Chapter 6 CONTINUOUS-TIME, IMPULSE-MODULATED, AND DISCRETE-TIME SIGNALS. 6.6 Sampling Theorem 6.7 Aliasing 6.8 Interrelations

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Chapter 6 CONTINUOUS-TIME, IMPULSE-MODULATED, AND DISCRETE-TIME SIGNALS 6.6 Sampling Theorem 6.7 Aliasing 6.8 Interrelations Copyright c 2005- Andreas Antoniou Victoria, BC, Canada Email: aantoniou@ieee.org September 12, 2008 Frame # 1 Slide # 1 A. Antoniou Digital Signal Processing Secs. 6.6-6.8

Introduction In order to process a continuous-time signal using digital signal processing methodologies, it is first necessary to convert the continuous-time signal into a discrete-time signal by applying sampling. Frame # 2 Slide # 2 A. Antoniou Digital Signal Processing Secs. 6.6-6.8

Introduction In order to process a continuous-time signal using digital signal processing methodologies, it is first necessary to convert the continuous-time signal into a discrete-time signal by applying sampling. Sampling obviously entails discarding part of the continuous-time signal and the question will immediately arise as to whether the sampling process will corrupt the signal. Frame # 2 Slide # 3 A. Antoniou Digital Signal Processing Secs. 6.6-6.8

Introduction In order to process a continuous-time signal using digital signal processing methodologies, it is first necessary to convert the continuous-time signal into a discrete-time signal by applying sampling. Sampling obviously entails discarding part of the continuous-time signal and the question will immediately arise as to whether the sampling process will corrupt the signal. It turns out that under a certain condition that is part of the sampling theorem, the information content of the continuous-time signal can be fully preserved. Frame # 2 Slide # 4 A. Antoniou Digital Signal Processing Secs. 6.6-6.8

The Sampling Theorem The sampling theorem states: A bandlimited signal x(t) forwhich X (jω) =0 for ω ω s 2 where ω s =2π/T, can be uniquely determined from its values x(nt ). Frame # 3 Slide # 5 A. Antoniou Digital Signal Processing Secs. 6.6-6.8

The Sampling Theorem The sampling theorem states: A bandlimited signal x(t) forwhich X (jω) =0 for ω ω s 2 where ω s =2π/T, can be uniquely determined from its values x(nt ). Alternatively, in what amounts to the same thing, a continuous-time signal whose spectrum is zero outside the baseband (i.e., ω s /2toω s /2) can, in theory, be recovered completely from an impulse-modulated version of the signal. Frame # 3 Slide # 6 A. Antoniou Digital Signal Processing Secs. 6.6-6.8

The Sampling Theorem Cont d Consider a two-sided bandlimited signal whose spectrum satisfies the condition of the sampling theorem. Frame # 4 Slide # 7 A. Antoniou Digital Signal Processing Secs. 6.6-6.8

The Sampling Theorem Cont d Consider a two-sided bandlimited signal whose spectrum satisfies the condition of the sampling theorem. By virtue of Poison s summation formula, i.e., ˆX (jω) = 1 T X (jω + jnω s ) impulse modulation will produce sidebands that are well separated from one another. Frame # 4 Slide # 8 A. Antoniou Digital Signal Processing Secs. 6.6-6.8

The Sampling Theorem Cont d X( jω) ω s 2 (a) ω s 2 ω ^ X( jω+ jω X( jω) T 1 X( jω) s ) T 1 X( jω jω s ) T 1 ω s ω s 2 (b) ω s 2 ω s ω Frame # 5 Slide # 9 A. Antoniou Digital Signal Processing Secs. 6.6-6.8

The Sampling Theorem Cont d Now if we pass the impulse-modulated signal through an ideal lowpass filter with a frequency response { T for ω<ω s /2 H(jω) = 0 otherwise then frequencies in the sidebands will be rejected and we will be left with the frequencies in the baseband, which constitute the original continuous-time signal. Frame# 6 Slide# 10 A. Antoniou Digital Signal Processing Secs. 6.6-6.8

