Analysis of Discrete-Time Signals: Sampling & Low-Pass Filtering, Assignments of Signals and Systems

Download Analysis of Discrete-Time Signals: Sampling & Low-Pass Filtering and more Assignments Signals and Systems in PDF only on Docsity! EEL3135: Homework #2 Solutions Problem 1: Consider the continuous-time signal: . (1) Assume that you sample such that, (2) where Hz, and then pass the sampled sequence through an ideal low-pass filter with cut-off frequencies at Hz to produce the signal . (a) Give an analytic expression for . Be sure to explain all the steps required to get your answer. Figure 1 illustrates the steps in deriving . The spectrum for is given by: (S-1) x1 t( ) 12πt( )cos= x1 t( ) x1 n[ ] x1 n fs⁄( )= fs 5= x1 n[ ] fs 2⁄± 2.5±= xr t( ) xr t( ) -10 -5 0 5 10 0 0.2 0.4 0.6 0.8 1 0 0.2 0.4 0.6 0.8 1 -1 -0.5 0 0.5 1 -10 -5 0 5 10 0 0.2 0.4 0.6 0.8 1 -10 -5 0 5 10 0 0.5 1 1.5 2 0 0.2 0.4 0.6 0.8 1 -1 -0.5 0 0.5 1 -10 -5 0 5 10 0 0.2 0.4 0.6 0.8 1 x1 t( ) 12πt( )cos= X1 f( ) f Xs f( ) Xr f( ) t t ff f low-pass filterfs 5Hz= xr t( ) 2πt( )cos= Figure 1 xr t( ) x1 t( ) X6 f( ) 1 2 --δ f 6+( ) 1 2 --δ f 6–( )+=- 1 - The spectrum of the discrete-time sequence in the continuous-time domain is given by, (S-2) The low-pass filter cuts off all frequencies outside of Hz leaving the following frequency components: (S-3) Therefore, the reconstructed waveform is given by, . (S-4) (b) Plot and for . (c) Give analytic expressions for at least two aliases of — that is, continuous-time signals not equal to that will yield the same discrete-time sequences as for Hz. One alias is since it produces an equivalent discrete-time sequence as for Hz. Other aliases (for the sam reason) include , , . The figure below plots (dashed orange line), and two of the above aliases ( , — pink lines); as can be observed, all three functions yield the same discrete time sequence for Hz. (d) What is the Nyquist sampling frequency corresponding to signal ? The Nyquist sampling frequency is given by, Hz. (S-5) Xs f( ) x n[ ] Xs f( ) X1 f kfs–( ) k ∞–= ∞ ∑= fs 2⁄± 2.5±= Xr f( ) 1 2 --δ f 1+( ) 1 2 --δ f 1–( )+= xr t( ) xr t( ) 2πt( )cos= xr n fs⁄( ) x1 n fs⁄( ) n 0 1 … 9 10, , , ,{ }∈ 0 2 4 6 8 10 -1.5 -1 -0.5 0 0.5 1 1.5 xr n fs⁄( ) x1 n fs⁄( )= n x1 t( ) x1 t( ) x1 n[ ] fs 5= xr t( ) 2πt( )cos= x1 t( ) fs 5= 8πt( )cos 18πt( )cos 32πt( )cos x1 t( ) 2πt( )cos 8πt( )cos fs 5= 0 0.5 1 1.5 2 -1 -0.5 0 0.5 1 t x1 t( ) fNyquist 2 12π 2π --------    12= =- 2 - See Figure 4 below for an illustration, where Hz. -4 -2 0 2 4 0 0.5 1 1.5 2 -4 -2 0 2 4 0 0.5 1 1.5 2 -4 -2 0 2 4 0 0.5 1 1.5 2 -4 -2 0 2 4 0 0.5 1 1.5 2X3 f( ) Xs f( ) Xr f( ) f low-pass filter fs 3Hz= Figure 3 f f f fmax 1= -4 -2 0 2 4 0 0.5 1 1.5 2 -4 -2 0 2 4 0 0.5 1 1.5 2 -4 -2 0 2 4 0 0.5 1 1.5 2 -4 -2 0 2 4 0 0.5 1 1.5 2X3 f( ) Xs f( ) Xr f( ) f low-pass filter fs 2 3⁄( )Hz= Figure 4 f f f- 5 - Problem 4: Assume that you sampled a piece of music at sampling frequency but encoded an incorrect sampling frequency in the resulting