# Polar form of the Fourier transform of $\sin(t)$

I came across the Fourier transform of sin(t).

It ends up being a purely imaginary (dirac delta) impulse pair. But when considering the frequency domain representation of a signal, we consider the graphs of (i) the magnitude of the signal w.r.t. frequency and (ii) the phase (i.e. argument or angle) of the signal w.r.t. frequency.

In the case of $\sin(t)$, the phase can be seen to be 90 degrees, treating the delta functions in manner analogous to real numbers.

But how would I find the magnitude? Could someone please provide a reference to a textbook?

As another example, what would be the magnitude and phase of

$$\operatorname{FourierTransform}(\cos(t)) \cdot (4 + 2j) \text{ ?}$$

• FT(sin(x))(w) is a complex number function of w. There is no trick to find the magnitude, just look at the definition Mar 23 '16 at 9:38
• But how do I interpret $\sqrt{\delta(w + 1)^2 + \delta(w - 1)^2}$?
– LMZ
Mar 23 '16 at 9:44
• $\left\lvert i\sqrt{\frac{\pi}{2}} \left(\delta(w+1) - \delta(w-1)\right)\right\rvert$ = $\sqrt{\frac{\pi}{2}} \left(\delta(w+1) - \delta(w-1)\right)$ Mar 23 '16 at 9:49
• Oh okay so you treat the delta function pair as a positive real number.
– LMZ
Mar 23 '16 at 9:55
• Are there any references that you could refer to that details that rule? (Consider posting as an answer if you do :) )
– LMZ
Mar 23 '16 at 9:57

I read two question here. One is about the graphical representation of the Fourier transform of a sinusoidal function, and the other is about the magnitude (and phase) of such a Fourier transform. For ordinary functions, these two would be equivalent, but for Fourier transforms containing Dirac delta impulses they are not. The reason is that mathematically you can't compute the magnitude of a Dirac delta impulse. It is not a function, so there are no function values to take the magnitude of.

A standard graphical representation of the Fourier transform of the function $\sin(\omega_0t)$ would usually just be two arrows at $\omega=\omega_0$ and $\omega=-\omega_0$, one pointing up, the other one pointing down, where the $y$-axis is labeled by $j$ (which makes a separate phase plot unnecessary). An example of such a graph is shown here.

Even though you can't compute the magnitude of a Dirac impulse, you could write down a representation of the Fourier transform of a sinusoid consisting of a sum of Dirac impulses with positive areas and a phase term. For the function $f(t)=\sin(\omega_0t)$ such a representation would be

$$F(\omega)=\pi[\delta(\omega-\omega_0)+\delta(\omega+\omega_0)]e^{-j\frac{\pi}{2}\text{sign}(\omega)}\tag{1}$$

(Note that I use the non-unitary definition of the Fourier transform, which is different from the unitary version used by Wolfram Alpha. This is why there's a factor of $\pi$ in $(1)$ instead of $\sqrt{\frac{\pi}{2}}$.)

A similar representation of your example $A\cdot\mathcal{F}\{\cos(\omega_0t)\}$ with complex $A=|A|e^{j\phi_A}$, where $\phi_A$ is the argument of the complex number $A$, is given by

$$A\cdot\mathcal{F}\{\cos(\omega_0t)\}=|A|\pi [\delta(\omega-\omega_0)+\delta(\omega+\omega_0)]e^{j\phi_A}\tag{2}$$

Regarding to the magnitude and phase representation of delta functions,

Now, for the second example, you may need to develop that function as:

$4\cdot\cos(t) + 2j\cdot\cos(t)$ and then, you will end up with some deltas as well. This is the answer for the $2j\cdot\cos(t)$ one: http://www.wolframalpha.com/input/?i=fourier+transform+2jcos(x)

Note that you don't really need to expand the second example, but if you want to find the polar representation, you are going to do it anyway.

• I'm not asking about the Fourier transform of $cos(t)(4 + 2j)$, and I'm comfortable with the polar form of complex numbers. I'm wondering what happens if you first compute FourierTransform(cos(t)), multiply it by (4 + 2j), and then try to express that in polar form. How would you compute the magnitude, and how would you compute the phase, considering that the spectrum FourierTransform(cos(t)) of the signal cos(t) is expressed in terms of delta functions. (you can consider 4 + 2j to be the rectangular frequency response of a filter in this example)
– LMZ
Mar 23 '16 at 7:55
• I have edited my answer, let me know Mar 23 '16 at 9:30
• Thank you very much, I would upvote you if I had voting privileges. I will have a read of the content you provided when I have the time (probably in 2 days), and probably accept your answer then, if it answers my question.
– LMZ
Mar 23 '16 at 9:47
• ok, I hope that helps you :) Mar 23 '16 at 17:54