The Sampling Theorem Cont d Now if we pass the impulse-modulated signal through an ideal lowpass filter with a frequency response { T for ω<ω s /2 H(jω) = 0 otherwise then frequencies in the sidebands will be rejected and we will be left with the frequencies in the baseband, which constitute the original continuous-time signal. A baseband gain of T is used to cancel out the scaling constant 1/T introduced by Poisson s summation formula. Frame# 6 Slide# 11 A. Antoniou Digital Signal Processing Secs. 6.6-6.8

The Sampling Theorem Cont d ω s 2 (a) ^ X( jω+ jω X( jω) T 1 X( jω) s ) T 1 X( jω jω s ) T 1 ω s 2 ω ω s ω s 2 (b) H( jω) ω s 2 ω s ω T ω s 2 (c) ω s 2 ω = X( jω) ^ TX( jω) ω s 2 (d) ω s 2 ω Frame# 7 Slide# 12 A. Antoniou Digital Signal Processing Secs. 6.6-6.8

The Sampling Theorem Cont d What has been done through a graphical illustration can now be repeated with mathematics. Frame# 8 Slide# 13 A. Antoniou Digital Signal Processing Secs. 6.6-6.8

The Sampling Theorem Cont d What has been done through a graphical illustration can now be repeated with mathematics. If the impulse-modulated signal is passed through a lowpass filter with a frequency response H(jω) as defined before, then the Fourier transform of the output of the filter will be Y (jω) =H(jω) ˆX (jω) where H(jω) = { T for ω<ω s /2 0 otherwise and ˆX (jω) = 1 T X (jω + jnω s ) Frame# 8 Slide# 14 A. Antoniou Digital Signal Processing Secs. 6.6-6.8

The Sampling Theorem Cont d Y (jω) =H(jω) ˆX(jω) If we apply the inverse Fourier transform, we get ] y(t) =F [H(jω) 1 x(nt )e jωnt = x(nt )F 1 [H(jω)e jωnt ] (A) Frame# 9 Slide# 15 A. Antoniou Digital Signal Processing Secs. 6.6-6.8

The Sampling Theorem Cont d Y (jω) =H(jω) ˆX(jω) If we apply the inverse Fourier transform, we get ] y(t) =F [H(jω) 1 x(nt )e jωnt = x(nt )F 1 [H(jω)e jωnt ] (A) The frequency response of a lowpass filter is actually a frequency-domain pulse of height T and base ω s, i.e., H(jω) =Tp ωs (ω) and hence from the table of Fourier transforms, we have T sin(ω s t/2) H(jω) (B) πt Frame# 9 Slide# 16 A. Antoniou Digital Signal Processing Secs. 6.6-6.8

The Sampling Theorem Cont d y(t) = x(nt )F 1 [H(jω)e jωnt ] T sin(ω s t/2) H(jω) πt From the time-shifting theorem of the Fourier transform T sin[ω s (t nt )/2] π(t nt ) H(jω)e jωnt (A) (B) (C) Frame # 10 Slide # 17 A. Antoniou Digital Signal Processing Secs. 6.6-6.8

The Sampling Theorem Cont d y(t) = x(nt )F 1 [H(jω)e jωnt ] T sin(ω s t/2) H(jω) πt From the time-shifting theorem of the Fourier transform T sin[ω s (t nt )/2] π(t nt ) H(jω)e jωnt (A) (B) (C) Therefore, from Eqs. (A) and (C), we conclude that y(t) = x(nt ) sin[ω s(t nt )/2] ω s (t nt )/2 Frame # 10 Slide # 18 A. Antoniou Digital Signal Processing Secs. 6.6-6.8

The Sampling Theorem Cont d y(t) = x(nt )F 1 [H(jω)e jωnt ] T sin(ω s t/2) H(jω) πt From the time-shifting theorem of the Fourier transform T sin[ω s (t nt )/2] π(t nt ) H(jω)e jωnt (A) (B) (C) Therefore, from Eqs. (A) and (C), we conclude that y(t) = x(nt ) sin[ω s(t nt )/2] ω s (t nt )/2 For t = nt, we have y(nt )=x(nt )forn =0, 1,..., kt, and for all other values of t the output of the lowpass filter is an interpolated version of x(t) according to the sampling theorem. Frame # 10 Slide # 19 A. Antoniou Digital Signal Processing Secs. 6.6-6.8

