Signals and Systems Test 5

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Signals and Systems Test 5

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Weekly Quiz Competition

Question 1
1 / -0

If the z-transform of a sequence x[n] = {1, 1, -1, -1} is X[z], then the value of X(1/2) is

A
-9

B
1.875

C
-1.125

D
15

Solution

\(X\left( z \right) = \mathop \sum \limits_{n = - \infty }^\infty x\left[ n \right]{z^{ - n}}\)
\(= \mathop \sum \limits_{n = 0}^3 x\left[ n \right]{z^{ - n}}\)
= z^-0 + z^-1 – z^-2 – z^-3
\(= 1 + \frac{1}{z} - \frac{1}{{{z^2}}} - \frac{1}{{{z^3}}}\)
\(x\left( {\frac{1}{2}} \right) = 1 + 2 - 4 - 8 = - 9\)
Question 2
1 / -0

It is given that \(H\left( z \right) = \frac{{11}}{{\left( {1 - \frac{1}{3}{z^{ - 1}}} \right)\left( {1 - 4{z^{ - 1}}} \right)}}\)
If H(z) is system function of LTI system and system is stable, then h(0) is given by

A
-1--1

B
-14--11

C
-17--16

D
-19--12

Solution

\(H\left( z \right) = \frac{{11}}{{\left( {1 - \frac{1}{3}{Z^{ - 1}}} \right)\left( {1 - 4{z^{ - 1}}} \right)}}\)
\(H\left( z \right) = \frac{{ - 1}}{{\left( {1 - \frac{1}{3}{Z^{ - 1}}} \right)}} + \frac{{12}}{{\left( {1 - 4{z^{ - 1}}} \right)}}\;\)
For stable system, ROC must include the unit circle.
\(ROC:\frac{1}{3} < \left| z \right| < 4\)
\(h\left( n \right) = - 1{\left( {\frac{1}{3}} \right)^n}u\left( n \right) - {124^n}\;u\left( { - n - 1} \right)\)
h(0) = - 1
Question 3
1 / -0

A sequence u[n] is defined as \(u\left[ n \right] = \left\{ {\begin{array}{*{20}{c}}{1,\;if\;n \ge 0}\\{0,\;if\;n < 0}\end{array}} \right.\) for n = {-∞, …, -1,0,1, …, ∞}. The z-transform of u[n] is U(z). The region of convergence for which the z-transform of u[n] exists is

A
|z| > 1

B
|z| < 1

C
|z| = 1

D
|z| > 0

Solution

Given signal is
\(u\left[ n \right] = \left\{ {\begin{array}{*{20}{c}}{1,\;if\;n \ge 0}\\{0,\;if\;n < 0}\end{array}} \right.\)
z transform for the given signal is
\(U\left( z \right) = \frac{1}{{1 - {z^{ - 1}}}}\)
The region of convergence will be
\(1 - {z^{ - 1}} > 0\)
⇒ |z| > 1
Question 4
1 / -0

The z-transform of a signal is given by \(H\left( z \right) = \frac{1}{8}\frac{{{z^{ - 1}}\left( {1 + {z^{ - 5}}} \right)}}{{\left( {1 + {z^{ - 1}}} \right)\left( {1 - {z^{ - 1}}} \right)}}\). It final value is –

