( ) ( ) numerically using the DFT. The DTFT is defined. [ ]e. [ ] = x n. [ ]e j 2π Fn and the DFT is defined by X k. [ ]e j 2π kn/n with N = 5.

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3 ( /13) in the Ω form. ind the DTT of 8rect 3 n 2 8rect ( 3( n 2) /13) 40drcl(,5)e j 4π Let = Ω / 2π. Then 8rect 3 n 2 40 drcl( Ω / 2π,5)e j 2Ω ( /13) ind the DTT of 8rect 3( n 2) /13 by X = x n numerically using the DT. The DTT is defined [ ]e j 2π n and the DT is defined by X k n= case the function 8rect 3 n 2 = x n X 4 n=0 [ ]e j 2π n Numerically, 8, 8, 8, 8, 8 = 8 j10π k/5 1 e j 2π k/5 1 e = N 1 [ ] = x n n=0 [ ]e j 2π kn/n ( /13) is non-zero only in the range 0 n < 5. So. It then follows that X k / N DT { } 8 e jπ k sin jπ k/5 e 5 N 1 = x n. In our [ ]e j 2π kn/n with N = 5. n=0 { 40,0,0,0,0}. Analytically X k / 5 = 40e j 4π k/5 drcl( k / 5,5) and 2πk sin 2πk / 5 k X( k / 5) which agrees with the numerical result but the resolution in is poor. 4 = 8 e n=0 j 2π kn/5

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5 ind the DTT of 3sinc( n / 24)cos( 2πn / 8). rom the DTT table of ourier pairs sinc( n / w) wrect ( w) δ 1 cos 2π 0 n 1/ 2 3sinc( n / 24) δ 1 ( 0 ) + δ 1 ( + 0 ) δ 1 ( 1/ 8) + δ 1 ( +1/ 8) 72rect ( 24) δ 1 cos 2πn / 8 1/ 2 Using x[ n]y[ n] X Y, 3sinc( n / 24)cos( 2πn / 8) 72rect ( 24) δ 1 1/ 2 Periodic convolution of two periodic functions with a common period is equivalent to δ 1 ( 1/ 8) + δ 1 ( +1/ 8) aperiodic convolution of either function with one period of the other function. Therefore 3sinc( n / 24)cos( 2πn / 8) 72rect ( 24) δ 1 ( 1/ 2) δ ( 1/ 8) + δ ( +1/ 8) 3sinc( n / 24)cos( 2πn / 8) 36 rect ( 24) δ 1 δ ( 1/ 8) + rect ( 24) δ 1 δ +1/ 8 3sinc( n / 24)cos( 2πn / 8) 36 rect ( 24) δ 1 ( 1/ 8) + rect ( 24) δ 1 ( +1/ 8) 3sinc( n / 24)cos( 2πn / 8) 36rect ( 24) δ 1 ( 1/ 8) + δ 1 ( +1/ 8) Convert to the Ω form. Using the scaling property of convolution, If y( t) = x( t) h( t), then y( at) = a x( at) h( at). 3sinc( n / 24)cos( 2πn / 8) ( 1/ 2π )36rect ( 24Ω / 2π ) δ 1 ( Ω / 2π 1/ 8) + δ 1 ( Ω / 2π +1/ 8) = bδ ab ( x) + δ 2π ( Ω + π / 4) + δ 2π ( Ω + π / 4) Using the scaling property of the periodic impulse, δ a x / b 3sinc( n / 24)cos( 2πn / 8) ( 1/ 2π )36rect ( 12Ω / π ) 2π δ 2π Ω π / 4 3sinc( n / 24)cos( 2πn / 8) 36rect ( 12Ω / π ) δ 2π Ω π / 4

6 ind the DTT of 7( 0.4) n 2 u[ n 2] δ 8 [ n]. rom the DTT table α n 1 u[ n] 1 αe, α < 1 and δ n jω N [ ] ( 1/ N )δ 1/N e and δ jω 8 [ n] ( 1/ 8)δ 1/8 or ( 1/ 8)δ 1/8 ( Ω / 2π ) = ( π / 4)δ π /4 Ω ( 0.4) n u[ n] Time shifting ( 0.4) n 2 u[ n 2] Linearity 7( 0.4) n 2 u[ n 2] Using the property x[ n] y n 7( 0.4) n 2 u[ n 2] δ 8 [ n] 7( 0.4) n 2 u[ n 2] δ 8 [ n] Convert to the form. 7( 0.4) n 2 u[ n 2] δ 8 [ n] 7( 0.4) n 2 u[ n 2] δ 8 [ n] 7( 0.4) n 2 u[ n 2] δ 8 [ n] [ ] e j 2Ω e jω 7e j 2Ω e jω X( e jω )Y( e jω ) ( e π / 4)δ jω π /4 Ω 7e j 2Ω e j 2Ω 7π / e δ jω π /4 Ω e j 4π 7π / e δ j 2π π /4 2π 7π / 4 ( 1/ 2π )δ e j 2π 1/8 e j 4π e j 4π 7 / e δ j 2π 1/8

