Utilizamos seu perfil e dados de atividades no LinkedIn para personalizar e exibir anúncios mais relevantes. Altere suas preferências de anúncios quando desejar.
Próximos SlideShares
Carregando em…5
×

# Digital Signal Processing[ECEG-3171]-Ch1_L05

266 visualizações

This Digital Signal Processing Lecture material is the property of the author (Rediet M.) . It is not for publication,nor is it to be sold or reproduced.
#Africa#Ethiopia

• Full Name
Comment goes here.

Are you sure you want to Yes No
• Hello! High Quality And Affordable Essays For You. Starting at \$4.99 per page - Check our website! https://vk.cc/82gJD2

Tem certeza que deseja  Sim  Não
Insira sua mensagem aqui

### Digital Signal Processing[ECEG-3171]-Ch1_L05

1. 1. Chapter Two Sampling and Reconstruction Lecture #5 Rediet Million AAiT, School Of Electrical and Computer Engineering rediet.million@aait.edu.et March,2018 (Rediet Million) DSP-Lecture #5 March,2018 1 / 22
2. 2. Introduction Under reasonable constraints,a continuous -time(analog) signals can be quite accurately represented by samples taken at discrete point in time. Basic digital processing of continuous-time signals perform in three stages: (Rediet Million) DSP-Lecture #5 March,2018 2 / 22
3. 3. 2.1 Periodic sampling Typically,discrete-time signals,x(n), are formed by periodically sampling of a continuous-time signals, xc(t), according to the relation x(n) = xc(nTs) ,−∞ < n < ∞ where Ts is the sampling period(interval) and fs = 1 Ts or Ωs = 2π Ts is the sampling frequency in sample/sec (Hz) or in rad/sec. It is convenient to represent an ideal sampling process mathematically in two steps: First, the CT signal, xc(t) , is multiplied by a periodic sequence of inﬁnite impulse train, s(t), to form the sampled signal xs(t). xs(t) = xc(t)s(t) (Rediet Million) DSP-Lecture #5 March,2018 3 / 22
4. 4. Periodic sampling Mathematically, s(t) = ∞ n=−∞ δ(t − nTs) xs(t) = xc(t)s(t) = xc(t) ∞ n=−∞ δ(t − nTs) xs(t) = ∞ n=−∞ xc(nTs)δ(t − nTs) Then the sampled signal, xs(t), is converted into a discrete-time signal by mapping the impulse that are spaced in time by Ts into a sequence x(n). x(n) = xs(nTs) = xc(nTs) (Rediet Million) DSP-Lecture #5 March,2018 4 / 22
5. 5. Periodic sampling (Rediet Million) DSP-Lecture #5 March,2018 5 / 22
6. 6. Periodic sampling Example (Rediet Million) DSP-Lecture #5 March,2018 6 / 22
7. 7. Periodic sampling Frequency domain representation of sampling The eﬀect of continuous to discrete time conversion maybe analyzed by considering the Fourier transform of xs(t). Since xs(t) is the product of xc(t) & s(t) ,the Fourier transform of xs(t) is the convolution of the Fourier transforms of Xc(jΩ) & S(jΩ). xs(t) = xc(t)s(t) ⇒ Xs(jΩ) = 1 2π Xc(jΩ) ∗ S(jΩ) s(t) is a periodic function of time and it is possible to compute its Fourier series representation s(t) = ∞ n=−∞ δ(t − nTs) S(jΩ) = 2π ∞ k=−∞ Ckδ(Ω − kΩs) Where Ckis Fourier series coeﬃcient and Ωs = 2π Ts (Rediet Million) DSP-Lecture #5 March,2018 7 / 22
8. 8. Periodic sampling Frequency domain representation of sampling Ck is determined from exponential Fourier series Ck = 1 Ts Ts /2 −Ts /2 δ(t)e−jkΩs tdt = 1 Ts Thus, S(jΩ) = 2π Ts ∞ k=−∞ δ(Ω − kΩs) ⇒ The Fourier transform of a periodic inﬁnite impulse train,s(t), is a periodic inﬁnite impulse train ,S(jΩ) ,in frequency domain. (Rediet Million) DSP-Lecture #5 March,2018 8 / 22
