Chapter 5 Frequency Domain Analysis of Systems

Transcription

1 Chapter 5 Frequency Domain Analysis of Systems

2 CT, LTI Systems Consider the following CT LTI system: xt () ht () yt () Assumption: the impulse response h(t) is absolutely integrable, i.e., ht ( ) dt< (this has to do with system stability (ECE 352))

3 Response of a CT, LTI System to a Sinusoidal Input What s the response y(t) of this system to the input signal xt () = Acos( ω t+ θ ), t? We start by looking for the response y c (t) of the same system to j( ω0t ) x () t Ae + θ t c 0 =

4 Response of a CT, LTI System to a Complex Exponential Input The output is obtained through convolution as y () t = h() t x () t = h( τ ) x ( t τ) dτ = c c c j( ω0 ( t τ) + θ) = h( τ) Ae dτ = j( ω0t+ θ) jω0τ = Ae h( τ) e dτ = = x c () t jωτ 0 x () t h( τ) e d c τ

5 The Frequency Response of a CT, LTI System By defining it is j H( ω) h( τ ωτ = ) e dτ y () t = H( ω ) x () t = c 0 j( 0t ) H( ω ω + θ ) Ae, t 0 Therefore, the response of the LTI system to a complex exponential is another complex exponential with the same frequency c = H ( ω) is the frequency response of the CT, LTI system = Fourier transform of h(t) ω 0

6 Analyzing the Output Signal y c (t) H ( ω ) Since 0 is in general a complex quantity, we can write y () t H( ) Ae c ω j( ω0t+ θ ) = = 0 jarg H( 0) j( 0t ) H( ω ω ω + θ = ) e Ae = 0 = A H ω0 ( ) e output signal s magnitude j( ω t+ θ+ arg H( ω )) 0 0 output signal s phase

7 Response of a CT, LTI System to a Sinusoidal Input With Euler s formulas we can express as xt () = Acos( ω t+ θ ) x( t) =R ( x ( t)) = ( x ( t) + x ( t)) 1 c 2 c c and, by exploiting linearity, it is yt ( ) =R ( y( t)) = ( y( t) + y( t)) = 1 c 2 c c 0 = A H( ω ) cos ω t+ θ + arg H( ω ) ( ) 0 0 0

8 Response of a CT, LTI System to a Sinusoidal Input Cont d Thus, the response to is xt () = Acos( ω t+ θ ) yt () = A H ( ω ) cosω t+ θ + arg H ( ω ) which is also a sinusoid with the same frequency ω 0 but with the amplitude scaled by the factor H ( ω0) and with the phase shifted by amount 0 ( ) arg H ( ω ) 0

9 DT, LTI Systems Consider the following DT, LTI system: xn [ ] hn [ ] yn [ ] The I/O relation is given by yn [ ] = hn [ ] xn [ ]

10 Response of a DT, LTI System to a Complex Exponential Input If the input signal is Then the output signal is given by where k j( ω0n ) x [ n] Ae + θ n c = y [ n] = H( ω ) x [ n] = c H( ω ) = h[ k] e jωk, ω 0 j( 0n ) H( ω ω + θ = ) Ae, n 0 c H ( ω) is the frequency response of the DT, LTI system = DT Fourier transform (DTFT) of h[n]

11 Response of a DT, LTI System to a Sinusoidal Input If the input signal is xn [ ] = Acos( ω n+ θ ) n Then the output signal is given by 0 yn [ ] = A H ( ω ) cos ω n+ θ + arg H ( ω ) ( ) 0 0 0

12 Example: Response of a CT, LTI System to Sinusoidal Inputs Suppose that the frequency response of a CT, LTI system is defined by the following specs: H ( ω) arg H ( ω) ω ω 1.5, 0 ω 20, H ( ω) = 0, ω > 20, arg H ( ω) = 60, ω

13 Example: Response of a CT, LTI System to Sinusoidal Inputs Cont d If the input to the system is xt ( ) = 2cos(10t + 90 ) + 5cos(25t+ 120 ) Then the output is yt () = 2 H (10) cos(10t arg H (10)) H (25) cos(25t arg H (25)) = = 3cos(10t + 30 )

14 Example: Frequency Analysis of an RC Circuit Consider the RC circuit shown in figure

15 Example: Frequency Analysis of an RC Circuit Cont d From ENGR 203, we know that: 1. The complex impedance of the capacitor is equal to 1/ sc where s = σ + jω st 2. If the input voltage is xc () t = e, then the output signal is given by 1/ sc st 1/ RC yc() t = e = e R + 1/ sc s + 1/ RC st

16 Example: Frequency Analysis of an RC Circuit Cont d Setting, it is x () t c whence we can write where s = jω 0 j 0t = e ω and y () t c = y () t = H( ω ) x () t c H ( ω) = 0 1/ c RC jω + 1/ RC jω 1/ 0 RC + 1/ RC e jω t 0

17 Example: Frequency Analysis of an RC Circuit Cont d H ( ω) = ω 1/ RC + (1/ RC) 2 2 arg H( ω) = arctan ( ωrc) 1/ RC = 1000

