size$($t$))$;

$\%$Since n $=1,3,5,7,...($Odd Harmonics$)$ until the number of harmonics wanted

$\%$Therefore, Loop from $1$ to X$,$ with increment of $2$ has syntax as shown

for n $=1$:$2$:$10$

$\%$The information below was copied directly from manual solution

bn $=2/($n $*$ pi$)$;

AC$\_$Component $=$ AC$\_$Component $+$ bn $*$ sin$($n $*$ omega$0*$ t$)$;

end

$\%$ Add the DC component here, because DC component is constant, and don't

$\%$ need to be looped! Only the AC component is looped

x$\_$t $=$ DC$\_$Component $+$ AC$\_$Component;

subplot$(3,1,2)$

plot$($t$,$ x$\_$t$,\text{'}$g$\text{'},$ 'LineWidth', $1\mathrm{.}5)$;

title$(\text{'}$Fourier Series Approximation of x$($t$)\text{'})$;

xlabel$(\text{'}$Time $($s$)\text{'})$;

ylabel$(\text{'}$Amplitude$\text{'})$;

ylim$([-1\mathrm{.}251\mathrm{.}25])$;

$\%------------------------20$ harmonics$------------------------------------$

AC$\_$Component $=$ zeros$($size$($t$))$;

$\%$Since n $=1,3,5,7,...($Odd Harmonics$)$ until the number of harmonics wanted

$\%$Therefore, Loop from $1$ to X$,$ with increment of $2$ has syntax as shown

for n $=1$:$2$:$20$

$\%$The information below was copied directly from manual solution

bn $=2/($n $*$ pi$)$;

AC$\_$Component $=$ AC$\_$Component $+$ bn $*$ sin$($n $*$ omega$0*$ t$)$;

end

$\%$ Add the DC component here, because DC component is constant, and don't

$\%$ need to be looped! Only the AC component is looped

x$\_$t $=$ DC$\_$Component $+$ AC$\_$Component;

subplot$(3,1,3)$

plot$($t$,$ x$\_$t$,\text{'}$b$\text{'},$ 'LineWidth', $1\mathrm{.}5)$;

title$(\text{'}$Fourier Series Approximation of x$($t$)\text{'})$;

xlabel$(\text{'}$Time $($s$)\text{'})$;

ylabel$(\text{'}$Amplitude$\text{'})$;

ylim$([-1\mathrm{.}251\mathrm{.}25])$;

$\%--------$Plotting the Low Pass Filter's Frequency Response Curve$----------$

$\%$ The Cutoff frequency given, in Hz

fc $=480$e$3$;

f$=0$: $10$e$3$ : $1200$e$3\%$This really need to trial and error to get the smooth curve

$\%$Inputting the transfer function derived manually,

$\%$Input log$10$ instead of log$10,$ it produces different result

gain $=20*$log$10(1./($sqrt$(1+($f$.^2/$fc$.^2))))$

figure;

plot$($f$,$ gain$)$;

title$(\text{'}$Frequency Response Curve of Low Pass Filter, of f$\_$c $=480$e$3\text{'})$;

xlabel$(\text{'}$Frequency $($kHz$)\text{'})$;

ylabel$(\text{'}$Gain $($dB$)\text{'})$;

grid on;

$\%--------------------------------------------------------------------------$

$\%$Since the fc should give a gain of $-3$ dB theoretically

$\%$To verify this, fc and $-3$dB line was plotted

$\%$ How to make matlab to display the gain, and fc calculated?

$\%\mathrm{.}6$f means $6$ floating decimal,

$\%"\%"$ is where the output would be

$\%$before $/$n means the unit, so that i don't need to type units

xline$($fc$,\text{'}--$r$\text{'},$ 'Cutoff Frequency $($f$\_$c$)\text{'},$ 'LabelVerticalAlignment', 'bottom'$)$;

yline$(-3,\text{'}--$r$\text{'},\text{'}-3$ dB$\text{'},$ 'LabelHorizontalAlignment' $,$'left'$)$;

$\%$Not used: below

$\%$fprintf$(\text{'}$Cutoff frequency $($fc$)$: $\%\mathrm{.}2$f Hz

$\text{'},$ fc$)$;

$\%$fprintf$(\text{'}$Time constant $($tau$)$: $\%\mathrm{.}6$f seconds

$\text{'},$ tau$)$;

$\%--------------------------------------------------------------------------$