digital music file (e.g. wav or mp3). (a) Explain qualitatively what the music would sound like when played back, if the incorrectly encoded sam- pling frequency is given by, . (7) The music will sound slow and muffled, since all the frequencies in the original piece of music will be played at three-fourth their original frequency. (b) Repeat part (a) for . The music will sound fast and high-pitched, since all the frequencies in the original piece of music will be played at four-thirds their original frequency. Problem 5: Compute for the following complex numbers: (a) (c) (b) (d) For part (a), lies in quadrant I; therefore . For part (b), lies in quadrant IV; therefore . For part (c), lies in quadrant III; therefore . For part (d), lies in quadrant II; therefore . Problem 6: Let , where , and are real numbers. Solve the following expressions in terms of , and . Your answers should not include the imaginary number . (a) (d) (b) (e) (c) (f) Let us first expand : (S-14) (S-15) Therefore: [part (a)] (S-16) [part (b)] (S-17) [part (c)] (S-18) For part (d), we know that , so that : fs 2fmax> fs' fs' 3fs( ) 4⁄= fs' 4fs( ) 3⁄= z( )arg z 1 j+= 1– j– z 1 j–= 1– j+ z z( )arg 1 1⁄( )atan π 4⁄= = z z( )arg 1– 1⁄( )atan π– 4⁄= = z z( )arg π– 1 1–⁄–( )atan+ 3π– 4⁄= = z z( )arg π 1 1–⁄( )atan+ 3π 4⁄= = z1 a jb+( )2ejc= a b c a b c j Re z1[ ] z1 Im z1[ ] z1( )arg Im jz1[ ] Re ejπ 4⁄ z1[ ] z1 z1 a 2 j2ab b2–+( ) c( )cos j c( )sin+( )= z1 a 2 b2–( ) c( )cos 2ab c( )sin– j 2ab c( )cos a2 b2–( ) c( )sin+( )+= Re z1[ ] a2 b2–( ) c( )cos 2ab c( )sin–= Im z1[ ] 2ab c( )cos a2 b2–( ) c( )sin+= Im jz1[ ] Re z1[ ] a2 b2–( ) c( )cos 2ab c( )sin–= = ejc 1= z1 a jb+( )2=- 6 - (S-19) [part (d)] (S-20) For part (e), we can rewrite the first part of as: (S-21) so that: (S-22) Therefore, [part (e)] (S-23) Finally, for part (f), we multiply equation (S-22) by : (S-24) (S-25) Therefore, [part (f)] (S-26) Problem 7: Let, and Solve the following expressions, giving your answers in both polar and rectangular form. You may verify your answers by computer, but must show all work required to arrive at the answers by hand calculation. Your answers should be exact (no numerical approximations). (a) (d) (b) (e) (c) (f) First, let us write in polar form: (S-27) and in rectangular form: (S-28) Part (a): (S-29) z1 a jb+( )2 a2 j2ab b2–+( )= = a2 b2–( )2 2ab( )2+= a4 2a2b2– b4 4a2b2+ += a4 2a2b2 b4+ += a2 b2+( )2 a2 b2+( )= = z1 a 2 b2+( )= z1 a jb+( )2 a jb+( )2 ej a jb+( )arg ej a jb+( )arg a2 b2+( )ej2 b a,( )atan= = z1 a 2 b2+( )ej2 b a,( )atan ejc a2 b2+( )ej 2 b a,( )atan c+[ ]= = z1( )arg 2 b a,( )atan c+= ejπ 4⁄ ejπ 4⁄ z1 a 2 b2+( )ej 2 b a,( )atan c+[ ]ejπ 4⁄= ejπ 4⁄ z1 a 2 b2+( )ej 2 b a,( )atan c π 4⁄+ +[ ]= Re ejπ 4⁄ z1[ ] a2 b2+( ) 2 b a,( )atan c π 4⁄+ +[ ]cos= z1 2– j2 3+= z2 5e jπ 3⁄–= 2z1 z2+ z1 z2⁄ z1z2( )2 e z2 z1z1∗ z1 z1 z1 2–( )2 2 3( ) 2+ ej 2 3 2–,( )atan 4ej2π 3⁄= = z2 z2 5 π– 3⁄( )cos j5 π– 3⁄( )sin+ 5 2⁄ j5 3 2⁄–= = 2z1 z2+ 2 2– j2 3+( ) 5 2⁄ j5 3 2⁄–( )+ 3 2⁄– j3 3 2⁄+= =- 7 -