# 7.1 Discrete time fourier series (dtfs)  (Page 2/2)

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## Complex sinusoids

If you are familiar with the basic sinusoid signal and with complex exponentials then you should not have any problem understanding this section. In most texts, you will see the the discrete-time,complex sinusoid noted as: $e^{i\omega n}$

## In the complex plane

The complex sinusoid can be directly mapped onto our complex plane , which allows us to easily visualize changes to the complexsinusoid and extract certain properties. The absolute value of our complex sinusoid has the followingcharacteristic:

$\forall n, n\in \mathbb{R}\colon \left|e^{i\omega n}\right|=1$
which tells that our complex sinusoid only takes values onthe unit circle. As for the angle, the following statement holds true:
$\angle (e^{i\omega n})=wn$
For more information, see the section on the Discrete Time Complex Exponential to learn about Aliasing , Negative Frequencies , and the formal definition of the Complex Conjugate .

Now that we have looked over the concepts of complex sinusoids, let us turn our attention back to finding a basisfor discrete-time, periodic signals. After looking at all the complex sinusoids, we must answer the question of whichdiscrete-time sinusoids do we need to represent periodic sequences with a period $N$ .

Find a set of vectors $\forall n, n=\{0, \dots , N-1\}\colon {b}_{k}=e^{i{\omega }_{k}n}$ such that $\{{b}_{k}\}$ are a basis for ${ℂ}^{n}$
In answer to the above question, let us try the "harmonic" sinusoids with a fundamental frequency ${\omega }_{0}=\frac{2\pi }{N}$ :

## Harmonic sinusoid

$e^{i\frac{2\pi }{N}kn}$

$e^{i\frac{2\pi }{N}kn}$ is periodic with period $N$ and has $k$ "cycles" between $n=0$ and $n=N-1$ .

If we let $\forall n, n=\{0, \dots , N-1\}\colon {b}_{k}(n)=\frac{1}{\sqrt{N}}e^{i\frac{2\pi }{N}kn}$ where the exponential term is a vector in $\mathbb{C}^{N}$ , then $(k=\{0, \dots , N-1\}, \{{b}_{k}\}())$ is an for $\mathbb{C}^{N}$ .

First of all, we must show $\{{b}_{k}\}()$ is orthonormal, i.e. ${b}_{k}\dot {b}_{l}={\delta }_{kl}$ ${b}_{k}\dot {b}_{l}=\sum_{n=0}^{N-1} {b}_{k}(n)\overline{{b}_{l}(n)}=\frac{1}{N}\sum_{n=0}^{N-1} e^{i\frac{2\pi }{N}kn}e^{-(i\frac{2\pi }{N}ln)}$

${b}_{k}\dot {b}_{l}=\frac{1}{N}\sum_{n=0}^{N-1} e^{i\frac{2\pi }{N}(l-k)n}$
If $l=k$ , then
${b}_{k}\dot {b}_{l}=\frac{1}{N}\sum_{n=0}^{N-1} 1()=1$
If $l\neq k$ , then we must use the "partial summation formula" shownbelow: $\sum_{n=0}^{N-1} \alpha ^{n}=-\sum$ 0 α n n N α n 1 1 α α N 1 α 1 α N 1 α ${b}_{k}\dot {b}_{l}=\frac{1}{N}\sum_{n=0}^{N-1} e^{i\frac{2\pi }{N}(l-k)n}$ where in the above equation we can say that $\alpha =e^{i\frac{2\pi }{N}(l-k)}$ , and thus we can see how this is in the form needed to utilize our partial summation formula. ${b}_{k}\dot {b}_{l}=\frac{1}{N}\left(\frac{1-e^{i\frac{2\pi }{N}(l-k)N}}{1-e^{i\frac{2\pi }{N}(l-k)}}\right)()=\frac{1}{N}\left(\frac{1-1}{1-e^{i\frac{2\pi }{N}(l-k)}}\right)()=0$ So,
${b}_{k}\dot {b}_{l}=\begin{cases}1 & \text{if k=l}\\ 0 & \text{if k\neq l}\end{cases}()$
Therefore: $\{{b}_{k}\}$ is an orthonormal set. $\{{b}_{k}\}$ is also a basis , since there are $N$ vectors which are linearly independent (orthogonality implies linear independence).

