Fourier Transform and the Delta Function

The Fourier transform of cosine is a pair of delta functions. The magnitude of both delta functions have infinite amplitude and infinitesimal width.

What I thought this meant: The cosine function can be constructed by the sum of two signals of infinite amplitude and corresponding frequencies.

I realised that only if I consider the 'weight' (area) of my delta functions will I obtain the correct result. Why is this?

Does the delta function in the frequency domain refer to it's weight?

• The delta function has amplitude of 1. Beyond that misconception, your question really doesn't make sense. Mar 7 '16 at 16:24
• The delta (impulse) function has amplitude that approaches infinity and duration that approaches zero, with constant area (weight). Mar 7 '16 at 16:32
• If you don't understand this concept you need a little more clarification wrt limits Mar 7 '16 at 16:34
• Sorry I can't give a definite answer, but I suspect you'll find the answer in Parseval's theorem, which says that the energy in the signal is equal whether you measure in time domain or frequency domain: $\int_{-\infty}^\infty{}\left|f(x)\right|^2\mathrm{d}x = \int_{-\infty}^\infty{}\left|F(\nu)\right|^2\mathrm{d}\nu$. To be more direct, the constant between the time and frequency domains is energy, not amplitude. Mar 7 '16 at 16:51
• @ThePhoton, i don't think Parseval's Theorem works with constant-amplitude sinusoids or with Dirac delta functions. Mar 7 '16 at 20:18

First you need to understand that the crucial property of the delta function is that it picks a single value of a function when it gets integrated

$$\int_{-\infty}^\infty f(x)\delta(x-a)\;dx = f(a)$$

Using this property to calculate the inverse fourier transform of $\pi \left[\delta(\omega+\omega_0) + \delta(\omega-\omega_0)\right]$ you get $$\frac 1{2\pi} \int_{-\infty}^\infty \pi\left[\delta(\omega-\omega_0) + \delta(\omega+\omega_0)\right] e^{j\omega t}d\omega = \frac 12 e^{j\omega_0 t} + \frac 12 e^{-j\omega_0 t} = \cos \omega_0 t$$ So in order to get a certain amplitude you have to multiply the delta function by some factor (weight), otherwise you get an amplitude of 1.

Since the amplitude of delta function is infinity by definition, the height is often used to indicate the weight.

The signal $x(t)=\cos(\omega_0t)$ can be written as

$$x(t)=\frac12\left(e^{j\omega_0 t}+e^{-j\omega_0t}\right)\tag{1}$$

And the Fourier transform of $e^{j\omega_0t}$ is a Dirac delta impulse:

$$e^{j\omega_0t}\Longleftrightarrow2\pi\delta(\omega-\omega_0)\tag{2}$$

This can be seen by considering its inverse Fourier transform:

$$\mathcal{F}^{-1}\{\delta(\omega-\omega_0)\}=\frac{1}{2\pi}\int_{-\infty}^{\infty}2\pi\delta(\omega-\omega_0)e^{j\omega t}d\omega=e^{j\omega_0t}\tag{3}$$

The last equality in (3) follows from this important property of the Dirac delta impulse:

$$\int_{-\infty}^{\infty}f(t)\delta(t-a)dt=f(a)\tag{4}$$

So you see that a cosine is not the sum of two signals of infinite amplitude. It's just that the Fourier transform of the complex exponentials in (1) is a Dirac delta.

• But deltas are infinite in amplitude Mar 7 '16 at 18:06
• @PDuarte, ... but the strength is unity.
– Chu
Mar 7 '16 at 18:15
• @Chu Yes, strenght (area) is unit, but amplitude is infinite. Mar 7 '16 at 18:17
• @PDuarte, if the delta functions had finite amplitude, yet still had width approaching zero in the limit, they would be indistinguishable from a function that is identically zero, as far as anything with an integral is concerned. Mar 7 '16 at 18:54
• @robert bristow-johnson, A Dirac delta function has infinite height, zero width and unit area. How can that be indistinguishable from a function that is identically zero? It's integral is unity.
– Chu
Mar 7 '16 at 19:36

First, let's make sure you understand the Fourier transform of a cosine. Start with the Euler identity:

$$e^{j\omega t} = \cos(\omega t) + j\sin(\omega t)$$ $$e^{j(-\omega t)} = e^{-j\omega t} = \cos(-\omega t) + j\sin(-\omega t) = \cos(\omega t) - j\sin(\omega t)$$

If you want to make a cosine out of complex exponentials, you need to get rid of the sine components:

$$\frac 1 2 (e^{j\omega t} + e^{-j\omega t}) = \frac 1 2 \big (\cos(\omega t) + j\sin(\omega t) + \cos(\omega t) - j\sin(\omega t) \big ) = \cos(\omega t)$$

Why do we want care about that? Because complex exponentials are the basis of the frequency domain! Each point of the Fourier transform represents a single complex exponential's magnitude and phase. A cosine is made of exactly two complex exponentials, so we'd expect there to be two non-zero points on the Fourier transform. That's what the delta functions are.

