Digital Processing of Continuous-Time Signals

Transcription

1 Chapter 4 Digital Processing of Continuous-Time Signals 清大電機系林嘉文 cwlin@ee.nthu.edu.tw Original PowerPoint slides prepared by S. K. Mitra Digital Processing of Continuous-Time Signals Digital processing of a continuous-time signal involves the following basic steps: 1. Conversion of the continuous-time signal into a discrete-time signal 2. Processing of the discrete-time signal 3. Conversion of the processed discrete-time signal back into a continuous-time signal Conversion of a continuous-time time signal into digital form is carried out by an analog-to-digital (A/D) converter The reverse operation of converting a digital signal into a continuous-time signal is performed by a digital-toanalog (D/A) converter Original PowerPoint slides prepared by S. K. Mitra

2 Digital Processing of Continuous-Time Signals Since the A/D conversion takes a finite amount of time, a sample-and-hold (S/H) circuit is used to ensure that the input analog signal remains constant in amplitude until the conversion is complete to minimize representation error To prevent aliasing, an analog anti-aliasing filter is employed before the S/H circuit To smooth the output signal of the D/A converter, which has a staircase-like waveform, an analog reconstruction filter is used Original PowerPoint slides prepared by S. K. Mitra Digital Processing of Continuous-Time Signals Both the anti-aliasing filter and the reconstruction filter are analog lowpass filters The most widely used IIR digital filter design method is based on the conversion of an analog lowpass prototype Discrete-time signals in many applications are generated by sampling continuous-time signals There exists an infinite number of continuous-time signals, which when sampled lead to the same discrete-timetime signal Under certain conditions, it is possible to relate a unique continuous-time signal to a given discrete-time signal If these conditions hold, then it is possible to recover the original continuous-time signal from its sampled values Original PowerPoint slides prepared by S. K. Mitra

3 Effect of Sampling in the Frequency Domain Let g a (t) be a continuous-time signal that is sampled uniformly at t = nt, generating g the sequence g[n] ] where g[n] = g a (nt), < n < with T being the sampling period The reciprocal of T is called the sampling frequency F T, F T = 1/T The frequency-domain representation of g a (t) is given by its continuous-time Fourier transform (CTFT): The frequency-domain representation of g[n] is given by its DTFT: Original PowerPoint slides prepared by S. K. Mitra Effect of Sampling in the Frequency Domain To establish the relation between G a (jω) and G(e jω ), we treat the sampling operation mathematically as a multiplication of g a (t) by a periodic impulse train p(t) p(t) consists of a train of ideal impulses with a period T as shown below The multiplication operation yields an impulse train: g () t = g () t p() t = g ( nt) δ ( t nt) p a a n= Original PowerPoint slides prepared by S. K. Mitra

4 Effect of Sampling in the Frequency Domain g p (t) is a continuous-time signal consisting of a train of uniformly spaced impulses with the impulse at t = nt weighted by the sampled value g a (nt) of g a (t) at that instant There are two different forms of G p (jω) One form is given by the weighted sum of the CTFTs of Original PowerPoint slides prepared by S. K. Mitra Effect of Sampling in the Frequency Domain To derive the second form, we make use of the Poisson s sum formula: where Ω T = 2π/T, and Φ(jΩ) is the CTFT of ϕ(t) For t = 0, the above equation reduces to From the frequency shifting property p of the CTFT, the CTFT of g a (t)e jψt is given by G a (j(ω + Ψ)) Substituting ϕ(t) = g a (t)e jψt into the above equation, we arrive at Original PowerPoint slides prepared by S. K. Mitra

5 Effect of Sampling in the Frequency Domain By replacing with Ω in the above equation we arrive at the alternative form of the CTFT G p (jω) of g p (t), given by Therefore, G p (jω) is a periodic function of Ω consisting of a sum of shifted and scaled replicas of G a (jω), shifted by integer multiples of Ω T and scaled by 1/T The term on the RHS of the previous equation for k = 0is the baseband portion of G p (jω), and each of the remaining terms are the frequency translated portions of G p (jω) The frequency range is called the baseband or Nyquist band Original PowerPoint slides prepared by S. K. Mitra Effect of Sampling in the Frequency Domain Assume g a (t) is a band-limited signal with a CTFT G a (jω) as shown below The spectrum, P(jΩ) of p(t) having a sampling period T = 2π/ΩT is indicated below Original PowerPoint slides prepared by S. K. Mitra