A
0.125-0.125

B
0.1254-0.1251

C
0.1257-0.1256

D
0.1259-0.1252

Solution

Final value theorem it states that
\(\begin{array}{l} x\left( \infty \right) = \begin{array}{*{20}{c}} {lim}\\ {z \to 1} \end{array}\left( {1 - {z^{ - 1}}} \right)x\left( z \right)\\ = \begin{array}{*{20}{c}} {lim}\\ {z \to 1} \end{array}\frac{1}{8}\left( {1 - {z^{ - 1}}} \right)\frac{{{z^{ - 1}}\left( {1 + {z^{ - 5}}} \right)}}{{\left( {1 - {z^{ - 1}}} \right)\left( {1 + {z^{ - 1}}} \right)}}\\ = \begin{array}{*{20}{c}} {lim}\\ {z \to 1} \end{array}\frac{1}{8}\frac{{{z^{ - 1}}\left( {1 + {z^{ - 5}}} \right)}}{{\left( {1 + {z^{ - 1}}} \right)}}\\ = \begin{array}{*{20}{c}} {lim}\\ {z \to 1} \end{array}\frac{1}{8}\frac{{\frac{1}{z}\left( {1 + \frac{1}{{{z^5}}}} \right)}}{{\left( {1 + \frac{1}{z}} \right)}} \end{array}\)
\(\begin{array}{l} = \begin{array}{*{20}{c}} {lim}\\ {z \to 1} \end{array}\frac{1}{8}.\frac{{{z^5} + 1}}{{\left( {z + 1} \right)}}\\ = \frac{1}{8} = 0.125 \end{array}\)
Question 5
1 / -0

The z-transform of a sequence x[n] is given as \(X\left( z \right) = \frac{{3{z^3} + 5{z^2} \pm 5z + 4}}{{{z^2}}}\). If y[n] is the second order backward difference (Δ²x_n), of x[n], then y(z) is given by

A
3z + 1 – 12z^-1 – 19z^-2 + 13z^-3 + 4z^-4

B
3z – 1 – 12z^-1 + 19z^-2 – 13z^-3+ 4z^-4

C
3z + 1 + 12z^-1 – 19z^-2 – 13z^-3 – 4z^-4

D
3z – 1 + 12z^-1 – 19z^-2 + 13z^-3 – 4z^-4

Solution

Concept:
The backward difference operator Δ is defined by the equation
Δx_n = x_n – x_n-1
Δ²x_n = Δ(Δx_n)
= Δ (x_n – x_n-1)
= Δx_n – Δx_n-1
= x_n – x_n-1 – (x_n-1 – x_n-2)
= x_n – 2x_n-1 + x_n-2
Calculation:
y[n] = Δ²x[n] = x_n – 2x_n-1 + x_n-2
= x[n] – 2x [n - 1] + x [n - 2]
By applying z transform,
y(z) = X(z) – 2z^-1 X(z) + z^-2 X(z)
= [1 – 2z^-1 + z^-2] X(z)
\(X\left( z \right) = \frac{{3{z^3} + 5{z^2} - 5z + 4}}{{{z^2}}} = 3z + 5 - 5{z^{ - 1}} + 4{z^{ - 2}}\)
y(z) = [1 – 2z^-1 + z^-2] [3z + 5 – 5z^-1 + 4z^-2]
= 3z + 5 – 5z^-1 + 4z^-2 – 6 – 10z^-1 + 10z^-2 – 8z^-3 + 3z^-1 + 5z^-2 – 5z^-3 + 4z^-4
= 3z – 1 – 12z^-1 + 19z^-2 – 13z^-3 + 4z^-4
Question 6
1 / -0

What is the signal corresponding to the following z-transform? (Where u[n] is the unit-step signal)
\(X\left( z \right) = \frac{{6{z^2}}}{{\left( {2z - 1} \right)\left( {3z - 1} \right)}},\frac{1}{3} < \left| z \right| < \frac{1}{2}\)

A
\({\left( {\frac{1}{2}} \right)^n}u\left[ { - n - 1} \right] + {\left( {\frac{1}{3}} \right)^n}u\left[ n \right]\)

B
\({\left( {\frac{1}{2}} \right)^n}u\left[ n \right] - {\left( {\frac{1}{3}} \right)^n}u\left[ { - n - 1} \right]\)

C
\( - 3{\left( {\frac{1}{2}} \right)^n}u\left[ { - n - 1} \right] - 2{\left( {\frac{1}{3}} \right)^n}u\left[ n \right]\)

D
\(- 3\;{\left( {\frac{1}{2}} \right)^n}u\left[ n \right] + 2{\left( {\frac{1}{3}} \right)^n}u\left[ { - n - 1} \right]\)