Aliasing If the spectrum of the continuous-time signal does not satisfy the condition imposed by the sampling theorem, i.e., if X (jω) 0 for ω ω s 2 then sideband frequencies will be aliased into baseband frequencies. Frame # 11 Slide # 20 A. Antoniou Digital Signal Processing Secs. 6.6-6.8

Aliasing If the spectrum of the continuous-time signal does not satisfy the condition imposed by the sampling theorem, i.e., if X (jω) 0 for ω ω s 2 then sideband frequencies will be aliased into baseband frequencies. As a result, ˆX (jω) will not be equal to X (jω)/t within the baseband. Frame # 11 Slide # 21 A. Antoniou Digital Signal Processing Secs. 6.6-6.8

Aliasing If the spectrum of the continuous-time signal does not satisfy the condition imposed by the sampling theorem, i.e., if X (jω) 0 for ω ω s 2 then sideband frequencies will be aliased into baseband frequencies. As a result, ˆX (jω) will not be equal to X (jω)/t within the baseband. Under these circumstances, the use of an ideal lowpass filter willyieldadistortedversionofx(t) atbest. Frame # 11 Slide # 22 A. Antoniou Digital Signal Processing Secs. 6.6-6.8

Aliasing Cont d Aliasing can be illustrated by examining an impulse-modulated signal generated by sampling the continuous-time signal x(t) =u(t)e at sin ω 0 t 0.2 0.1 X( jω) 0 30 20 10 0 10 20 ω 30 (a) Frame # 12 Slide # 23 A. Antoniou Digital Signal Processing Secs. 6.6-6.8

Aliasing Cont d Aliasing can be illustrated by examining an impulse-modulated signal generated by sampling the continuous-time signal x(t) =u(t)e at sin ω 0 t The frequency spectrum of x(t), X (jω), extends over the infinite range <ω<. 0.2 0.1 X( jω) 0 30 20 10 0 10 20 ω 30 (a) Frame # 12 Slide # 24 A. Antoniou Digital Signal Processing Secs. 6.6-6.8

Aliasing Cont d The frequency spectrum of impulse-modulated signal ˆx(t) can be obtained as ˆX (jω) = 1 T X (jω + jnω s ) by using Poisson s summation formula. Frame # 13 Slide # 25 A. Antoniou Digital Signal Processing Secs. 6.6-6.8

Aliasing Cont d ˆX (jω) = 1 T X (jω + jnω s ) The shifted copies of X (jω) or sidebands, namely,..., X (jω j2ω s ), X (jω jω s ), X (jω + jω s ), X (jω + j2ω s ),... overlap with the baseband ω s /2 <ω<ω s /2 and, therefore, the above sum can be expressed as where ˆX (jω) = 1 [X (jω)+e(jω)] T E (jω) = k= k 0 X (jω + jkω s ) is the contribution of the sidebands to the baseband. Frame # 14 Slide # 26 A. Antoniou Digital Signal Processing Secs. 6.6-6.8

Aliasing Cont d Now if we filter the impulse-modulated signal, ˆx(t), using an ideal lowpass filter with a frequency response { T for ω s /2 <ω<ω s /2 H(jω) = 0 otherwise we will get a signal y(t) whose frequency spectrum is given by Y (jω) =H(jω) ˆX (jω) = H(jω) 1 T = X (jω)+e(jω) X (jω + jnω s ) Frame # 15 Slide # 27 A. Antoniou Digital Signal Processing Secs. 6.6-6.8

Aliasing Cont d Y (jω) =X (jω)+e(jω) In other words, the output of the filter will be signal x(t) plus an error e(t) =F 1 E(jω) which is commonly referred to as the aliasing error. Frame # 16 Slide # 28 A. Antoniou Digital Signal Processing Secs. 6.6-6.8

Aliasing Cont d With a sampling frequency of 12.5 rad/s, E (jω), i.e., the discrepancy between the solid and dashed curves in the figure is quite large. 0.2 0.1 X(jω) 0 40 30 20 10 0 10 20 30 40 X D (e jωt ) 0.2 0.1 0 40 30 20 10 0 10 20 30 40 0.2 0.1 Filtered X D (e jωt ) 0 40 30 20 10 0 10 20 30 40 Frame # 17 Slide # 29 A. Antoniou Digital Signal Processing Secs. 6.6-6.8

