Signals and Systems Test 6

Question 1
1 / -0

Let x₁[n], x₂[n] and x₃[n] be three discrete time signals and X₁(z), X₂(z) and X₃(z) be their z-transforms given as:
\({X_1}\left( z \right) = \frac{{{z^2}}}{{\left( {z - 1} \right)\left( {z - 0.5} \right)}}\;\),
\({X_2}\left( z \right) = \frac{z}{{\left( {z - 1} \right)\left( {z - 0.5} \right)}}\)
\({X_3}\left( z \right) = \frac{1}{{\left( {z - 1} \right)\left( {z - 0.5} \right)}}\)
Then x₁[n], x₂[n] and x₃[n] are related as:

A
x₁[n - 2] = x₂[n - 1] = x₃[n]

B
x₁[n + 2] = x₂[n + 1] = x₃[n]

C
x₁[n ] = x₂[n - 1] = x₃[n - 2]

D
x₁[n + 1] = x₂[n - 1] = x₃[n]

Question 2
1 / -0

Let 3 + 4j be a zero of a fourth order linear-phase FIR filter. The complex number which is NOT a zero of this filter is

A
3 – 4j

B
\(\frac{3}{{25}} + \frac{4}{{25}}j\)

C
\(\frac{3}{{25}} - \frac{4}{{25}}j\)

D
\(\frac{1}{3} - \frac{1}{4}j\)

Question 3
1 / -0

H(z) is discrete rational transfer function. To ensure that both H(z) and its inverse are stable its

A
Poles must be inside the unit circle and zeros must be outside the unit circle.

B
Poles and zeros must be inside the unit circle.

C
Poles and zeros must be outside the unit circle.

D
Poles must be outside the unit circle and the zeros should be inside the unit circle.

Question 4
1 / -0

A discrete-time causal signal x[n] has the z-transform:
\(X\left( z \right) = \frac{z}{{z - 0.4}},\;ROC:\left| z \right| > 0.4\)
The ROC for z-transform of the even part of x[n] will be:

A
same as the ROC of X(z)

B
0.4 < |z| < 2.5

C
|z| > 0.2

D
|z| > 0.8

Question 5
1 / -0

The signal x[n] = (0.5)ⁿ u[n], when applied to a digital filter yields the following output:
y[n] = δ[n] – 2δ[n - 1].
If the impulse response of the filter is h[n], what will be the value of the sample h[1]?

A
-2.5--2.5

B
-2.54--2.51

C
-2.57--2.56

D
-2.59--2.52

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Signals and Sys...

Question 1

x₁[n - 2] = x₂[n - 1] = x₃[n]

x₁[n + 2] = x₂[n + 1] = x₃[n]

x₁[n ] = x₂[n - 1] = x₃[n - 2]

x₁[n + 1] = x₂[n - 1] = x₃[n]

Question 2

3 – 4j

\(\frac{3}{{25}} + \frac{4}{{25}}j\)

\(\frac{3}{{25}} - \frac{4}{{25}}j\)

\(\frac{1}{3} - \frac{1}{4}j\)

Question 3

Poles must be inside the unit circle and zeros must be outside the unit circle.

Poles and zeros must be inside the unit circle.

Poles and zeros must be outside the unit circle.

Poles must be outside the unit circle and the zeros should be inside the unit circle.

Question 4

same as the ROC of X(z)

0.4 < |z| < 2.5

|z| > 0.2

|z| > 0.8

Question 5

-2.5--2.5

-2.54--2.51

-2.57--2.56

-2.59--2.52

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Question Analysis

Signals and Sys...

Question 1

x1[n - 2] = x2[n - 1] = x3[n]

x1[n + 2] = x2[n + 1] = x3[n]

x1[n ] = x2[n - 1] = x3[n - 2]

x1[n + 1] = x2[n - 1] = x3[n]

Question 2

3 – 4j

\(\frac{3}{{25}} + \frac{4}{{25}}j\)

\(\frac{3}{{25}} - \frac{4}{{25}}j\)

\(\frac{1}{3} - \frac{1}{4}j\)

Question 3

Poles must be inside the unit circle and zeros must be outside the unit circle.

Poles and zeros must be inside the unit circle.

Poles and zeros must be outside the unit circle.

Poles must be outside the unit circle and the zeros should be inside the unit circle.

Question 4

same as the ROC of X(z)

0.4 < |z| < 2.5

|z| > 0.2

|z| > 0.8

Question 5

-2.5--2.5

-2.54--2.51

-2.57--2.56

-2.59--2.52

User

Question Analysis

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x₁[n - 2] = x₂[n - 1] = x₃[n]

x₁[n + 2] = x₂[n + 1] = x₃[n]

x₁[n ] = x₂[n - 1] = x₃[n - 2]

x₁[n + 1] = x₂[n - 1] = x₃[n]