ECE4330 Lecture 23 Sampling and Signal Reconstruction Prof. Mohamad Hassoun Sampling of Continuous-Time Analog Signals Discrete-time system implementation is very convenient because it takes advantage of the flexibility and efficiency of digital processors. In practice, most discrete-time systems require us to discretize the (physical) continuous input signal. A sequence 푓[푛] (or 푓[푘]) can be obtained from a continuous-time signal 푓(푡) through sampling. Here, the value of 푓[푛] for some integer 푛 is equal to 푓(푛푇푠), the value of 푓(푡) at time 푡 = 푛푇푠. So, we sample at regular intervals of width 푇푠 seconds or, equivalently, with a 1 sampling rate of 푓푠 = Hz. It is kind of intuitive to argue that sampling 푇푠 generally leads to loss of signal information. But we may also argue that this loss of information about the signal 푓(푡) can be made insignificant if we sample at a very high rate (i.e., select very small 푇푠). The question now is whether or not, for a finite sampling rate, 푓푠, we can perfectly reconstruct 푓(푡) from its samples, 푓[푛]. If the answer is yes, then we need to answer the following questions: (1) What is a sufficient sampling rate that allows for perfect signal reconstruction? (2) Are there restrictions on the nature of the continuous-time signal 푓(푡) being sampled? (3) Can we derive a formula to reconstruct 푓(푡) from its samples 푓[푛]? (4) Can we realize a physical (or simulated) system that accepts 푓[푛] and generates 푓(푡)? Surprisingly, the answer to the above questions is affirmative. This is captured by Shannon’s Sampling Theorem: Let 푓(푡) be a real-valued, continuous-time, bandlimited (finite bandwidth) signal with bandwidth 퐵. And let 푓[푛] be the sequence of numbers obtained by sampling 푓(푡) at a sampling rate of 푓푠 samples per second (Hz); i.e., a sample is taken every 1 푇푠 = seconds. Then, 푓(푡) can be perfectly reconstructed from its 푓푠 samples 푓[푛] if and only if 푓푠 > 2퐵. The sampling rate must exceed twice the bandwidth of the signal (2퐵 is referred to as Nyquist sampling rate). Furthermore, 푓(푡) can be uniquely reconstructed by the following interpolation formula: ∞ ∞ sin [휋푓푠(푡 − 푛푇푠)] 푓(푡) = ∑ 푓(푛푇푠) = ∑ 푓[푛]sinc [휋푓푠(푡 − 푛푇푠)] 휋푓푠(푡 − 푛푇푠) 푛=−∞ 푛=−∞ In the remainder of this lecture, we will prove the Sampling Theorem and discuss its practical applications to storage and reconstruction of digital data. Again, we will witness the importance of Fourier methods (transform and series) in the analysis and design of sampling systems. The Sampling Process Today, digital computers are used to store and process physical and synthetic signals in the form of digital data. For example, an audio signal picked up by the microphone of your laptop or smart phone is first sampled and then digitized (amplitude converted into binary code) prior to processing or storage on disk. Later, before the digital audio signal can drive a speaker, the digital amplitudes (binary code) must be converted back into analog values and then reconstructed as a continuous-time signal. The analog to digital conversion is performed using an analog-to- digital convertor (ADC) circuit. Similarly, the digital to analog conversion is performed using a digital-to-analog (DAC) circuit. The following figure illustrates such steps. In the remainder of this lecture, we will use 푓(푛푇푠) to signify the sampled signal. We will also use the symbol 푓[푛] to represent the value of the 푛th sample of 푓(푡) at time 푛푇푠. It should be stressed here that 푓[푛] is a sequence of numbers. It does not retain any information about the sampling rate or time (i.e., it can’t be displayed on an oscilloscope). On the other hand, 푓(푛푇푠) is the true sampled signal that position its non-zero values at exact time instances, 푛푇푠 (this signal will display physically on an oscilloscope). We will have more to say about this in the next lecture. As we will see, the real signal 푓(푡) is generally not bandlimited, therefore the reconstructed signal will have some degree of distortion. Another source of distortion is the approximate nature of the reconstruction filter (a low-pass filter) that is used to reconstruct (interpolate) the signal. Sampling, in theory, consists of simply modulating (multiplying) the signal 푓(푡) by another signal, 훿푠(푡), which consists of a train of unit- impulses separated (in time) by 푇푠 seconds. We will refer to the sampled signal as 푔(푡), and express it analytically as ∞ 푔(푡) = 푓(푡)훿푠(푡) = 푓(푡) ∑ 훿(푡 − 푛푇푠) 푛=−∞ Employing the sampling property of the delta function [namely, 