# 6.3 Properties of the ctfs

 Page 1 / 1
An introduction to the general properties of the Fourier series

## Introduction

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

$f(t)=\sum_{n=()}$ c n ω 0 n t
${c}_{n}=\frac{1}{T}\int_{0}^{T} f(t)e^{-(i{\omega }_{0}nt)}\,d t$
Let $ℱ(·)$ denote the transformation from $f(t)$ to the Fourier coefficients $ℱ(f(t))=\forall n, n\in \mathbb{Z}\colon {c}_{n}$ $ℱ(·)$ maps complex valued functions to sequences of complex numbers .

## Linearity

$ℱ(·)$ is a linear transformation .

If $ℱ(f(t))={c}_{n}$ and $ℱ(g(t))={d}_{n}$ . Then $\forall \alpha , \alpha \in \mathbb{C}\colon ℱ(\alpha f(t))=\alpha {c}_{n}$ and $ℱ(f(t)+g(t))={c}_{n}+{d}_{n}$

Easy. Just linearity of integral.

$ℱ(f(t)+g(t))=\forall n, n\in \mathbb{Z}\colon \int_{0}^{T} (f(t)+g(t))e^{-(i{\omega }_{0}nt)}\,d t=\forall n, n\in \mathbb{Z}\colon \frac{1}{T}\int_{0}^{T} f(t)e^{-(i{\omega }_{0}nt)}\,d t+\frac{1}{T}\int_{0}^{T} g(t)e^{-(i{\omega }_{0}nt)}\,d t=\forall n, n\in \mathbb{Z}\colon {c}_{n}+{d}_{n}={c}_{n}+{d}_{n}$

## Shifting

Shifting in time equals a phase shift of Fourier coefficients

$ℱ(f(t-{t}_{0}))=e^{-(i{\omega }_{0}n{t}_{0})}{c}_{n}$ if ${c}_{n}=\left|{c}_{n}\right|e^{i\angle ({c}_{n})}$ , then $\left|e^{-(i{\omega }_{0}n{t}_{0})}{c}_{n}\right|=\left|e^{-(i{\omega }_{0}n{t}_{0})}\right|\left|{c}_{n}\right|=\left|{c}_{n}\right|$ $\angle (e^{-(i{\omega }_{0}{t}_{0}n)})=\angle ({c}_{n})-{\omega }_{0}{t}_{0}n$

$ℱ(f(t-{t}_{0}))=\forall n, n\in \mathbb{Z}\colon \frac{1}{T}\int_{0}^{T} f(t-{t}_{0})e^{-(i{\omega }_{0}nt)}\,d t=\forall n, n\in \mathbb{Z}\colon \frac{1}{T}\int_{-{t}_{0}}^{T-{t}_{0}} f(t-{t}_{0})e^{-(i{\omega }_{0}n(t-{t}_{0}))}e^{-(i{\omega }_{0}n{t}_{0})}\,d t=\forall n, n\in \mathbb{Z}\colon \frac{1}{T}\int_{-{t}_{0}}^{T-{t}_{0}} f(\stackrel{~}{t}())e^{-(i{\omega }_{0}n\stackrel{~}{t})}e^{-(i{\omega }_{0}n{t}_{0})}\,d t=\forall n, n\in \mathbb{Z}\colon e^{-(i{\omega }_{0}n\stackrel{~}{t})}{c}_{n}$

## Parseval's relation

$\int_{0}^{T} \left|f(t)\right|^{2}\,d t=T\sum_{n=()}$ c n 2
Parseval's relation tells us that the energy of a signal is equal to the energy of its Fourier transform.
Parseval tells us that the Fourier series maps $L(\left[0 , T\right])^{2}$ to $l(\mathbb{Z})^{2}$ .

For $f(t)$ to have "finite energy," what do the ${c}_{n}$ do as $n\to$ ?

$\left|{c}_{n}\right|^{2}$ for $f(t)$ to have finite energy.

If $\forall n, \left|n\right|> 0\colon {c}_{n}=\frac{1}{n}$ , is $f\in L(\left[0 , T\right])^{2}$ ?

Yes, because $\left|{c}_{n}\right|^{2}=\frac{1}{n^{2}}$ , which is summable.

