Digital Signal Processing

Transcription

1 Digital Signal Processing Introduction Moslem Amiri, Václav Přenosil Embedded Systems Laboratory Faculty of Informatics, Masaryk University Brno, Czech Republic February 25, 2012

2 The Concept of Frequency Frequency is closely related to a specific type of periodic motion called harmonic oscillation Described by sinusoidal functions Frequency has dimension of inverse time Nature of time (continuous or discrete) would affect nature of frequency accordingly 2 / 45

3 Continuous-Time Sinusoidal Signals A simple harmonic oscillation x a (t) = A cos(ωt + θ), < t < Subscript a = analog signal A = amplitude Ω = frequency (in rad/s) θ = phase (in radians) Rewriting above equation using frequency F in cycles per second or hertz (Hz) x a (t) = A cos(2πft + θ), < t < 3 / 45

4 Continuous-Time Sinusoidal Signals 4 / 45

5 Continuous-Time Sinusoidal Signals Properties of x a (t) = A cos(2πft + θ), < t < 1 For every fixed F, x a (t) is periodic x a (t + T p ) = x a (t) T p = 1/F = fundamental period of sinusoidal signal 2 Signals with distinct frequencies are themselves distinct 3 Increasing F results in an increase in rate of oscillation of signal Using Euler identity e ±jφ = cos φ ± j sin φ and introducing negative frequencies x a (t) = A cos(ωt + θ) = A 2 ej(ωt+θ) + A 2 e j(ωt+θ) A sinusoidal signal can be obtained by adding two equal-amplitude complex-conjugate exponential signals, called phasors As time progresses, phasors rotate in opposite directions with angular frequencies ±Ω radians/second 5 / 45

6 Discrete-Time Sinusoidal Signals A discrete-time sinusoidal signal x(n) = A cos(ωn + θ), < n < n = an integer called sample number A = amplitude ω = frequency in radians/sample θ = phase in radians Using ω = 2πf frequency f is in cycles/sample x(n) = A cos(2πfn + θ), < n < 6 / 45

7 Discrete-Time Sinusoidal Signals 7 / 45

8 Discrete-Time Sinusoidal Signals Properties of discrete-time sinusoids 1 A discrete-time sinusoid is periodic only if its frequency f is a rational number x(n) is periodic with period N(N > 0) if and only if x(n + N) = x(n) for all n Smallest value of N for which this equation is true is called fundamental period Proof of this property cos[2πf 0(N + n) + θ] = cos(2πf 0n + θ) 2πf 0N = 2kπ f 0 = k/n To determine fundamental period N, express its frequency as f 0 = k/n and cancel common factors so that k and N are relatively prime, then N is answer 8 / 45

9 Discrete-Time Sinusoidal Signals Properties of discrete-time sinusoids (continued) 2 Discrete-time sinusoids whose frequencies are separated by an integer multiple of 2π are identical cos[(ω 0 + 2kπ)n + θ] = cos(ω 0 n + 2πkn + θ) = cos(ω 0 n + θ) where π ω 0 π Discrete-time sinusoids with ω π or f 1 are unique 2 Any sequence resulting from a sinusoid with ω > π or f > 1 is 2 identical to a sequence obtained from a sinusoid with ω < π Sinusoid having ω > π is called an alias of a corresponding sinusoid with ω < π 9 / 45

10 Discrete-Time Sinusoidal Signals Properties of discrete-time sinusoids (continued) 3 The highest rate of oscillation in a discrete-time sinusoid is attained when ω = π (or ω = π) or, equivalently, f = 1 2 (or f = 1 2 ) 2 x(n) = cos ω 0n, ω 0 = 0 = N = x(n) x(n) = cos 0.5 ω 0n, ω 0 = π = N = 16 1 x(n) n n / 45

11 Discrete-Time Sinusoidal Signals Properties of discrete-time sinusoids (continued) 3 The highest rate of oscillation is when ω = π 2 x(n) = cos ω 0n, ω 0 = π = N = x(n) n x(n) = cos 1.5 ω 0n, ω 0 = π = N = x(n) n / 45

