Chapter 4 The Fourier Series and Fourier Transform Representation of Signals in Terms of Frequency Components Consider the CT signal defined by N xt () = Acos( ω t+ θ ), t k = 1 k k k The frequencies `present in the signal are the frequency of the component sinusoids ω k The signal x(t) is completely characterized by the set of frequencies ω k, the set of amplitudes A, and the set of phases θ k k 1
Example: Sum of Sinusoids Consider the CT signal given by x( t) = A1cos( t) + A2cos(4 t+ π / 3) + A3cos(8 t+ π / 2), t The signal has only three frequency components at 1,4, and 8 rad/sec, amplitudes A and phases 1, A2, A3 0, π / 3, π / 2 The shape of the signal x(t) depends on the relative magnitudes of the frequency components, specified in terms of the amplitudes A, A, A 1 2 3 Example: Sum of Sinusoids Cont d A1 = 0.5 A2 = 1 A3 = 0 A1 = 1 A2 = 0.5 A3 = 0 A1 = 1 A2 = 1 A3 = 0 2
Example: Sum of Sinusoids Cont d A1 = 0.5 A2 = 1 A3 = 0.5 A1 = 1 A2 = 0.5 A3 = 0.5 A1 = 1 A2 = 1 A3 = 1 Amplitude Spectrum Plot of the amplitudes making up x(t) vs. ω A k of the sinusoids Example: ω 3
Phase Spectrum Plot of the phases θk making up x(t) vs. ω of the sinusoids Example: ω jα Euler formula: e = cos( α) + jsin( α) Thus j( kt k) Ak cos( kt k) Ake ω + ω + θ =R θ whence Complex Exponential Form real part N j( ωkt θk) () + k, k = 1 xt = R Ae t 4
Complex Exponential Form Cont d And, recalling that R ( z) = ( z+ z )/2 where z = a+ jb, we can also write N 1 j( ωkt θk) j( ωkt θk) xt () = Ae + k + Ae + k, t 2 k = 1 This signal contains both positive and negative frequencies The negative frequencies ω k stem from writing the cosine in terms of complex exponentials and have no physical meaning Complex Exponential Form Cont d By defining Ak ck = 2 it is also N k = 1 j k e θ c k = Ak 2 j k e θ N jωkt jωkt jωkt k k k k= N k 0 xt () = ce + c e = ce, t complex exponential form of the signal x(t) 5
Line Spectra The amplitude spectrum of x(t) is defined as the plot of the magnitudes c k versus ω The phase spectrum of x(t) is defined as the plot of the angles c versus ω k = arg( ck) This results in line spectra which are defined for both positive and negative frequencies Notice: for k = 1, 2, c k = c k ck = c k ck = c k arg( ) arg( ) Example: Line Spectra xt ( ) = cos( t) + 0.5cos(4 t+ π / 3) + cos(8 t+ π / 2) 0. 0. 6
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 jkω0t xt () ce, t = k = where ω0 = 2 π /T is the fundamental frequency (rad/sec) of the signal and T /2 T /2 k 1 jkωot ck = x( t) e dt, k = 0, ± 1, ± 2, T c 0 is called the constant or dc component of x(t) 7
The Fourier Series Cont d The frequencies kω 0 present in x(t) are integer multiples of the fundamental frequencyω 0 Notice that, if the dc term c 0 is added to N j kt x() t = cke ω k= N k 0 and we set N =, the Fourier series is a special case of the above equation where all the frequencies are integer multiples of ω 0 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 8
Example: The Rectangular Pulse Train ω = π = π From figure, T = 2 whence 0 2 /2 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 kπ k = kodd 9
By using Euler s formula, we can rewrite as Trigonometric Fourier Series jkω0t xt () ce, t = k = k xt () = c + 2 c cos( kω t+ c), t k k 0 0 k = 1 dc component k-th harmonic 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 = kodd can be rewritten as 1 2 ( k 1)/2 π xt () = + cos kπt+ ( 1) 1, t 2 k = 1 kπ 2 kodd 10
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 k odd According to Fourier s s theorem, it should be lim x () t x() t = 0 N N Gibbs Phenomenon Cont d x () t 3 x () t 9 11
Gibbs Phenomenon Cont d x t x () t 45 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 = ck k = jkω0t xt () ce, t 2 = k = k 12
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 13
Frequency Content of the Rectangular Pulse Cont d Since xt () t is periodic with period T, we can write where jkω0t x () t c e, t T T /2 T /2 = k = k 1 jkωot ck = x( t) e dt, k = 0, ± 1, ± 2, T Frequency Content of the Rectangular Pulse Cont d What happens to the frequency components of xt () t as T? For k = 0 1 c0 = T For k 0 c k 2 kω 1 kω = = =± ± kω T 2 kπ 2 0 0 0 sin sin, k 1, 2, ω0 = 2 π /T 14
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 ω lim Tck = sinc, ω T 2π where sin( πλ sinc( λ) ) πλ 15
Fourier Transform of the Rectangular Pulse The Fourier transform of the rectangular pulse x(t) is defined to be the limit of Tc k as T, i.e., ω X( ω) = limtck = sinc, ω T 2π X ( ω ) arg( X ( ω )) Fourier Transform of the Rectangular Pulse Cont d The Fourier transform X ( ω) of the rectangular pulse x(t) can be expressed in terms of x(t) as follows: 1 jkωot ck = x( t) e dt, k = 0, ± 1, ± 2, T whence jkωot Tc = x( t) e dt, k = 0, ± 1, ± 2, k x() t = 0 for t < T /2and t > T /2 16
Fourier Transform of the Rectangular Pulse Cont d Now, by definition X( ω) = limtc k and, T since kω0 ω as T jωt X( ω) = x( t) e dt, ω The inverse Fourier transform of X ( ω) is 1 jωt xt () = X( ω) e dω, t 2π The Fourier Transform in the General Case Given a signal x(t), its Fourier transform X ( ω) is defined as jωt X( 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 17
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 18
Example: The Exponential Signal bt Consider the signal xt ( ) = e ut ( ), b Its Fourier transform is given by bt jωt X( ω) = e u( t) e dt 1 = e dt = e b+ jω ( b+ jω) t ( b+ jω) t t= 0 t= 0 Example: The Exponential Signal Cont d If b < 0, X ( ω) does not exist If b = 0, xt () = ut () and X ( ω) does not exist either in the ordinary sense If b > 0, it is 1 X ( ω) = b + jω amplitude spectrum phase spectrum 1 X ( ω) = ω arg( X ( ω)) = arctan 2 2 b + ω b 19
Example: Amplitude and Phase Spectra of the Exponential Signal 10 t xt () = e ut () Rectangular Form of the Fourier Transform Consider jωt X( ω) = x( t) e dt, ω Since X ( ω) in general is a complex function, by using Euler s formula X( ω) = x( t)cos( ωt) dt+ j xt ( )sin( ωtdt ) R( ω ) I ( ω ) X( ω) = R( ω) + ji( ω) 20
Polar Form of the Fourier Transform X( ω) = R( ω) + ji( ω) can be expressed in a polar form as X( ω) = X( ω) exp( j arg( 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 X( ω) = X ( ω) Moreover Hermitian symmetry X ( ω) = X( ω) exp( jarg( X( ω))) whence X( ω) = X( ω) and arg( X( ω)) = arg( X( ω)) 21
Fourier Transforms of Signals with Even or Odd Symmetry Even signal: xt () = x( t) X( ω) = 2 x( t)cos( ωt) dt Odd signal: xt () = x( t) 0 X ( ω) = j2 xt ( )sin( ωtdt ) 0 Example: Fourier Transform of the Rectangular Pulse Consider the even signal It is τ /2 2 t= τ /2 2 ωτ X( ω) = 2 (1)cos( ωt) dt = [ sin( ωt) ] = sin t= 0 ω ω 2 0 ωτ = τ sinc 2 π 22
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 23
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 24
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 1 jωt xt () = X( ω) e dω, t 2π Transform pair xt () Xω ( ) Properties of the Fourier Transform Linearity: xt () Xω ( ) y() t Y( ω) α xt () + βy() t αx( ω) + βy( ω) Left or Right Shift in Time: Time Scaling: j t xt ( t) X( ω) e ω 0 xat ( ) 1 ω X a a 0 25
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 26
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 ( ) y( t) X( ω) Y( ω) Multiplication in the Time Domain: xt () y() t X( ω) Y( ω) Properties of the Fourier Transform Parseval s Theorem: xt 1 () y () tdt ( ) ( ) 2 X ω Y ω d ω π if y() t = x() t Duality: 1 () ( ) 2π 2 2 xt dt Xω d X() t 2 π x( ω ) ω 27
Properties of the Fourier Transform - Summary Example: Linearity x() t = p () t + p () t 4 2 2ω ω X ( ω) = 4sinc + 2sinc π π 28
Example: Time Shift xt () = p( t 1) 2 ω X( ω) = 2sinc π j e ω Example: Time Scaling p () t 2 2sinc ω π p (2 t) 2 ω sinc 2 π a > 1 0< a < 1 time compression time expansion frequency expansion frequency compression 29
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 2 ω 30
Example: Multiplication by a Sinusoid xt () p()cos( t t) = τ ω sinusoidal 0 burst 1 τ ( ω + ω0) τ( ω ω0) X ( ω) = τsinc τsinc 2 + 2π 2π Example: Multiplication by a Sinusoid Cont d 1 τ ( ω + ω0) τ( ω ω0) X ( ω) = τsinc τsinc 2 + 2π 2π ω 0 = 60 rad / sec τ = 0.5 31
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 4π 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 32
Example: Integration in the Time Domain Cont d τ 2 τω V ( ω) = sinc 2 4π Generalized Fourier Transform Fourier transform of δ () t jωt δ () te dt= 1 Applying the duality property xt () = 1, t 2 πδ( ω) δ () t 1 generalized Fourier transform of the constant signal x() t = 1, t 33
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 0t Since e ω 2 πδω ( ω), it is k X( ω) = 2 πc δ( ω kω ) k = k 0 0 34
Since Fourier Transform of the Unit-Step Function ut () = δ ( τ) dτ using the integration property, it is t t 1 ut () = δ ( τ) dτ + πδ( ω) jω Common Fourier Transform Pairs 35