And finally, we have shown that the harmonic sinusoids $\{\frac{1}{\sqrt{N}}e^{i\frac{2\pi }{N}kn}\}$ form an orthonormal basis for ${ℂ}^{n}$

## Periodic extension to dtfs

Now that we have an understanding of the discrete-time Fourier series (DTFS) , we can consider the periodic extension of $c(k)$ (the Discrete-time Fourier coefficients). [link] shows a simple illustration of how we can represent a sequence as a periodic signal mapped over an infinite numberof intervals.

Why does a periodic extension to the DTFS coefficients $c(k)$ make sense?

Aliasing: ${b}_{k}=e^{i\frac{2\pi }{N}kn}$

${b}_{k+N}=e^{i\frac{2\pi }{N}(k+N)n}=e^{i\frac{2\pi }{N}kn}e^{i\times 2\pi n}=e^{i\frac{2\pi }{N}n}={b}_{k}$
→ DTFS coefficients are also periodic with period $N$ .

## Discrete time square wave

Calculate the DTFS $c(k)$ using:

$c(k)=\frac{1}{N}\sum_{n=0}^{N-1} f(n)e^{-(i\frac{2\pi }{N}kn)}$
Just like continuous time Fourier series, we can take the summation over any interval, so we have
${c}_{k}=\frac{1}{N}\sum_{n=-{N}_{1}}^{{N}_{1}} e^{-(i\frac{2\pi }{N}kn)}$
Let $m=n+{N}_{1}$ (so we can get a geometric series starting at 0)
${c}_{k}=\frac{1}{N}\sum_{m=0}^{2{N}_{1}} e^{-(i\frac{2\pi }{N}(m-{N}_{1})k)}=\frac{1}{N}e^{i\frac{2\pi }{N}k}\sum_{m=0}^{2{N}_{1}} e^{-(i\frac{2\pi }{N}mk)}$
Now, using the "partial summation formula"
$\sum_{n=0}^{M} a^{n}=\frac{1-a^{(M+1)}}{1-a}$
${c}_{k}=\frac{1}{N}e^{i\frac{2\pi }{N}{N}_{1}k}\sum_{m=0}^{2{N}_{1}} e^{-(i\frac{2\pi }{N}k)}^{m}=\frac{1}{N}e^{i\frac{2\pi }{N}{N}_{1}k}\frac{1-e^{-(i\frac{2\pi }{N}(2{N}_{1}+1))}}{1-e^{-(ik\frac{2\pi }{N})}}$
Manipulate to make this look like a sinc function (distribute):
${c}_{k}=\frac{1}{N}\frac{e^{-(ik\frac{2\pi }{2N})}(e^{ik\frac{2\pi }{N}({N}_{1}+\frac{1}{2})}-e^{-(ik\frac{2\pi }{N}({N}_{1}+\frac{1}{2}))})}{e^{-(ik\frac{2\pi }{2N})}(e^{ik\frac{2\pi }{N}\frac{1}{2}}-e^{-(ik\frac{2\pi }{N}\frac{1}{2})})}=\frac{1}{N}\frac{\sin \left(\frac{2\pi k({N}_{1}+\frac{1}{2})}{N}\right)}{\sin \left(\frac{\pi k}{N}\right)}=\text{digital sinc}$
It's periodic! [link] , [link] , and [link] show our above function and coefficients for various values of ${N}_{1}$ .

## Dtfs conclusion

Using the steps shown above in the derivation and our previous understanding of Hilbert Spaces and Orthogonal Expansions , the rest of the derivation is automatic. Given a discrete-time, periodicsignal (vector in ${ℂ}^{n}$ ) $f(n)$ , we can write:

$f(n)=\frac{1}{\sqrt{N}}\sum_{k=0}^{N-1} {c}_{k}e^{i\frac{2\pi }{N}kn}$
${c}_{k}=\frac{1}{\sqrt{N}}\sum_{n=0}^{N-1} f(n)e^{-(i\frac{2\pi }{N}kn)}$
Note: Most people collect both the $\frac{1}{\sqrt{N}}$ terms into the expression for ${c}_{k}$ .
Here is the common form of the DTFS with the above note taken into account: $f(n)=\sum_{k=0}^{N-1} {c}_{k}e^{i\frac{2\pi }{N}kn}$ ${c}_{k}=\frac{1}{N}\sum_{n=0}^{N-1} f(n)e^{-(i\frac{2\pi }{N}kn)}$ This is what the fft command in MATLAB does.

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