12 Discrete-Time Sinusoidal Signals Properties of discrete-time sinusoids (continued) 3 The highest rate of oscillation is when ω = π 2 x(n) = cos ω 0n, ω 0 = π = N = x(n) n For π ω 0 2π, if consider sinusoids with ω 1 = ω 0 and ω 2 = 2π ω x 1(n) = A cos ω 1n = A cos ω 0n x 2(n) = A cos ω 2n = A cos(2π ω 0)n = A cos( ω 0n) = x 1(n) Hence, ω 2 is an alias of ω 1 Using a sine function, result would be same, except phase difference would be π between x 1(n) and x 2(n) 12 / 45

13 Discrete-Time Sinusoidal Signals Negative frequencies for discrete-time signals x(n) = A cos(ωn + θ) = A 2 ej(ωn+θ) + A 2 e j(ωn+θ) Since discrete-time sinusoids with frequencies separated by 2kπ are identical Frequency range for discrete-time sinusoids is finite with duration 2π Usually 0 ω 2π or π ω π is called fundamental range 13 / 45

14 Harmonically Related Complex Exponentials Harmonically related complex exponentials Sets of periodic complex exponentials with fundamental frequencies that are multiples of a single positive frequency Properties which hold for complex exponentials, also hold for sinusoidal signals We confine our discussion to complex exponentials 14 / 45

15 Continuous-Time Exponentials Basic signals for continuous-time, harmonically related exponentials s k (t) = e jkω 0t = e j2πkf 0t k = 0, ±1, ±2,... For each k, s k (t) is periodic with fundamental period 1/(kF 0 ) = T p /k or fundamental frequency kf 0 A signal that is periodic with period T p /k is also periodic with period k(t p /k) = T p for any positive integer k Hence all s k (t) have a common period of T p A linear combination of harmonically related complex exponentials x a (t) = k= c ks k (t) = k= c ke jkω 0t This is Fourier series expansion for x a (t) c k, k = 0, ±1, ±2,... are arbitrary complex constants (Fourier series coefficients) s k (t) is kth harmonic of x a (t) x a (t) is periodic with fundamental period T p = 1/F 0 15 / 45

16 Discrete-Time Exponentials A discrete-time complex exponential is periodic if its frequency is a rational number Hence, we choose f 0 = 1/N Sets of harmonically related complex exponentials s k (n) = e j2πkf 0n, k = 0, ±1, ±2,... Since s k+n (n) = e j2πn(k+n)/n = e j2πn s k (n) = s k (n) there are only N distinct periodic complex exponentials in the set All members of the set have a common period of N samples We can choose any consecutive N complex exponentials to form a set For convenience s k (n) = e j2πkn/n, k = 0, 1, 2,..., N 1 Fourier series representation for a periodic discrete-time sequence x(n) = N 1 k=0 c ks k (n) = N 1 k=0 c ke j2πkn/n Fundamental period = N Fourier coefficients = {c k } Sequence s k (n) is called kth harmonic of x(n) 16 / 45

17 Discrete-Time Exponentials Example Stored in memory is one cycle of sinusoidal signal x(n) = sin ( 2πn N + θ) where θ = 2πq/N, where q and N are integers Obtain values of harmonically related sinusoids having the same phase x k (n) = sin ( ( ) 2πnk N + θ) = sin 2π(kn) N + θ = x(kn) Thus x k (0) = x(0), x k (1) = x(k), x k (2) = x(2k),... Obtain sinusoids of the same frequency but different phase We control phase θ of sinusoid with f k = k/n by taking first value of sequence from memory location q = θn/2π, where q is an integer We wrap around table each time index (kn) exceeds N 17 / 45

18 Analog-to-Digital and Digital-to-Analog Conversion Analog-to-digital (A/D) conversion Converting analog signals to a sequence of numbers having finite precision Corresponding devices are called A/D converters (ADCs) A/D conversion is a three-step process 18 / 45

