Question

QUESTION 1 The output of the LSI system y[n] = 0.5(x[n]- x[n - 1]) to the input x[n] = 1 is: Oy[n] = cos(it) Oy[n] = 1 Oy[n] = 0 y[n] = Tt QUESTION 2 The output of an LSI system with frequency response Halw)= wejsinw, [W] < it to the input x[n] = 10 cos cos (Th 1 + 45 45°) is Oy[n] = 0 y[n] = 5jt sin 2 πη 4 O 1 πη y[n] = 10 cosi 4 + TT 4 + 11 / + = πη TT vin] = 51 sin 4 4 2 QUESTION 3 The impulse response of the system described by y[n] = 0.5(x[n] - x[n - 10]) is real-valued. O False O True

Answer 1

Solution:

The DC value is not affected by shifting hence

x(n) = x(n-1) =1

By substituting the above values we will get answer as 0.

For LSI system we got the value of y[n] with reference to the x[n]

In the last Q. 3 the impulse response is real valued hence h[n] is a real valued signal.

It's TRUE.

1. 
   yinla 0 s [«Cup-u (n-1) 1 [n] = 1 [oc value wont (oc value by shifting Y [n 0.5 [~[n] -1 [n 1] n[n] = n(n-1) = 1 yen] = 0.5 [ren] -4 (0-1] = 0.5 (1-1) = 0.5X o Y[n] = 0 option 3rd y [n] = 0 is correct
   
2. 
   0.2 jsinw Hd 1w) = we I Hd In 1 = W <tblul - sinu yin = | Halo | XM(W) x <Hd () W-000 Wo = tlu from nch] =1000s = 1000s | Th tus my tus) I Hd (wo) 1 = <Hd wla sinaly 4 yil. = <1 x to rol Costust! yon = numerous [***** yin) an sin + +1+1 ( У(и) jala sin (AM + I +) e porno 5 2 4 ylu) = sin sin ( * * . + tl
   
3. 
   yan) Sin sin an + +1 2 4 4 By the observations all the given options are incorrect . The (0 rrect answer should be sia sin( the helt y[n T +1 1) 2 4. be In option (d) it must of I instead A. √2 The correct answer is y[n] = sir sin an at 2 4 u
   
4. 
   0.3 yln ÖS [ a[n] - [n+c]] Taking a transform on both sides - 10 712) 0.5 X12) - 0.57 X12) y 12) = x12) [0. X12) 0.5-05 2 70 -10 Y12) X12) 0.5-0.52 yiz) X12) - Impulse Response H(2) = 0.5-0.5210 Taking inverse Z transport hin) 0.5 -0.5 g (1-10) Impulse response и (и) h[n] = 0.5-0.5 8 [1-10 4 The obtained h[n] is real valued It is a True statement "TRUE"