Mathematically, the Dirac delta function is a strange thing. It's defined only by its integral:

$$\int_a^b {\delta(x - c) dx} = \begin{cases} 1, & a < c < b \\ 0, & \text{otherwise} \end{cases}$$

$$\int_{-\infty}^{\infty} {f(t) \delta(x - c) dx} = f(c)$$

In other words, there's a "spike" at $x = c$, and the area under the "spike" is 1. This is sloppy math, but mathematicians prefer not to call the delta function a function at all, so let's not get too hung up on formalities. We have enough problems. :-)

It's easy enough to see how the delta function works with the inverse Fourier transform:

$$x(t) = \cos(\omega_0 t)$$

$$X(\omega) = \pi \big( \delta(\omega - \omega_0) + \delta(\omega + \omega_0) \big)$$

\begin{align} \mathcal{F}^{-1}\{X(\omega)\} = & \frac 1 {2\pi} \int_{-\infty}^{\infty} X(\omega) e^{j\omega t} d\omega \\ & = \frac 1 {2\pi} \big(\int_{-\infty}^{\infty} \pi\delta(\omega - \omega_0) e^{j\omega t} d\omega + \int_{-\infty}^{\infty} \pi\delta(\omega + \omega_0) e^{j\omega t} d\omega \big) \\ & = \frac 1 2 \big( e^{j\omega_0 t} + e^{-j\omega_0 t}\big) \\ & = \cos(\omega_0 t) \end{align}

But what does all this mean? It turns out that the Fourier transform is actually very similar to taking the dot product of two vectors. The "vectors" in this case are the function $x(t)$ and the "unit vector" $e^{j\omega t}$. You've probably seen the dot product defined like this before:

$$\overline A \cdot \overline B = \sum_i A_iB_i = A_xB_x + A_yB_y + A_zB_z$$

If you want to get a single component, you dot the vector with a unit vector:

$$\overline A \cdot \hat k = A_z$$

Well, the math is a lot more complicated, and you have an infinite number of components ("dimensions") instead of three, but otherwise that's exactly what this formula is doing:

$$X(\omega_0) = \overline{x(t)} \cdot \widehat{e^{j\omega_0 t}} = \int_{-\infty}^{\infty} x(t) e^{-j\omega_0 t} dt$$

We change the sum to an integral, the vectors to continuous functions, and (for some reason) take the complex conjugate of the second vector. Poof! Dot product. Pretty neat, huh?

So how do you do the inverse transform? Multiply each component by the unit vector, then add them!

$$\overline A = \sum_i A_i \hat e_i = A_x \hat i + A_y \hat j + A_z \hat k$$

$$x(t) = \sum_\omega X(w) \widehat {e^{j\omega t}} = \frac 1 {2\pi} \int_{-\infty}^\infty X(\omega) e^{j\omega t}d\omega$$

We made the same changes as before -- sum to integral and vectors to functions. The factor of $2\pi$ is there because of the complex exponentials, but it's not important.

In this context, a Dirac delta function represents a single component of a function. As a rule of thumb, when you go continuous, you start having to work with infinitesimal quantities. This is no exception.

Dirac's delta-"function" is interesting, because it deals with yet another form of infinity; and one that is hard to comprehend. You see, it is not a function in the regular sense. It represents a limit of functions. An analogy would be $\pi$, which is the limit of any valid approximation of the area of a unit circle.

There are many ways to create Dirac's function. One way is to define a sequence of functions like this one:

$$\delta_k(t) = \cases{k/2,\;\text{when |t|<\frac{1}{k}}\\0,\;\text{otherwise}}$$

You can see that $\delta_k$ is a function and that

$$\int_{-\infty}^{\infty} \delta_k dt = \frac{k}{2}\int_{-1/k}^{1/k}dt = \frac{k}{2}\cdot \frac{2}{k} = 1$$

for every $k$. Because of this, and because $\delta_k(0)\to\infty$, as $k\to\infty$ we call the resulting "function" Dirac's Delta Function: $$\delta(t) = \lim_{k\to\infty}\delta_k(t)$$

A neat property of a sequence of functions like the above is that: $$\lim_{k\to\infty} \int \delta_k(t-t_0) f(t) dt = f(t_0),$$ as long the function $f$ is reasonably nice (all continuous compactly supported functions). Using Dirac's generalized function, we can write this as $$\int \delta(t-t_0) f(t) dt = f(t_0),$$

If you pause to think about it, that's a very nice property. Calculating an integral is hard, but here you have a formula where you can just skip it and evaluate a function! Quite a win in my book.

But the main takeaway from all of that is that while Dirac's delta function is infinite at 0, it still has a bounded area (the integral is one). Whenever you see Dirac's function, think in your head that it is the limit of a sequence of functions. In other words, to make it a "real" calculation, you have to pick an arbitrarily large $k$ and use $\delta_k$ instead of $\delta$.

• normally you see "$\delta(t)$" in the context of a signal that is proportional to voltage. power and energy are proportional to the square of voltage. the energy of a Dirac delta is not bounded. the point you're making might be valid, but maybe you should just revert to the common terms like "area" or "strength". Mar 8 '16 at 1:22
• Yes, I agree. Especially since it can quickly lead to innocent little expressions such as $\delta^2(t)$, which do not make sense. So, I've edited it. Thanks! Mar 8 '16 at 3:12