Question

QUESTION 1 (a) For a given mathematical equation, sketch the signals of: (1) x1 [n] = (n − 2)(u[n + 2] – u[n – 2]) + 48[n + 1] + 8[n – 2] (ii) x2 [n] = 3u [n + 1] - 3u[n – 1] (b) (8 marks) Based on the signal that you sketch in Qla(i), illustrate a new transformed signal yı[n] by performing (1), discuss the difference between original signal in Q(a)(i) and new signal in term of amplitude and time. (c) y. [n] = 3 – 2x[n] (1) (5 marks) By replacing n with (2n – 1) in the signal of x2[n], illustrate a new transformed signal of yz[n]. Explain the difference between the transformed signal yz[n] in (b) with y, [n] in (c). (4 marks) Using the signal of y. [n] and yz [n] that you obtained in Q1(b) and Q1(c), determine the equation of z[n] in Fiqure Q1(d) and then plot the signal of z[n]. (d) (3 marks) + yı[n] 2[n] y2[n] Figure Q1(d): Connection of the Discrete Time systems (e) Determine and draw the odd components of z[n] based on the signal you obtained in Qi(d). (5 marks) [25 Marks

Answer 1

Here i am uploading pictures of my solution.

2,Chๆ 2 (1-2)(21 (142) - 1( 1-2ว) + 66 (0+10+8C-2) T12(( หt7) - H (h-22 (1) การ)(1) 0 น - -4 1 2 (h-?) 36 ht ) 2 9

(b) Now, y (n)= 3-22, (2) 22,(n)=2 -2 6 2 2 -4 Yon )

(b) Now, y (n)= 3-22, (2) 22,(n)=2 -2 6 2 2 -4 Yon )

=) Here these is no chande in time when operation 3-20 is ressesmed. chile, after Perbsming the operation, amplitude of cell component has become positive (c) reflucing n with (22-12 in a no azcan-1= 3u Can-1 +1) 34 (2n-1-12 Y₂n2=3u Can - 30 C 2n-22 3 304 am 32(22-22 3 3. 3 3 2 2 4 ay ang Buana -34 (22-2 Sich comparing y ch und Y2(02, pese is no similarity between both sidnad us their components and Amplitude of Components che different.

(ы) culla 51 () T Culth 20h - 2 - 2 2,C)+ 3 и 5 ) - 3 (2n-2 ) jo 5 Cult о 2 | -- Бит); 3- (1-4) (un+2 ) - (hr)) +8 S (7) + c ch-) + 3 иһ) - зус-1)
Here i have uploaded the solution.

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