( ) ( ) ( ) () () Signals And Systems Exam#1. 1. Given x(t) and y(t) below: x(t) y(t) (A) Give the expression of x(t) in terms of step functions.

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1 Signals And Sysems Exam#. Given x() and y() below: x() y() 4 4 (A) Give he expression of x() in erms of sep funcions. (%) x () = q() q( ) + q( 4) (B) Plo x(.5). (%) x() g() = x( ) h() = g(. 5) = x(. 5) p( ) = h( ) = x(. 5) x() 6 (C) Plo y(+). (%) y () g() = y( ) h() = g( +. 5) = y( + ) 4 y().5.5 ( ) ( ) ( ) (D) Compue x. 5 y + δ d. (%) 4 x(. 5) y( + ) δ ( ) d = x(. 5) y( ) δ ( ) d = δ ( ) d = 4 4 (E) Compue () y() x() y() x (%) () () () << x y = x(τ) y(τ) 4 4 τ

2 () () () <<4 x y = x(τ) y(τ) 4 4 τ () () () 4<<6 x y =4 () () (4) 6<<8 x y = 6 x(τ) y(τ) x(τ) y(τ) 4 4 τ τ (5) >8 () y() x = Hence, from () o (5) we have x ( ) y( ) Consider v[n] and w[n] below: v[n] w[n] n n (A) Plo v e [n] and v o [n], which are he even and odd pars of v[n]. (4%) ( v[] n + v[ n] ), ( v[] n v[ n] ) v e = v o = v e [n] / / / 4 5 / n

3 [] [] (B) Compue v n w n (8%) v o [n] / / 4 5 / / n Le h[ n] = v[ n]* w[ n] = v[ k] w[ n k] h[] = v[ k] w[ k] = k = k = h[] = v[ k] w[ k] = k = h[] = v[ k] w[ k] = k = h[] = v[ k] w[ k] = k = h[4] = v[ k] w[4 k] = k = h[5] = v[ k] w[5 k] = k = h[6] = v[ k] w[6 k] = k = Because here is no overlapping beween v[k] and w[n-k] when n< or n>6 So hn [ ] =, when n< or n> 6. Consider he following periodic signal: z = sin. 8 + sin. 7 + cos. 8 () ( ) ( ) ( ) (A) Find he fundamenal period. (%) π π n.8 π π π Le n = n = = = = = fundamenal period.8.7 n

4 (B) Express z() using complex exponenials. (5%) e e e e e + e j j j.8 j.8 j.7 j.7 j.8 j.8 j () = e + + z = e + ( j) e + (.5.5 j) e + ( j) e + ( j) e j j.8 j.8 j.7 j.7 = ( e ) e + ( e ) e + ( e ) e + ( e ) e + ( e ) e j π j π j π π j j j 4 j.8 4 j.8 j. 7 j.7 (C) Wha are he Fourier series coefficiens? (%) j π j π j π j 4 4 Fourier series coefficien: e, e, e, e, e j 4. Consider g()=sin(). (A) Le he sampling period be T=. sec. Is he sampled sequence sin(nt), n=,,,, periodic? Why? (%) Le sin( nt ) = sin(( N + n) T), N Z NT = kπ N.=.6N = kπ, k Z kπ N = Z.6 i is no periodic (B) Wha are he frequencies of he sampled sequences sin(nt), n=,,,, for T=.4, and.8 sec? (4%) π ω = when T=.4s ω Ny π π = = T he freqency of he sample sequence = when T=.8s ω Ny π π = =.97 T.8.97 he freqency of he sample sequence = rad s rad s

5 (C) Under wha condiion will he frequency of sin(nt) equal rad/sec? (4%) π π if ω ( ωny, ωny ] = (, ], hen he freqency of he sample sequence = ω T T π π π π (, ] T.47 T T T 5. Le y () and y () be he oupu responses of u () and u (), respecively. They are depiced as below: u () y () u () y () Find he oupu responses corresponding o u () and u 4 () given as u () u 4 () (8%, 8%) (a) u () = u () + u ( ) + u () y () = y () + y ( ) + y () ( graphic in Fig 5.) ( b) u () = u ( ) + u ( ) u () 4 y () = y ( ) + y ( ) y () 4 ( graphic in Fig 5.)

6 Fig 5. Fig Compue he impulse responses of he following wo sysems: y n = u n + u n u n + u n 5 (%) (A) [] [] [ ] [ ] [ ] n= : h[] = δ[] + δ[ ] δ[ ] + δ[ 5] = n= : h[] = δ[] + δ[ ] δ[ ] + δ[ 4] = n= : h[] = δ[] + δ[] δ[ ] + δ[ ] = n= : h[] = δ[] + δ[] δ[] + δ[ ] = n= 4 : h[4] = δ[4] + δ[] δ[] + δ[] = n= 5 : h[5] = δ[5] + δ[] δ[] + δ[] = for n< or n> 5, h[ n] = i is an FIR filer (B) [ n] y[ n ] = u[ n ] u[ n ] y (5%) n= : h[] = h[ ] + δ[ ] δ[ ] = n= : h[] = h[] + δ[] δ[ ] = n= : h[] = h[] + δ[] δ[] = = n= : h[] = h[] = n= 4: h[4] = h[] = 4 n= 5: h[5] = h[4] = 8 for n <, h[ n] = for n > 5, h[ n] = i is an IIR filer n

7 7. (A) Find a differenial equaion o describe he following circui. (8%) (B) Deermine he seady-sae response y() as for u()=sin(). (4%) u + H +.5Ω F y ( A) dil () u () = Vc () + L d Vc() dvc() il () = + C.5 d d Vc() dvc() u () = Vc () + ( + ) d.5 d dvc() d Vc() = Vc ( ) + + d d V () = y() c u () = y () + y'() + y''() ( B) i is in seady sae y() = y () + y () = y () ( y () = ) u () = sin() h p p h Le y ( ) = Acos( ) + Bsin( ) p u () = y () + y'() + y''() u() = y () + y '() + y '() p p p sin( ) = ( Acos( ) + Bsin( )) + ( Asin( ) + Bcos( )) + ( Acos( ) Bsin( ) ) sin() = ( A+ B A)cos( ) + ( B A B)sin( ) B=, A= B=, A= y () = yp () = cos()

8 8. (A) Find a differenial equaion o describe he following circui. (6%) (B) Deermine is sep response wih zero iniial condiion. (4%) F u Ω Ω y ( A) u () iin() = dy() y() iin() + C + = d u () dy () + + y () = d u () = y'() y () ( B) Le u() = q(), y() = y () + y () dy() for, + y( ) = d c find y (): le y () = Ae h char equ : c + = c = y () = Ae h h find y p(): y p() = y () = yh() + yp() = Ae + for zero iniial condiion : y( ) = + y( ) = A = A= y () = yh() + yp() = e for seady sae: y( ) = h p