18 Example: Frequency Analysis of an RC Circuit Cont d The knowledge of the frequency response H ( ω) allows us to compute the response y(t) of the system to any sinusoidal input signal since xt () = Acos( ω t+ θ ) yt () = A H ( ω ) cosω t+ θ + arg H ( ω ) 0 ( ) 0 0 0

19 Example: Frequency Analysis of an RC Circuit Cont d 1/ RC = 1000 Suppose that and that xt ( ) = cos(100 t) + cos(3000 t) Then, the output signal is y( t) = H (100) cos(100 t+ arg H (100)) + + H (3000) cos(3000 t + arg H (3000)) = = cos(100t 5.71 ) cos(3000t )

20 Example: Frequency Analysis of an RC Circuit Cont d xt () yt ()

21 Example: Frequency Analysis of an RC Circuit Cont d Suppose now that = + xt ( ) cos(100 t) cos(50,000 t) Then, the output signal is y( t) = H (100) cos(100 t+ arg H (100)) + + H (50,000) cos(50,000 t + arg H (50,000)) = = cos(100t 5.71 ) cos(50,000t )

22 Example: Frequency Analysis of an RC Circuit Cont d xt () yt () The RC circuit behaves as a lowpass filter, by letting lowfrequency sinusoidal signals pass with little attenuation and by significantly attenuating high-frequency sinusoidal signals

23 Response of a CT, LTI System to Periodic Inputs Suppose that the input to the CT, LTI system is a periodic signal x(t) having period T This signal can be represented through its Fourier series as where = k k = x jkω0t xt () ce, t t + T x 1 ck x t e dt k T 0 jkω0t = (), t 0

24 Response of a CT, LTI System to Periodic Inputs Cont d By exploiting the previous results and the linearity of the system, the output of the system is x yt () = H( kω ) ce k = 0 k jkω t 0 x x j( kω0t+ arg( ck ) + arg H ( kω0 )) = H ( kω0) ck e = k = arg c y y k ck k= y y j( kω0t+ arg( ck )) y jkω0t, = c e = ce t k k= k

25 Example: Response of an RC Circuit to a Rectangular Pulse Train Consider the RC circuit with input xt ( ) = rect( t 2 n) n

26 Example: Response of an RC Circuit to a Rectangular Pulse Train Cont d xt ( ) = rect( t 2 n) n We have found its Fourier series to be with x jkπt xt () = ce k, t c x k k 1 k = sinc 2 2

27 Example: Response of an RC Circuit to a Rectangular Pulse Train Cont d x Magnitude spectrum of input signal x(t) c k

28 Example: Response of an RC Circuit to a Rectangular Pulse Train Cont d The frequency response of the RC circuit was found to be H ( ω) = Thus, the Fourier series of the output signal is given by 1/ RC jω + 1/ RC x jkω0t y jkω0t = ω0 k = k k= k= yt () H( k ) ce ce

29 Example: Response of an RC Circuit to a Rectangular Pulse Train Cont d H( ω) ( db) 1/ RC = 100 filter more selective 1/ RC = 10 1/ RC = 1 ω

30 Example: Response of an RC Circuit to a Rectangular Pulse Train Cont d 1/ RC = 1 y c k y c k 1/ RC = 10 filter more selective y c k 1/ RC = 100

31 Example: Response of an RC Circuit to a Rectangular Pulse Train Cont d 1/ RC = 1 y() t 1/ RC = 10 y() t filter more selective 1/ RC = 100 y() t

32 Response of a CT, LTI System to Aperiodic Inputs Consider the following CT, LTI system xt () ht () yt () Its I/O relation is given by yt () = ht () xt () which, in the frequency domain, becomes Y( ω) = H( ω) X( ω)

33 Response of a CT, LTI System to Aperiodic Inputs Cont d Y( ) = H( ) X( ) From ω ω ω, the magnitude spectrum of the output signal y(t) is given by Y( ω) = H ( ω) X( ω) and its phase spectrum is given by arg Y( ω) = arg H ( ω) + arg X( ω)

34 Example: Response of an RC Circuit to a Rectangular Pulse Consider the RC circuit with input xt ( ) = rect( t)

35 Example: Response of an RC Circuit to a Rectangular Pulse Cont d xt ( ) = rect( t) The Fourier transform of x(t) is ω X ( ω) = sinc 2 π

36 Example: Response of an RC Circuit to a Rectangular Pulse Cont d X ( ω) arg X ( ω)

37 Example: Response of an RC Circuit to a Rectangular Pulse Cont d 1/ RC = 1 Y ( ω) arg Y ( ω)

38 Example: Response of an RC Circuit to a Rectangular Pulse Cont d 1/ RC = 10 Y ( ω) arg Y ( ω)

39 Example: Response of an RC Circuit to a Rectangular Pulse Cont d The response of the system in the time domain can be found by computing the convolution where yt () = ht () xt () ht RCe ut xt ( ) = rect( t) (1/ RC) t () = (1/ ) ()

40 Example: Response of an RC Circuit to a Rectangular Pulse Cont d yt () 1/ RC = 1 yt () 1/ RC = 10 filter more selective