6 Effect of Sampling in the Frequency Domain Two possible spectra of G p (jω) are shown below aliasing If Ω T 2Ω m, there is no overlap between the shifted replicas of G a (jω) generating G p (jω) If Ω T < 2Ω m, there is an overlap of the spectra of the shifted replicas of G a (jω) generating G p (jω) (Aliasing) Original PowerPoint slides prepared by S. K. Mitra Effect of Sampling in the Frequency Domain If Ω T > 2Ω m, g a (t) can be recovered exactly from g p (t) by passing it through an ideal lowpass filter H r (jω) with a gain T and a cutoff frequency Ω c greater than Ω m and less than Ω c Ω m as shown below Original PowerPoint slides prepared by S. K. Mitra

7 Effect of Sampling in the Frequency Domain The spectra of the filter and pertinent signals are as below If Ω T < 2Ω m, due to the overlap of the shifted replicas of G a (jω), the spectrum G a (jω) cannot be separated by filtering to recover G a (jω) because of the distortion caused by a part of the replicas immediately outside the baseband folded back or aliased into the baseband Original PowerPoint slides prepared by S. K. Mitra Sampling Theorem Sampling theorem - Let g a (t) be a band-limited signal with CTFT G a (jω) = 0 for Ω > Ω m. Then g a (t) is uniquely determined by its samples g a (nt), n if Ω T 2Ω m where Ω T = 2π/T The condition Ω T 2Ω m is often referred to as the Nyquist condition The frequency Ω T /2 is usually referred to as the folding frequency Original PowerPoint slides prepared by S. K. Mitra

8 Sampling Theorem Given {g a (nt)}, we can recover exactly g a (t) by generating an impulse train and then passing it through an ideal lowpass filter H r (jω) with a gain T and a cutoff frequency Ω c satisfying Ω m < Ω c < (Ω T Ω m ) The highest frequency Ω m contained in g a (t) is usually called the Nyquist frequency since it determines the minimum sampling frequency Ω T =2Ω m that must be used to fully recover g a (t) from its sampled version The frequency 2Ω m is called the Nyquist rate Original PowerPoint slides prepared by S. K. Mitra Sampling Theorem Oversampling - The sampling frequency is higher than the Nyquist rate Undersampling - The sampling frequency is lower than the Nyquist rate Critical sampling - The sampling frequency is equal to the Nyquist rate In digital telephony, a 3.4 khz signal bandwidth is acceptable for telephone conversation a sampling rate of 8 khz is used in telecommunication In high-quality analog music signal processing, a bandwidth of 20 khz has been determined to preserve the fidelity a sampling rate of 44.1 khz is used in CD music systems Original PowerPoint slides prepared by S. K. Mitra

9 Sampling Theorem Example - Consider the three sinusoidal signals: g 1 (t) = cos(6πt) g 2 (t) = cos(14πt) g 3 (t) = cos(26πt) Their corresponding CTFTs are: G 1 ( jω) = π[δ(ω 6π) + δ(ω + 6π)] G 2 ( jω) = π[δ(ω 14π) + δ(ω + 14π)] G 3 ( jω) = π[δ(ω 26π) + δ(ω + 26π)] Original PowerPoint slides prepared by S. K. Mitra Sampling Theorem These continuous-time signals sampled at a rate of T = 0.1 sec, i.e., with a sampling frequency rad/sec The sampling process generates the continuous-time time impulse trains g 1p (t), g 2p (t), and g 3p (t) Their corresponding CTFTs are given by aliasing Original PowerPoint slides prepared by S. K. Mitra aliasing 9

10 Sampling Theorem We now derive the relation between the DTFT of g[n] and the CTFT of g p (t) To this end we compare with and make use of g[n] = g a (nt), < n < Observation: We have or, equivalently, As a result Original PowerPoint slides prepared by S. K. Mitra Sampling Theorem We arrive at the desired result given by The relation derived above can be alternately expressed as From or from it follows that G(e jω ) is obtained from G p (jω) by applying the mapping Ω = ω/t (note, both G(e jω ) are G p (jω) are periodic) Original PowerPoint slides prepared by S. K. Mitra

11 Recovery of the Analog Signal (1/3) We now derive the expression for the output of the ideal lowpass reconstruction filter H r (jω) as a function of the samples g[n] The impulse response h r (t) of the lowpass reconstruction filter is obtained by taking the inverse DTFT of H r (jω) Thus, the impulse response is given by The input to the lowpass filter is the impulse train g p (t) Original PowerPoint slides prepared by S. K. Mitra Recovery of the Analog Signal (2/3) Therefore, the output of the ideal lowpass filter is given by: Substituting h r (t) = sin(ω c t) /(Ω T t/2) in the above and assuming for simplicity Ω c = Ω T /2 = π/t, we get sin [ π ( t nt) / T] ga () t = g[ n] n= π ( t nt)/ T The ideal band-limited interpolation process is shown below Amplitud de Original PowerPoint slides prepared by S. K. Mitra Time