9. 9. Periodic sampling Frequency domain representation of sampling Since Xs(jΩ) = 1 2π Xc(jΩ) ∗ S(jΩ) , it follows that Xs(jΩ) = 1 2π Xc(jΩ) ∗ 2π Ts ∞ k=−∞ δ(Ω − kΩs) Xs(jΩ) = 1 Ts ∞ k=−∞ Xc(jΩ − jkΩs) Finally the discrete Fourier transform of x(n) can be obtained as follows: - We know that x(ejw ) = ∞ n=−∞ x(n)e−jnw = ∞ n=−∞ Xc(nTs)e−jnw and Xs(jΩ) = 1 Ts ∞ k=−∞ Xc(jΩ − jkΩs) = ∞ n=−∞ Xc(nTs)e−jnΩTs -Combining the two equations: x(ejw ) = Xs(jΩ)|Ω=w/Ts = 1 Ts ∞ k=−∞ Xc( jw Ts − 2πk Ts ) (Rediet Million) DSP-Lecture #5 March,2018 9 / 22
10. 10. Periodic sampling Frequency domain representation of sampling Example: Suppose that Xc(jΩ) is strictly band-limited so that Xc(jΩ) = 0 for |(Ω)| > ΩN as shown in ﬁgure bellow. If xc(t) is sampled with a sampling frequency Ωs > 2ΩN , the Fourier transform of xs(t) is formed by periodically replicating Xc(jΩ) as illustrated bellow. (Rediet Million) DSP-Lecture #5 March,2018 10 / 22
11. 11. Periodic sampling Frequency domain representation of sampling However ,if Ωs < 2ΩN ,the shifted spectra Xc(jΩ − jkΩs) overlap, and when these spectra are summed to form Xs(jΩ) ,the result is as shown bellow. This overlapping of spectral components is called aliasing.When aliasing occurs,the frequency of xc(t) is corrupted, and Xc(jΩ) cannot recovered from Xs(jΩ). Aliasing can be avoided by using Nyquist sampling theorem. (Rediet Million) DSP-Lecture #5 March,2018 11 / 22
12. 12. Periodic sampling Nyquist sampling theorem If xc(t) is strictly band-limited, Xc(jΩ) = 0 for |Ω| > ΩN then xc(t) may be uniquely recovered from its sample xc(nTs) if Ωs ≥ 2ΩN = 2π Ts The frequency ΩN = Ωs/2 is called the Nyquist frequency, and the minimum sampling frequency, Ωs = 2ΩN, is called the Nyquist rate. Thus ,in order to avoid aliasing, the operating frequency of the signal should be in Nyquist interval [− Ωs 2 , Ωs 2 ]. This is done by analog low-pass pre-ﬁlter known as an anti-aliasing pre-ﬁlter. The cutoﬀ frequency of this pre-ﬁlter,Ωc, is taken to be the Nyquist frequency, Ωs 2 . (Rediet Million) DSP-Lecture #5 March,2018 12 / 22
13. 13. Periodic sampling Aliased frequency components The unique frequency components that appear within the Nyquist interval maybe obtained by reducing the original frequency f modulo fs, i.e fa = f mod (fs) -This modulo-fs operation is obtained by adding or subtracting from f enough multiples of fs until it lies within the Nyquist interval [−fs/2, fs/2]. Thus, the sinusoid that is extracted by the analog reconstructor will be xa(t) = ej2πfat fa = f if |f | ≤ fs 2 (Rediet Million) DSP-Lecture #5 March,2018 13 / 22
14. 14. Periodic sampling Aliased frequency components Example: Let x(t) = 4 + 3cos(πt) + 2cos(2πt) + cos(3πt). Determine the minimum sampling rate that will not cause any aliasing eﬀects, If x(t) is sampled at half the Nyquist rate determine xa(t) that would be aliased with x(t). -The frequencies in x(t) are f1 = 0Hz f2 = 1 2 Hz f3 = 1Hz f4 = 1.5Hz Thus,fmax = f4 = 1.5Hz and the minimum sampling rate is fs = 2fmax = 3Hz - If x(t) is sampled at half of this rate i.e fs = fs 2 = 1.5Hz ,then aliasing will occur. The Nyquist interval of this case is [− fs 2 , fs 2 ] i.e [−0.75, 0.75]. (Rediet Million) DSP-Lecture #5 March,2018 14 / 22
15. 15. Aliased frequency components -The aliased frequencies will be only due to f3 & f4 since they are outside the Nyquist interval. - The aliased components will have frequencies f3a = f3mod(fs ) = 1mod(1.5) = 1 − 1.5 = −0.5Hz f4a = f4mod(fs ) = 1.5mod(1.5) = 1.5 − 1.5 = 0Hz - Thus,x(t) will be aliased with xa(t) = 4cos(2πf1t) + 3cos(2πf2t) + 2cos(2πf3at) + cos(2πf4at) = 4 + 3cos(πt) + 2cos(−πt) + cos(0) = 5 + 5cos(πt) -The signal x(t)& xa(t) are shown below.They agree only at their sampled values i.e x(nT) = xa(nT) (Rediet Million) DSP-Lecture #5 March,2018 15 / 22