Solution

Concept:
The Z-transform of standard signals with their ROC are:
\({a^n}u\left( n \right) \leftrightarrow \frac{z}{{z - a}};\left| z \right| > \left| a \right|\)
\( - {a^n}u\left( { - n - 1} \right) \leftrightarrow \frac{z}{{z - a}};\left| z \right| < \left| a \right|\)
Application:
\(X\left( z \right) = \frac{{6{z^2}}}{{\left( {2z - 1} \right)\left( {3z - 1} \right)}}\)
\( = \frac{{{z^2}}}{{\left( {z - \frac{1}{2}} \right)\left( {z - \frac{1}{3}} \right)}}\)
\(= z[\frac{z}{{\left( {z - \frac{1}{2}} \right)\left( {z - \frac{1}{3}} \right)}}\)
\(= z\left[ {\frac{3}{{\left( {z - \frac{1}{2}} \right)}} - \frac{2}{{\left( {z - \frac{1}{3}} \right)}}} \right]\)
\(= \frac{{3z}}{{\left( {z - \frac{1}{2}} \right)}} - \frac{{2z}}{{\left( {z - \frac{1}{3}} \right)}}\)
For \(\left| z \right| < \frac{1}{2},\;\frac{{3z}}{{\left( {z - \frac{1}{2}} \right)}} \leftrightarrow \; - 3{\left( {\frac{1}{2}} \right)^n}u\left[ { - n - 1} \right]\)
For \(\left| z \right| > \frac{1}{3},\;\frac{{2z}}{{\left( {z - \frac{1}{3}} \right)}} \leftrightarrow 2{\left( {\frac{1}{3}} \right)^n}u\left[ n \right]\)
\(x\left[ n \right] = - 3{\left( {\frac{1}{2}} \right)^n}u\left[ { - n - 1} \right] + 2{\left( {\frac{1}{3}} \right)^n}u\left[ n \right]\)
Question 7
1 / -0

The z transform of the signal \(x\left[ n \right] = n{\left( {\frac{1}{2}} \right)^{\left| n \right|}}\) is

A
\(\frac{{ - 3z\left( {z - 1} \right)\left( {z + 1} \right)}}{{2{{\left( {z - 3} \right)}^2}{{\left( {z - \frac{1}{2}} \right)}^2}}},\;\frac{1}{2} < \left| z \right| < 2\)

B
\(\frac{{ - 3z\left( {z - 1} \right)\left( {z + 1} \right)}}{{{{\left( {z - 2} \right)}^2}{{\left( {z - \frac{1}{2}} \right)}^2}}},\;\frac{1}{2} < \left| z \right| < 2\)

C
\(\frac{{ - 2z\left( {z - 1} \right)\left( {z + 1} \right)}}{{3{{\left( {z - 2} \right)}^2}{{\left( {z - \frac{1}{2}} \right)}^2}}},\;\frac{1}{2} < \left| z \right| < 2\)

D
\(\frac{{ - z\left( {z - 1} \right)\left( {z + 1} \right)}}{{2{{\left( {z - 2} \right)}^2}{{\left( {z - \frac{1}{2}} \right)}^2}}},\;\frac{1}{2} < \left| z \right| < 2\)