12 Recovery of the Analog Signal (3/3) It can be shown that when Ω c = Ω T /2 in h r (0) =1 and h r (nt) = 0 for n 0 As a result, from we observe g a ( rt ) = g [ r ] = ga ( rt ) for all integer values of r in the range < r < The above relation holds whether or not the condition of the sampling theorem is satisfied However, for all values of t only if the sampling frequency Ω T satisfies the condition of the sampling theorem Original PowerPoint slides prepared by S. K. Mitra Implication of the Sampling Process Consider again the three continuous-time signals: g 1 (t) = cos(6πt), g 2 (t) = cos(14πt), and g 3 (t) = cos(26πt) The plot of the CTFT G 1p (jω) ofg g 1p (t) is shown below It is apparent that we can recover any of its frequencytranslated versions cos[(20k ± 6)πt] outside the baseband by passing g 1p (t) through an ideal analog bandpass filter with a passband centered at Ω = (20k ± 6)π For example, to recover the signal cos(34πt), it will be necessary to use a bandpass filter with a frequency response Original PowerPoint slides prepared by S. K. Mitra

13 Implication of the Sampling Process Likewise, we can recover the aliased baseband component cos(6πt) from the sampled version of either g 2p (t) or g 3p (t) by passing it through an ideal lowpass filter with a frequency response: There is no aliasing distortion unless the original continuous-time signal also contains the component cos(6πt) Similarly, from either g 2p (t) or g 3p (t) we can recover any one of the frequency translated versions, including the parent continuous-time signal g 2 (t) or g 3 (t) as the case may be, by employing suitable filters Original PowerPoint slides prepared by S. K. Mitra Sampling of Bandpass Signals (1/3) The conditions for the unique representation of a continuous-time signal by the discrete-time signal obtained by uniform sampling assumed that the continuous-time time signal is bandlimited in the frequency range from dc to some frequency Ω m Such a continuous-time signal is commonly referred to as a lowpass signal There are applications where the continuous-time signal is bandlimited to a higher frequency range Ω L Ω Ω H Such a signal is usually referred to as a bandpass signal To prevent aliasing a bandpass signal can of course be sampled at a rate greater than twice the highest frequency, i.e. by ensuring Ω L 2Ω H Original PowerPoint slides prepared by S. K. Mitra

14 Sampling of Bandpass Signals (2/3) The spectrum of the discrete-time signal obtained by sampling a bandpass signal will have spectral gaps with no signal components present in these gaps Moreover, if Ω H is very large, the sampling rate also has to be very large which may not be practical in some situations A more practical approach is to use under sampling Let Ω = Ω H Ω L define the bandwidth of the bandpass signal Assume that the highest frequency contained in the signal is an integer multiple of the bandwidth, i.e., Ω H = M( Ω) Choose the sampling frequency Ω T to satisfy the condition Ω T = 2( Ω) = 2Ω H /M which is smaller than 2Ω H, the Nyquist rate Original PowerPoint slides prepared by S. K. Mitra Sampling of Bandpass Signals (3/3) Substitute the above expression for Ω T in This leads to As before, G p (jω) consists of a sum of G a (jω) and replicas of G a (jω) shifted by integer multiples of twice the bandwidth Ω and scaled by 1/T The amount of shift for each value of k ensures that there will be no overlap between all shifted replicas no aliasing Original PowerPoint slides prepared by S. K. Mitra

15 Sampling of Bandpass Signals As shown above, g a (t) can be recovered from g p (t) by passing it through an ideal bandpass filter with a passband given by Ω L Ω Ω H and a gain of T Note: Any of the replicas in the lower frequency bands can be retained by passing g p (t) through bandpass filters with passbands Ω L k( Ω) Ω Ω H k( Ω), 1 k M 1 Original PowerPoint slides prepared by S. K. Mitra Analog Lowpass Filter Specifications Typical magnitude response H a (jω) of analog lowpass filter: In the passband, defined by 0 Ω Ω p, we require 1 δ p H a (jω) 1+ δ p, Ω Ω p, i.e., H a (jω) approximates unity within an error of ± δ p In the stopband, defined by Ω s Ω <, we require H a (jω) δ s, Ω s Ω <, i.e., H a (jω) approximates zero within an error of δ s Original PowerPoint slides prepared by S. K. Mitra