19 Analog-to-Digital and Digital-to-Analog Conversion A/D conversion process 1 Sampling Taking samples of continuous-time signal at discrete-time instants x a(t) is input x a(nt ) x(n) is output T = sampling interval 2 Quantization Conversion of a discrete-time continuous-valued signal into a discrete-time, discrete-valued signal Value of each sample is selected from a finite set of possible values Quantization error: Difference between unquantized sample x(n) and quantized output x q(n) 3 Coding Each discrete value x q(n) is represented by a b-bit binary sequence 19 / 45

20 Analog-to-Digital and Digital-to-Analog Conversion Digital-to-analog (D/A) conversion Process of converting a digital signal into an analog signal Interpolation Connecting dots in a digital signal Approximations: zero-order hold (staircase), linear, quadratic, and so on 20 / 45

21 Sampling of Analog Signals Periodic or uniform sampling x(n) = x a (nt ), < n < T = sampling period or sample interval 1/T = F s = sampling rate (samples/second) or sampling frequency (hertz) 21 / 45

22 Sampling of Analog Signals Relationship between t of continuous-time and n of discrete-time signals t = nt = n F s To establish a relationship between F (or Ω) and f (or ω) x a (t) = A cos(2πft + θ) ( x a (nt ) x(n) = A cos(2πfnt + θ) = A cos 2πnF F s f = F /F s ω = ΩT Substituting f = F /F s and ω = ΩT into following range 1 2 < f < 1 2 π < ω < π we find that F and Ω must fall in the range 1 2T = Fs 2 F Fs 2 = 1 2T π T = πf s Ω πf s = π T ) + θ 22 / 45

23 Sampling of Analog Signals Summary of relations among frequency variables Continuous-time signals Discrete-time signals Ω = 2πF ω = 2πf radians radians sec Hz sample ω=ωt,f =F /F s cycles sample π ω π 1 2 f 1 2 Ω=ω/T,F =f.f s < Ω < π/t Ω π/t < F < F s /2 F F s /2 Since the highest frequency in a discrete-time signal is ω = π or f = 1 2 F max = Fs 2T Ω max = πf s = π T 2 = 1 23 / 45

24 Sampling of Analog Signals Example Consider these analog sinusoids sampled at F s = 40 Hz x 1 (t) = cos 2π(10)t x 2 (t) = cos 2π(50)t Corresponding discrete-time signals x 1 (n) = cos 2π ( ) n = cos π 2 n x 2 (n) = cos 2π ( ) n = cos 5π 2 n However cos 5πn/2 = cos(2πn + πn/2) = cos πn/2 x 2 (n) = x 1 (n) Given sampled values generated by cos(π/2)n, there is ambiguity as to whether they correspond to x 1 (t) or x 2 (t) F 2 = 50 Hz is an alias of F 1 = 10 Hz at F s = 40 All cos 2π(F k)t, k = 1, 2,... sampled at F s = 40 are aliases of F 1 = / 45

25 Sampling of Analog Signals If sinusoids where x a (t) = A cos(2πf k t + θ) F k = F 0 + kf s, k = ±1, ±2,... are sampled at F s, then F k is outside the range F s /2 F F s /2 ( x(n) x a (nt ) = A cos 2π F ) 0 + kf s n + θ F s = A cos(2πnf 0 /F s + θ + 2πkn) = A cos(2πf 0 n + θ) Frequencies F k = F 0 + kf s, are aliases of F 0 after sampling < k < 25 / 45

26 Sampling of Analog Signals 26 / 45

27 Sampling of Analog Signals Example Two sinusoids: F 1 = 7 8 Hz and F 2 = 1 8 Hz with F s = 1 Hz F 1 = F 2 + kf s, k = ±1, ±2,... k = 1, F 2 = F 1 + F s = ( )Hz = 1 8 Hz 27 / 45

