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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,

If you know please help :(

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  • $\begingroup$ where have read your definitions? $\endgroup$ – Stanley Pawlukiewicz Nov 8 '18 at 13:18
  • $\begingroup$ 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? $\endgroup$ – Natalie Johnson Nov 8 '18 at 13:21
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    $\begingroup$ 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 $s=j\Omega$. $\endgroup$ – robert bristow-johnson Nov 8 '18 at 22:35
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    $\begingroup$ @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. $\endgroup$ – Matt L. Nov 9 '18 at 10:43
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    $\begingroup$ 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. $\endgroup$ – Natalie Johnson Nov 13 '18 at 20:33
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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.

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  • $\begingroup$ not sure this answers the question? $\endgroup$ – Natalie Johnson Nov 20 '18 at 14:08
  • $\begingroup$ 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. $\endgroup$ – Fat32 Dec 3 '18 at 15:44
  • $\begingroup$ I was on holiday, sorry. accepted. Thankyou for your kind explanation. $\endgroup$ – Natalie Johnson Dec 10 '18 at 17:06

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