Exam (elaborations) TEST BANK FOR Signals and Systems Analysis of Sign

Exam (elaborations) TEST BANK FOR Signals and Systems Analysis of Sign If g t e t ( ) = − − 7 2 3 write out and simplify (a) g3 7 9 ( ) = − e (b) g2 7 7 22 3 7 2 ( ) − = = − − ( )− −+ te e t t (c) g t e t 10 4 7 5 11 +       = − − (d) g jt e j t ( ) = − − 7 2 3 (e) g g cos jt jt e e e e t jt jt ( ) + −( ) = + = ( ) − − − 2 7 2 7 2 3 2 2 3 (f) g g cos jt jt e e t jt jt  −      +  − −      = + = ( ) − 3 2 3 2 2 7 2 7 2. If g( ) xx x =− + 2 4 4 write out and simplify (a) g( )zz z =− + 2 4 4 (b) g( ) u v u v u v u v uv u v + = + ( ) − + ( ) += + + − − + 2 2 2 4 4 2 444 (c) g ee e e e e jt jt jt j t jt jt ( ) = ( ) − += − += − ( ) 2 2 2 44 44 2 (d) gg g ( ) ( )t tt tt tt = −+ ( ) = −+ ( ) − −+ ( ) + 2 2 2 2 44 44 4 444 g g( ) ( )tt t t t =− + − + 4 (e) g2 4 8 4 0 ( ) =−+= 3. What would be the numerical value of “g” in each of the following MATLAB instructions? (a) t = 3 ; g = sin(t) ; 0.1411 (b) x = 1:5 ; g = cos(pi*x) ; [-1,1,-1,1,-1] (c) f = -1:0.5:1 ; w = 2*pi*f ; g = 1./(1+j*w) ; M. J. Roberts - 7/12/03 Solutions 2-2 1 . . . . . . . . + + − −                 j j j j 4. Let two functions be defined by x , sin , sin 1 1 20 0 1 20 0 t t t ( ) = ( ) ≥ − ( ) <    π π and x , sin , sin 2 2 0 2 0 t t t t t ( ) = ( ) ≥ − ( ) <    π π . Graph the product of these two functions versus time over the time range, −<< 2 2 t . t -2 2 x(t) -2 2 5. For each function, g( )t , sketch g( ) −t , −g( )t , g( ) t −1 , and g 2( )t . (a) (b) t g(t) 2 4 t g(t) 1 -1 3 -3 t g(-t) -2 4 t g(-t) 1 -1 3 -3 t -g(t) 2 4 t -g(t) 1 -1 3 -3 t g(t-1) 1 3 4 t g(t-1) 1 2 3 -3 t g(2t) 1 4 t g(2t) 1 3 -3 2 1 2 - 6. A function, G( )f , is defined by M. J. Roberts - 7/12/03 Solutions 2-3 G rect f e f j f ( ) =       − 2 2 π . Graph the magnitude and phase of G G ( ) f f −10 10 over the range, + + ( ) −<< 20 20 f . G G rect rect f fe f e f jf jf ( ) − + + ( ) =  −      +  +      − − ( ) − + ( ) 10 10 10 2 10 2 2 10 2 10 π π f -20 20 |G( f )| 1 f -20 20 Phase of G( f ) -π π 7. Sketch the derivatives of these functions. (All sketches at end.) (a) g sinc ( )t t = ( ) ′( ) = ( ) − ( ) ( ) = ( ) − ( ) g cos sin cos sin t tt t t tt t t π ππ π π ππ π π 2 2 2 (b) g u t et t ( ) = − ( ) ( ) − 1 ′( ) = ≥ <       = ( ) − − g , , t u e t t e t t t 0 0 0 t -4 4 x(t) -1 1 t -4 4 dx/dt -1 1 t -1 4 x(t) -1 1 t -1 4 dx/dt -1 1 (a) (b) 8. Sketch the integral from negative infinity to time, t, of these functions which are zero for all time before time, t = 0. M. J. Roberts - 7/12/03 Solutions 2-4 g(t) t 1 12 3 1 2 g(t) t 1 123 ∫ g(t) dt ∫ g(t) dt t 1 123 1 2 t 1 123 9. Find the even and odd parts of these