Solution

Concept:
The Z-transform of standard signals with their ROC are:
\({a^n}u\left( n \right) \leftrightarrow \frac{z}{{z - a}};\left| z \right| > \left| a \right|\)
\( - {a^n}u\left( { - n - 1} \right) \leftrightarrow \frac{z}{{z - a}};\left| z \right| < \left| a \right|\)
Application:
\(x\left[ n \right] = n{\left( {\frac{1}{2}} \right)^{\left| n \right|}}\)
\( = n{\left( {\frac{1}{2}} \right)^n}u\left[ n \right] + n{\left( {\frac{1}{2}} \right)^{ - n}}u\left[ { - n - 1} \right]\)
\({\left( {\frac{1}{2}} \right)^n}u\left[ n \right] \leftrightarrow \frac{z}{{\left( {z - \frac{1}{2}} \right)}}\;\;;\;\;\left| z \right| > \frac{1}{2}\)
\(n{\left( {\frac{1}{2}} \right)^n}u\left[ n \right] \leftrightarrow \frac{z}{{2{{\left( {z - \frac{1}{2}} \right)}^2}}};\left| z \right| > \frac{1}{2}\)
\({\left( {\frac{1}{2}} \right)^{ - n}}u\left[ { - n - 1} \right] \leftrightarrow \frac{{ - z}}{{\left( {z - 2} \right)}}\;\;;\;\;\left| z \right| < 2\)
\(n{\left( {\frac{1}{2}} \right)^{ - n}}u\left[ { - n - 1} \right] \leftrightarrow \frac{{ - 2z}}{{{{\left( {z - 2} \right)}^2}}}\)
\(X\left( z \right) = \frac{z}{{2{{\left( {z - \frac{1}{2}} \right)}^2}}} - \frac{{2z}}{{{{\left( {z - 2} \right)}^2}}}\;\;;\;\frac{1}{2} < \left| z \right| < 2\)
\(= \frac{{3z\left( {z - 1} \right)\left( {z + 1} \right)}}{{2{{\left( {z - 2} \right)}^2}{{\left( {z - \frac{1}{2}} \right)}^2}}}\;\;;\;\frac{1}{2} < \left| z \right| < 2\)
Question 8
1 / -0

The following is known about a discrete-time LTI system with input x[n] and output y[n]:
1. If x[n] = (-2)ⁿ for all n, then y[n] = 0 for all n.
2. If \(x\left[ n \right] = {\left( {\frac{1}{2}} \right)^n}u\left[ n \right]\) for all n, then y[n] for all n is of the form \(y\left[ n \right] = \delta \left[ n \right] + a{\left( {\frac{1}{4}} \right)^n}u\left[ n \right]\).
The value of constant ‘a’ is

A
-1.2--1.1

B
-1.24--1.11

C
-1.27--1.16

D
-1.29--1.12

Solution

\(x\left[ n \right] = {\left( {\frac{1}{2}} \right)^n}u\left[ n \right]\)
\(y\left[ n \right] = \delta \left[ n \right] + a{\left( {\frac{1}{4}} \right)^n}u\left[ n \right]\)
\(H\left( z \right) = \frac{{Y\left( z \right)}}{{X\left( z \right)}}\)
\(= \frac{{1 + \left( {\frac{a}{{1 - \frac{1}{4}{z^{ - 1}}}}} \right)}}{{\frac{1}{{1 - \left( {\frac{1}{2}{z^{ - 1}}} \right)}}}}\)
\(= \frac{{\left( {1 + a} \right) - \frac{1}{4}{z^{ - 1}}}}{{\left( {1 - \frac{1}{4}{z^{ - 1}}} \right)\left( {1 - \frac{1}{2}{z^{ - 1}}} \right)}}\)
ROC of \(Y\left( z \right)\;:\left| z \right| > \frac{1}{4}\)
ROC of \(X\left( z \right)\;:\left| z \right| > \frac{1}{2}\)
ROC of \(H\left( z \right)\;:\left| z \right| > \frac{1}{2}\)
It is given that the output of the system to the input x[n] = (-2)ⁿ is y[n] = 0
So, H(-2) = 0
\(H\left( z \right) = \frac{{\left( {1 + a} \right) - \frac{1}{4}{z^{ - 1}}}}{{\left( {1 - \frac{1}{4}{z^{ - 1}}} \right)\left( {1 - \frac{1}{2}{z^{ - 1}}} \right)}}\)
H(-2) = 0
\( \Rightarrow \left( {1 + a} \right) - \frac{1}{4}\left( {\frac{1}{{ - 2}}} \right) = 0\)
\(\Rightarrow 1 + a = \frac{{ - 1}}{8}\)
\(\Rightarrow a = \frac{{ - 9}}{8} = - 1.125\)
Question 9
1 / -0

The system represented by the difference equation \(y\left[ n \right] - 2\;y\left[ {n - 2} \right] = x\left[ n \right] - \frac{1}{2}x\left[ {n - 1} \right]\) is

