LECTURE 12 Sections Introduction to the Fourier series of periodic signals

Signals and Systems I Wednesday, February 11, 29 LECURE 12 Sections 3.1-3.3 Introduction to the Fourier series of periodic signals

Chapter 3: Fourier Series of periodic signals 3. Introduction 3.1 Historical perspective 3.2 he response of LI systems to complex exponentials 3.3 Fourier series representation of continuous-time periodic signals 3.4 Convergence of Fourier series 3.5 Properties of continuous-time Fourier series 3.8 Fourier series and LI systems 3.9 Filtering

Q. How to mae a rectangular shape out of circles?

he Fourier series By analogy, could we plot a square function using sinusoidal functions? f(t) t In the 18 s, Jean Baptiste Joseph Fourier found that most periodic functions with finite average power could be represented by a sum of sines and cosines xt () + + jω t ( ω ) sin ( ω ) a cos t + j t ae + ( ω ) sin ( ω ) a + A cos t + B t J.B. Fourier 1768-183

Example,

f 1Hz f 1Hz + f 3Hz + f 5Hz +... f 7, 9...Hz

With three sin(2πf t) components f 1, 3, 5 Hz With twentyfive sin(2πf t) components f 1, 3, 5...51 Hz

o represent any arbitrary periodic function, what frequency component must be chosen? What is the amplitude coefficient of each term?

Eigenfunctions of LI Difference and Differential Systems We already now that the complex exponentials of the type Ae st and Cz n remain basically invariant under the action of time shifts (difference systems) and derivatives (differential systems). he response of an LI system to a complex exponential input is the same complex exponential with only a change in (complex) amplitude. Continuous-time LI system: Discrete-time LI system: e z st n H() s e H( z) z st n Where the complex amplitude factors H(s), H(z) are functions of the complex variable s or z. 1

Input signals lie x(t)e st and x[n]z n for which the system output is a constant times the input signal are called eigenfunctions of the LI system, and the complex gains are the system's eigenvalues corresponding to the eigenfunctions. 11

o show that x(t)e st is indeed an eigenfunction of any LI system of impulse response h(t), we loo at the following convolution integral: () ( ) ( ) y t h τ x t τ dτ st ( τ ) ( τ ) h e dτ + st sτ e h e d ( τ ) he system's response has the form y(t)h(s)e st, where is an eigenvalue and e st is an eigenfunction + + τ z + sτ Hs () h() τ e dτ 12

Similarly for LI discrete-time systems, the complex exponential x[n]z n is an eigenfunction: [ ] [ ] [ ] yn hxn [ ] he system's response has the form z + + n hz + [ ] n h z yn [ ] Hzz ( ) n Where + H( z) h[ ] z 13

he eigenvalues of LI systems in continuous and discrete-time + s H() s h() τ τ e dτ + H( z) h[ ] z Are respectively nown as the Laplace transform and the z- transform of the system's impulse response (also called transfer function, more of this later in the course and Signals 2). 14

Linear Combinations of Harmonically- Related Complex Exponentials Periodic signals satisfy for some positive value of. ( ) ( + ) { < <+ } xt xt t he smallest such is the fundamental period and ω 2π/ (radians/s) is the fundamental frequency A periodic signal x(t) is completely nown from prior nowledge of only one period A special case of periodic signals, the harmonically-related complex exponentials have frequencies that are integer multiples of ω jω t φ t e ± ± (),, 1, 2, 16

Harmonically related exponentials form an orthogonal set. hat is, 2π 2π ω ω * jωt jmωt φ () t φ () t dt e e dt m 2π ω e j( m) ω t 1 j ( m) ω j( m)2π 2 π, ω, m dt e m 1 Use l Hopital s rule 17

A linear combination (i.e. summation) of harmonically-related complex exponentials φ (t) + + + ω xt () aφ ae ae is also periodic with fundamental period. erms with ±1 he fundamental frequency component or the first harmonic of the signal erms with ±2 second harmonic components (at frequency 2ω ) erms with ±N N th harmonic components. 2π j( ) t j t he representation of periodic signals in the form given above is referred to as the Fourier series representation 18

Example 4.1: Consider the periodic signal with fundamental frequency ω π/2 rad/s made up of the sum of five harmonic components: 5 π j t 2 () ae, 5 xt a a a a a a ± 1 ± 2 ± 3 ± 4 ± 5.226.225.81 19