Aliasing Cont d As the sampling frequency is increased to 25, the sidebands are spread out and E (jω) will be decreased quite a bit as shown. 0.2 0.1 0 40 30 20 10 0 10 20 30 40 0.2 X(jω+jωs)/T 0.1 0 40 30 20 10 0 10 20 30 40 0.2 0.1 0 40 30 20 10 0 10 20 30 40 Frame # 18 Slide # 30 A. Antoniou Digital Signal Processing Secs. 6.6-6.8

Aliasing Cont d A further increase to 40 rad/s will render E (jω) for all practical purposes negligible as can be seen. 0.2 0.1 0 40 30 20 10 0 10 20 30 40 0.2 0.1 0 40 30 20 10 0 10 20 30 40 0.2 0.1 0 40 30 20 10 0 10 20 30 40 Frame # 19 Slide # 31 A. Antoniou Digital Signal Processing Secs. 6.6-6.8

Summary of Interrelations Impulse-modulate signal: ˆx(t) = x(nt )δ(t nt ) (6.42c) Frame # 20 Slide # 32 A. Antoniou Digital Signal Processing Secs. 6.6-6.8

Summary of Interrelations Impulse-modulate signal: ˆx(t) = x(nt )δ(t nt ) (6.42c) Spectrum of impulse modulated signal or discrete-time signal in terms of the spectrum of the original continuous-time signal: ˆX(jω) =X D (e jωt )= 1 T X (jω + jnω s ) (6.45a) where X D (e jωt )= x(nt )e jωnt Frame # 20 Slide # 33 A. Antoniou Digital Signal Processing Secs. 6.6-6.8

Summary of Interrelations Cont d Spectrum of impulse-modulated signal (or discrete-time signal) in terms of the spectrum of the original continuous-time signal for a right-sided signal: ˆX (jω) =X D (e jωt )= x(0+) + 1 2 T X (jω+jnω s )(6.45b) Frame # 21 Slide # 34 A. Antoniou Digital Signal Processing Secs. 6.6-6.8

Summary of Interrelations Cont d Spectrum of impulse-modulated signal (or discrete-time signal) in terms of the spectrum of the original continuous-time signal for a right-sided signal: ˆX (jω) =X D (e jωt )= x(0+) + 1 2 T X (jω+jnω s )(6.45b) Laplace transform of impulse-modulated signal in terms of the Laplace transform of the original continuous-time signal for a right-sided signal: ˆX(s) =X D (z) = x(0+) 2 where z = e st. + 1 T X (s + jnω s ) (6.46a) Frame # 21 Slide # 35 A. Antoniou Digital Signal Processing Secs. 6.6-6.8

Summary of Interrelations Cont d Recovery of a continuous-time signal by lowpass filtering an impulse-modulated filter frequency domain: Y (jω) =H(jω) ˆX (jω) (6.48) where H(jω) = { T for ω <ω s /2 0 for ω ω s /2 Frame # 22 Slide # 36 A. Antoniou Digital Signal Processing Secs. 6.6-6.8

Summary of Interrelations Cont d Recovery of a continuous-time signal by lowpass filtering an impulse-modulated filter frequency domain: Y (jω) =H(jω) ˆX (jω) (6.48) where H(jω) = { T for ω <ω s /2 0 for ω ω s /2 Recovery of a continuous-time signal by lowpass filtering an impulse-modulated filter time-domain: y(t) = x(nt ) sin[ω s(t nt )/2] ω s (t nt )/2 (6.51) Frame # 22 Slide # 37 A. Antoniou Digital Signal Processing Secs. 6.6-6.8

Graphical Representation of Interrelations X(s) L L 1 jω s s jω F x(t) X( jω) F 1 Eq. (6.51) Eq. (6.42d) or (6.42e) Eq. (6.48) Eq. (6.45a) or (6.45b) x(t) ˆ F F 1 X( ˆ jω) Replace numbers by impulses Replace impulses by numbers z e jωt 1 jω ln z T x(nt ) Z Z 1 X D (z) Frame # 23 Slide # 38 A. Antoniou Digital Signal Processing Secs. 6.6-6.8

This slide concludes the presentation. Thank you for your attention. Frame # 24 Slide # 39 A. Antoniou Digital Signal Processing Secs. 6.6-6.8