41 Example: Attenuation of High- Frequency Components H ( ω) Y ( ω) = X ( ω)

42 Example: Attenuation of High- Frequency Components xt () y() t

43 Filtering Signals The response of a CT, LTI system with frequency response H ( ω) to a sinusoidal signal is xt () = Acos( ω t+ θ ) yt () = A H ( ω ) cosω t+ θ + arg H ( ω ) Filtering: if 0 or then or 0 ( ) H ( ω ) = 0 H ( ω0) 0 yt () = 0 yt () 0, t

44 Four Basic Types of Filters lowpass H ( ω) passband highpass H ( ω) stopband stopband cutoff frequency bandpass H ( ω) bandstop H ( ω) (many more details about filter design in ECE 464/564 and ECE 567)

45 Phase Function Filters are usually designed based on specifications on the magnitude response The phase response arg H ( ω) has to be taken into account too in order to prevent signal distortion as the signal goes through the system If the filter has linear phase in its passband(s), then there is no distortion H ( ω)

46 A filter is said to have linear phase if If ω 0 is in passband of a linear phase filter, its response to is Linear-Phase Filters H ( ω) arg H( ω) = ωt d, ω passband xt () = Acos( ω t) yt () = A H( ω ) cos( ω t ω t) = = A H( ω ) cos( ω ( t t )) 0 0 d d

47 Ideal Linear-Phase Lowpass The frequency response of an ideal lowpass filter is defined by H ( ω) jωt e d, ω [ B, B] = 0, ω [ B, B] arg H ( ω)

48 Ideal Linear-Phase Lowpass Cont d H ( ω) can be written as ω H( ω) = rect e 2B jωt d whose inverse Fourier transform is B B ht () sinc = ( t td ) π π

49 Ideal Linear-Phase Lowpass Cont d B B ht () = sinc ( t td ) π π ht () t < 0 Notice: the filter is noncausal since is not zero for

50 Ideal Sampling Consider the ideal sampler:.. xt () xn [ ] = xt ( ) t= nt= xnt ( ) t T n It is convenient to express the sampled signal as xt pt where xnt ( ) () () pt () = δ ( t nt) n

51 Ideal Sampling Cont d Thus, the sampled waveform is = = xt () pt () xt () δ ( t nt) xnt ( ) δ ( t nt) xt () pt () n n xt () pt () is an impulse train whose weights (areas) are the sample values xnt ( ) of the original signal x(t)

52 Ideal Sampling Cont d Since p(t) is periodic with period T, it can be represented by its Fourier series where jkωst pt () = ce k, ωs = k T /2 /2 /2 /2 2π 1 jkωst ck = p() t e dt, k T T T T 1 jkω 1 st = δ () te dt= T T T sampling frequency (rad/sec)

53 Therefore and Ideal Sampling Cont d 1 jkωs pt () = e T k 1 ω 1 s xs () t = xt () pt () = xte () = xte () T T k whose Fourier transform is 1 X ( ω s ) = X( ω kωs) T k t jk t jkω t k s

54 Ideal Sampling Cont d X ( ω) 1 Xs( ω) = X( ω kωs) T k

55 Signal Reconstruction Suppose that the signal x(t) is bandlimited with bandwidth B, i.e., X ω Then, if the replicas of in ( ) = 0,for ω > ωs 2 B, X ( ω) 1 X ( ω s ) = X( ω kωs) T k X ( ω) do not overlap and can be recovered by applying an ideal lowpass filter to X ( ω s ) (interpolation filter) B

56 Interpolation Filter for Signal Reconstruction H ( ω) T, ω [ B, B] = 0, ω [ BB, ]

57 Interpolation Formula The impulse response h(t) of the interpolation filter is BT B ht () = sinc t π π and the output y(t) of the interpolation filter is given by yt () = ht () x() t s

58 But Interpolation Formula Cont d xs () t = x() t p() t = x( nt) δ ( t nt) whence Moreover, n yt () = ht () x() t = xntht ( ) ( nt) = s n BT B = xnt ( )sinc ( t nt) π π n yt () = xt ()

59 Shannon s Sampling Theorem A CT bandlimited signal x(t) with frequencies no higher than B can be reconstructed from its samples xn [ ] = xnt ( ) if the samples are taken at a rate ωs = 2 π / T 2B The reconstruction of x(t) from its samples xn [ ] = x( nt ) is provided by the interpolation formula BT B xt ( ) = x( nt ) sinc ( t nt ) π π n

60 Nyquist Rate The minimum sampling rate is called the Nyquist rate ωs = 2 π / T = 2B Question: Why do CD s adopt a sampling rate of 44.1 khz? Answer: Since the highest frequency perceived by humans is about 20 khz, 44.1 khz is slightly more than twice this upper bound

61 Aliasing X ( ω) 1 Xs( ω) = X( ω kωs) T k

62 Aliasing Cont d Because of aliasing, it is not possible to reconstruct x(t) exactly by lowpass filtering the sampled signal xs () t = x() t p() t Aliasing results in a distorted version of the original signal x(t) It can be eliminated (theoretically) by lowpass filtering x(t) before sampling it so that for X ( ω ) = 0 ω B