Collecting the harmonic components together, we obtain xt () 5 5 ae π j t 2 π π 3π 3π 5π 5π j t j t j t j t j t j t 2 2 2 2 2 2.226( e + e ).225( e + e ).81( e + e ) π 3π 5π.452cos( t).45cos( t).162cos( t) 2 2 2 2

π 3π 5π x( t).452cos( t).45cos( t).162cos( t) 2 2 2 Relative amplitude ime [s] 21

Given a periodic function, how to represent it with a summation of harmonic components? xt () + j t ae ω We use the orthogonality property of the harmonically related complex exponentials. Multiplying both sides of the above equation with another complex exponential and integrating over one fundamental period of the signal jnωt jωt jnωt xte () dt ae e dt + jωt jnωt a e e dt + { } a n 1 jn t () ω a x t e dt 22

In conclusion, if a periodic signal x(t) has a Fourier series representation, then we have the Fourier series equation pair ω x t ae ae () + + 2π j t j t 2π j t j t 1 () ω 1 () a x t e dt x t e dt 23

+ + 2π j t j t () ω x t ae ae he first equation provides a time-domain representation of the signal as a sum of periodic complex exponential signals. his is the synthesis equation. 2π j t 1 1 j t () ω () a x t e dt x t e dt he second equation provides a frequency-domain representation of the signal. he a represent the Fourier series coefficients, or the spectral coefficients of x(t). 24

Example: he periodic sawtooth signal he fundamental period is 1s hence 2π ω 2π rad/s he first a coefficient with. 1 1 jωt a x(t)e dt a x(t)dt Since the average over one period is, we find a. 25

jωt a x t e dt 1 1 1 xte () j 2πt (1 2 te ) () dt j 2πt { } dt j 2πt 1 j 2πt 1 e e (1 2 t) 2 j2π j2π 1 1 1 + j2π j2π jπ j π ( ) 26

2π j t + j x() t ae 1, a, π ±1 ±1,2 ±1,2,3 27

2π j t + x() t ae 1, a ± j π 28

Signals and Systems I Wednesday, February 11, 29 LECURE 13 Section 3.3-More examples of Fourier series Section 3.4-Convergence of the Fourier series

Example Consider the following periodic rectangular wave of fundamental period and fundamental frequency ω 2π. 3

he Fourier series coefficients are, for a t 1 1dt t and for 1 t 2t jωt a e dt t 1 jω t t e jω t 1 jω 2sin( ωt) ω 2t sin π π a : average value of x(t) jω t jω t ( e e ) () x t 1 () jω t jωt a x t e dt + a e 31

he a s are scaled samples of the continuous sinc function defined as sinc( x) sinπ x π x his function equals to one at x and has zeros at x±n, n1, 2, 3 1 sinc( x) -3-2 -1 1 2 3 x 32

he spectral coefficients then become 2t sin π a a { } π 2t sin 2t 2t 2t sinc π 2t We define the duty cycle of the rectangular wave π η 2t sinc ( x) sinπ x π x and therefore a sinc( ) η η 33

For a 5% duty cycle, η.5 a sinc( ) η η 1 2 2 sinc -/4 /4 a t 1 1dt t 2t a : is also taen into account in the sinc function 34

Remember that is a multiple of the fundamental frequency ω, so for a 1 Hz (2π rad/s) square wave, a ±1 represents the fundamental components at ω 2π rad/sec, f 1Hz, a ±2 represents the second harmonic components at ω 4π rad/sec, f 2Hz 35

a sinc( ) η η For smaller values of duty cycle, the sinc envelope of the spectral coefficients expands, and a coefficients are more densely paced under in each lobe of the sinc enveloppe. 36

Example: Find the Fourier series coefficients of x t 2+ cosω t ( ) Clearly here, because this function can be rewritten as () x t 1 () jω t jωt a x t e dt + a e x t () the Fourier coefficients are 2 + e + e 2 jω t jω t a a 2 1 a 1 1 2 For a real function x(t), a real coefficient a involves a cosinus function 37

Example: Find the Fourier series coefficients of his function can be rewritten as x t sinω t+ 2sin 2ω t ( ) () x t 1 () jω t jωt a x t e dt + a e x t () e e e e + 2 2j 2j jω t jω t j2ω t j2ω t the Fourier coefficients are a a a 1 2 j a 1 j 1 1 2 2 For a real function x(t), an imaginary coefficient a involves a sinus function 38

Real form of the Fourier series If and only if the signal x(t) is real, xt () + ae + 1 jω t jωt jωt ( ) a + a e + a e + ( cos( ω ) sin ( ω ) cos( ω ) sin ( ω ) ) a + a t + j t + a t j t 1 + ([ ] cos( ω ) [ ] sin ( ω )) a + a + a t + a a j t 1 then we must have a - a * in order to get a +a - and j(a -a - ) to be real If we now represent the Fourier series coefficients as a B +jc, we obtain the real form of the Fourier series + ( ( ω ) ( ω )) xt () a + B cos t C sin t 1 39

Graph of the Fourier Series Coefficients: he Line Spectrum he set of complex Fourier series coefficients a, {- <<+ } of a signal can be plotted in two distinct graphs: magnitude vs (or ω ) and phase vs (or ω ). he combination of both plots is called the line spectrum of the signal. his representation contains all the information required to recover x(t). 4

Consider the Fourier series coefficients of the sawtooth signal heir magnitude is given by a, a a j a, π 1 π { } heir phase is given by ( a ) ( a ) Im a arctan Re π, > a 2 π, 2 < a 41

ry this http://ptolemy.eecs.bereley.edu/eecs2/wee8/examples.html 42

Convergence of the Fourier Series he Fourier coefficients a enable a signal approximation that becomes more precise as the number of Fourier coefficients increase. How precise is a Fourier series to represent a periodic signal? o answer this, consider the Fourier series truncated to N th coefficient We define the error signal as N j x () t a e ω N + N N e t x t x t ( ) ( ) ( ) N + N xt () j N t ae ω t 44

he error signal at each point can be integrated under the form of the energy of the error signal evaluated over one period Expanding the equation above * N N N E e t e t dt () () E () N en t dt + N + N () ω () N N j t jωt xt ae x t a e dt 2 45

+ N + N + jωt jnωt E x() t x () t dt a e a e dt N n N n N + N + N jωt jωt x() t a e dt x () t a e dt N N + N N + N 2 ( ) N + n N n N N j n ωt jωt E x() t dt a a e dt a x() t e dt + N N a x ( te ) jω t dt + N + N + N 2 + jωt jωt E xt () dt aa a xte () dt a x() te dt N N N N 46

We express the Fourier coefficients component in rectangular coordinates to highlight the real and imaginary parts a Α +jβ, + N + N 2 2 2 N N N jωt E xt () dt+ ( A + B ) ( A jb) xte () dt + N ( + ) ( ) N jωt A jb x t e dt () + ( + ) 2 Re () + N jωt 2 B Im x( t) e dt N + N + N 2 2 2 N N N jωt E xt dt A B A xte dt hen, we want to find the function that minimizes the energy E N (α, β ) by taing the partial derivative with respect to Α and Β and setting it equal to zero. 47

Differentiating with respect to Α, we obtain + N + N EN A x t e dt A N N jωt 2 2 Re ( ) his equation is satisfied with 1 Re ( ) jωt A xte dt 48

Similarly, minimizing the energy of the approximation error with respect to Β yields herefore, the complex coefficients minimize the approximation error. 1 Im ( ) jωt B x t e dt 1 + jωt a A jb x t e dt his shows that the Fourier series is the most efficient way to express a periodic signal with a sum of harmonic components. () 49

Moreover, if the signal x(t) has a Fourier series representation, then the limit of the approximation error as N tends to infinity is zero. z EN en() t 2 dt For this reason, the Fourier series converges to the signal x(t) with increasing N. 5

Existence of a Fourier Series Representation What classes of periodic signals have Fourier series representation? We will distinguish between two classes: 1) Periodic signals with finite energy over one period (finite total average power), i.e., signals with z xt () 2 dt< hese signals have Fourier series that converge in the sense that the energy in the error between the signal and its Fourier series decreases as N increases. However, this does not implies that a signal x(t) and its Fourier representation are equal at every value of t. 51

2) More restricted class of signals--signals that satisfy the Dirichlet conditions. hese signals equal their Fourier series representation, except at isolated values of t where x(t) is discontinuous (finite jumps). At these values, the Fourier series converges to the average of the values on either side of the discontinuity. What are the Dirichlet conditions? 52

Dirichlet Condition 1 x(t) must be absolutely integrable over one period z xt () dt< For example, the function x t is not integrable over one period 1 t < 1 t () { } 53

Dirichlet Condition 2 In any finite interval of time, x(t) must be of bounded variations. his means that x(t) must have a finite number of maxima and minima during any single period. For example, F xt () sin H 2πI, t t K < 1 does not meet this requirement because it generates an infinite number of oscillations as t approaches from +. 54