7 A digital filter has a frequency response H( e jω ) = 1+ e jω + e j 2Ω x[ n] = 15sin( 2πn /12). The DTT of the response is Y( e jω ) = 1+ e jω + e j 2Ω 3 = j15π 1+ e jω + e j 2Ω 3 = H( e jω ) X( e jω ) j15π δ 2π Ω + π / 6 δ 2π ( Ω π / 6) 3. ind its response to δ 2π ( Ω + π / 6) 1+ e jω + e j 2Ω δ 2π ( Ω π / 6) 3 Using the equivalence property of the impulse and the periodicity of the complex sinusoid e jω Y( e jω ) = j5π 1+ e jπ /6 jπ /3 ( + e )δ 2π ( Ω + π / 6) 1+ e jπ /6 jπ /3 ( + e )δ 2π ( Ω π / 6) Using Euler's identity Y( e jω ) ( 1+ cos ( π / 6 ) = j5π + cos( π / 3) ) δ 2π ( Ω + π / 6) ( 1+ cos( π / 6) + cos( π / 3) )δ 2π Ω π / 6 + ( j sin( π / 6) + j sin( π / 3) )δ 2π ( Ω + π / 6) ( j sin( π / 6) + j sin( π / 3) )δ 2π Ω π / 6 = j5π { δ 2π ( Ω + π / 6) δ 2π ( Ω π / 6) + j1.366 δ 2π ( Ω + π / 6) + δ 2π ( Ω π / 6) } [ ] = 11.83sin( πn / 6) 6.83cos( πn / 6) = cos( πn / ) y n

8 The signal x[ n] = sinc( n / 21) δ 100 [ n] excites a system with transfer function H( z) = = H z z e jω 5. ind and graph the response y 1 [ n] of the system z X e jω Using sinc( n / w) wrect ( w) δ 1 and δ N [ n] ( 1/ N )δ 1/N and x[ n] y[ n] XY sinc( n / 21) δ 100 n 21rect ( 21) δ 1 ( 1 /100)δ 1/100 [ ] Using the scaling properties of convolution and the periodic impulse, sinc( n / 21) δ 100 [ n] ( 1/ 2π )21rect ( 21Ω / 2π ) δ 1 ( Ω / 2π ) ( 1/100)δ 1/100 Ω / 2π sinc( n / 21) δ 100 [ n] ( 1/ 2π )21rect ( 21Ω / 2π ) 2πδ 2π ( Ω) ( 2π /100 )δ 2π /100 ( Ω) sinc( n / 21) δ 100 n ( 21π / 50) rect ( 21Ω / 2π ) δ 2π ( Ω) δ π /50 ( Ω) [ ] This result is periodic with period 2π, one period of which consists of set of impulses of strength 21π / 50 at radian frequencies πk / 50 in the range 1/ 2 < 21πk /100π < 1/ 2 or 50 / 21 < k < 50 / 21 or, since k must be an integer, 2 k 2. This can be written [ ] in the form sinc( n / 21) δ 100 n 21π / 50 2 δ ( Ω πk / 50) k= 2 δ 2π ( Ω)

9 [ ] sinc( n / 21) δ 100 n 21π / 50 2 δ ( Ω πk / 50) k= Then Y( e jω ) = 21π / e jω k= 2 Using the equivalence property of the impulse, = 21π /10 = 2.1π = 2.1 [ ] = 2.1 y n δ ( Ω πk / 50) 2 δ ( Ω πk / 50) jπ k/50 k= e δ 2π δ ( Ω + 2π / 50) + δ Ω + π / e j 2π /50 jπ / e + δ ( Ω π / 50) 1 0.8e + δ Ω 2π / 50 jπ /50 j 2π / e 3.921π δ Ω + 2π / 50 + j1.906π δ Ω + 2π / π δ Ω π / 50 + j1.164π δ Ω π / cos 2πn / cos πn / 50 ( Ω) δ ( Ω) δ ( Ω 2π / 50) δ ( Ω 2π / 50) + δ ( Ω π / 50) δ ( Ω π / 50) 1.906sin ( 2πn / 50 ) 1.164sin( πn / 50) δ 2π ( Ω) δ ( Ω) 2π δ 2π ( Ω) + 5πδ Ω δ 2π ( Ω)

10 [ ] = 2.1 y n 3.921cos 2πn / cos πn / sin ( 2πn / 50 ) 1.164sin( πn / 50) Using Acos( x) + Bsin( x) = A 2 + B 2 cos x tan 1 ( B / A) y[ n] = cos( 2πn / ) cos( πn / ) y[ n] = 9.156cos( 2πn / ) cos( πn / )