I came across this Fourier Transform question when revising for an upcoming test. The question asked me to determine the Fourier transform of the 'on-off' pulse as seen in the attached photo. Fourier Transform Question

It can be seen that Euler's formula was used to simplify the original expression in the second line as it is an odd function, hence you could just integrate one side and multiply it by two.

Euler's Formula states that,

$Ae^{-j\omega t}=A(\cos (\omega t)-j \sin( \omega t))$ and

$-Ae^{-jωt}=A(-\cos⁡(\omega t)+j \sin( \omega t))$

However, what I am confused about is why is the cosine term negated in the second step of integration? Is it because the on-off pulse is an odd function, hence there are no cosine terms? (Is it similar to how you do a Fourier series question, where an even function means that $b_n=0$, while an odd function means that $a_0=a_n=0$.

P.S (Apologies about the image, I can't embed images yet)

  • $\begingroup$ Hint: what is $A\,e^{j \omega t}-A\,e^{-j \omega t}$? $\endgroup$ Commented Apr 10, 2022 at 14:30

1 Answer 1


Here's the linked image of the problem in the question above.

Image of problem

It's not "just integrate one side and multiply it by two".

The Fourier transform of

$$f(t)=\left\{ \begin{array}{cc} A & -T<t<0 \\ -A & 0<t<T \\ \end{array}\right.\tag{1}$$

can be evaluated as

$$F(\omega)=\mathcal{F}_t[f(t)](\omega)=\int\limits_{-\infty}^{\infty} f(t)\,e^{-j \omega t}\,dt=\int\limits_{-T}^0 A\,e^{-j \omega t}\,dt+\int_\limits0^T -A\,e^{-j \omega t}\,dt.\tag{2}$$

With the variable substitution $t=-t$ the integral

$$\int\limits_{-T}^0 A\,e^{-j \omega t}\,dt\tag{3}$$


$$-\int\limits_T^0 A\,e^{j \omega t}\,dt\tag{4}$$

and reversing the limits this becomes

$$\int\limits_0^T A\,e^{j \omega t}\,dt\tag{5}$$


$$F(\omega)=\int\limits_0^T A\,e^{j \omega t}\,dt+\int_\limits0^T -A\,e^{-j \omega t}\,dt=\int\limits_0^T A \left(e^{j \omega t}-e^{-j \omega t}\right)\,dt=\int\limits_0^T 2 j A \sin (\omega t)\,dt\tag{6}$$

With respect to the misconception "just integrate one side and multiply it by two" note that

$$2 \int\limits_0^T -A e^{-j \omega t} \, dt=\frac{2 A j \left(1-e^{-j T \omega }\right)}{\omega }=\frac{2 A j (1+i \sin(T \omega)-\cos(T \omega))}{\omega}\tag{7}$$

is not the same as

$$\int\limits_0^T 2 A j \sin(\omega t)\,dt=\frac{2 A j (1-\cos(T \omega))}{\omega}\tag{8}$$


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