Dirichlet Condition 3 In any finite interval of time, x(t) must have a finite number of discontinuities. Furthermore, each of these discontinuities must be finite. For example, the function ( ) x t u t n () [ ] n 1 does not meet this requirement as it has an infinite number of discontinuities over one period 55

Example: Square wave N () + N x t a e N N jω t ( ω ) a + 2 a cos t 1 he Fourier series reaches a limitation named the Gibbs Phenomenon when x(t) has a discontinuity. For example, the periodic square wave 56

With the Dirichlet conditions, the Fourier series leads to pointwise convergence that is, x t x t ( ) ( ) N s s N + for any point t s except at a discontinuity where the approximation converges to half of the jump. he Gibbs phenomenon eeps the amplitude of the first overshoot on both sides of the discontinuity to remain at 9% of the height of the discontinuity. 57

Suggested reading/problems Read sections 3.1-3.4 Problems: 3.1, 3.3, 3.4, 3.6, 3.22 58

Question () x t 1 () jω t jωt a x t e dt + a e A periodic signal x(t) is expressed in volts and has a 1 A/2j and a -1 -A/2j. What are the units of a? 59

Signals and Systems I Wednesday, February 11, 29 LECURE 14 Section 3.5-Properties of continuous-time Fourier series Assignment 2 will be posted after class

Properties of Continuous-ime Fourier Series he set of spectral coefficients a, [-, ] determines x(t) completely. he conversion between the signal and its spectral representation is denoted as xt () a he following properties of the Fourier series are easy to show from () x t 1 FS () You may prove them as an exercise. jω t jωt a x t e dt + a e 62

Linearity he operation of calculating the Fourier series of a periodic signal is linear. Considering at first if we form the linear combination then we have FS FS xt () a y() t b z() t Ax() t + By() t FS z() t Aa + Bb 63

ime Shifting ime shifting leads to a multiplication by a complex exponential. xt () a jω t xt ( t) e a FS FS he magnitudes of the Fourier series coefficients are not changed, but only their phases is changed. 64

Example: 2 a, + π j t x() t ae j a, π Delay of t /2 j π 2 j e jπ π j ( 1) π jω t a e t 65

ime Reversal ime reversal leads to a "sequence reversal" of the corresponding sequence of Fourier series coefficients: FS x( t) a 66

Example: 2π j( )t + x( t ) ae 1, a j π ime reversal a becomes a - a a j π j π 67

ime Scaling ime scaling applied on a periodic signal changes the fundamental frequency of the signal, but it remains periodic with the same profile. For example x(αt) has fundamental frequency αω and fundamental period /α. he Fourier series coefficients do not change FS x( αt) a but the harmonic components now have frequencies ± ω ± αω ± 2ω ± 2αω ± 3ω ± 3αω 68

Multiplication of wo Signals A multiplication of two function in the time domain converts into a convolution of their Fourier coefficients in the frequency domain. Suppose that x(t) and y(t) are both periodic with period x t () y t () FS a FS b then + x t y t c a b () () FS l l 69

Proof ω () () () () + + j t jzωt z z xt ae yt be xt yt + + z abe z ( + ) j zω t Changing for variable lz+, (and zl-), therefore l also spreads from - to () () + + jlωt l l + + jlωt ab l e l + jlωt ( al bl) e l xt yt ab e 7

Conjugate Symmetry of real signals If and only if x(t) is real, the Fourier coefficients must be conjugate symmetric his implies a () * x t real a a a a Re{ a } Re{ a } a Im{ a } Im{ a } 71

In the particular case where x(t) is real and even, the sequence of Fourier coefficients is also real and even x() t even () * x t real a a a a his happens because x(t) real and even can be expressed as a sum of cosines and the Fourier coefficients arising from cosines are also real and even jωt jωt e + e B cos( ωt) B 2 B jωt B jωt B e + e a a 2 2 2 72

riangular wave Real and even function a ( ) ( π ) 1 1 2 a 2 2 a ( π ) ( π ) 1 2 1 2 a a 73

Square wave Real and even function a π sin 2 π 2 a π π sin sin 2 2 2 2 a π π π π 2 2 1 1 a a 74

In the special case where x(t) is real and odd, the sequence of coefficients is purely imaginary and odd () his happens because x(t) real and odd can be expressed as a sum of sinuses and the Fourier coefficients arising from a sinus are also imaginary and odd () xtodd * x t real a a a a imaginary ( ω ) C sin t C e e 2 j jω t jω t C jωt C jωt C e e a a 2j 2j 2j 75

a a j π j j a π π 1 1 a a Imaginary a a j π 1 1 ( ) 1 j j a π π a Imaginary a Delay of t /2 76

Average power he average power of a complex signal x(t) is equal to the integral over one period of the signal squared in amplitude 1 2 P x() t dt 1 2 1 r r 1 jω t 2 jωt * jrωt ae ae dt * r 1 + r ( ) j rω t ( ) * * j rωt aa aae dt r r r r a ae aae dt dt 77

Parseval heorem he average power of a periodic signal x(t) is equal to the sum of the average powers in all of its harmonic components. he average power in the th harmonic components of a signal is: 1 jωt P ae dt 1 a 2 a P 2 dt 2 Note that for a real signal, P P - so the total power of the th harmonic of the signal is 2P. a 2 78

Example: Let us compute the average power in the unit-amplitude square wave of period and 5% duty cycle. -/4 /4 We already now the value of its Fourier spectral coefficients: a sinc( ) η η 1 2 2 sinc 79

Using Parseval s relation, we compute the average signal power from the frequency domain P a 2 1 sinc 2 2 1 1 + 2 sinc 4 4 2 1 π sin 1 1 + 2 4 2 π 1,3,5,... 2 2 2 2 P 1 1 + 2 2 2 4 π 1,3,5,... 2 1 2 π + 2 4 π 8 1 2 8

and compare this result against the average power in the time domain 1 2 P x() t dt 1 4 4 1 2 dt 1 + 4 4 1 2 -/4 In conclusion, Parseval s theorem provides the average power contained in individual spectral components P /4 a 2 81

Power Spectrum he power spectrum of a signal is the sequence of average powers in each harmonic component a For real periodic signals, the power spectrum is a real even sequence as 2 2 2 a a a Example: Power spectrum of the rectangular wave η1/8. 2 2 2 ( ) a η sinc η ( ) 1 2 sin c 64 8 All power is mostly contained in the main lobe, that is, the DC part + six harmonic components ω ω 82

Example: he power spectrum of ( ) ( ω ) x t Asin t A e 2 j ( jω ) t e jωt A A a1 j, a 1 j 2 2 2 a.25a 2-1 1 83

otal Harmonic Distortion Suppose that a signal that was expected to be a pure sine wave of amplitude A is distorted One way to quantify the level of distortion is by using the total harmonic distortion 84

otal harmonic distortion (HD) HD + 2 a a 1 2 2 1% he HD represents the total average power from all the harmonics that should not be there, that is, for >1, divided by the average power of the pure sine wave, that is, for ±1. his ratio of powers is taen in square root to reduce it in units of amplitude ratio 85

Suggested problems Demonstrate the properties of the Fourier series Linearity ime-shifting ime reversal ime-scaling Multiplication Conjugate symmetry of real signals 87

Signals and Systems I Wednesday, February 11, 29 LECURE 15 More Properties of Fourier Series Section 3.8-Fourier series and LI systems Section 3.9-Filtering

We have seen Linearity ime shifting FS FS jω t xt () a xt ( t) e a ime reversal ime scaling Multiplication of two signals Conjugate symmetry of real signals FS FS x() t a y() t b zt () Axt () + Byt () z t Aa Bb () FS + FS x( t) a FS x( αt) a ± ω ± αω, ± 2ω ± 2 αω, FS FS FS () () () () xt a yt b xt yt c ab l l * () x t real a a + 89

More properties of Fourier Series ime differentiation Given a signal xt () FS a he time derivative of x(t) has Fourier coefficients given by d xt j a dt () FS ω Proof: () xt ae jω t d xt () j ω ae dt jω t { ω } jωt be b j a 9

ime integration Given a signal xt () FS a he running integral of x(t) has Fourier coefficients given by t FS a { } x( τ) dτ if a jω a Non periodic integral Proof: () xt ae jω t t x d e ( τ) τ a jω jω t ω be b j t a jω 91

Fourier Series of a Periodic rain of Impulses It is useful to have the Fourier series representation of an impulse train. 1 + vt () δ( t n) n -2-2 3 t Such a signal falls into the first category of signals that can be represented by Fourier series. hat is, it contains a finite amount of energy in one period and will thus converge towards the fonction with increasing number of Fourier coefficients. However, this signal does not meet the Dirichlet conditions. he Fourier representation will thus not coincide point to point with the original function. 92

Reminder Dirichlet Condition 1 x(t) must be absolutely integrable over one period z xt () dt< Dirichlet Condition 2 In any finite interval of time, x(t) must be of bounded variations. his means that x(t) must have a finite number of maxima and minima during any single period. Dirichlet Condition 3 In any finite interval of time, x(t) has a finite number of discontinuities. Furthermore, each of these discontinuities is finite. 93

1 jωt a v t e dt 1 1 /2 /2 () δ () te jω t dt he spectrum of an impulse train is a real, constant sequence. a 1/ 1 + () δ ( ) vt t n n -2-2 3 t -16-8 8 16 94

Example A periodic signal y(t) can be expressed as a convolution of a finite duration signal h(t) with a train of impulses v(t). vt ( ) y h( t ) ( t) + + + yt () δ ( t n) ht () δτ ( nht ) ( τ) dτ n n + + + δτ ( n ) h( t τ) dτ h( t n ) n n 95

Eigenfunctions of LI Differential Systems We already now from section 3.2 that the complex exponentials of the type Ae st remain basically invariant under the action of time derivatives in differential systems. As a result, the response of an LI system to a complex exponential input is the same complex exponential with only a change in (complex) amplitude. Continuous-time LI system: e st H() s e st Where the complex amplitude factors H(s), is a function of the complex variable s. 97

Response of LI Systems to Periodic Signals A special case of the general s transformation (Laplace transform) z + sτ Hs () h() τ e dτ is the Fourier transform for which sjω (purely imaginary) + j ( ) ( ) ωτ H jω h τ e dτ Referring here to h(t) as the impulse response of a system, then the complex function H(jω) is called the system's frequency response. aing again the result from a convolution, the output of a system to an input x(t)e jωt would then have the form ω ω τ τ + jωτ y t e H j H j h e d jωt ( ) ( ) ( ) ( ) 99

Consider a periodic input signal which has a Fourier series representation xt () + ae ω j t his signal is made of a summation of exponentials, all eigenfunctions of LI systems. As a consequence, the response of an LI system to such input is the same sum of complex exponentials multiplied by their eigenvalue, the system s frequency response + yt () ah( jω ) e jω t hus, the output signal y(t) is periodic with Fourier series coefficients b given by b a H( j ) ω 1

Example Input: Periodic rectangular wave Fourier coefficients following a sinc LI system s impulse response: A sinc function Square frequency response a xt () y() t ht () H( jω ) ( ) b a H jω Application: Filtering Filtering periodic signals with an LI system involves the design of a filter with a desirable frequency spectrum H(jω o ) that retains some frequencies and cuts off others. 11

Example A periodic signal y(t) can be expressed as a convolution of a finite duration signal h(t) with a train of impulses v(t). vt ( ) y h( t ) ( t) () FS a v t F () ( ) h t H jω F () ( ) y t a H jω A periodic signal y(t) can ALSO be expressed as a multiplication of the frequency response of a signal h(t) (H(jω)) with the Fourier coefficients of a train of impulses v(t) (a ). 12

RC FILER A first order low-pass filter with impulse response 1 t RC () () h t e u t RC attenuates high-frequency harmonics in a periodic input signal. he frequency response of this filter is: + jωτ H j h e d ( ) ( ) ω τ τ + τ RC e e RC 1 1 + jωrc jωτ dτ 14

H ( jω ) 1 + 1 jωrc Woring with Fourier series, we have ωω where ω is the fundamental frequency of the signal H ( jω ) 1 + 1 jω RC As the frequency order increases, the magnitude of the frequency response of the filter decreases H ( jω ) 1+ 1 ( ω RC) 2 herefore high frequencies are attenuated. 15

Example: If the input signal is a rectangular wavefunction with zero average a 2t η ( η ) a η sinc he output signal will have Fourier series coefficients given by b b ah( jω) sin( ω t ) 1 π (1 + jωrc) 16

1 t RC () () ht e ut RC RC.5 ( jω ) 1 (. π ) 1+ 1 2 Input Output Attenuation on high frequencies smoother output 17

1 t RC () () ht e ut RC RC.1 ( jω ) 1 (. π ) 1+ 2 2 Input Output 18

1 t RC h( t ) e u( t ) RC RC.2 ( jω ) 1 (. π ) 1+ 4 2 Input Output 19

1 t RC h( t ) e u( t ) RC RC 1 H ( jω ) 1 ( π ) 1+ 2 2 Input Output 11