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# Introduction

In this module we will discuss the basic properties of the Discrete-Time Fourier Series. We will begin by refreshing your memory of our basic Fourier series equations:

$f⁢n=∑ k =0N−1 c k ⁢ei⁢ ω 0 ⁢k⁢n f n k N 1 0 c k ω 0 k n$
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$c k =1N⁢∑ n =0N−1f⁢n⁢e−(i⁢2⁢πN⁢k⁢n) c k 1 N n N 1 0 f n 2 N k n$
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Let $ℱ⁢· ℱ ·$ denote the transformation from $f⁢n f n$ to the Fourier coefficients $ℱ⁢f⁢n=∀ k ,k∈Z: c k ℱ f n k k c k$ $ℱ⁢· ℱ ·$ maps complex valued functions to sequences of complex numbers.

# Linearity

$ℱ⁢· ℱ ·$ is a linear transformation.

Theorem 1

If $ℱ⁢f⁢n= c k ℱ f n c k$ and $ℱ⁢g⁢n= d k ℱ g n d k$. Then $∀ α ,α∈C:ℱ⁢α⁢f⁢n=α⁢ c k α α ℱ α f n α c k$ and $ℱ⁢f⁢n+g⁢n= c k + d k ℱ f n g n c k d k$

Proof

Easy. Just linearity of integral.

$ℱ⁢f⁢n+g⁢n=∀ k ,k∈Z:∑ n =0N(f⁢n+g⁢n)⁢e−(i⁢ ω 0 ⁢k⁢n)=∀ k ,k∈Z:1N⁢∑ n =0Nf⁢n⁢e−(i⁢ ω 0 ⁢k⁢n)+1N⁢∑ n =0Ng⁢n⁢e−(i⁢ ω 0 ⁢k⁢n)=∀ k ,k∈Z: c k + d k = c k + d k ℱ f n g n k k n 0 N f n g n ω 0 k n k k 1 N n 0 N f n ω 0 k n 1 N n 0 N g n ω 0 k n k k c k d k c k d k$
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# Shifting

Shifting in time equals a phase shift of Fourier coefficients

Theorem 2

$ℱ⁢f⁢n− n 0 =e−(i⁢ ω 0 ⁢k⁢ n 0 )⁢ c k ℱ f n n 0 ω 0 k n 0 c k$ if $c k =| c k |⁢ei⁢∠⁢ c k c k c k ∠ c k$, then $|e−(i⁢ ω 0 ⁢k⁢ n 0 )⁢ c k |=|e−(i⁢ ω 0 ⁢k⁢ n 0 )|⁢| c k |=| c k | ω 0 k n 0 c k ω 0 k n 0 c k c k$ $∠⁢e−(i⁢ ω 0 ⁢ n 0 ⁢k)=∠⁢ c k − ω 0 ⁢ n 0 ⁢k ∠ ω 0 n 0 k ∠ c k ω 0 n 0 k$

Proof

$ℱ⁢f⁢n− n 0 =∀ k ,k∈Z:1N⁢∑ n =0Nf⁢n− n 0 ⁢e−(i⁢ ω 0 ⁢k⁢n)=∀ k ,k∈Z:1N⁢∑ n =− n 0 N− n 0 f⁢n− n 0 ⁢e−(i⁢ ω 0 ⁢k⁢(n− n 0 ))⁢e−(i⁢ ω 0 ⁢k⁢ n 0 )=∀ k ,k∈Z:1N⁢∑ n =− n 0 N− n 0 f⁢ n ~ ⁢e−(i⁢ ω 0 ⁢k⁢ n ~ )⁢e−(i⁢ ω 0 ⁢k⁢ n 0 )=∀ k ,k∈Z:e−(i⁢ ω 0 ⁢k⁢ n ~ )⁢ c k ℱ f n n 0 k k 1 N n 0 N f n n 0 ω 0 k n k k 1 N n n 0 N n 0 f n n 0 ω 0 k n n 0 ω 0 k n 0 k k 1 N n n 0 N n 0 f n ~ ω 0 k n ~ ω 0 k n 0 k k ω 0 k n ~ c k$
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# Parseval's Relation

$∑ n =0N|f⁢n|2=N⁢∑ k =0N−1| c k |2 n 0 N f n 2 N k N 1 0 c k 2$
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Parseval's relation tells us that the energy of a signal is equal to the energy of its Fourier transform.
Note:
Parseval tells us that the Fourier series maps $L20N L 0 N 2$ to $l2Z l 2$.

Exercise 1

For $f⁢n f n$ to have "finite energy," what do the $c k c k$ do as $k→∞ k$?

Solution

$| c k |2<∞ c k 2$ for $f⁢n f n$ to have finite energy.

Exercise 2

If $∀ k ,|k|>0: c k =1k k k 0 c k 1 k$, is $f∈L2 0 N f L 0 N 2$?

Solution

Yes, because $| c k |2=1k2 c k 2 1 k 2$, which is summable.

