# Discrete Harmonics - Why multiplying digital frequency by k does not get next harmonic

For continuous time $$e^{jk\Omega_0t}$$ gives a complete set of orthogonal harmonics for fourier decomposition but for discrete $$e^{jk\omega_0n}$$ does not form a complete set orthogonal basis set to decompose a signal because... ?

No Answer yet on 21 Nov 2018,

• where have read your definitions? – Stanley Pawlukiewicz Nov 8 '18 at 13:18
• This book - complextoreal.com/fftguide and chapter 3 which is free to view here...complextoreal.com/tutorials/… its a portion of the chapter from pg 86. Is my approach above wrong? – Natalie Johnson Nov 8 '18 at 13:21
• i would reverse the roles of your symbols $\Omega_0$ and $\omega_0$ to be consistent with the convention we see in DSP books. i know that in the old analog days we use $\omega=2\pi f$ for continuous-time signals (like $s=j\omega$) but in DSP we now say $z=e^{j\omega}$ and when we need to reference an analog context, we say $s=j\Omega$. – robert bristow-johnson Nov 8 '18 at 22:35
• @NatalieJohnson: The set $\phi_k[n]=e^{j2\pi nk/N}$ is an orthogonal set, and that's what you use to find the Fourier series of a discrete-time signal. – Matt L. Nov 9 '18 at 10:43
• I know that. But I am confused how we get there - I edited my post to make it crystal clear what I am confused about. – Natalie Johnson Nov 13 '18 at 20:33

First the continuous-time:

Consider the complex exponential: $$e^{j \Omega_0 t}$$ where $$\Omega_0$$ is the continuous-time radian frequency, $$t$$ is the continuous time itself. Now from elementary calculus, the period of this complex exponential can be found from $$x(t) = x(t+T)$$ as :

$$e^{j \Omega_0 t} = e^{j \Omega_0 (t+T)} = e^{j \Omega_0 t} e^{j \Omega_0 T}$$

Now in order for this equality to hold, the last exponential term must be $$1$$. From the algebra of complex-exponentials it's known that it will be $$1$$ whenever $$\Omega_0 T = 2\pi m$$ for some integer $$m$$ and real $$T,\Omega_0$$ which indicates that

$$e^{j \Omega_0 T} = 1 \implies \Omega_0 T = 2\pi m$$ from which we find the period of the continuous-time complex exponential as:

$$T = \frac{2 \pi m}{\Omega_0} = \frac{2 \pi}{\Omega_0} = T_0$$

where in the left we used the fact that the fundamental period $$T_0$$ is the minimum (nonzero) real number $$T$$ that satisfied the equality for $$m=1$$.

Conclusion: any continous-time complex exponential $$e^{j \Omega_0 t}$$ is periodic for any real value of $$\Omega_0$$... This is so because $$t$$ is a continuous variable which admits the period $$T_0$$ to be continuous as well. Since the period is allowed to be continous, it can always be found to make $$\Omega_0 T$$ an integer multiple of $$2\pi$$.

Furthermore, in the continous-time case the harmonic family of the complex exponential $$e^{j \Omega_0 t}$$ is defined to be: $$\phi_k(t) = e^{j k \Omega_0 t}$$ for integer $$k = 1,2,...,\infty$$.

The particular period associated with the k-th harmonic $$\phi_k(t) = e^{j k \Omega_0 t}$$ is $$T_k = \frac{T_0}{k}$$, nevertheless its fundamental period is $$T_0$$.