16. 16. Sampling and reconstruction The above example shows that the sample values x(nT) don’t uniquely determine the continuous-time signal they can form. Hence, an ideal reconstructor is supposed to extract from a sampled signal,all the frequency components that lie within the Nyquist interval, [− Ωs 2 , Ωs 2 ], & removes all frequencies outside the interval. This shows that an ideal reconstructor,Hr (jΩ), acts as an ideal low-pass ﬁlter with cutoﬀ frequency equal to the Nyquist frequency Ωs 2 . (Rediet Million) DSP-Lecture #5 March,2018 16 / 22
17. 17. Sampling and reconstruction Example: Sampling and reconstruction of sinusoidal signal Suppose we sample the CT signal xc(t) = cos(4000πt) with sampling period Ts = 1 6000 sec. -We can obtain x(n) = xc(nTs) = cos(4000πTsn) = cos(ω0n) ,where ω0 = 4000πTs = 2π 3 - The highest frequency of xc(t) is Ω0 = 4000π and the sampling frequency is Ωs = 2π Ts = 12000π Ωs = 12000π ≥ 2Ω0 = 8000π ⇒ there is no aliasing. - The Fourier transform of xc(t) is Xc(jΩ) = πδ(Ω − 4000π) + πδ(Ω + 4000π) (Rediet Million) DSP-Lecture #5 March,2018 17 / 22
18. 18. Sampling and reconstruction and the sampled signal for sampling frequency of Ωs = 12000π is Xs(jΩ) = 1 Ts ∞ k=−∞ Xc(jΩ − jkΩs) = π Ts ∞ k=−∞ δ(Ω − 4000π − kΩs) + δ(Ω + 4000π + kΩs) (Rediet Million) DSP-Lecture #5 March,2018 18 / 22
19. 19. Sampling and reconstruction the above ﬁgure(a) shows Xc(jΩ) is a pair of impulse at Ω = ±4000π and the shifted copies of is FT centered on ±Ωs, ±2Ωs , etc Plotting x(ejω) = xs( jω Ts ) as a function of the normalized frequency ω = ΩTs gives in ﬁgure(b). The above ﬁgure also shows the frequency response of the ideal reconstructor ﬁlter Hr (jΩ) for Ωs = 12000π ⇒ the signal that would be reconstructed would have frequency Ω0 = 4000π which is the frequency of the original signal. (Rediet Million) DSP-Lecture #5 March,2018 19 / 22
20. 20. Sampling and reconstruction Example: Aliasing in the reconstruction of under-sampled sinusoidal signal Suppose the signal is again xc(t) = cos(4000πt).However now the sampling period is increased to Ts = 1 1500 sec - This sampling period fails to satisfy the Nyquist criterion ,since Ωs = 3000π 2Ω0 = 8000π ⇒ there is will be aliasing. - the sampled signal for sampling frequency of Ωs = 3000π is Xs(jΩ) = π Ts ∞ k=−∞ δ(Ω − 4000π − kΩs) + δ(Ω + 4000π + kΩs) (Rediet Million) DSP-Lecture #5 March,2018 20 / 22
21. 21. Sampling and reconstruction This time the impulse is located at Ω = −1000π is from δ(Ω + 4000π − kΩs) and the impulse at Ω = 1000π is from δ(Ω + 4000π + kΩs). It is clear that the signal would be reconstructed using sampling rate Ts = 1 1500 sec would have frequency Ω0 = 1000π and not 4000π. (Rediet Million) DSP-Lecture #5 March,2018 21 / 22
22. 22. Periodic sampling (#1 ) Class exercises & Assignment 1) The sequence x(n) = cos( πn 4 ) was obtained by sampling a CT signal xc(t) = cos(Ω0t) at a sampling rate of 1000 sample/sec. Determine the two possible values of Ω0 that could have resulted in sequence x(n). 2 Consider the discrete-time sequence x(n) = sin( πn 8 ) .Find the two diﬀerent continuous-time signals that would produce this sequence when sampled at a frequency of fs = 20Hz. 3) For the following continuous signals xc(t) and the corresponding sampling frequency fs a.xc(t) = 10sin(2πt) + 10sin(8πt) + 5sin(12πt) ,fs = 5Hz b.xc(t) = sin(6πt)[1 + 2cos(4πt)] ,fs = 4Hz i.Determine xa(t) aliased with xc(t) and show that the two signals have the same sample values i.e xc(nTs) = xa(nTs). ii. Find the condition for Ts to correctly sample xc(t) and avoid aliasing. (Rediet Million) DSP-Lecture #5 March,2018 22 / 22