I've the following matlab code written by myself, in the course of Signal and Systems Fourier Series. How to get the input x$($t$),$ in frequency domain with the theory of coefficients of fourier series being the frequency domain, $($Don$\text{'}$t use fourier transform to obtain the frequency domain because there are two methods to get frequency domain right?$).$ And obtain the low pass filter's output in frequency domain, Y$($f$)$ by using theory of output$\_$coefficient $=$ c$\_$k $*$ H$($k$*$$0),$ where c$\_$k is the coefficient of input signal. And then get the output in time domain, Y$($t$)$ back by using inverse fourier transform $($y$($t$)=[$c$\_$k $*$ H$($k$*$$0)]*$ e$^($j$*$k$*$$0*$t$))?$ Use MATBLAB. Will upvote if give good answer, and vice versa. Thank you.

$\%-----------------------$Defining all the parameters$------------------------$

$\%$ 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 $=1/2$;

$\%----------$Computing the input,in time domain, x$($t$)$ and its plot$-----------$

$\%$ Initialize the signal

AC$\_$Component$\_5=$ zeros$($size$($t$))$;

AC$\_$Component$\_10=$ zeros$($size$($t$))$;

AC$\_$Component$\_20=$ 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 $=1$:$2$:$50$

$\%5$ harmonics

if n $<=5$

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

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

AC$\_$Component$\_5=$ AC$\_$Component$\_5+$ bn $*$ sin$($n $*$ omega$0*$ t$)$;

end

if n $<=10$

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

AC$\_$Component$\_10=$ AC$\_$Component$\_10+$ bn $*$ sin$($n $*$ omega$0*$ t$)$;

end

if n $<=20$

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

AC$\_$Component$\_20=$ AC$\_$Component$\_20+$ bn $*$ sin$($n $*$ omega$0*$ t$)$;

end

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$\_5=$ DC$\_$Component $+$ AC$\_$Component$\_5$;

x$\_$t$\_10=$ DC$\_$Component $+$ AC$\_$Component$\_10$;

x$\_$t$\_20=$ DC$\_$Component $+$ AC$\_$Component$\_20$;

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 black, and line width to be $1\mathrm{.}5?$

$\%$so that it is beautiful?

plot$($t$,$ x$\_$t$\_5,\text{'}$k$\text{'},$ 'LineWidth', $1\mathrm{.}5)$;

title$(\text{'}$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 behind of each subplot or it

$\%$won't work!

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

subplot$(3,1,2)$

plot$($t$,$ x$\_$t$\_10,\text{'}$k$\text{'},$ 'LineWidth', $1\mathrm{.}5)$;

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

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

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

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

subplot$(3,1,3)$

plot$($t$,$ x$\_$t$\_20,\text{'}$k$\text{'},$ '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 input, in frequency domain, X$($w$)$ and its plot$--------$

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

$\%$Define the Low$-$Pass Filter Transfer function derived from my manual method

$\%$Magnitude only

H$\_$f$\_$magitude $=1./$ sqrt$(1+($f $/$ fc$).^2)$;

$\%$Conver to gain $($dB$)$

gain $=20*$log$10($H$\_$f$\_$magitude$)$;

$\%$Start another figure

figure;

$\%$I want the x$-$axis to have unit of scale of kHz$,$ so I make all the f to be

$\%$devided by $1000,$ or $"1$e$3"$

$\%$Since I want the curve to be black colour, black is $"$k$"$

plot$($f$/1$e$3,$ gain,$\text{'}$k$\text{'},$ 'LineWidth', $1\mathrm{.}5)$;

title$(\text{'}$Frequency Response Curve of Low Pass Filter'$)$;

$\%$This curve is actually a hard fact, every f$\_$c gives different curve!

$\%$But just depend on where is my f$\_0$

$\%$f$\_$c only, just plotting the transfer function only

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

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

$\%$Make the axis show from $0$ to $600$ khz only

xlim$([0600])$;

grid on;

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

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

$\%$Dashed line, so use $--,$ solid line, use $-,$ followed by the code of colour

$\%$Want the label alligned vertically $(90$ degree rotated$)?$ then put

$\%$'LabelVerticalAlignment' at behind,

$\%$Plot my f$\_$c line

xline$($fc $/1$e$3,\text{'}--$b$\text{'},$ 'Cutoff Frequency $($f$\_$c$)\text{'},$ 'LabelVerticalAlignment', 'bottom'$)$;

$\%$Plot my f$\_0$ line

xline$(20,\text{'}--$b$\text{'},$ 'Input Frequency $($f$\_0)\text{'},$ 'LabelVerticalAlignment', 'top'$)$;

$\%$Plot the $-3$ dB horizontal line

yline$(-3,\text{'}--$b$\text{'},\text{'}-3$ dB$\text{'},$ 'LabelHorizontalAlignment' $,$'center'$)$;

$\%-------------------$Computing the output and its plot$----------------------$