Now, since the period in continous-time is a real variable it can take any value possible, as you can see, as the harmonic member index $$k$$ increases the particular period decreases like $$T_k = \frac{T_0}{k}$$. As the member index $$k$$ goes to infinity, the member period goes to zero but is a valid value. Hence we see that there are an infinite number of such members; i.e., the harmonic family in continuous-time has infinite members. for each $$k=1,2,... , \infty$$

Part-II: The discrete-time case :

The complex exponential is $$e^{j \omega_0 n}$$, where $$n$$ is an integer and $$\omega_0$$ is the discrete-time radian frequency (real variable).The headache comes beacuse the index $$n$$ is not continous but an integer, and therefore, admits only integer periods $$N=1,2,...$$ . The smallest allowed period is $$1$$ (unlike in the CT case where the smallest period goes to zero taking any real value)

Now for a discrete-time complex expoeential to be periodic, you should have $$x[n] = x[n+N]$$ for some integer $$N$$ as shown:

$$e^{j \omega_0 n} = e^{j \omega_0 (n+N)} = e^{j \omega_0 n} e^{j \omega_0 N}$$

Again from the complex exponential algebra it's seen that : $$e^{j \omega_0 N} = 1 \implies \omega_0 N = 2\pi m$$ for some integer $$N$$ and $$m$$ . This implies that :

$$\omega_0 N = 2\pi m \implies N = \frac{ 2 \pi m }{\omega_0}$$

Now, in order for a discrete-time complex exponential $$e^{j \omega_0 n}$$ to be periodic with period $$N$$, it must be true that its frequency $$w_0$$ must be a rational multiple of $$\pi$$ as indicated by :

$$\omega_0 = \frac{2 \pi m}{N}$$

Hence if $$\omega_0$$ does not satisfy the above condition that that complex exponential cannot be periodic at all. This is never the case with continuous-time complex exponentials which are always periodic for any $$\Omega_0$$.

After finding the condition on the periodicity of the discrete-time complex exponential, let's also observe their harmonic family:

In the discrete-time case, the harmonic family of the periodic complex exponential $$e^{j \omega n}$$ with a period of $$N$$ is defined to be: $$\phi_k[n] = e^{j k \omega_0 n}$$ for integer $$k = 0,1,...,N-1$$.

Another difference between the CT and DT harmonic families occur on the particular period $$N_k$$ associated with the k-th harmonic: it is not $$N_k = \frac{N}{k}$$, as that won't be an integer for any $$N$$ and $$k$$. The correct period for the k-th member is $$N_k = \frac{N m}{k}$$ where integer $$m$$ is chosen to make $$N_k$$ the minimum integer for given integer $$N$$ and $$k$$

Now the last point, why are there just a finite number of harmonics in DT unlike the CT where there are infinite harmonics? This can most easily explained by the following: Let integer $$k = mN + r$$ for some integers $$r,m$$ and let $$r < N$$ where $$N$$ is the period of the fundamental member $$e^{j \omega_0 n}$$. Then we have the k-the member as: $$\phi_{k}[n] = \phi_{m N+r}[n] = e^{j \omega_0 (mN + r ) n} = e^{j \omega_0 m N n } e^{j \omega_0 rn }$$

Now since $$N$$ is the period, we have $$e^{j \omega_0 m N n } = e^{j 2\pi l m N n} = 1$$ and therefore we see that

$$\phi_{k}[n] = \phi_{m N+r}[n] = e^{j \omega_0 rn } = \phi_{r}[n]$$

The $$k = mN + r$$ th member is identical to the $$0\leq r \leq N-1$$ st member. Hence we conlude that in discrete-time, for a periodic complex exponential $$e^{j\omega_0 n}$$ of period $$N$$ there are only $$N$$ distinguishable members of the harmonic family.

I hope this clarified it.

• not sure this answers the question? – Natalie Johnson Nov 20 '18 at 14:08
• Hi @NatalieJohnson I have provided this extremely long and detailed answer just to relieve your confusion about discrete complex exponential. I will be quite glad if you take your time, read it and take action to upvote / accept or return a feedback. – Fat32 Dec 3 '18 at 15:44
• I was on holiday, sorry. accepted. Thankyou for your kind explanation. – Natalie Johnson Dec 10 '18 at 17:06