16 Analog Lowpass Filter Specifications Ω p - passband edge frequency Ω s - stopband edge frequency δ p - peak ripple value in the passband δ s - peak ripple value in the stopband Peak passband ripple α p = 20log 10 (1 δ p ) Minimum stopband attenuation α s = 20log 10 (δ s ) Original PowerPoint slides prepared by S. K. Mitra Analog Lowpass Filter Specifications Magnitude specifications may alternately be given in a normalized form as indicated below Here, the maximum value of the magnitude in the passband assumed to be unity - Maximum passband deviation, given by the minimum value of the magnitude in the passband 1/A -Maximum stopband magnitude Original PowerPoint slides prepared by S. K. Mitra

17 Analog Lowpass Filter Design Two additional parameters are defined: 1) Transition ratio k = Ω p /Ω s for a lowpass filter k < 1 2) Discrimination parameter usually k 1 << 1 Original PowerPoint slides prepared by S. K. Mitra Butterworth Approximation (1/5) The magnitude-square response of an N-th order analog lowpass Butterworth filter is given by First 2N 1 derivatives of H a (jω) 2 at Ω = 0 are equal to zero The Butterworth lowpass filter thus is said to have a maximally-flat magnitude at Ω = 0 Gain in db is G(Ω) = 10log 10 H a (jω) 2 As G(0) = 0 and G(Ω c ) =10log 10 (0.5) = dB Ω c is called the 3-dB cutoff frequency Original PowerPoint slides prepared by S. K. Mitra

18 Butterworth Approximation (2/5) Typical magnitude responses with Ω c = 1 Two parameters completely characterizing a Butterworth lowpass filter are Ω c and N These are determined from the specified banedges Ω p and Ω s, and minimum passband magnitude, and maximum stopband ripple 1/A Original PowerPoint slides prepared by S. K. Mitra Butterworth Approximation (3/5) Ω c and N are thus determined from Solving the above we get Since order N must be an integer, value obtained is rounded up to the next highest integer N is used next to determine Ω c by satisfying either the stopband edge or the passband edge spec exactly If the stopband edge spec is satisfied, then the passband edge spec is exceeded providing a safety margin Original PowerPoint slides prepared by S. K. Mitra

19 Butterworth Approximation (4/5) Transfer function of an analog Butterworth lowpass filter is given by where Denominator D N N(s) is known aste the Butterworth t polynomial of order N Original PowerPoint slides prepared by S. K. Mitra Butterworth Approximation (5/5) Example- Determine the lowest order of a Butterworth lowpass filter with a 1-dB cutoff frequency at 1 khz and a minimum attenuation of 40 db at 5 khz Now which yields ε 2 = , and 10log10(1/A 2 ) = 40 Which yields A 2 = 10,000 Therefore and Hence We choose N = 4 Original PowerPoint slides prepared by S. K. Mitra

20 Chebyshev Approximation (1/3) The magnitude-square response of an N-th order analog lowpass Type 1 Chebyshev filter is given by where T N (Ω) is the Chebyshev polynomial of order N: Typical magnitude response plots of the analog lowpass Type 1 Chebyshev filter are shown below Original PowerPoint slides prepared by S. K. Mitra Chebyshev Approximation (2/3) If Ω = Ω s at the magnitude is equal to 1/A, then Solving the above we get Order N is chosen as the nearest integer greater than or equal to the above value The magnitude-square response of an N-th order analog lowpass Type 2 Chebyshev (also called inverse Chebyshev) filter is given by where T N (Ω) is the Chebyshev polynomial of order N Original PowerPoint slides prepared by S. K. Mitra

21 Chebyshev Approximation (3/3) Typical response plots of Type 2 Chebyshev lowpass filter: The order N of the Type 2 Chebyshev filter is determined by Example - Determine the lowest order of a Chebyshev lowpass filter with an 1-dB cut-off frequency at 1 KHz and a maximum attenuation of 40 db at 5 KHz Original PowerPoint slides prepared by S. K. Mitra Elliptic Approximation (1/2) The square-magnitude response of an elliptic lowpass filter is given by where R N (Ω) is a rational function of order N satisfying R N (1/Ω) = 1/R N (Ω), with the roots of its numerator lying in the interval 0 < Ω <1 and the roots of its denominator lying in the interval 1< Ω < For given Ω p, ε, Ω s, and A, the filter order can be estimated using where Original PowerPoint slides prepared by S. K. Mitra