Exercise 3

Now, if $∀ k ,|k|>0: c k =1k k k 0 c k 1 k$, is $f∈L2 0 N f L 0 N 2$?

Solution

No, because $| c k |2=1k c k 2 1 k$, which is not summable.

The rate of decay of the Fourier series determines if $f⁢n f n$ has finite energy.

# Symmetry Properties

Rule 1: Even Signals
Even Signals
• $f[n]=f[-n]f[n]=f[-n]$
• $∥ck∥=∥c-k∥∥ck∥=∥c-k∥$
Proof
• $c k = 1 N Σ 0 N f [ n ] exp [ - ı ω 0 k n ] c k = 1 N Σ 0 N f [ n ] exp [ - ı ω 0 k n ]$
• $= 1 N Σ 0 N 2 f [ n ] exp [ - ı ω 0 k n ] + 1 N Σ N 2 N f [ n ] exp [ - ı ω 0 k n ] = 1 N Σ 0 N 2 f [ n ] exp [ - ı ω 0 k n ] + 1 N Σ N 2 N f [ n ] exp [ - ı ω 0 k n ]$
• $= 1 N Σ 0 N 2 f [ - n ] exp [ - ı ω 0 k n ] + 1 N Σ N 2 N f [ - n ] exp [ - ı ω 0 k n ] = 1 N Σ 0 N 2 f [ - n ] exp [ - ı ω 0 k n ] + 1 N Σ N 2 N f [ - n ] exp [ - ı ω 0 k n ]$
• $= 1 N Σ 0 N f [ n ] exp [ ı ω 0 k n ] + exp [ - ı ω 0 k n ] = 1 N Σ 0 N f [ n ] exp [ ı ω 0 k n ] + exp [ - ı ω 0 k n ]$
• $= 1 N Σ 0 N f [ n ] 2 cos [ ω 0 k n ] = 1 N Σ 0 N f [ n ] 2 cos [ ω 0 k n ]$
Rule 2: Odd Signals
Odd Signals
• $f [ n ] = -f [ -n ] f [ n ] = -f [ -n ]$
• $ck=c-kck=c-k$*
Proof
• $c k = 1 N Σ 0 N f [ n ] exp [ - ı ω 0 k n ] c k = 1 N Σ 0 N f [ n ] exp [ - ı ω 0 k n ]$
• $= 1 N Σ 0 N 2 f [ n ] exp [ - ı ω 0 k n ] + 1 N Σ N 2 N f [ n ] exp [ - ı ω 0 k n ] = 1 N Σ 0 N 2 f [ n ] exp [ - ı ω 0 k n ] + 1 N Σ N 2 N f [ n ] exp [ - ı ω 0 k n ]$
• $= 1 N Σ 0 N 2 f [ n ] exp [ - ı ω 0 k n ] - 1 N Σ N 2 N f [ - n ] exp [ ı ω 0 k n ] = 1 N Σ 0 N 2 f [ n ] exp [ - ı ω 0 k n ] - 1 N Σ N 2 N f [ - n ] exp [ ı ω 0 k n ]$
• $= - 1 N Σ 0 N f [ n ] exp [ ı ω 0 k n ] - exp [ - ı ω 0 k n ] = - 1 N Σ 0 N f [ n ] exp [ ı ω 0 k n ] - exp [ - ı ω 0 k n ]$
• $= - 1 N Σ 0 N f [ n ] 2 ı sin [ ω 0 k n ] = - 1 N Σ 0 N f [ n ] 2 ı sin [ ω 0 k n ]$
Rule 3: Real Signals
Real Signals
• $f[n]=ff[n]=f$*$[n][n]$
• $ck=c-kck=c-k$*
Proof
• $c k = 1 N Σ 0 N f [ n ] exp [ - ı ω 0 k n ] c k = 1 N Σ 0 N f [ n ] exp [ - ı ω 0 k n ]$
• $= 1 N Σ 0 N 2 f [ n ] exp [ - ı ω 0 k n ] + 1 N Σ N 2 N f [ n ] exp [ - ı ω 0 k n ] = 1 N Σ 0 N 2 f [ n ] exp [ - ı ω 0 k n ] + 1 N Σ N 2 N f [ n ] exp [ - ı ω 0 k n ]$
• $= 1 N Σ 0 N 2 f [ - n ] exp [ - ı ω 0 k n ] + 1 N Σ N 2 N f [ - n ] exp [ - ı ω 0 k n ] = 1 N Σ 0 N 2 f [ - n ] exp [ - ı ω 0 k n ] + 1 N Σ N 2 N f [ - n ] exp [ - ı ω 0 k n ]$
• $= 1 N Σ 0 N f [ n ] exp [ ı ω 0 k n ] + exp [ - ı ω 0 k n ] = 1 N Σ 0 N f [ n ] exp [ ı ω 0 k n ] + exp [ - ı ω 0 k n ]$
• $= 1 N Σ 0 N f [ n ] 2 cos [ ω 0 k n ] = 1 N Σ 0 N f [ n ] 2 cos [ ω 0 k n ]$