A
Causal, stable and minimum phase system

B
No causal and stable but minimum phase system

C
No causal, not stable and not a minimum phase system

D
Causal but not stable and not a minimum phase system

Solution

\(y\left[ n \right] - 2\;y\left[ {n - 2} \right] = x\left[ n \right] - \frac{1}{2}x\left[ {n - 1} \right]\)
By applying z transform
\(H\left( z \right) = \frac{{z\left( {z - \frac{1}{2}} \right)}}{{{z^2} - 2}}\)
Zeros at, \(z = 0,\frac{1}{2}\)
Poles at, \(z = \pm \sqrt 2 \)
Not all poles are inside |z| = 1, the system is not stable
Not all poles and zeros and inside |z| = 1, the system is not minimum phase
As the system is depending upon only past values but not the future inputs, the system is causal.

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Re-Attempt Weekly Quiz Competition

Signals and Systems Test 5

Result

Signals and Systems Test 5

Score

-

Rank

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SHARING IS CARING

Question 1

-9

1.875

-1.125

15

Solution

Question 2

-1--1

-14--11

-17--16

-19--12

Solution

Question 3

|z| > 1

|z| < 1

|z| = 1

|z| > 0

Solution

Question 4

0.125-0.125

0.1254-0.1251

0.1257-0.1256

0.1259-0.1252

Solution

Question 5

3z + 1 – 12z-1 – 19z-2 + 13z-3 + 4z-4

3z – 1 – 12z-1 + 19z-2 – 13z-3 + 4z-4

3z + 1 + 12z-1 – 19z-2 – 13z-3 – 4z-4

3z – 1 + 12z-1 – 19z-2 + 13z-3 – 4z-4

Solution

Question 6

\({\left( {\frac{1}{2}} \right)^n}u\left[ { - n - 1} \right] + {\left( {\frac{1}{3}} \right)^n}u\left[ n \right]\)

\({\left( {\frac{1}{2}} \right)^n}u\left[ n \right] - {\left( {\frac{1}{3}} \right)^n}u\left[ { - n - 1} \right]\)

\( - 3{\left( {\frac{1}{2}} \right)^n}u\left[ { - n - 1} \right] - 2{\left( {\frac{1}{3}} \right)^n}u\left[ n \right]\)

\(- 3\;{\left( {\frac{1}{2}} \right)^n}u\left[ n \right] + 2{\left( {\frac{1}{3}} \right)^n}u\left[ { - n - 1} \right]\)

Solution

Question 7

\(\frac{{ - 3z\left( {z - 1} \right)\left( {z + 1} \right)}}{{2{{\left( {z - 3} \right)}^2}{{\left( {z - \frac{1}{2}} \right)}^2}}},\;\frac{1}{2} < \left| z \right| < 2\)

\(\frac{{ - 3z\left( {z - 1} \right)\left( {z + 1} \right)}}{{{{\left( {z - 2} \right)}^2}{{\left( {z - \frac{1}{2}} \right)}^2}}},\;\frac{1}{2} < \left| z \right| < 2\)

\(\frac{{ - 2z\left( {z - 1} \right)\left( {z + 1} \right)}}{{3{{\left( {z - 2} \right)}^2}{{\left( {z - \frac{1}{2}} \right)}^2}}},\;\frac{1}{2} < \left| z \right| < 2\)

\(\frac{{ - z\left( {z - 1} \right)\left( {z + 1} \right)}}{{2{{\left( {z - 2} \right)}^2}{{\left( {z - \frac{1}{2}} \right)}^2}}},\;\frac{1}{2} < \left| z \right| < 2\)

Solution

Question 8

-1.2--1.1

-1.24--1.11

-1.27--1.16

-1.29--1.12

Solution

Question 9

Causal, stable and minimum phase system

No causal and stable but minimum phase system

No causal, not stable and not a minimum phase system

Causal but not stable and not a minimum phase system

Solution

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3z + 1 – 12z^-1 – 19z^-2 + 13z^-3 + 4z^-4

3z – 1 – 12z^-1 + 19z^-2 – 13z^-3+ 4z^-4

3z + 1 + 12z^-1 – 19z^-2 – 13z^-3 – 4z^-4

3z – 1 + 12z^-1 – 19z^-2 + 13z^-3 – 4z^-4