Example: Telephone line is a bandpass lter which passes only Hz thus in the

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1 CHAPTER 3 FILTERING AND SIGNAL DISTORTION page 3.1 We can think of a lter in both frequency and time domain Example: Telephone line is a bandpass lter which passes only Hz thus in the frequency domain Thus a telephone line is not good for Hi-Fi music, since the low and high frequencies are cut out. lter H(f) 300-3,400 Hz (telephone line frequency response) hi- music input signal X(f), 20-20,000 Hz (CD frequency response) output Y (f) =X(f)H(f) lo- ltered music Y (f), missing Hz and 3,400-20,000 Hz 15

2 page 3.2 Read text p Recall Convolution y(t) = h()x(t, )d = h(t) x(t) text notation? same as, convolution symbol Filter impulse response h(t) Filter frequency reponse H(f) Y (f) =H(f)X(f) A lter is also called a system Causal lter h(t) = 0 t<0 Stable lter Section 3.2 p. 91 jh(t)jdt < 1 Frequency response H(f) Also called transfer function if y(t) =x(t)h(t) Then Y (f) =X(f)H(f) 16

3 page 3.3 Text Page 93 Amplitude and Phase Response H(f) = jh(f)j exp[j(f)] where (f) = arg H(f) jh(f)j = Amplitude Response (f) = Phase Response H(f) = Frequency Response Example 3 p H(f) =,jsgn f = 8 >< >:,j f>0 0 f=0 j f<0 j=e j H(f) produces phase shift of,90 o for + Frequencies +90 o for, Frequencies Amplitude is constant for all frequencies H(f) called Hilbert Transformer We will use Hilbert transform for single sideband signals (text chapter 7) 17

4 page 3.4 From tables if H(f) =,jsgnf then h(t) = 1 t ^x(t) = x(t) h(t) =? = x(t) 1 t = x()h(t, )dt = 1 x() t, d ^x(t) is Hilbert transform of x(t) System Bandwidth See notes p

5 page Linear Distortion p Distortionless transmission Output signal is exact replica of input signal except for Change in amplitude K Constant delay t0 y(t) =Kx(t, t0) From tables property 4 g(t,t o )$G(f) exp(,j2ft o ) Y (f) =KX(f) exp(,j2ft o )! H(f) = Y(f) X(f) =Kexp(,j2ft o) using (t, t o ) $ exp(,j2ft o ), we obtain h(t) =K(t, t o )!jh(f)j=k Constant Amplitude (f) =,2ft o Linear Phase 19

6 page 3.6 For Distortionless Transmission: Constant delay in time domain same as linear phase in frequency domain For practical systems there is always a certain amount of linear distortion - Amplitude Distortion - Phase Distortion (Delay Distortion) Note distinction between - constant phase shift - constant delay They are not the same - see text p. 100 top. 20

7 page Ideal Low Pass Filter Transmits all frequencies in passband jf j < B Completely rejects all frequencies in stopband jf j > B Recall distortionless transmission H(f) = ( Kexp(,j2fto ) jfj B 0 jfj>b Impulse Response h(t) = F fh(f)g = = = h(t) = Z B,B Z B,B H(f) exp(j2ft)df exp(,j2ft o ) exp(j2ft)df exp(j2f(t, t o ))df 1 j2(t, t o ) [exp(j2b(t, t o), exp(,j2b(t, t o )] 21

8 page 3.8 h(t) = 1 (t, t o ) sin 2B(t, t o) =2Bsinc [2B(t, t o )] where sinc x = sin x x Thus impulse repsonse of ideal lowpass lter is non-causal because for any nite t o there is a response for t<0. Exercise 6 p. 106 Problems p. 123: 1, 4, 11 22

9 page Bandpass Transmission Deferred until we consider single sideband in section 7.4 p. 284 Omit CHAPTER 4 SPECTRAL DENSITY AND CORRELATION Energy and power signals - Text 1.2 p. 3-4 Energy spectral density -Text 4.1, p Correlation of energy signals 4.2 Power spectral density 4.3 Correlation of power signals 23

10 page 3.10 Energy and Power Signals p. 3-4 Instantaneous Power P = v2 (t) R = i2 (t)r If = R = 1 ohm P = g 2 (t) Total Energy Z T Z E = lim g 2 1 (t)dt = g 2 (t)dt T!1,T Average Power P = lim T!1 1 2T Z T,T g 2 (t)dt Energy Signal 0 <E<1 Power Signal 0 <P <1 24

11 page 3.11 Spectral Density -Text 128 Energy Spectral Density g(f) If FT of energy signal g(t)! G(f), then total energy E = jg(t)j 2 dt = g(t)g (t)dt = G()G ()d Proof: using results of text page 49 exercise 9, solution in notes p E = jg(f)j 2 df = jg(t)j 2 dt Rayleigh Energy Theorm Dene energy spectral density g(f) such that E = g(f)df! {z } g(f) =jg(f)j 2 " Density in Joules/Hertz 25

12 page 3.12 Example 1 - p. 129 Evaluate energy in g(t) = Asinc (2wt) (textp:25) E = jg(t)j 2 dt = A 2 sinc 2 (2wt)dt =? This is a dicult integral, but from Table, note G(f) = A 2w rect f 2w and use Rayleigh Energy Theorm where ( rect 1 jxj < 1=2 gate x = 0 jxj > 1=2 E = A 2w 2 rect 2 f 2w df = jg(f)j 2 df = A2 4w 2 Z w = A2 2w,w df 26