# Differentiation in Fourier Domain

$(ℱ⁢f⁢n= c k )⇒(ℱ⁢df⁢nd n =i⁢k⁢ ω 0 ⁢ c k ) ℱ f n c k ℱ n f n k ω 0 c k$
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Since

$f⁢n=∑ k =0N c k ⁢ei⁢ ω 0 ⁢k⁢n f n k 0 N c k ω 0 k n$
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then
$d d n f⁢n=∑ k =0N c k ⁢dei⁢ ω 0 ⁢k⁢nd n =∑ k =0N c k ⁢i⁢ ω 0 ⁢k⁢ei⁢ ω 0 ⁢k⁢n n f n k 0 N c k n ω 0 k n k 0 N c k ω 0 k ω 0 k n$
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A differentiator attenuates the low frequencies in $f⁢n f n$ and accentuates the high frequencies. It removes general trends and accentuates areas of sharp variation.
Note:
A common way to mathematically measure the smoothness of a function $f⁢n f n$ is to see how many derivatives are finite energy.
This is done by looking at the Fourier coefficients of the signal, specifically how fast they decay as $k→∞ k$. If $ℱ⁢f⁢n= c k ℱ f n c k$ and $| c k | c k$ has the form $1kl 1 k l$, then $ℱ⁢d m f⁢nd n m =i⁢k⁢ ω 0 m⁢ c k ℱ n m f n k ω 0 m c k$ and has the form $kmkl k m k l$. So for the $m th m th$ derivative to have finite energy, we need $∑ k |kmkl|2<∞ k k m k l 2$ thus $kmkl k m k l$ decays faster than $1k 1 k$ which implies that $2⁢l−2⁢m>1 2 l 2 m 1$ or $l>2⁢m+12 l 2 m 1 2$ Thus the decay rate of the Fourier series dictates smoothness.

# Integration in the Fourier Domain

If

$ℱ⁢f⁢n= c k ℱ f n c k$
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then
$ℱ⁢∑ η =0nf⁢η=1i⁢ ω 0 ⁢k⁢ c k ℱ η 0 n f η 1 ω 0 k c k$
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Note:
If $c 0 ≠0 c 0 0$, this expression doesn't make sense.

Integration accentuates low frequencies and attenuates high frequencies. Integrators bring out the general trends in signals and suppress short term variation (which is noise in many cases). Integrators are much nicer than differentiators.

# Signal Multiplication

Given a signal $f⁢n f n$ with Fourier coefficients $c k c k$ and a signal $g⁢n g n$ with Fourier coefficients $d k d k$, we can define a new signal, $y⁢n y n$, where $y⁢n=f⁢n⁢g⁢n y n f n g n$. We find that the Fourier Series representation of $y⁢n y n$, $e k e k$, is such that $e k =∑ l =0N c l ⁢ d k - l e k l 0 N c l d k - l$. This is to say that signal multiplication in the time domain is equivalent to discrete-time circular convolution in the frequency domain. The proof of this is as follows

$e k =1N⁢∑ n =0Nf⁢n⁢g⁢n⁢e−(i⁢ ω 0 ⁢k⁢n)=1N⁢∑ n =0N∑ l =0N c l ⁢ei⁢ ω 0 ⁢l⁢n⁢g⁢n⁢e−(i⁢ ω 0 ⁢k⁢n)=∑ l =0N c l ⁢(1N⁢∑ n =0Ng⁢n⁢e−(i⁢ ω 0 ⁢(k−l)⁢n))=∑ l =0N c l ⁢ d k - l e k 1 N n 0 N f n g n ω 0 k n 1 N n 0 N l 0 N c l ω 0 l n g n ω 0 k n l 0 N c l 1 N n 0 N g n ω 0 k l n l 0 N c l d k - l$
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# Conclusion

Like other Fourier transforms, the DTFS has many useful properties, including linearity, equal energy in the time and frequency domains, and analogs for shifting, differentation, and integration.

 Property Signal DTFS Linearity $a x ( n ) + b y ( n ) a x ( n ) + b y ( n )$ $a X ( k ) + b Y ( k ) a X ( k ) + b Y ( k )$ Time Shifting $x ( n - m ) x ( n - m )$ $X ( k ) e - j 2 π m k / N X ( k ) e - j 2 π m k / N$ Time Modulation $x ( n ) e j 2 π m n / N x ( n ) e j 2 π m n / N$ $X ( k - m ) X ( k - m )$ Multiplication $x ( n ) y ( n ) x ( n ) y ( n )$ $X ( k ) * Y ( k ) X ( k ) * Y ( k )$ Circular Convolution $x ( n ) * y ( n ) x ( n ) * y ( n )$ $X ( k ) Y ( K ) X ( k ) Y ( K )$