Chapter 4 The Fourier Series and Fourier Transform

Chapter 4 The Fourier Series and Fourier Transform

Fourier Series Representation of Periodic Signals Let x(t) be a CT periodic signal with period T, i.e., xt ( + T) = xt ( ), t R Example: the rectangular pulse train

The Fourier Series Then, x(t) can be expressed as = k k = jkω0t xt () ce, t where is the fundamental frequency (rad/sec) of the signal and T /2 1 jkωot ck = x( t) e dt, k = 0, ± 1, ± 2, T c 0 ω0 = 2 π /T T /2 is called the constant or dc component of x(t)

Dirichlet Conditions A periodic signal x(t), has a Fourier series if it satisfies the following conditions: 1. x(t) is absolutely integrable over any period, namely a+ T a xt ( ) dt<, a 2. x(t) has only a finite number of maxima and minima over any period 3. x(t) has only a finite number of discontinuities over any period

Example: The Rectangular Pulse Train From figure, so T = 2 ω0 = 2 π /2= π Clearly x(t) satisfies the Dirichlet conditions and thus has a Fourier series representation

Example: The Rectangular Pulse Train Cont d 1 1 ( k 1) / 2 jkπt xt () = + ( 1) e, t 2 k k = k odd π

Trigonometric Fourier Series By using Euler s formula, we can rewrite as = k k = jkω0t xt () ce, t xt () = c + 2 c cos( kω t+ ck), t 0 k 0 k = 1 dc component k-th harmonic as long as x(t) is real This expression is called the trigonometric Fourier series of x(t)

Example: Trigonometric Fourier Series of the Rectangular Pulse Train The expression 1 1 ( k 1) / 2 jkπt xt () = + ( 1) e, t 2 kπ k = k odd can be rewritten as 1 2 ( k 1)/2 π xt () = + cos kπt+ ( 1) 1, t 2 k = 1 kπ 2 kodd

Gibbs Phenomenon Given an odd positive integer N, define the N-th partial sum of the previous series N 1 2 ( k 1)/2 π xn () t = + cos kπt+ ( 1) 1, t 2 k = 1 kπ 2 kodd According to Fourier s s theorem, it should be lim x () t x() t = 0 N N

Gibbs Phenomenon Cont d x3( t) x9( t)

Gibbs Phenomenon Cont d x t x45( t) 21( ) overshoot: about 9 % of the signal magnitude (present even if ) N

Parseval s Theorem Let x(t) be a periodic signal with period T The average power P of the signal is defined as T /2 1 2 P = x () t dt T T /2 Expressing the signal as it is also P = k = c k = k k = jkω0t xt () ce, t 2

Fourier Transform We have seen that periodic signals can be represented with the Fourier series Can aperiodic signals be analyzed in terms of frequency components? Yes, and the Fourier transform provides the tool for this analysis The major difference w.r.t. the line spectra of periodic signals is that the spectra of aperiodic signals are defined for all real values of the frequency variable ω not just for a discrete set of values

Frequency Content of the Rectangular Pulse xt () xt () t xt () = lim x() t T T

Frequency Content of the Rectangular Pulse Cont d x () t Since T is periodic with period T, we can write where jkω0t x () t c e, t T T /2 /2 = k k = 1 jkωot ck = x( t) e dt, k = 0, ± 1, ± 2, T T

What happens to the frequency components of xt () t as T? For k = 0: For Frequency Content of the Rectangular Pulse Cont d c0 = 1/ T k =± 1, ± 2, : c k 2 kω0 1 kω0 = sin = sin kω T 2 kπ 2 0 ω0 = 2 π /T

Frequency Content of the Rectangular Pulse Cont d plots of T c k vs. ω = kω0 for T = 2,5,10

Frequency Content of the Rectangular Pulse Cont d It can be easily shown that where ω lim Tck = sinc, ω T 2π sin( πλ sinc( ) ) λ πλ

Fourier Transform of the Rectangular Pulse The Fourier transform of the rectangular pulse x(t) is defined to be the limit of as, i.e., T Tc k ω X( ω) = limtck = sinc, T 2π ω X ( ω ) arg( X ( ω ))

The Fourier Transform in the General Case Given a signal x(t), its Fourier transform is defined as X ( ω) j t X( ω ω ) = x( t) e dt, ω A signal x(t) is said to have a Fourier transform in the ordinary sense if the above integral converges

The Fourier Transform in the General Case Cont d The integral does converge if 1. the signal x(t) is well-behaved 2. and x(t) is absolutely integrable, namely, xt ( ) dt < Note: well behaved means that the signal has a finite number of discontinuities, maxima, and minima within any finite time interval

Example: The DC or Constant Signal Consider the signal xt () 1, t Clearly x(t) does not satisfy the first requirement since = xt ( ) dt= dt= Therefore, the constant signal does not have a Fourier transform in the ordinary sense Later on, we ll see that it has however a Fourier transform in a generalized sense

Example: The Exponential Signal bt xt () = e ut (), b Consider the signal Its Fourier transform is given by bt j t X( ω ω ) = e u( t) e dt 1 = e dt = e b+ j ω ( b+ jω) t ( b+ jω) t t= 0 t= 0

Example: The Exponential Signal Cont d b < 0 X ( ω) b = 0 xt () = ut () X ( ω) If, does not exist If, and does not exist either in the ordinary sense b > 0 If, it is X ( ω) amplitude spectrum 1 X ( ω) = 2 2 b + ω 1 = b + jω phase spectrum ω arg( X ( ω)) = arctan b

Example: Amplitude and Phase Spectra of the Exponential Signal xt () = e ut () () = 10 t ()

Rectangular Form of the Fourier Transform Consider j t X( ω ω ) = x( t) e dt, ω X ( ω) Since in general is a complex function, by using Euler s formula X( ω) = x( t)cos( ωt) dt+ j x( t)sin( ωt) dt R( ω ) I ( ω ) X( ω) = R( ω) + ji( ω)

Polar Form of the Fourier Transform X( ) = R( ) + ji( ) ω ω ω can be expressed in a polar form as X( ω) = X( ω) exp( jarg( X( ω))) where 2 2 X( ω) = R ( ω) + I ( ω) I( ω) arg( X ( ω)) = arctan R( ω)

Fourier Transform of Real-Valued Signals If x(t) is real-valued, it is Moreover whence X( ω) = X ( ω) X ( ω) = X( ω) exp( jarg( X( ω))) X( ω) = X( ω) and arg( X( ω)) = arg( X( ω)) Hermitian symmetry

Example: Fourier Transform of the Rectangular Pulse Consider the even signal It is τ /2 ω ω ω ω 2 [ ] t= τ /2 2 ωτ X( ) = 2 (1)cos( t) dt = sin( t) = sin t= 0 ω 2 0 ωτ = τ sinc 2 π

Example: Fourier Transform of the Rectangular Pulse Cont d ωτ X ( ω) = τsinc 2 π

Example: Fourier Transform of the Rectangular Pulse Cont d amplitude spectrum phase spectrum

Bandlimited Signals A signal x(t) is said to be bandlimited if its Fourier transform X ( ω) is zero for all ω > B where B is some positive number, called the bandwidth of the signal It turns out that any bandlimited signal must have an infinite duration in time, i.e., bandlimited signals cannot be time limited

Bandlimited Signals Cont d If a signal x(t) is not bandlimited, it is said to have infinite bandwidth or an infinite spectrum Time-limited signals cannot be bandlimited and thus all time-limited signals have infinite bandwidth However, for any well-behaved signal x(t) it can be proven that lim X ( ω) = 0 ω whence it can be assumed that X( ω) 0 ω > B B being a convenient large number

Inverse Fourier Transform Given a signal x(t) with Fourier transform X ( ω), x(t) can be recomputed from X ( ω) by applying the inverse Fourier transform given by Transform pair 1 jωt xt () = X( ω) e dω, t 2π xt () Xω ( )

Properties of the Fourier Transform Linearity: xt () Xω ( ) yt () Yω ( ) α xt () + βyt () αx( ω) + βy( ω) Left or Right Shift in Time: j t xt ( t) X( ω) e ω 0 0 Time Scaling: xat ( ) 1 ω X a a

Properties of the Fourier Transform Time Reversal: x( t) X( ω ) Multiplication by a Power of t: n n n d txt () ( j) X( ω) n dω Multiplication by a Complex Exponential: xte jω0t () X( ω ω ) 0

Properties of the Fourier Transform Multiplication by a Sinusoid (Modulation): j xt ()sin( ω0t) X( ω + ω0) X( ω ω0) 2 1 xt ()cos( ω0t) X( ω + ω0) + X( ω ω0) 2 [ ] [ ] Differentiation in the Time Domain: d n xt n () ( j ω) X ( ω) n dt

Properties of the Fourier Transform Integration in the Time Domain: t 1 x( τ ) dτ X( ω) + πx(0) δ( ω) jω Convolution in the Time Domain: xt () yt () X( ω) Y( ω) Multiplication in the Time Domain: xt () yt () X( ω) Y( ω)

Properties of the Fourier Transform Parseval s Theorem: if Duality: 1 xt () ytdt () X ( ω) Y( ω) dω 2π yt () = xt () 1 ( ) ( ) 2 2 xt dt Xω dω 2π X() t 2 π x( ω )

Properties of the Fourier Transform - Summary

Example: Linearity x() t = p () t + p () t 4 2 2ω ω X ( ω) = 4sinc + 2sinc π π

Example: Time Shift xt () = p( t 1) 2 X ω ( ω) = 2sinc π j e ω

Example: Time Scaling p2( t) 2sinc ω π p2(2 t) ω sinc 2 π a > 1 0< a < 1 time compression frequency expansion time expansion frequency compression

Example: Multiplication in Time xt () = tp() t 2 d ω d sinω ωcosω sinω X( ω) = j 2sinc j2 j2 2 dω π = = dω ω ω

Example: Multiplication in Time Cont d ω cosω sinω X( ω) = j2 ω 2

Example: Multiplication by a Sinusoid xt () p()cos( t ω t) = sinusoidal τ 0 burst X 1 τ ( ω + ω0) τ( ω ω0) ( ω) = τsinc τsinc 2 + 2π 2π

Example: Multiplication by a Sinusoid Cont d X 1 τ ( ω + ω0) τ( ω ω0) ( ω) = τsinc τsinc 2 + 2π 2π ω 0 = 60 rad / sec τ = 0.5

Example: Integration in the Time Domain 2 t vt () = 1 pτ () t τ xt () = dv() t dt

Example: Integration in the Time Domain Cont d The Fourier transform of x(t) can be easily found to be τω τω X( ω) = sinc j2sin 4 π 4 Now, by using the integration property, it is 1 τ 2 τω V( ω) = X( ω) + πx(0) δ( ω) = sinc jω 2 4π

Example: Integration in the Time Domain Cont d τ 2 τω V ( ω) = sinc 2 4π

Generalized Fourier Transform Fourier transform of δ jωt () te dt= 1 δ () t Applying the duality property xt () = 1, t 2 πδ( ω) δ () t 1 generalized Fourier transform of the constant signal xt () = 1, t

Generalized Fourier Transform of Sinusoidal Signals [ ] cos( ω t) π δ( ω + ω ) + δ( ω ω ) 0 0 0 [ ] sin( ω t) jπ δ( ω + ω ) δ( ω ω ) 0 0 0

Fourier Transform of Periodic Signals Let x(t) be a periodic signal with period T; as such, it can be represented with its Fourier transform xt () jk 0t = ce ω ω0 = 2 π /T k = j t e ω 2 πδω ( ω ) 0 Since, it is k X( ω) = 2 πc δ k ( ω kω0) k = 0

Fourier Transform of the Unit-Step Function Since t ut () = δ ( τ) dτ using the integration property, it is t 1 ut () = δ ( τ) dτ + πδ( ω) jω

Common Fourier Transform Pairs