DEPARTMENT OF ELECTRICAL AND ELECTRONIC ENGINEERING EXAMINATIONS 2008

Transcription

1 [E5] IMPERIAL COLLEGE LONDON DEPARTMENT OF ELECTRICAL AND ELECTRONIC ENGINEERING EXAMINATIONS 008 EEE/ISE PART II MEng BEng and ACGI SIGNALS AND LINEAR SYSTEMS Time allowed: :00 hours There are FOUR quesions on his paper Q is compulsory Answer Q and any wo of quesions -4 Q carries 40% of he marks Quesions o 4 carry equal marks (30% each) Any special insrucions for invigilaors and informaion for candidaes are on page Firs Marker: Second Marker: Peer Y K Cheung M M Draief

2 [E5] Special insrucions for invigilaors: None Informaion for candidaes: None Page of 7

3 [E5] [Quesion is compulsory] a) Wih a single equaion, define he characerisic of a linear sysem j b) Find he even and odd componens of he signal x () = e θ [] [] c) A coninuous-ime signal x () is shown in Figure Skech he signals i) x ()[ u () u ( )] ii) 3 x () δ ( ) [3] [3] Figure d) Consider he RC circui shown in Figure Find he relaionship beween he inpu x () = v() and he oupu y () = i () in he form of: s i) a differenial equaion; [3] ii) a ransfer funcion [3] R + + v s () i() C V c () Figure 3 e) The uni impulse response of an LTI sysem is h () = e e u () Find he sysem s zero-sae response y () if he inpu x () = e u () Noe ha e e () () = () for λ λ λ λ λ λ λ λ e u e u u Page 3 of 7

4 [E5] f) Using he graphical mehod, find y () = x () h () where x () and h () are shown in Figure 3 Figure 3 g) Find he pole and zero locaions for a n sysem wih he ransfer funcion s s+ 5/ H() s = s + 5s+ 4 h) Given ha he Fourier ransform of he signal x () is X ( ω ), ie x () Xω ( ), prove from firs principle ha x e X jω0 ( 0) ( ω) i) Using he z-ransform pairs uk [ ] z k z and γ uk [ ] z z γ inverse z-ransform of zz ( 7) Fz [ ] = z 5z+ 4 j) A TV signal has a bandwidh of 45 MHz This signal is sampled and quanized wih an analogue-o-digial converer i) Deermine he sampling rae if he signal is o be sampled a a rae 0% above he Nyquis rae [] ii) If he samples are quanized ino 04 levels, deermine he bi-rae (ie bis/second) of he binary coded signal [] Page 4 of 7

5 [E5] a) Given he iniial condiions y (0) = 0 and y (0) = 0 0, find he uni impulse response of an LTI sysem specified by he equaion d y dy dx y ( ) = + 9 x ( ) d d d [5] b) An inpu signal f() is expressed in erms of sep componens as shown in Figure The sep componen a ime = τ has a heigh of Δ f which can be expressed as Δf Δ f = Δ τ = f ( τ) Δτ Δτ If g () is he uni sep response of an LTI sysem o he sep inpu u (), show ha he zero-sae response y () of he sysem o he inpu f() can be expressed as y () = f ( τ) g ( τ) dτ = f () g () [5] f() df dτ Δτ τ = nδτ Figure Page 5 of 7

6 [E5] 3 a) Find he Fourier ransform of he signal shown in Figure 3 using wo differen mehods: i) By direc inegraion using he definiion of he Fourier ransform [0] ii) Using only he ime-shifing propery and he Fourier ransform pair rec τ sinc ωτ τ [0] b) Given ha x / e dx= π, show ha he energy f E of a Gaussian pulse f() = e σ σ π is given by E f = σ π You should derive he energy E f from F( ω ) using he Parseval s heorem and he following Fourier ransform pair e / σ σ ω / σ π e [0] Figure 3 Page 6 of 7

7 [E5] 4 A discree-ime LTI sysem is specified by he difference equaion yk [ + ] 05 yk [ ] = f[ k+ ] + 08 f[ k] a) Derive is ransfer funcion in he z-domain [6] b) Find he ampliude and phase response of he sysem π c) Find he sysem response yk [ ] for he inpu f[ k] = cos(05 k ) 3 [0] [THE END] Page 7 of 7

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9 E5 Signals and Linear Sysems Soluions 008 All quesions are unseen Quesion is compulsory Answer o Quesion a) If x y and x y, for a linear sysem, kx + kx ky + k y where k and k are consans [] b) Therefore, Even: cosθ Odd: j sinθ j x () = e θ = cosθ + jsinθ [] c) i) ii) [3] [3] Page of 7

10 d) i) v () = Ri() + v () s vc () = i( τ) dτ C x () = v(), y () = i () s Ry() + y( τ) dτ = x () C c Differeniae boh sides wr : dy dx R + y () = d C d dy dx + y () = d RC R d ii) Take Laplace ransform on boh sides: ( s+ ) Y() s = sx() s RC R [3] Y() s s H() s = = C X() s RCs+ [3] e) y () = h () x () 3 = ( e e ) u( ) e u( ) 3 = e u( ) e u( ) e u( ) e u( ) 3 ( e e ) ( e e ) = u () = e e u() 3 ( ) f) Page of 7

