How is a continuous spectrum for the DTFT possible?

So we that a complex sinusoid of the form $e^{j\omega_0n}$ is periodic over $N=2\pi/\omega_0$ only if $\omega_0$ is a rational multiple of $\pi$, otherwise the exponential is not periodic. (see EDIT!)

Then we got to the DTFT, specifically the synthesis equation:

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

Here $X\left(e^{j\omega}\right)$ if a function that's continuous over $\omega$, and $X\left(e^{j\omega_0}\right)$ is how much a complex sinusoid $e^{j\omega_0n}$ contributes to $x[n]$.

But that confuses me, because I feel like not all sinusoids should be able to contribute to $x[n]$ since not all complex sinusoids are periodic. Or are aperiodic sinusoids actually the key to the DTFT?

EDIT: I should clarify that in that first bit where I talk about complex sinusoids being periodic, I'm talking about discrete complex sinusoids, $x[n] = e^{j\omega_0n}$ where $n\in\mathbb{Z}$. For $x[n]$ to be considered periodic there must be some $N\in\mathbb{Z} \mid x[n+N]=x[n]$. For such an $N$ to exist, $\omega_0$ must be a rational multiple of $pi$.

My confusion comes from the fact that the synthesis equation includes values of $\omega$ that are not rational multiples of $\pi$, so aperiodic discrete sinusoids are contributing to synthesizing $x[n]$.

• I'm not actually sure where your confusion lies. I would argue that all complex sinusoids are periodic because I would argue that all complex sinusoids are continuous signals. On the other hand, there are complex sinusoids that will result in aperiodic sequences when sampled as you describe. Still, the underlying continuous signal that was sampled to obtain the sequence is periodic. This point of view depends on viewing a discrete time system as a sampled version of a continuous time system. – hops May 9 '17 at 3:42
• @hops Sorry in that first bit I was referring to discrete complex sinusoids. I just edited my post! – rcplusplus May 9 '17 at 4:08

The reason the DTFT has values at frequencies that result in discrete aperiodic sinusoids is because we are interested in representing all possible discrete time signals including (linear combinations of) discrete aperiodic sinusoids. Let's say that we have a signal $x[n] = e^{j \omega_0 n}$ where $\omega_0$ is not a rational multiple of $\pi$. Would you like this signal to have a DTFT? Of course. It would not have one if we did not allow the existence of the DTFT at all frequencies (including irrational multiples of $\pi$).

Put another way, periodicity is not a requirement for the existence of the DTFT of a signal.

• Oh shoot, when you put it that way it's fairly obvious I guess haha. Followup though, the DTFS coefficients are defined by $a_k=\sum_{k=<N>}x[n]e^{-jk\omega_0n}$. Say $x[n]$ is periodic with $\omega_0=3$. That isn't a rational multiple of $\pi$, and so all harmonics of $e^{j\omega_0n}$ will not be periodic as a result. How does this all factor in? – rcplusplus May 9 '17 at 4:33
• By that I mean, how can a periodic signal $x[n]$ be composed of a finite number of aperiodic signals? – rcplusplus May 9 '17 at 4:42
• I'm not sure I follow all of your notation. My definition of the DTFT does not sum over a single period, but from $-\infty$ to $+\infty$. I think I'm going to need a little bit more clarity to respond to this properly. – hops May 9 '17 at 4:43
• Oh I was talking about the Discrete Time Fourier Series, that's what the notation in my comment was referring to. – rcplusplus May 9 '17 at 4:46
• Okay, well in that case, the signal is viewed as being extended periodically and therefore it is periodic with period $N$ regardless. – hops May 9 '17 at 4:49

$e^{j\omega n}$ in the synthesis equation

$$\displaystyle x[n]=\frac{1}{2\pi}\int_{2\pi}X\left(e^{j\omega}\right)e^{j\omega n}d\omega$$ is NOT discrete since $\omega$ is the independent variable. Hence, it is periodic over $[0, 2\pi]$ for any $n$. Expand it by Euler's formula for better understanding.

Same applies to the analysis equation: $$X(e^{j\omega})=\sum_{n=-\infty}^{\infty}x[n]e^{-j\omega n}$$ Again, $e^{j\omega n}$ is continuous with respect to $\color{red}\omega$.

What you are mixing up with this is the DTFS:

$$x[n] = \sum_{k=0}^{N-1}a_ke^{-jk\omega_0 n}$$ Here, there are only $N$ periodic bases that are defined by $e^{-jk\omega_0 n}$ for $k=0,1,\cdots,N-1$:

$$e^{jk\omega_0n}=e^{j(k+N)\omega_0n}$$

They are discrete with respect to $\color{red}n$.

• Yeah $\omega$ is continuous and that makes sense but what was boggling me was that non-periodic discrete sinusoids in the form of $a_ke^{-jk\omega_0n}$ would contribute to the synthesis of $x[n]$. I guess I was just getting confused that a non-periodic discrete sinusoid could factor in, but $x[n]$ is a periodic, so I guess it isn't really a stretch that its components won't all be. – rcplusplus May 9 '17 at 4:38
• I understand your confusion. You should note that there is no contradiction when the weighted sum of an infinite number of complex exponential functions constitutes another periodic signal. – msm May 9 '17 at 4:48