# What is the meaning of the DTFT of the unit impulse sequence?

In an exercice, I'm asked to draw the $$X_{imp}(\omega)$$ Discrete-Time Fourier Transform (DTFT) of the $$x_{imp}(n)$$ unit impulse sequence defined as: $$x_{imp}(n) = \begin{cases} 1 & \text{if } n = 0 \\ 0 & \text{otherwise} \end{cases}$$

So, given $$X(\omega) = \sum_{-\infty}^{\infty} x(n)\cdot{\rm e}^{-j\omega n}$$, I end up with $$X_{imp}(\omega) = 1$$ because all terms of the summation would be $$0$$, except at $$n=0$$ where $$x(n)\cdot{\rm e}^{-j\omega n} = 1$$.

If my answer is correct, does that mean the unit impulse sequence would have the same amount of "energy" (← is this the right word?) for all frequencies, including audio, radio frequencies X-Rays, light or whatever else. That seems somewhat irrealistic to me. So, where is the flaw in my reasoning?

• Your problem is in the interpretation of the discrete-time frequency $\omega$... But your concern is more realistic if you would consider the continuous-time Fourier transform of the continuous-time Dirac impulse $\delta(t)$ which indeed has the "same" unit magnitude accross the whole range of physical frequencies... – Fat32 Dec 17 '19 at 1:50
• What is equally unrealistic is the impulse in the time domain itself, so a good direction as Fat32 indicated to consider that case: What can possibly be 0 everywhere except when time = 0, and then only at that time rise to infinity? – Dan Boschen Dec 17 '19 at 4:06
• However what is really interesting, and I believe @Fat32 can answer it well is the comparison of Parseval's theorem in the continuous time domain for a Dirac impulse versus in the discrete time case with the unit sample function and the related DTFT which is one for $\omega = -\pi$ to $\pi$. – Dan Boschen Dec 17 '19 at 4:19
• Thanks for the comments. @Fat32, indeed, it seems I do not properly understand the meaning of "the discrete-time frequency 𝜔". I've made a similar remark while reading Juancho's answer below. Any hint or suggestion that could help me? I read about the continuous-time Dirac impulse--which seems to be more realistic to me (it makes me think to a physical impulse like a transient). But I'm not sure I understand all the maths and their implications. – Sylvain Leroux Dec 17 '19 at 10:44

Your signal has its energy equally distributed in all frequencies.

But note that in discrete-time, "all frequencies" means from $$\omega = -\pi$$ to $$\omega = +\pi$$. Frequencies beyond that range exist but result in the same signal as one with a frequency within the range.

For example, $$\cos(0\times n) = \cos(2\pi \times n)$$. Similarly $$\cos(0.1 \pi \times n) = \cos(2.1 \pi \times n)$$, and so on.

So the energy of the signal is calculated integrating in a range of $$2\pi$$.

The formal result is known as Parseval's theorem, which states that energy measured in time domain is equal to energy measured in frequency domain:

$$\sum_n |x[n]|^2 = \frac{1}{2\pi} \int_{-\pi}^{\pi} |X(e^{j\omega})|^2 d\omega$$

Note: this is a particular version of Parseval. The more general version states that the projection (think cross product) of one vector onto another is the same calculated in time domain or frequency domain, which is reasonable since Fourier can be seen as a change of base, and a change of base should not alter projections.

• @Juancho "But note that in discrete-time, 'all frequencies' means from 𝜔=−𝜋 to 𝜔=+𝜋." I think I failed to understand that notion of discrete-time. What does 𝜔 represent exactly? Does it have a physical meaning or is it a purely abstract mathematical artifact? – Sylvain Leroux Dec 17 '19 at 10:27
• @SylvainLeroux: For a (real-valued) discrete-time signal, the maximum frequency is the Nyquist frequency, equal to half the sampling frequency. That would be an alternating sequence of numbers. So a unit impulse can be represented by the integral over all frequencies in the interval $[-f_s/2,f_s/2]$. – Matt L. Dec 17 '19 at 11:28
• Deleting my prior comment because I misread the question (d'oh)! – TimWescott Dec 17 '19 at 16:19
• $\omega$ is the frequency of the sinusoids underlying the Fourier transform. $e^{j \theta} = \cos \theta + j \sin \theta$; so in a sense it's just shorthand for a pair of sinusoids. – TimWescott Dec 17 '19 at 17:01

You're right that the DTFT of a unit impulse is constant:

$$\textrm{DTFT}\{\delta[n]\}=\sum_{n=-\infty}^{\infty}\delta[n]e^{-jn\omega}=1$$

That means that a unit impulse can be represented by the integral over all frequencies from $$-f_s/2$$ to $$f_s/2$$, where $$f_s$$ is the sampling frequency. So the problem with your reasoning is that you thought the frequency axis extends from $$-\infty$$ to $$\infty$$. For discrete-time signals it doesn't. The spectrum of a discrete-time signal is periodic with period $$f_s$$.

It's straightforward to show that $$\delta[n]$$ can be obtained by the inverse DTFT of a constant spectrum:

$$\textrm{IDTFT}\{1\}=\frac{1}{2\pi}\int_{-\pi}^{\pi}e^{jn\omega}d\omega=\begin{cases}1,&n=0\\\frac{1}{2\pi jn}\left[e^{jn\pi}-e^{-jn\pi}\right]=0,& n\neq 0\end{cases}$$

Note that $$\omega$$ is the normalized frequency in radians:

$$\omega=\frac{2\pi f}{f_s}$$

So the integration limits $$\pm\pi$$ correspond to the frequencies $$\pm f_s/2$$.