11 g) The complex zeros are given by: z z+ = 0 Therefore he zeros are a: ± -0 3 z= = ± j The poles are given by: p + 5p+ 5 = ( p+ 4)( p+ ) = 0 Therefore he poles are a: p = and p = 4 5 h) By definiion of Fourier ransform, Le τ = 0, jω FT of x ( ) = x ( ) e d 0 0 jω jωτ ( + 0 ) x ( 0) e d= x( τ) e d jω0 jωτ jω0 τ = e x( τ) e dτ = e X( ω) i) Divide F[z] by z, and perform parial fracion: Fz [ ] z 7 z 7 = = = z z 5z+ 4 ( z )( z 4) z z 4 z z Fz [ ] = z z 4 k f[ k] = [ 4 ] u[ k] j) 6 i) Nyquis rae is 45 0 = 9 MHz Therefore he acual sampling rae = 9 MHz =08 MHz ii) 04 levels require 0 bis per sample Therefore bi-rae is: = 08 Mbis/sec [] [] Page 3 of 7

12 Answer o Quesion Express he differenial equaion in erms of D operaors: ( D + 6D+ 9) y() = (D+ 9) x() Q( D) y() = P( D) x() QD D D PD D ( ) = ( ), ( ) = ( + 9) The characerisic equaion is herefore: ( λ + 6λ+ 9) = 0 ( λ+ 3) = 0 y ( ) = ( c + c ) e and y ( ) = [ 3( c + c ) + c ] e Seing = 0, and subsiuing 3 e y0 y 0 (0) = 0 and (0) =, gives 0= c c = 0 = 3c+ c c = y ( ) = e and y ( ) = ( 3+ ) e Now he impulse response can be calculaed: h () = [ PD ( ) y()] u () 0 = [ y ( ) + 9 y ( )] u( ) 0 0 = ( 6e e 9 e ) u( ) = + 3 ( 3 e ) u ( ) [5] a) The sysem response o u () is g (), and he response o he sep u ( τ ) is g ( τ ) (ime-invarian propery) Δf I is given ha Δ f = Δ τ = f ( τ) Δτ The sep componen a = nδ τ herefore has a Δτ heigh of f ( nδτ) Δτ, and can be expressed as f ( n τ ) τ Δ Δ u ( nδτ ) This gives a response Δ y () a he oupu, where Δ y () = f ( nδτ) Δτ g ( nδτ) Therefore, he oal response due o ALL sep componens is: y () = lim f ( nδτ) g ( nδτ) Δτ Δτ 0 n = = f ( τ) g( τ) dτ = f ( τ) g( τ) = f () g() [5] Page 4 of 7

13 Answer o Quesion 3 a) i) From definiion of Fourier ransform, jω F( ω) = f( ) e d 0 τ jω jω e d e d τ 0 = jω = e e jω jω 0 τ τ jω 0 jω jω = + e + e jω jω jω jω = + cos ωτ jω jω 4 sin ωτ = j ω ii) Express f() as sum of wo recangular funcions: + τ / τ / f () = rec rec τ τ Given ha rec ωτ τ sinc τ, apply ime-shifing propery gives Therefore ± τ / ωτ rec τ sinc τ ± j / e ωτ ωτ ωτ F( ω) = τ sinc e τ sinc e ωτ ωτ = j τ sinc sin 4 sin ωτ = j ω + jωτ / jωτ / [0] [0] Page 5 of 7

14 b) and f() = e σ σ π e σ π σ / e σ ω Parseval s Theorem saes: Given we obain: Ef = F( ω) dω π F( ω) = / e σ ω Ef e d π σ ω = ω x x Le σω =, hen σ ω = and dω = dx σ Therefore Ef = e dx= π σ σ π x / [0] Page 6 of 7

15 Answer o Quesion 4 a) Taking z-ransform of boh sides: zyz [ ] 05 Yz [ ] = zfz [ ] + 08 Fz [ ] Therefore Yz [ ] z+ 08 Hz [ ] = = Fz [ ] z 05 b) The frequency response is given by: jω jω e + 08 (cos Ω+ 08) + jsin Ω He [ ] = = jω e 05 (cosω 05) + jsin Ω Therefore, he ampliude response is jω jω jω He [ ] = He [ ] He [ ] jω jω ( e + 08)( e + 08) = jω jω ( e 05)( e 05) cosΩ = 5 cosω The phase response is j sin sin He [ Ω ] an Ω an Ω = cosω+ 08 cosω 05 π c) Since f[ k] = cos(05 k ), Ω= 05 3 Therefore j cos 05 He [ Ω ] = = cos 05 j He [ Ω ] = 86 j sin 05 sin 05 He [ Ω ] an an = cos cos05 05 = = 0653 radian or 3583 Therfore, he sysem response is π yk [ ] = 86 cos(05k 0653) = 86 cos(05k 675) 3 Page 7 of 7

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