28 Sampling of Analog Signals F s /2 (which corresponds to ω = π) is highest frequency that can be represented uniquely with F s Use F s /2 or ω = π as pivotal point and fold alias frequency to range 0 ω π F s /2 (ω = π) is called folding frequency 28 / 45

29 Sampling of Analog Signals Example x a (t) = 3 cos 100πt 1 Minimum sampling rate required to avoid aliasing F = 50 Hz F s = 100 Hz 2 Suppose F s = 200 Hz. Discrete-time signal obtained after sampling x(n) = 3 cos 100π 200 n = 3 cos π 2 n 3 Suppose F s = 75 Hz. Discrete-time signal obtained after sampling x(n) = 3 cos 100π 75 n = 3 cos 4π 3 n = 3 cos ( 2π 2π ) 3 n = 3 cos 2π 3 n 4 Frequency 0 < F < F s /2 of a sinusoid that yields samples identical to those obtained in part (3) For F s = 75 Hz, F = ff s = 75f In part (3), f = 1 3 F = 25 Hz y a (t) = 3 cos 2πFt = 3 cos 50πt Hence F = 50 Hz is an alias of F = 25 Hz for F s = 75 Hz 29 / 45

30 The Sampling Theorem Given any analog signal, how T or equivalently F s should be selected Knowing F max of general class of signals (e.g., class of speech signals), we can specify F s Suppose any analog signal can be represented as x a (t) = N i=1 A i cos(2πf i t + θ i ) N = number of frequency components F max may vary slightly from different realizations among signals of any given class (e.g., from speaker to speaker) To ensure F max does not exceed some predetermined value, pass analog signal through a filter that attenuates frequency components above F max Any frequency outside F s /2 F F s /2 results in samples identical with a corresponding frequency inside this range To avoid aliasing F s > 2F max This condition ensures that any frequency component ( F i < F max) in analog signal is mapped into a discrete-time sinusoid with a frequency 1 2 f i = F i F s / 45

31 The Sampling Theorem Sampling theorem If F max = B for x a (t) and F s > 2F max 2B, then x a (t) can be exactly recovered from its sample values using the interpolation function x a (t) = sin 2πBt g(t) = 2πBt ( ) ) n x a g (t nfs F s n= where x a (n/f s ) = x a (nt ) x(n) are samples of x a (t) Nyquist rate = F N = 2B = 2F max 31 / 45

32 The Sampling Theorem Example x a (t) = 3 cos 50πt + 10 sin 300πt + cos 100πt Nyquist rate for this signal F 1 = 25 Hz, F 2 = 150 Hz, F 3 = 50 Hz F max = 150 Hz F s > 2F max = 300 Hz F N = 2F max = 300 Hz Component 10 sin 300πt sampled at F N = 300 results in 10 sin πn which are identically zero If θ 0 or π, samples taken at Nyquist rate are not all zero 10 sin(πn + θ) = 10(sin πn cos θ + cos πn sin θ) = 10 sin θ cos πn = ( 1) n 10 sin θ To avoid this uncertain situation, sample analog signal at a rate higher than Nyquist 32 / 45

33 The Sampling Theorem Example x a (t) = 3 cos 2000πt + 5 sin 6000πt + 10 cos 12000πt Nyquist rate for this signal F 1 = 1 khz, F 2 = 3 khz, F 3 = 6 khz F max = 6 khz F s > 2F max = 12 khz F N = 12 khz Assume F s = 5000 samples/s. Signal obtained after sampling Folding frequency ( ) = Fs 2 = 2.5 khz x(n) = x a (nt ) = x n a F s = 3 cos 2π( 1 5 )n + 5 sin 2π( 3 5 )n + 10 cos 2π( 6 5 )n = 3 cos 2π( 1 5 )n + 5 sin 2π(1 2 5 )n + 10 cos 2π( )n = 3 cos 2π( 1 5 )n + 5 sin 2π( 2 5 )n + 10 cos 2π( 1 5 )n = 13 cos 2π( 1 5 )n 5 sin 2π( 2 5 )n 33 / 45

