# Find Fourier transform and plot spectrum by hand & MATLAB

## Homework Equations

##\omega=2\pi t##
Fourier: ## Y(f)=\int ^{\infty}_{-\infty}y(t)\mathrm{exp}(-j\omega t)dt##
Linearity Property: ##ay_1(t)+by_2(t)=aY_1(f)+bY_2(f)##, where a and b are constants
Scaling Property: ##y(at)=\frac{1}{|a|}G(\frac{f}{a})##, where a is constant

## The Attempt at a Solution

From the hint, I think I need to find the Fourier transform of the function for each region:
##(-2\leq t\leq -1)##: ##Y(f)=te^{-j\omega t}+\frac{j}{\omega}(e^{j\omega}-e^{j2\omega})##
##(-1\leq t\leq 1)##: ##Y(f)=\frac{j}{\omega}(e^{-j\omega}-e^{j\omega})##
##(1\leq t\leq 2)##: ##Y(f)=-te^{-j\omega t}+\frac{3j}{\omega}(e^{-j2\omega}-e^{-j\omega})##
Please let me know if you want to see my work.

Why do I need to use the scaling property and how do I use it?

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Delta2
Homework Helper
Gold Member
First of all, fourier transform is an integral from (-infinity) to (+infinity) and because you function is defined in that way and it is zero in "most" of real line, the fourier transform will be the sum of 3 different integrals (one integral from -2 to -1, one from -1 to +1, and one from +1 to +2, the integral from -infinity to -2 is zero, and zero is also the integral from +2 to +infinity). So , even when our function is defined in a segmented way, the fourier transform is not segmented (we wont have 3 different fourier transforms as you wrote, it is one transform (integral) that is the sum of 3 smaller integrals as I said above), so it doesn't mean that we find the fourier transform in each segment, rather we work on the whole fourier transform and break up the big integral as necessary.

Second the hint says to use the triangular function as basis. How is the triangular function being defined in your textbook?

How is the triangular function being defined in your textbook?
The book I am using is "Introduction to Analog & Digital Communications" by Simon Haykin & Michael Moher, 2nd ed. They define the Fourier transform of a triangular pulse as ##AT^2\mathrm{sinc}^2(fT)##. That is all I could find in the book. However, I found the following definition of the unit triangle function online: ##\mathrm{tri}(t) = \begin{cases}1-|t| & |t| <1\\0 & |t|\geq 1\end{cases}##.

Delta2
Homework Helper
Gold Member
Ok i believe it is ##y(t)=2tri(\frac{t}{2})-tri(t)##. Can you verify this? If it is indeed true then all you have to use is linear and scaling properties of fourier transform, to get the fourier transform of y(t), knowing the transform of tri(t).

Last edited:
Ok i believe it is ##y(t)=2tri(\frac{t}{2})-tri(t)##. Can you verify this?
Well, ##y(t)=2\mathrm{tri}(\frac{t}{2})-\mathrm{tri}(t)## is equivalent to a constant +1. So, that matches the given function over the region ##-1\leq t\leq 1## only.

To be honest, I'm stuck. I have not found a transform that agrees with the given function.

Is the following correct for defining y(t) in terms of the triangular function? ##y(t) = \begin{cases}2(\mathrm{tri}(\frac{t}{2})) & -2\leq t\leq -1\\2(\mathrm{tri}(\frac{t}{2}))-\mathrm{tri}(t) & -1\leq t\leq 1\\2(\mathrm{tri}(\frac{t}{2})) & 1\leq t\leq 2\end{cases}##

Delta2
Homework Helper
Gold Member
Well, ##y(t)=2\mathrm{tri}(\frac{t}{2})-\mathrm{tri}(t)## is equivalent to a constant +1. So, that matches the given function over the region ##-1\leq t\leq 1## only.
it is +1 in [-1,1]. But look what happens for example in [-2,-1]

##-2\leq t \leq-1\implies tri(t)=0 , -1\leq \frac{t}{2}\leq\frac{-1}{2} \implies tri(t)=0 , tri(\frac{t}{2})=1-\frac{|t|}{2}\implies 2tri(\frac{t}{2})-tri(t)=2+t-0##
I believe if u take the other cases (-infinity,-2],[1,2],[2,+infinity] you ll get the desired result.
To see it more clearly make the graph for ##2tri(t/2)## and on the same graph also draw ##tri(t)##.

Delta2
Homework Helper
Gold Member
Is the following correct for defining y(t) in terms of the triangular function? ##y(t) = \begin{cases}2(\mathrm{tri}(\frac{t}{2})) & -2\leq t\leq -1\\2(\mathrm{tri}(\frac{t}{2}))-\mathrm{tri}(t) & -1\leq t\leq 1\\2(\mathrm{tri}(\frac{t}{2})) & 1\leq t\leq 2\end{cases}##
this is correct, but then again no need to take cases because tri(t)=0 outside [-1,1].

tri(t)=0 outside [-1,1].
I think I realize my mistake. When defined in terms of the triangular function, the given function is indeed ##y(t)=2tri(\frac{t}{2})-tri(t)##.
FT of ##\mathrm{tri}(t)##: ##\mathrm{sinc}^2(f)##
FT of ##\mathrm{tri}(\frac{t}{2})##: ##\mathrm{sinc}^2(2f)##
Therefore FT of y(t): ##2\mathrm{sinc}^2(2f)-\mathrm{sinc}^2(f)##
Which produces the following graph: