I am reading a book by Zoran Gajic titled Linear Dynamic Systems and Signals.

There were two problems in chapter 3 that I was curious about.

The first question asks for the Fourier Transform of a function $x^2(t-5)$, given the Fourier Transform of $x(t) = \mathrm{X}(jw)$. Is my approach correct?

$$\mathrm{X}(jw) = \frac{1}{jw}$$

Then, from the tables, we know:

$$ x(t) = \frac{1}{2} \text{sgn} (t)$$

Then, $$ x^2(t) = \begin{cases} \dfrac{1}{4}, t \neq 0 \\ \\ 0, t = 0 \end{cases}$$

The Fourier Transform of two different functions $f(t)$ and $g(t)$ are the same if they are both bounded and differ at a finite number of points. Thus, the Fourier Transform of $x^2(t)$ is the same as the Fourier Transform of $g(t) = \frac{1}{4}$.

$$ \mathfrak{F}\left(\frac{1}{4}\right) = \frac{\pi}{2} \delta(w)$$

Then, applying the time-shifting property,

$$ \mathfrak{F}\left(x^2(t-5)\right) = e^{-jw5} \frac{\pi}{2} \delta(w) = \frac{\pi}{2} \delta(w)$$

My second question is, what is the Inverse Fourier Transform (or how do I approach the Inverse Fourier Transform) of the following?

$$ \mathrm{Y}(jw) = \frac{jw}{5 + jw}$$

Specifically, it is the $jw$ term in the numerator that is making the problem difficult for me. I know the inverse of the denominator, and I am aware of the Fourier Transform Property that convolution in the time domain corresponds to multiplication in the frequency domain.

Thank you for your assistance.

  • $\begingroup$ I just realized I also need to use the property $\frac{\mathrm{d}^n x(t)}{\mathrm{d}t^n} \leftrightarrow (jw)^n \mathrm{X} (jw)$ to obtain the inverse Fourier transform of the numerator. $\endgroup$
    – jrand
    Commented Jul 15, 2012 at 6:01
  • 1
    $\begingroup$ Try using $\frac{i\omega}{5+i\omega} = 1-\frac{5}{5+i\omega}$. $\endgroup$
    – copper.hat
    Commented Jul 15, 2012 at 6:25
  • $\begingroup$ Hi copper.hat: Thank you for your advice. I used it to verify my answer. $\endgroup$
    – jrand
    Commented Jul 16, 2012 at 2:59

1 Answer 1


Given that $$x(t)\Longleftrightarrow X(j\omega)$$ the given signal can be written as $$x^2(t-5) = x^2(t)\star \delta(t-5)$$ where $\star$ denotes convolution and $\delta(t)$ is Dirac delta function. Thus, by using multiplication in time-domain, convolution in time-domain and time-shifting properties we easily find that $$x^2(t-5)\Longleftrightarrow \dfrac{1}{2\pi}\left[X(j\omega)\star X(j\omega)\right]e^{-j5\omega}$$

  • $\begingroup$ Hi Bhupala; I agree with your answer. I also cannot find any errors in my answer. Therefore, our answers should be the same. However, I am having trouble showing equality in our answers. Specifically, I am trying to show $\frac{\pi}{2} \delta(w) = \frac{1}{2\pi} \left[\frac{1}{jw} \star \frac{1}{jw}\right] e^{-jw5}$. On the right side, I am ending up with the integral $\frac{1}{2\pi}e^{-jw5} \int_{-\infty}^{\infty} \frac{1}{-\lambda w + \lambda^2} \, \mathrm{d} \lambda$. How do I show that the two quantities are equal? $\endgroup$
    – jrand
    Commented Jul 18, 2012 at 21:26
  • $\begingroup$ I already tried Wolfram Integrator but it is not obvious where the Dirac delta is coming from. Thanks. $\endgroup$
    – jrand
    Commented Jul 18, 2012 at 21:26

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