22 Elliptic Approximation (2/2) Example - Determine the lowest order of an Elliptic lowpass filter with an 1-dB cut-off frequency at 1 KHz and a maximum attenuation of 40 db at 5 KHz Note k = 0.2 and 1/ k 1 = Substituting these values we get k = , ρ 0 = , and ρ = Hence N = (choosing N = 3) Typical magnitude response plots with Ω p =1 are shown below Original PowerPoint slides prepared by S. K. Mitra Analog Lowpass Filter Design Example - Determine the lowest order of an Elliptic lowpass filter with an 1-dB cut-off frequency at 1 KHz and a maximum attenuation of 40 db at 5 KHz Code fragments used [N, Wn] = ellipord(wp, Ws, Rp, Rs, s ); [b, a] = ellip(n, Rp, Rs, Wn, s ); with Wp = 2*pi*1000; Ws = 2*pi*5000; Rp = 1; Rs = 40; Original PowerPoint slides prepared by S. K. Mitra

23 Design of Analog Highpass, Bandpass, and Bandstop Filters Steps involved in the design process: Step 1 Develop specifications of a prototype analog lowpass filter H LP (s) from specifications of desired analog filter H D (s) using a frequency transformation Step 2 Design the prototype analog lowpass filter Step 3 Determine the transfer function H D (s) of desired filter by applying inverse frequency transformation to H LP (s) Let s denote the Laplace transform variable of prototype analog lowpass filter H LP (s) and denote the Laplace transform variable ŝ of desired analog filter H D (ŝ) Then Original PowerPoint slides prepared by S. K. Mitra Analog Highpass Filter Design (1/2) Spectral Transformation: where Ω p is the passband edge frequency of H LP (s) and is the passband edge frequency of H HP (ŝ) On the imaginary axis the transformation is Original PowerPoint slides prepared by S. K. Mitra

24 Analog Highpass Filter Design (2/2) Example - Determine the lowest order of an Butterworth lowpass filter with the specifications: Choose Ω p = 1, then Analog lowpass filter specifications: Ω p = 1, Ω s = 1, α p = 0.1 db, α s = 40 db, Code fragments used [N, Wn] = buttord(1, 4, 0.1, 40, s ); [B, A] = butter(n, Wn, s ); [num, den] = lp2hp(b, A, 2*pi*4000); Original PowerPoint slides prepared by S. K. Mitra Analog Bandpass Filter Design (1/7) Spectral Transformation: where Ω p is the passband edge frequency of H LP (s), and are the lower and upper passband edge frequencies of desired bandpass filter H BP (ŝ) On the imaginary axis the transformation is Original PowerPoint slides prepared by S. K. Mitra

25 Analog Bandpass Filter Design (2/7) Case 1: To make we can either increase any one of the stopband edges or decrease any one of the passband edges as shown below 1) Decrease to larger passband and shorter leftmost transition band 2) Increase to No change in passband and shorter leftmost transition band Original PowerPoint slides prepared by S. K. Mitra Analog Bandpass Filter Design (3/7) Note the condition can also be satisfied by decreasing which is acceptable as the passband is reduced from the desired value Alternately, the condition can be satisfied by increasing which is not acceptable as the upper stop band is reduced from the desired value Original PowerPoint slides prepared by S. K. Mitra

26 Analog Bandpass Filter Design (4/7) Case 2: To make we can either increase any one of the stopband edges or decrease any one of the passband edges as shown below 1) Increase to larger passband and shorter rightmost transition band 2) Decrease to No change in passband and shorter rightmost transition band Original PowerPoint slides prepared by S. K. Mitra Analog Bandpass Filter Design (5/7) Note the condition can also be satisfied by increasing which is acceptable as the passband is reduced from the desired value Alternately, the condition can be satisfied by decreasing which is not acceptable as the lower stop band is reduced from the desired value Original PowerPoint slides prepared by S. K. Mitra

27 Analog Bandpass Filter Design (6/7) Example - Determine the lowest order of an Elliptic bandpass filter with the specifications: Now and, since we choose We choose Ω p = 1 Hence Analog lowpass filter specifications: Ω p = 1, Ω s = 1.4, α p = 1 db, α s = 22 db Original PowerPoint slides prepared by S. K. Mitra Analog Bandpass Filter Design (7/7) Code fragments used: [N, Wn] = ellipord(1, 1.4, 1, 22, s ); [B, A] = ellip(n, 1, 22, Wn, s ); s); [num, den] = lp2bp(b, A, 2*pi* , 2*pi*25/7); Gain plot: Frequency KHz Original PowerPoint slides prepared by S. K. Mitra