Exercise Questions - Introduction to Digital Signal Processing

Exercise Questions - Introduction to Digital Signal Processing Notes | Study Digital Signal Processing - Electrical Engineering (EE)

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Exercise Questions

Q. y(n)-0.4 y(n-1) = x(n). find the anti-causal impulse response? h(n)=0 for n≥ 0

h(n-1) = 2.5 [h(n)- δ (n) ]

h(-1) =  2.5 [h(0)- δ (0) ] = -2.5

h(-2) = -2.52  . …….. h(n) = -2.5valid for n ≤ -1

Q. x(n)={1,2,3} y(n)={3,4} Obtain difference equation from i/p & o/p information

y(n) + 2 y(n-1) + 3 y(n-2) = 3 x(n) + 4 x(n-1) (Ans)

Q. x(n) = {4,4,}, y(n)= x(n)- 0.5x(n-1). Find the difference equation of the inverse system. Sketch the realization of each system and find the output of each system.

Solution:

The original system is y(n)=x(n)-0.5 x(n-1)

The inverse system is x(n)= y(n)-0.5 y(n-1)

y (n) = x (n) – 0.5 x(n-1)

Y (z) = X (z) [1-0.5Z-1]

System

Inverse System

y (n) – 0.5 y(n-1) =x(n)

Y (z) [1-0.5 Z-1] = X (z)

g (n) = 4 δ (n) - 2δ(n-1) + 4δ (n-1) - 2δ (n-2) = 4δ (n) + 2δ (n-1) - 2δ (n-2)

y (n) = 0.5 y(n-1) + 4δ (n) + 2δ (n-1) – 2δ (n-2)

y (0) = 0.5y(-1) + 4δ (0) = 4

y(1) = 4

y(2) = 0.5 y(1) - 2δ (0) = 0

y(n) = {4, 4} same as i/p

 Non Recursive filters Recursive filters y(n) for causal systemFor causal i/p sequence Present response is a function of the present and past N values of the excitation as well as the past N values of response. It gives IIR o/p but not y(n) =   ak x(n-k)Present response depends only on present i/p & previous i/ps but not future i/ps. It gives FIR o/p always.(n) – y(n-1) = x(n) – x(n-3)

Q. y(n) = 1/3[x (n+1) + x (n) + x (n-1)]    Find the given system is stable or not?

Let x(n) = δ (n)

h(n) = 1/3 [ δ (n+1) + δ (n) + δ(n-1)]

h(0) = 1/3

h(-1) = 1/3

h(1) = 1/3

s=∑h(n) <∞ therefore Stable.

Q. y(n) = a y(n-1) + x(n)   given y(-1) = 0

Let x(n) = δ (n)

h(n) = y(n) = a y(n-1) + δ (n)

h(0) = a y(-1) +δ (0) = 1 = y(0)

h(1) = a y(0) + δ (1) = a

h(2) = a y(1) + δ (2) = a2 . . . . . . . h(n) = an u(n)   stable if a<1.

y(n-1) =1/a [ y(n) – x(n)]

y(n) = 1/a [ y(n+1) – x(n+1)]

y(-1) = 1/a [ y(0) – x(0)]=0

y(-2) = 0

Q. y(n) = 1/n+1 y(n-1) + x(n) for n ≥ 0

= 0        otherwise. Find whether given system is time variant or not?

Let x(n) = δ (n)

h (0) = 1 y(-1) +δ (0) = 1

h(1) = ½ y(0) + δ(1) = ½

h(2) = 1/6

h(3) = 1/24

if x(n) = δ (n-1)

y(n) = h(n-1)

h(n-1) = y(n) =1/n+1 h(n-2) + δ (n-1)

n=0 h(-1) = y(0) = 1 x 0+0 =0

n=1 h(0) = y(1) = ½ x 0 +δ (0)= 1

n=2 h(1) = y(2) = 1/3 x 1 + 0 = 1/3

h(2) = 1/12

∴ h (n, 0) ≠ h (n,1)  ∴ TV

Q. y (n) = 2n x(n)    Time varying

Q. y (n) =1/3 [x (n+1) + x (n) + x (n-1)] Linear

Q. y (n) = 12 x (n-1) + 11 x(n-2) TIV

Q. y (n) = 7 x2(n-1) non linear

Q. y (n) = x2(n) non linear

Q. y (n) =  n2 x (n+2) linear

Q. y (n) = x (n2) linear

Q. y (n) =  ex(n) non linear

Q. y (n) =  2x(n) x (n) non linear, TIV

(If the roots of characteristics equation are a magnitude less than unity. It is a necessary & sufficient condition) Non recursive system, or FIR filter are always stable

Q. y (n) + 2 y2(n) = 2 x(n) – x(n-1)  non linear, TIV

Q. y (n) - 2 y (n-1) = 2x(n) x (n) non linear, TIV

Q. y (n) + 4 y (n) y (2n) = x (n) non linear, TIV

Q. y (n+1) – y (n) = x (n+1) is causal

Q. y (n) - 2 y (n-2) = x (n) causal

Q. y (n) - 2 y (n-2) = x (n+1) non causal

Q. y (n+1) – y (n) = x (n+2) non causal

Q. y (n-2) = 3 x (n-2) is static or Instantaneous.
Q. y (n) = 3 x (n-2) dynamic

Q. y (n+4) + y (n+3) = x (n+2) causal & dynamic

Q. y (n) = 2 x (αη )

If α=1 causal, static

α <1 causal, dynamic

α >1 non causal, dynamic

α≠ 1 TV

Q. y (n) = 2(n+1) x (n) is causal & static but TV.

Q. y (n) = x (-n) TV

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