푓(푡)훿(푡 − 푎) = 푓(푎)훿(푡 − 푎)] we may express the above expression as ∞ 푔(푡) = ∑ 푓(푛푇푠)훿(푡 − 푛푇푠) 푛=−∞ The following figure depicts the sampling process for an arbitrary, bandlimited signal. We may now gain a better picture of the sampling process by considering the signals in the frequency domain. Let us apply the Fourier transform to the sampled signal ∞ 푔(푡) = 푓(푡)훿푠(푡) = 푓(푡) ∑ 훿(푡 − 푛푇푠) (1) 푛=−∞ We will employ the frequency-convolution property 푓1(푡)푓2(푡) ↔ 1 퐹 (휔) ∗ 퐹 (휔). We may also note that 훿 (푡) is a periodic signal (of 2휋 1 2 푠 2휋 fundamental frequency 휔푠 = = 2휋푓푠) having the following exponential 푇푠 Fourier series representation (derive it! This was assigned in Lecture 15) ∞ ∞ ∞ 1 휔푠 푗푛휔푠푡 푗푛휔푠푡 훿푠(푡) = ∑ 훿(푡 − 푛푇푠) = ∑ 푒 = ∑ 푒 푇푠 2휋 푛=−∞ 푛=−∞ 푛=−∞ 푗푛휔표푡 By employing the Fourier Pair 푒 ↔ 2휋훿(휔 − 푛휔표) and the superposition property, then the Fourier transform of 훿푠(푡) would be ∞ 퐹{훿푠(푡)} = 휔푠 ∑ 훿(휔 − 푛휔푠) 푛=−∞ Transforming the signal 푔(푡) = 푓(푡)훿푠(푡) we obtain ∞ ∞ 1 휔 퐺(휔) = 퐹(휔) ∗ [휔 ∑ 훿(휔 − 푛휔 )] = 푠 ∑ 퐹(휔 − 푛휔 ) 2휋 푠 푠 2휋 푠 푛=−∞ 푛=−∞ where the distributive property and the convolution with a shifted impulse, 퐹(휔) ∗ 훿(휔 − 푛휔푠) = 퐹(휔 − 푛휔푠) were applied. We may also, for convenience, express the above result in terms of the Hz frequency, 푓, as 휔 (recall that 푓 = 푠) 푠 2휋 ∞ 퐺(푓) = ∑ 푓푠퐹(푓 − 푛푓푠) 푛=−∞ [Do not confuse the Hz frequencies 푓 and 푓푠 with the signal 푓(푡). ] The above equation represents a very important and elegant result. It states that, in the frequency domain, the sampled signal is essentially a superposition of shifted (by integer multiples of the sampling frequency, 푓푠) versions of the spectrum of the signal 푓(푡). Also, the amplitude of the spectrum 퐺(푓) is scaled by the sampling rate, 푓푠. The following figure depicts, graphically, the magnitude spectra |퐹(푓)| and |퐺(푓)| as a function of frequency, 푓. Note how the magnitude spectrum |퐹(푓)| is duplicated every 푓푠 (Hz) in the spectrum for |퐺(푓)|. These duplicates are known as the image spectra. It shows that if the sampling frequency 푓푠 is twice the bandwidth (i.e., 푓푠 = 2퐵) of the bandlimited signal 푓(푡), then there will be no overlap between |퐹(푓)| and its images, therefore an ideal (brick-wall) low-pass filter with a 푓 cutoff frequency 푓 = 푠 = 퐵 (and dc gain 푇 ) can be used to perfectly 표 2 푠 reconstruct 푓(푡) (refer to the following figure, Part 푐). Increasing the sampling rate beyond 2퐵 (푓푠 > 2퐵) allows for further separation between |퐹(푓)| and its adjacent image spectra, and may allow for a practical filter (say a 4th or higher order Butterworth filter) to properly reconstruct the signal 푓(푡) [refer to (푑) in the following figure]. On the other hand if we under sample, 푓푠 < 2퐵, then there will be overlap with the image spectra and the reconstructed signal will always have distortion (aliasing) [refer to (푏) in the following figure]. The 푓푠 = 2퐵 threshold value on the sampling frequency is known as the Nyquist sampling rate. There are some special bandlimited signals that require the sampling frequency to strictly exceed this sampling rate for perfect reconstruction. Also, for non-bandlimited signals with effective bandwidth 퐵, 푓푠 would have to significantly exceed 2퐵. Your turn: Consider 푓(푡) = sin(2휋푡) sampled at the Nyquist rate, 푓푠 = 2Hz. Can the reconstruction lead to a unique signal? Explain. Your turn: Compute the Nyquist sampling rate [think (95%) effective bandwidth 퐵; Lecture 18] for reconstructing the signal 푓(푡) = 0.5푒−|푡| from its samples. Sketch the spectrum, 퐺(푓), of the sampled signal assuming: (1) 푓푠 = 2퐵, (2) 푓푠 = 4퐵 and (3) 푓푠 = 8퐵. What is the smallest sampling rate that is appropriate for this non bandlimited signal? Explain. Your turn: Repeat the above problem assuming the signal (휎 = 0.5) 푡2 − 푓(푡) = 푒 2휎2 The Sinc Interpolation Formula Next, we derive an analytical formula for reconstructing the signal 푓(푡) from its samples. Assuming the sampling condition 푓푠 ≥ 2퐵 and a brick- wall low-pass reconstruction filter, we may solve for the (zero-state) output 푦(푡) of the filter by convoluting 푔(푡) with the unit-impulse response, ℎ(푡), of the filter, 푦(푡) = 푔(푡) ∗ ℎ(푡). The transfer function of 푓 the filter (a rectangular, brick-wall spectrum with cutoff frequency 푠 and a 2 1 푓 dc gain 푇푠) can be expressed as 퐻(푓) = rect ( ), or 퐻(휔) = 푓푠 푓푠 2휋 휔 rect ( ). We may employ Fourier Pair #18 which is reproduced below 휔푠 휔푠 to find the filter’s impulse response, 2휋 휔푠 휔푠 ℎ(푡) = ( ) ( ) sinc ( 푡) = sinc(휋푓푠푡) 휔푠 2휋 2 sin(휋푓푠푡) where sinc(휋푓푠푡) = .