Now, if $\forall n, \left|n\right|> 0\colon {c}_{n}=\frac{1}{\sqrt{n}}$ , is $f\in L(\left[0 , T\right])^{2}$ ?

No, because $\left|{c}_{n}\right|^{2}=\frac{1}{n}$ , which is not summable.

The rate of decay of the Fourier series determines if $f(t)$ has finite energy .

## Even signals

• $f\left(t\right)=f\left(-t\right)$
• $\parallel {c}_{n}\parallel =\parallel {c}_{-n}\parallel$
• ${c}_{n}=\frac{1}{T}{\int }_{0}^{T}\phantom{\rule{-0.166667em}{0ex}}f\left(t\right)exp\left(-ı{\omega }_{0}nt\right)\phantom{\rule{0.166667em}{0ex}}dt$
• $=\frac{1}{T}{\int }_{0}^{\frac{T}{2}}\phantom{\rule{-0.166667em}{0ex}}f\left(t\right)exp\left(-ı{\omega }_{0}nt\right)\phantom{\rule{0.166667em}{0ex}}dt+\frac{1}{T}{\int }_{\frac{T}{2}}^{T}\phantom{\rule{-0.166667em}{0ex}}f\left(t\right)exp\left(-ı{\omega }_{0}nt\right)\phantom{\rule{0.166667em}{0ex}}dt$
• $=\frac{1}{T}{\int }_{0}^{\frac{T}{2}}\phantom{\rule{-0.166667em}{0ex}}f\left(-t\right)exp\left(-ı{\omega }_{0}nt\right)\phantom{\rule{0.166667em}{0ex}}dt+\frac{1}{T}{\int }_{\frac{T}{2}}^{T}\phantom{\rule{-0.166667em}{0ex}}f\left(-t\right)exp\left(-ı{\omega }_{0}nt\right)\phantom{\rule{0.166667em}{0ex}}dt$
• $=\frac{1}{T}{\int }_{0}^{T}\phantom{\rule{-0.166667em}{0ex}}f\left(t\right)\left[exp,\left(ı{\omega }_{0}nt\right),\phantom{\rule{0.166667em}{0ex}},d,t,+,exp,\left(-ı{\omega }_{0}nt\right)\right]\phantom{\rule{0.166667em}{0ex}}dt$
• $=\frac{1}{T}{\int }_{0}^{T}\phantom{\rule{-0.166667em}{0ex}}f\left(t\right)2cos\left({\omega }_{0}nt\right)\phantom{\rule{0.166667em}{0ex}}dt$

## Odd signals

• $f\left(t\right)=\mathrm{-f}\left(\mathrm{-t}\right)$
• ${c}_{n}={c}_{-n}$ *
• ${c}_{n}=\frac{1}{T}{\int }_{0}^{T}\phantom{\rule{-0.166667em}{0ex}}f\left(t\right)exp\left(-ı{\omega }_{0}nt\right)\phantom{\rule{0.166667em}{0ex}}dt$
• $=\frac{1}{T}{\int }_{0}^{\frac{T}{2}}\phantom{\rule{-0.166667em}{0ex}}f\left(t\right)exp\left(-ı{\omega }_{0}nt\right)\phantom{\rule{0.166667em}{0ex}}dt+\frac{1}{T}{\int }_{\frac{T}{2}}^{T}\phantom{\rule{-0.166667em}{0ex}}f\left(t\right)exp\left(-ı{\omega }_{0}nt\right)\phantom{\rule{0.166667em}{0ex}}dt$
• $=\frac{1}{T}{\int }_{0}^{\frac{T}{2}}\phantom{\rule{-0.166667em}{0ex}}f\left(t\right)exp\left(-ı{\omega }_{0}nt\right)\phantom{\rule{0.166667em}{0ex}}dt-\frac{1}{T}{\int }_{\frac{T}{2}}^{T}\phantom{\rule{-0.166667em}{0ex}}f\left(-t\right)exp\left(ı{\omega }_{0}nt\right)\phantom{\rule{0.166667em}{0ex}}dt$
• $=-\frac{1}{T}{\int }_{0}^{T}\phantom{\rule{-0.166667em}{0ex}}f\left(t\right)\left[exp,\left(ı{\omega }_{0}nt\right),\phantom{\rule{0.166667em}{0ex}},d,t,-,exp,\left(-ı{\omega }_{0}nt\right)\right]\phantom{\rule{0.166667em}{0ex}}dt$
• $=-\frac{1}{T}{\int }_{0}^{T}\phantom{\rule{-0.166667em}{0ex}}f\left(t\right)2ısin\left({\omega }_{0}nt\right)\phantom{\rule{0.166667em}{0ex}}dt$