functions. (a) g( )ttt = −+ 2 36 2 ge t tt t t t ( ) = t − ++−( ) − −( ) + = + = + 2 3 62 3 6 2 4 12 2 2 6 2 2 2 2 go t tt t t t ( ) = t − +− −( ) + −( ) − = − = − 2 3 62 3 6 2 6 2 3 2 2 (b) g cos ( )t t = −      20 40  4 π π g cos cos e t t t ( ) = −       + −−      20 40  4 20 40 4 2 π π π π Using cos cos cos sin sin zz z z z z ( ) 12 1 2 1 2 + = () ( ) − () ( ) g cos cos sin sin cos cos sin sin e t t t t t ( ) = ( ) −       − ( ) −             + −( ) −       − −( ) −                           20 40 4 40 4 20 40 4 40 4 2 π π π π π π π π g cos cos sin sin cos cos sin sin e t t t t t ( ) = ( )       + ( )             + ( )       − ( )                           20 40 4 40 4 20 40 4 40 4 2 π π π π π π π π M. J. Roberts - 7/12/03 Solutions 2-5 g cos cos cos e ( )t tt =      20  ( ) = ( ) 4 40 20 2 40 π π π g cos cos o t t t ( ) = −       − −−      20 40  4 20 40 4 2 π π π π Using cos cos cos sin sin zz z z z z ( ) 12 1 2 1 2 + = () ( ) − () ( ) g cos cos sin sin cos cos sin sin o t t t t t ( ) = ( ) −       − ( ) −             − −( ) −       − −( ) −                           20 40 4 40 4 20 40 4 40 4 2 π π π π π π π π g cos cos sin sin cos cos sin sin o t t t t t ( ) = ( )       + ( )             − ( )       − ( )                           20 40 4 40 4 20 40 4 40 4 2 π π π π π π π π g sin sin sin o ( )t tt =      20  ( ) = ( ) 4 40 20 2 40 π π π (c) g t t t t ( ) = − + + 2 36 1 2 ge t t t t t t t ( ) = − + + + + + − 2 36 1 2 36 1 2 2 2 ge t tt t tt t t t ( ) = ( ) − + ( ) − + ++ ( )( ) + ( ) + ( ) − 2 3 61 2 3 61 1 1 2 2 2 ge t t t t t t ( ) = + + ( ) − = + − 4 12 6 2 1 6 5 1 2 2 2 2 2 go t t t t t t t ( ) = − + + − + + − 2 36 1 2 36 1 2 2 2 M. J. Roberts - 7/12/03 Solutions 2-6 go t tt t tt t t t ( ) = ( ) − + ( ) − − ++ ( )( ) + ( ) + ( ) − 2 3 61 2 3 61 1 1 2 2 2 go t tt t t t t t ( ) = −− − ( ) − = − + − 6 4 12 2 1 2 9 1 3 2 2 2 (d) g sinc ( )t t = ( ) g sin sin sin e t t t t t t t ( ) = ( ) + ( ) − − = ( ) π π π π π 2 π go ( )t = 0 (e) g( )tt t t = − ( ) 2 14 ( ) + 2 2 g( )tt t t = − ( )( ) + odd even even { 34 2 14 2 2 Therefore g( )t is odd, g g e o ( )t tt t t = 0 2 14 ( ) = − ( )( ) + 2 2 and (f) g( )tt t t = − ( ) 2 14 ( ) + ge t tt t t t t ( ) = ( ) 2 14 2 14 − ( ) + + −( )( ) + ( ) − 2 ge ( )t t = 7 2 go t tt t t t t ( ) = ( ) 2 14 2 14 − ( ) + − −( )( ) + ( ) − 2 go ( )tt t = − ( ) 2 4 2 10. Sketch the even and odd parts of these functions. M. J. Roberts - 7/12/03 Solutions 2-7 t g(t) 1 1 t g(t) 1 2 1 -1 t g (t) 1 1 t g (t) 1 2 1 -1 t g (t) 1 1 t g (t) 1 2 1 -1 e e o o (a) (b) 11. Sketch the indicated product or quotient, g( )t , of these functions. t 1 -1 1 -1 t -1 1 1 g(t) Multiplication t 1 -1 1 -1 t 1 -1 -1 1 g(t) g(t) g(t) Multiplication (a) (b) t 1 -1 1 -1 t -1 1 1 -1 M. J. Roberts - 7/12/03 Solutions 2-8 t 1 1 g(t) Multiplication (c) t -1 1 t 1 1 g(t) g(t) Multiplication (d) t 1 1 t -1 -1 1 g(t) t -1 1 1 (e) (f) t -1 1 1 -1 t -1 1 1 g(t) Multiplication ... ... t 1 1 -1 t 1 -1 1 g(t) Multiplication g(t) t -1 1 1 -1 ... ... g(t) t 1 1 -1 t 1 1 g(t) Division Division (g) t -1 -1 -1 1 1 1 g(t) g(t) (h) t 1 t π -1 1 1 t g(t) t -1 12. Use the properties of integrals of even and odd functions to evaluate these integrals in the quickest way. M. J. Roberts - 7/12/03 Solutions 2-9 (a) 2 2 22 4 1 1 1 1 1 1 0 1 ( ) + = +== − −− ∫ ∫∫ ∫ t dt dt t dt dt even odd { { (b) 4 10 8 5 4 10 8 5 8 10 8 10 1 20 1 20 1 20 1 20 1 20 1 20 0 1 20 cos sin cos sin cos ππ π π π π [ ] ( )t t dt t dt t dt t dt + ( ) = ( ) + ( ) = ( ) = − −− ∫ ∫ ∫∫ even odd (c) 4 10 0 1 20 1 20 t t dt odd even odd { 1 24 34 1 24 34 cos( ) π = − ∫ (d) t t dt t t dt t t t dt odd odd even { 1 24 34 1 24 34 sin sin cos cos 10 10 10 1 10 1 10 0 1 10 0 1 10 0 1 10 π π π π π π ( ) = ( ) = − ( ) + ( )          −  ∫∫ ∫ t t dt t odd odd even { 1 24 34 1 24 34 sin sin 10 2 1 100 10 10 1 50 1 10 1 10 2 0 1 10 π π π π π ( ) == + ( ) ( )           = − ∫ (e) e dt e dt e dt e e − t tt t − −−− − ∫∫∫ = = =−[ ] = − ( ) ≈ even { 1 1 0 1 0 1 0 1 1 2 2 2 2 1 1 264 . (f) t e dt t odd even odd { { 123 − − ∫ = 1 1 0 13. Find the fundamental period and fundamental frequency of each of these functions. (a) g cos ( )t t = 10 50 ( ) π f T 0 0 25 1 25 = = Hz s , (b) g cos ( )t t = +      10 50  4 π π f T 0 0 25 1 25 = = Hz s , (c) g cos sin ( )t tt = ( ) 50 15 π π + ( ) f T 0 0 25 15 2 2 5 1 2 5 = 0 4      GCD , . ,  = == . Hz s . (d) g cos sin cos ( )t tt t = ( ) + ( ) + −      23 5  3 4 ππ π π M. J. Roberts - 7/12/03 Solutions 2-10 f T 0 0 1 3 2 5 2 1 2 1 1 2 = 2      GCD , , Hz , s  = == 14. Find the fundamental period and fundamental frequency of g( )t . g(t) t ... ... 1 t ... ... 1 g(t) t ... ... 1 + (a) (b) t ... ... 1 g(t) t ... ... 1 + (c) (a) f T 0 0 3 1 3 = = Hz and s (b) f T 0 0 64 2 1 2 = GCD , Hz and s ( ) = = (c) f T 0 0 = GCD , Hz and s ( ) 65 1 1 = = 15. Plot these DT functions. (a) x cos sin n n n [ ] =       −  ( ) −     4  2 12 3 2 2 8 π π , − ≤< 24 24 n n -24 24 x[n] -7 7 (b) x n ne n [ ] = − 3 5 , − ≤< 20 20 n n -20 20 x[n] -6 6 (c) x n n [ ] = n      21  + 2 14 2 3 , −≤ < 5 5 n M. J. Roberts - 7/12/03 Solutions 2-11 n -5 5 x[n] -2000 2000 16. Let x cos 1 5 2 8 n n [ ] =       π and x2 6 8 2 n e n [ ] = − −      . Plot the following combinations of those two signals over the DT range, − ≤< 20 20 n . If a signal has some defined and some undefined values, just plot the d