34 The Sampling Theorem Example (continued) Second solution: Aliases of F 0 : F k = F 0 + kf s F 0 = F k kf s such that F s /2 F 0 F s /2 F 1 is less than F s /2 F 2 = F 2 F s = 2 khz F 3 = F 3 F s = 1 khz f = F F s f 1 = 1 5, f 2 = 2 5, f 3 = 1 5 Analog signal y a (t) reconstructed from samples using ideal interpolation Only frequency components at 1 khz and 2 khz are present in sampled signal y a (t) = 13 cos 2000πt 5 sin 4000πt This distortion was caused by aliasing effect due to low F s used 34 / 45

35 Quantization of Continuous-Amplitude Signals Quantization Process of converting a discrete-time continuous-amplitude signal into a digital signal x q (n) = Q[x(n)] = sequence of quantized samples Each sample value is expressed as a finite number of digits Error introduced is called quantization error or quantization noise e q (n) = x q (n) x(n) Example Consider discrete-time signal x(n) = { 0.9 n, n 0 0, n < 0 obtained by sampling x a (t) = 0.9 t, t 0 with F s = 1 Hz 35 / 45

36 Quantization of Continuous-Amplitude Signals Example (continued) Following table shows values of first 10 samples of x(n) Description of sample value x(n) requires n significant digits 36 / 45

37 Quantization of Continuous-Amplitude Signals Example (continued) x(n) x q (n) x q (n) e q (n) = x q (n) x(n) n Discrete-time signal (Truncation) (Rounding) (Rounding) Assume using one significant digit. To eliminate excess digits do truncation or do rounding Rounding process is illustrated in next figure 37 / 45

38 Quantization of Continuous-Amplitude Signals Example (continued) Quantization step size (or resolution) = = = / 45

39 Quantization of Continuous-Amplitude Signals Range of quantization error e q (n) in rounding 2 e q(n) 2 = x max x min L 1 x min and x max = minimum and maximum value of x(n) L = number of quantization levels Dynamic range of signal = x max x min Quantization of analog signals always results in a loss of information 39 / 45

40 Quantization of Sinusoidal Signals Sampling and quantization of x a (t) = A cos Ω 0 t 40 / 45

41 Quantization of Sinusoidal Signals If F s satisfies sampling theorem, quantization is the only error in A/D process Thus we can evaluate quantization error by quantizing x a (t) instead of x(n) = x a (nt ) x a (t) is almost linear between quantization levels Quantization error = e q (t) = x a (t) x q (t) e q(t) Δ Δ/2 Δ/2 -Δ/2 -τ 0 τ t -τ 0 τ t τ denotes time that x a (t) stays within quantization levels 41 / 45

42 Quantization of Sinusoidal Signals Mean-square error power P q = 1 2τ τ τ since e q (t) = ( /2τ)t, τ t τ e 2 q(t) dt = 1 τ τ 0 e 2 q(t) dt P q = 1 τ τ 0 ( ) 2 t 2 dt = 2 2τ 12 If quantizer has b bits of accuracy and covers range 2A Average power of x a (t) = 2A 2 b P q = A2 /3 2 2b P x = 1 T p Tp 0 (A cos Ω 0 t) 2 dt = A / 45

43 Quantization of Sinusoidal Signals Quality of output of A/D converter is measured by signal-to-quantization noise ratio (SQNR) Provides ratio of signal power to noise power SQNR = P x P q = b SQNR(dB) = 10 log 10 SQNR = b 43 / 45

44 Coding of Quantized Samples Coding process assigns a unique binary number to each quantization level L levels need at least L different binary numbers b bits 2 b different binary numbers 2 b L b log 2 L 44 / 45

45 References John G. Proakis, Dimitris G. Manolakis, Digital Signal Processing: Principles, Algorithms, and Applications, Prentice Hall, / 45