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

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: $$e^{i\omega n}$$

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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:

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 $N$ .

$e^{i\frac{2\pi }{N}kn}$ 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 orthonormal basis for $\mathbb{C}^N$ .

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.

Examples

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)

$c_k=\frac{1}{N}\sum_{m=0}^{2N_1} e^{-(i\frac{2\pi }{N}(m-N_1)k)}=\frac{1}{N}e^{i\frac{2\pi }{N}k}\sum_{m=0}^{2N_1} e^{-(i\frac{2\pi }{N}mk)}$

Now, using the "partial summation formula"

$\sum_{n=0}^M a^n=\frac{1-a^{(M+1)}}{1-a}$

$c_k=\frac{1}{N}e^{i\frac{2\pi }{N}{N}_{1}k}\sum_{m=0}^{2N_1} e^{-(i\frac{2\pi }{N}k)}^m=\frac{1}{N}e^{i\frac{2\pi }{N}{N}_{1}k}\frac{1-e^{-(i\frac{2\pi }{N}(2N_1+1))}}{1-e^{-(ik\frac{2\pi }{N})}}$

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 $N_1$ .

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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: