Signals Represented in Terms of Magnitude Phase The physical interpretation of the Fourier transform of a time signal is its decomposition into a linear combination of frequency components with different magnitudes phase shifts. A continuous signal x(t) can be written as x(t) = Z Z X(!)e j!t d! = X(f)e j2ßft df 2ß Moreover, if the signal is real, the above can be written as: x(t) = 2ß Z 2jX(!)jcos(6 X(!)+!t)d! 0 where X(!) is the complex weight for the frequency component e j!t If the signal is periodic with period T, it can be written as x T (t) = X X[k]e jk! 0t k= where X[k] is the complex weight for the kth frequency component e jk! 0t. Again, if x T (t) is real, it can be written as x T (t) =X[0] + X k= 2jX[k]j cos(6 X[k]+k! 0 t) Whether the signal is periodic or not, its waveform is determined by the magnitude (jx(!)j or jx[k]j) the phase (6 X(!) or6 X[k]) of each of the frequency components at frequency! or k! 0. Example : Waveform determined by phase shift The waveform of a signal is determined by the phase shifts as well as the magnitudes of its frequency components. The three waveforms shown in the figure below have the same frequency composition (sin(k Λ x), k = ; 3; 5) but different phase shifts for different frequency components. For linear phase shift, i.e., the phase shift is proportional to the frequency, the waveform (blue) is a shifted version of the original waveform (red). However, if the phase shift is not linear, e.g., a constant phase for all frequencies, the resulting waveform (green) is drastically different. Example 2: Human auditory system
The different frequency components in a sound signal are processed first by the sensory system in the ear then some specialized cortical areas in the brain. The auditory system is sensitive to the magnitude, but not the phase, of the frequency components in a signal of relatively short duration, such as the sound of a vowel. But the phase information in a longer signal, such as a spoken sentence, certainly matters. For example, represent a recorded sentence as a real signal x(t), then x( t) represents the recording played backward. In frequency domain, the signal's spectrum can be written as F[x(t)] = X(!) =jx(!)je j 6 X(!) According to the Fourier transform property, the spectrum of the time reversed signal x( t) is Moreover, due to the property F[x( t)] = X(!) F[x Λ (t)] = X Λ (!) the fact that x(t) is also real, i.e., x( t) =x Λ ( t), we have F[x( t)] = F[x Λ ( t)] = X Λ (!) =jx(!)je j 6 X(!) i.e., the forward backward signals differ only in phase, as they have identical magnitude jx(!)j for all frequencies. Example 3: Human visual system A visual scene projected on the retina can be considered as a two-dimensional (2D) spatial signal f(x; y) whose 2D Fourier transform F (! x ;! y ) can be obtained by one-dimensional Fourier transform of its rows then its columns. Here! x! y are the spatial frequencies in x y directions, respectively. Unlike the auditory system, the human visual system is sensitive to both magnitude phase of the visual signal, System Represented in Terms of Magnitude Phase An LTI system with impulse response function h(t) can also be represented in frequency domain by its transfer function F[h(t)] = H(!) =jh(!)je j 6 H(!) After passing through the system, an input signal x(t) becomes or in frequency domain y(t) =h(t) Λ x(t) Y (!) =jy (!)je j 6 Y (!) = H(!)X(!)=jH(!)j jx(!)je j( 6 H(!)+6 X(!)) The effect the LTI system exerts on the signal x(t) is the modification of the complex weights of the frequency components: jy (!) =jh(!)j jx(!)j 6 Y (!) =6 H(!)+6 X(!) The waveform of a signal x(t) is determined by the phase shift 6 X(!) aswell as the magnitude jx(!)j of its frequency components. After passing through an LTI system H(!), the input X(!) becomes output Y (!) =H(!)X(!) withchanged waveforms. 2
Group Delay In particular, if the transfer function of the system is H(!) = e jt 0!, i.e., jh(!)j = is a constant 6 H(!) = t 0! is a linear function of! (where t 0 is a constant), then Y (!) =H(!)X(!) =X(!)e j!t 0 according to the time-delay property of Fourier transform, we have y(t) =F [Y (!)] = F [X(!)e j!t 0 ]=x(t t 0 ) Similarly, a periodic signal x(t) = x(t + T ) passing through a linear-phase LTI system H[k] = e jkt 0! 0 (jh[k]j =,6 H[k] = t 0 k! 0 ) becomes y(t) =F [Y [k]] = F [X[k]e jk! 0t 0 ]=x(t t 0 ) A time delay by t 0 of the waveform of a signal x(t) will be caused by a linear phase LTI system which exerts a phase shift t 0! proportional to the frequency! of each component). The amount of time delay t 0 is the negative slope of the phase function: d d! 6 H(!) = d d! ( t 0!) =t 0 An example of a signal delayed by a linear phase LTI system is shown in these figures: The linear phase delay can also be seen intuitively. For a waveform to be time shifted by t 0 without distortion, all of its frequency components have to be shifted in time by the same amount. In terms of phase shift (2ß per period T = 2ß=!), a high frequency component (large! or small T ) needs to be phase shifted more, while a low frequency component (small! or large T ) needs to phase shifted less. In other words, for the output of an LTI to be a delayed version of the input, the amount of phase shift for each component of the signal should be proportional to its frequency, the proportional constant is the amount of delay. 3
A nonlinear phase shift 6 H(!) of an arbitrary LTI system can be approximately linearized within a narrow frequency b centered around! =! c : where 6 H(!)j j!!cj<ffi ß ffi t 0! ffi = 6 H(!)j!=!c ; t 0 = d d! 6 H(!)j!=!c In frequency domain, the response of the system to the input within this narrow b is Y (!) =X(!)H(!)=X(!)jH(!)je j(ffi t 0!) = X(!)jH(!)je jffi e jt 0! i.e., the output is a time delayed version of X(!)jH(!)je jffi, where the amount of delay, called the group delay at frequency! c,isthe negative of the slope of the phase at frequency! c, as shown above. 4
Continuous-Time Systems Bode Plots First-order system A first-order system is described by this differential equation fi d y(t)+y(t) =x(t) dt The stability requires that the time constant fi > 0. The impulse response of the system is the step response of the system is h(t) = fi e t=fi u(t) s(t) =h(t) Λ u(t) = fi Z 0 e t0 =fi u(t t 0 )dt 0 =( e t=tau )u(t) The frequency response of the system is H(j!)=F[h(t)] = fi Z e t=fi e j! dt = 0 j!fi + 5
Second-order system Example : An RCL circuit with the voltage x(t) across all three elements in series as the input x(t) =v L (t)+v R (t)+v C (t) =L d dt i(t)+ri(t)+ C the voltage across one of the elements, e.g., the capacitor, as the output Z t i(fi)dfi Since y(t) =v C (t) = C i(t) =C d dt y(t) Z t i(fi)dfi the differential equation above becomes d 2 dt 2 y(t)+ R L d dt y(t)+ LC y(t) = LC x(t) Example 2: A mechanical system containing a mass m attached to a wall by a spring k a dashpot b in parallel with applied force x(t) to the mass as the input the displacement of the mass y(t) as the output. The equation of motion of the system is m d2 dt y(t)+b d y(t)+ky(t) =x(t) 2 dt or d 2 dt y(t)+ b d 2 m dt y(t)+ k m y(t) = m x(t) In general, a second-order system is described by this differential equation dt y(t)+a d 2 dt y(t)+a 0y(t) =b 0 x(t) d 2 Conventionally, thecoefficients a a 0 are represented in terms of two other constants! n : ( a =2! n the above equation is rewritten as where for the first example d 2 a 0 =! 2 dt y(t)+2! d 2 n dt y(t)+!2y(t) n =b 0x(t) s! n = LC ; = R C 2 L s k! n = m ; = b 2 p km for the second example. The stability of the system requires > 0. By Fourier transform, the frequency response of the system can be found to be H(j!)= Y (j!) X(j!) = b 0 (j!) 2 +2! n (j!)+! 2 n 6 = b 0 (j! c )(j! c 2 )
where c ;2 are the two roots of the denominator: ( c =( + p 2 )! n c 2 =( p 2 )! n If we assume b 0 =! 2 (the RCL example), H(j!) canbe written by partial fraction expansion as H(j!)= the impulse response can be found to be In particular, when =,we have h(t) =! n 2 p 2 [ ] j! c j! c 2! n 2 p 2 [ec t e c 2t ] H(j!)=! 2 n (j! +! n ) 2 h(t) =! 2 n te!nt u(t) Note that h(t) takes different forms depending on the value of, astobediscussed in Laplace transform. If 0 < <, we have h(t) =! p p p n 2j 2 e!nt [e j 2!nt e j 2!nt ]u(t) =! ne p!nt q sin( 2! 2 n t)u(t) =! ne p!nt 2 dt)u(t) where! d = p 2! n is damped natural frequency. The step response can be found by s(t) =h(t) Λ u(t) =! n 2 p 2 Z t 0! n [e c t 0 e c 2t 0 ]dt 0 =[+ 2 p 2 (e c t c e c 2 t )]u(t) c 2 when =,we have s(t) =[ e!nt! n te!nt ]u(t) Bode Plot In general the impulse response h(t) is real, therefore its Fourier transform, the frequency response H(j!) is symmetric (even symmetry of the real part odd symmetry for the imaginary part) with respect to! = 0 can be plotted in the range 0»! <. Conventionally, the complex frequency response is plotted not as real imaginary parts, but in terms of it magnitude phase. Moreover, logarithmic scale is used for the frequency axis, 20log 0 jh(j!)j with unit decibel (db) is used for the magnitude of H(j!). Such a plot is called a Bode plot. For a first-order system, the phase magnitude of the frequency response H(j!) = =(j!fi + ) can be found to be 6 H(j!)= tan (!fi) or jh(j!)j = p!2 fi 2 + 20log 0 jh(j!)j = 0 log 0 [(!fi) 2 +] 7
In particular, when! = =fi (corner frequency), we have jh(j!)j = p 2=0:707; 20 log 0 jh(j!)j = 0 log 0 2 ß 3 db 6 H(j!)= tan = ß=4 be For a second-order system, the phase magnitude of the frequency response H(j!) can be found to 2 (!=! 6 H(j!)= tan n ) [ (!=! n ) ]= 2 8 >< >: 0! =0 ß=2! =! n ß!! ρ ff 20 log 0 jh(j!)j = 0 log 0 [ (! ) 2 ] 2 +4 2 (! ) 2! n! n ß 8 >< >: 0! fi! n 20 log 0 (2 ) =20log 0 Q! =! n 40 log 0 (!=! n )! fl! n where Q = 4 =2 is the quality factor of the system. Note that when p < 2=2 ß 0:707, H(j!) has a maximum value jh(j! max )j = p 2 2 where! max =! n q 2 2 8
Discrete-Time Systems Bode Plots First-order system A first-order system is described by this difference equation y[n] ay[n ] = x[n] The stability of this system requires that jaj <. The impulse response of the system is the step response of the system is s[n] =h[n] Λ u[n] = The frequency response of the system is X k= H(e j! )=Ffh[n]g = which can be expressed as magnitude jh(e j! )j = h[n] =a n u[n] a k u[k]u[n k] = nx k=0 a k = an+ a u[n] ae = j! ( acos!)+ja sin! p 2a cos!+ a 2 phase 6 H(e j! a sin! )= tan [ acos! ] Both H(e j! ) 6 H(e j! ) are periodic functions of frequency! with period 2ß. Second-order system A second-order system is described by this difference equation y[n]+a y[n ] + a 2 y[n 2] = b 0 x[n] Conventionally, thecoefficients a a 2 are replaced by some two other constants ( a = 2r cos a 2 = r 2 the above equation can be rewritten as y[n] 2r cos y[n ] + r 2 y[n 2] = b 0 x[n] where 0»» ß, the stability of the system requires 0 < r <. By discrete Fourier transform, the frequency response of the system can be found to be H(e j! )= 2r cos e j! + r 2 e = j2! [ (re j )e j! ][ (re j )e j! ] When a 6= 0 (i.e., 6= 0 or ß), the two roots of the denominator are different by partial fraction expansion, H(e j! ) can be written as H(e j! )= j j 2j sin [ e (re j )e e j! (re j )e j!] 9
The impulse response of the system can be found to be h[n] =F [H(e j! )] = rn 2j sin [(ej ) n+ (e j ) n+ ] u[n] = sin[(n +) ] sin r n u[n] When =0,we have When = ß, we have H(e j! )= ( re j! ) 2 h[n] =F [H(e j! )] = (n +)r n u[n] H(e j! )= ( + re j! ) 2 h[n] =F [H(e j! )] = (n + )( r) n u[n] From these expressions of h[n] we see that the impulse response of the system is a sinusoidal function which decays (as 0 < r < ) over time (n! ). r determines the rate of decay determines the frequency of oscillation. When a 6=0(i.e., 6= 0orß), the step response of the system is s[n] =h[n] Λ u[n] = 2j sin j (re ) n+ j [ej e j (re ) n+ ] u[n] re j re j When =0 When = ß s[n] =[ (r ) r n+ 2 (r ) + r n+ (n +) ] u[n] 2 r s[n] =[ (r +) ( r)n+ 2 (r +) ( r)n+ (n +) ] u[n] 2 r + 0