## Real signals

• $f\left(t\right)=f$ * $\left(t\right)$
• ${c}_{n}={c}_{-n}$ *
• ${c}_{n}=\frac{1}{T}{\int }_{0}^{T}\phantom{\rule{-0.166667em}{0ex}}f\left(t\right)exp\left(-ı{\omega }_{0}nt\right)\phantom{\rule{0.166667em}{0ex}}dt$
• $=\frac{1}{T}{\int }_{0}^{\frac{T}{2}}\phantom{\rule{-0.166667em}{0ex}}f\left(t\right)exp\left(-ı{\omega }_{0}nt\right)\phantom{\rule{0.166667em}{0ex}}dt+\frac{1}{T}{\int }_{\frac{T}{2}}^{T}\phantom{\rule{-0.166667em}{0ex}}f\left(t\right)exp\left(-ı{\omega }_{0}nt\right)\phantom{\rule{0.166667em}{0ex}}dt$
• $=\frac{1}{T}{\int }_{0}^{\frac{T}{2}}\phantom{\rule{-0.166667em}{0ex}}f\left(-t\right)exp\left(-ı{\omega }_{0}nt\right)\phantom{\rule{0.166667em}{0ex}}dt+\frac{1}{T}{\int }_{\frac{T}{2}}^{T}\phantom{\rule{-0.166667em}{0ex}}f\left(-t\right)exp\left(-ı{\omega }_{0}nt\right)\phantom{\rule{0.166667em}{0ex}}dt$
• $=\frac{1}{T}{\int }_{0}^{T}\phantom{\rule{-0.166667em}{0ex}}f\left(t\right)\left[exp,\left(ı{\omega }_{0}nt\right),\phantom{\rule{0.166667em}{0ex}},d,t,+,exp,\left(-ı{\omega }_{0}nt\right)\right]\phantom{\rule{0.166667em}{0ex}}dt$
• $=\frac{1}{T}{\int }_{0}^{T}\phantom{\rule{-0.166667em}{0ex}}f\left(t\right)2cos\left({\omega }_{0}nt\right)\phantom{\rule{0.166667em}{0ex}}dt$

## Differentiation in fourier domain

$(ℱ(f(t))={c}_{n})\implies (ℱ(\frac{d f(t)}{d t}})=in{\omega }_{0}{c}_{n})$

Since

$f(t)=\sum_{n=()}$ c n ω 0 n t
then
$\frac{d f(t)}{d t}}=\sum_{n=()}$ c n t ω 0 n t n c n ω 0 n ω 0 n t
A differentiator attenuates the low frequencies in $f(t)$ and accentuates the high frequencies. It removes general trends and accentuates areas of sharpvariation.
A common way to mathematically measure the smoothness of a function $f(t)$ is to see how many derivatives are finite energy.
This is done by looking at the Fourier coefficients of thesignal, specifically how fast they decay as $n\to$ .If $ℱ(f(t))={c}_{n}$ and $\left|{c}_{n}\right|$ has the form $\frac{1}{n^{k}}$ , then $ℱ(\frac{d^{m}f(t)}{dt^{m}})=(in{\omega }_{0})^{m}{c}_{n}$ and has the form $\frac{n^{m}}{n^{k}}$ .So for the ${m}^{\mathrm{th}}$ derivative to have finite energy, we need $\sum \left|\frac{n^{m}}{n^{k}}\right|^{2}$ thus $\frac{n^{m}}{n^{k}}$ decays faster than $\frac{1}{n}$ which implies that $2k-2m> 1$ or $k> \frac{2m+1}{2}$ Thus the decay rate of the Fourier series dictates smoothness.