13 page 3.13 Skip ahead to p Power Spectral Density S g (f) Recall average power of g(t) 1 P = lim T!1 2T Z T,T g 2 (t)dt If P 6= 0, then g(t) has innite energy and thus we must be careful about nding G(f). Text p. 142 considers a truncated version of g(t) g T (t) = ( g(t) jtjt 0 jtj>t and shows that the power spectral density (PSD) 1 S g (f) = lim T!1 2T jg T (f)j 2 units are watts/hz For periodic signals g p (t) =g p (t,nt o ), we nd the PSD S g p(f) contains -functions with weights equal to the Fourier coecients. 27

14 page 3.14 Text. p. 144 Properties of PSD Total Power P = S g (f)df Units: watts/hz times Hz = watts y(t) =h(t)x(t) Y(f) = H(f)X(f) S y (f) = jh(f)j 2 S x (f) Interpretation of PSD Assume H(f) narrowband centered at f c. Then S y (f) = ( Sx (f c ) in passband 0 in stopband Average Power P y =2S x (f c )4f To measure PSD at a particular f c : Choose 4f, measure P y, calculate S x (f c )= Py 24f watts/hz 28

15 page Spectrum analyzer can measure PSD directly by sweeping across a range of frequencies and displaying the result - Usually 4f is selectable on front panel - For small 4f, sweep time is longer and resolution is better Text p. 152 PSD of periodic signal g p (t) S g p(f) = 1X n= jc n j 2 f, nto where c n = Complex Fourier Coecients The spectrum analyzer will spread out -functions to lter shape (notes page 2.9). The spectrum analyser will display the -functions as peaks of nite width (depending on the bandwidth setting) and height proportional to jc n j 2 or 20logjc n j, depending on the scale setting (linear or log). Total power for periodic signal is same as power over one period Z T o=2 P =,To=2 g2 p(t)dt 29

16 page 3.16 Text p Exercise 9 a) Use g1(t)g2(t) $ to show that G1()G2(f, ) d g1(t)g2(t)dt = Solution: From Line 1 We know that G1(f)G2(,f)df g1(t)g2(t) exp(,j2ft)dt = G1()G2(f, )d Now relabel! f f! g 1(t)g2(t) exp(,j2t)dt = Now set =0 To set the desired result G 1(f)G2(, f)df b) If replace G2(,f) with G 2(f) then, using FT property 10, this is the same as replacing g2(t) with g2(t) Thus g1(t)g2(t)dt = G1(f)G 2(f)df 30

17 page 3.17 Text p.40 - Exercise 6 Show that Recall exp(,t 2 )dt =1 exp(,t 2 ) $ exp(,f 2 ) therefore Set f = 0 on both sides exp(,t 2 ) exp(,j2ft)dt = exp(,f 2 ) exp(,t 2 )dt =1 from table g(t)dt = G(0) g(t) = exp(,t 2 ) G(f =0) = exp(,f 2 ) f=0 = 1 31

18 page 3.18 Text P Exercise 4 G(f) = 8 >< >: exp(,f) f > 0 1=2 f=0 0 f<0 Find g(t) using duality, and exp(t)u(,t) $ 1 1, j2f Solution: recall u(f) = 8 >< >: 1 f>0 1=2 f=0 0 f<0 Write G(f) = exp(,f)u(f) Duality: If h(t) $ H(f) then H(t) $ h(,f) let h(t) = exp(t)u(,t) $ H(f) = 1 1,j2f Then h(,f) $ H(t) from Duality 32

19 Write h(,f) =exp(,f)u(f) o = G(f) $ H(t) = 1 1,j2t o = g(t) Thus we have found g(t). 33

20 page X.1 Communications Signals Convey a message m(t) using a carrier wave at some radio frequency f c with amplitude A c. f c may be in range 10 KHz to 300 GHz for radio, or much higher (infrared, lightwave) for optical ber. The signal is written in one of two general formats (polar or rectangular) s(t) = r(t) {z} cos[2f c t + (t)] {z } Time Varying Amplitude Time Varying Phase s(t) = x(t) cos 2f c t, y(t) sin 2f c t Use cos( + ) =cos cos, sin sin with =2f c t, = (t). x(t) = r(t) cos (t) r(t) = q x 2 (t)+y 2 (t) y(t) = r(t) sin (t) (t) = arctan y(t) x(t) If we are given the modulation type, e.g. amplitude modulation (AM), then we can nd x(t) y(t) r(t) (t) in terms of the message m(t) and the carrier frequency f c. Example: AM : s(t) =A c (1 + k a m(t)] cos 2f c t x(t) =A c [1 + k a m(t)] y(t) =0 r(t)=a c [1 + k a m(t)] (t) =0 34

21 page X.2 PREVIEW: Other modulation types, to be explained later DSB-SC: SSB-SC: PM: Narrowband Noise s(t) =m(t) cos 2f c t x(t) =m(t) y(t)=0 r(t)=m(t) (t)=0 s(t) =m(t) cos 2f c t, ^m(t) sin 2f c t ^m(t) = Hilbert Transform of m(t) x(t) =m(t) y(t)= ^m(t) r(t)=[m 2 (t)+ ^m 2 (t)] 1=2 (t) = arctan [ ^m(t)=m(t)] s(t) =A c cos[2f c t +2k p m(t)] r(t) =A c (t)=2k p m(t) x(t) =A c cos 2k p m(t) y(t)=a c sin 2k p m(t) s(t) =n I (t) cos 2f c t ),n Q (t) sin 2f c t n I (t) Bandlimited White Noise n Q (t) x(t) =n I (t) y(t)=n Q (t) r(t)=::: (t) =::: 35

University of Toronto Electrical & Computer Engineering ECE 316, Winter 2015 Thursday, February 12, Test #1

Name: Student No.: University of Toronto Electrical & Computer Engineering ECE 36, Winter 205 Thursday, February 2, 205 Test # Professor Dimitrios Hatzinakos Professor Deepa Kundur Duration: 50 minutes