## Integration in the fourier domain

If

$ℱ(f(t))={c}_{n}$
then
$ℱ(\int_{()} \,d \tau )$ t f τ 1 ω 0 n c n
If ${c}_{0}\neq 0$ , 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 and convolution

Given a signal $f(t)$ with Fourier coefficients ${c}_{n}$ and a signal $g(t)$ with Fourier coefficients ${d}_{n}$ , we can define a new signal, $y(t)$ , where $y(t)=f(t)g(t)$ . We find that the Fourier Series representation of $y(t)$ , ${e}_{n}$ , is such that ${e}_{n}=\sum_{k=()}$ c k d n - k . This is to say that signal multiplication in the time domainis equivalent to signal convolution in the frequency domain, and vice-versa: signal multiplication in the frequency domain is equivalent to signal convolution in the time domain.The proof of this is as follows

${e}_{n}=\frac{1}{T}\int_{0}^{T} f(t)g(t)e^{-(i{\omega }_{0}nt)}\,d t=\frac{1}{T}\int_{0}^{T} \sum_{k=()} \,d t$ c k ω 0 k t g t ω 0 n t k c k 1 T t 0 T g t ω 0 n k t k c k d n - k
for more details, see the section on Signal convolution and the CTFS

## Conclusion

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

 Property Signal CTFS Linearity $ax\left(t\right)+by\left(t\right)$ $aX\left(f\right)+bY\left(f\right)$ Time Shifting $x\left(t-\tau \right)$ $X\left(f\right){e}^{-j2\pi f\tau /T}$ Time Modulation $x\left(t\right){e}^{j2\pi f\tau /T}$ $X\left(f-k\right)$ Multiplication $x\left(t\right)y\left(t\right)$ $X\left(f\right)*Y\left(f\right)$ Continuous Convolution $x\left(t\right)*y\left(t\right)$ $X\left(f\right)Y\left(f\right)$

how do you translate this in Algebraic Expressions
why surface tension is zero at critical temperature
Shanjida
Need to simplify the expresin. 3/7 (x+y)-1/7 (x-1)=
. After 3 months on a diet, Lisa had lost 12% of her original weight. She lost 21 pounds. What was Lisa's original weight?
what is biological synthesis of nanoparticles
what's the easiest and fastest way to the synthesize AgNP?
China
Cied
types of nano material
I start with an easy one. carbon nanotubes woven into a long filament like a string
Porter
many many of nanotubes
Porter
what is the k.e before it land
Yasmin
what is the function of carbon nanotubes?
Cesar
I'm interested in nanotube
Uday
what is nanomaterials​ and their applications of sensors.
what is nano technology
what is system testing?
preparation of nanomaterial
Yes, Nanotechnology has a very fast field of applications and their is always something new to do with it...
what is system testing
what is the application of nanotechnology?
Stotaw
In this morden time nanotechnology used in many field . 1-Electronics-manufacturad IC ,RAM,MRAM,solar panel etc 2-Helth and Medical-Nanomedicine,Drug Dilivery for cancer treatment etc 3- Atomobile -MEMS, Coating on car etc. and may other field for details you can check at Google
Azam
anybody can imagine what will be happen after 100 years from now in nano tech world
Prasenjit
after 100 year this will be not nanotechnology maybe this technology name will be change . maybe aftet 100 year . we work on electron lable practically about its properties and behaviour by the different instruments
Azam
name doesn't matter , whatever it will be change... I'm taking about effect on circumstances of the microscopic world
Prasenjit
how hard could it be to apply nanotechnology against viral infections such HIV or Ebola?
Damian
silver nanoparticles could handle the job?
Damian
not now but maybe in future only AgNP maybe any other nanomaterials
Azam
Hello
Uday
I'm interested in Nanotube
Uday
this technology will not going on for the long time , so I'm thinking about femtotechnology 10^-15
Prasenjit
can nanotechnology change the direction of the face of the world
At high concentrations (>0.01 M), the relation between absorptivity coefficient and absorbance is no longer linear. This is due to the electrostatic interactions between the quantum dots in close proximity. If the concentration of the solution is high, another effect that is seen is the scattering of light from the large number of quantum dots. This assumption only works at low concentrations of the analyte. Presence of stray light.
the Beer law works very well for dilute solutions but fails for very high concentrations. why?
how did you get the value of 2000N.What calculations are needed to arrive at it
Privacy Information Security Software Version 1.1a
Good
Got questions? Join the online conversation and get instant answers!