1. The condition for memory-less system is given by _____
a) h[k] = cδ[k]
b) h[k] = cδ[n-k]
c) h[k] = ch[k]δ[k]
d) h[k] = ch[n-k]δ[k]
Explanation: The LTI discrete-time system is said to be memory-less if and only if it satisfies the condition h[k]=cδ[k]. All memory-less LTI systems perform scalar multiplication on the input
2. The causal continuous system with impulse response should satisfy ____ equation.
a) h(t)=0,t<0
b) h(t)=0,t>0
c) h(t)≠0,t<0
d) h(t)≠0,t≤0
Explanation: To the continuous system to be causal, the impulse response should satisfy the equation h(t)=0,t < 0 and convolution integral is given by y(t)=∫0∞ h(τ)x(t-τ)dτ
3. Causal systems are ______
a) Anticipative
b) Non anticipative
c) For certain cases anticipative
d) For certain cases anticipative and non anticipative
Explanation: Causal systems are non anticipative. They cannot generate an output before the input is applied. Which indicates the impulse response is zero for negative time.
4. Which of the following is true for discrete-time stable systems?
a) \(∑_{k=-∞}^∞\) |h[k]| < ∞
b) \(∑_{k=0}^∞\) |h[k]|< ∞
c) \(∑_{k=-∞}^∞\) |h[k]| ≤ ∞
d) \(∑_{k=-∞}^0\) |h[k]|< ∞
Explanation: If the condition ∑∞k=-∞|h[k]|<∞ is satisfied by an LTI system then it is said to be stable and for continuous time signal the condition is given by integral
∫∞-∞ |h(τ)|dτ<∞.
5. The impulse response of discrete-time signal is given by h [n] = u [n+3]. Whether the system is causal or not?
a) Causal
b) Non-causal
c) Insufficient information
d) The system cannot be classified
Explanation: The given impulse response h [n] = u [n+3] is not causal because of the term u [n+3] which implies it is non zero for n= -1, -2, -3.
6. Find the values of x, y, z and w from the below condition.
\(5\begin{bmatrix}
x & z \\
y & w \\
\end{bmatrix} = \begin{bmatrix}
2 & 10 \\
3 & 2x+y \\
\end{bmatrix} + \begin{bmatrix}
z & 5 \\
7 & w \\
\end{bmatrix} \).
a) x=1, y=3, z=4, w=0
b) x=2, y=3, z=8, w=1
c) x=1, y=2, z=3, w=1
d) x=1, y=2, z=4, w=1
Explanation: 5z=10+5 => 5z=15 => z=3
5x=2+z => 5x=5 => x=1
5y=3+7 => 5y=10 => y=2
5w=2+2+w => 4w=4 => w=1.
7. The matrix A is represented as \(\begin{bmatrix}
1 & 4 \\
-2 & 9 \\
-3 & -8 \\
\end{bmatrix}\). The transpose of the matrix of this matrix is represented as?
a) \(\begin{bmatrix}
1 & 4 \\
-2 & 9 \\
\end{bmatrix}\)
b) \(\begin{bmatrix}
1 & 4 \\
-2 & 9 \\
-3 & 8 \\
\end{bmatrix}\)
c) \(\begin{bmatrix}
1 & -2 & -3\\
4 & 9 & 8\\
\end{bmatrix}\)
d) \(\begin{bmatrix}
-1 & 2 & 3\\
-4 & -9 & 8\\
\end{bmatrix}\)
Explanation:Given matrix is a 3×2 matrix and the transpose of the matrix is 3×2 matrix.
The values of matrix are not changed but, the elements are interchanged, as row elements of a given matrix to the column elements of the transpose matrix and vice versa but the polarities of the elements remains same
8. Find the inverse of the matrix \(A = \begin{bmatrix}
8 & 5 & 2\\
4 & 6 & 3\\
7 & 4 & 2\\
\end{bmatrix}\).
a) \(\frac{1}{13}*\begin{bmatrix}
90 & 65 & 80\\
65 & 61 & 54\\
80 & 58 & 69\\
\end{bmatrix}\)
b) \(\frac{1}{14}*\begin{bmatrix}
93 & 68 & 80\\
68 & 61 & 58\\
80 & 58 & 69\\
\end{bmatrix}\)
c) \(\frac{1}{13}*\begin{bmatrix}
94 & 67 & 80\\
67 & 60 & 56\\
80 & 58 & 69\\
\end{bmatrix}\)
d) \(\frac{1}{13}*\begin{bmatrix}
93 & 68 & 80\\
68 & 61 & 58\\
80 & 58 & 69\\
\end{bmatrix}\)
Explanation: The inverse of matrix A = \(\frac{adjA}{|A|}\),
adjA=AA-1,
adjA = \(\frac{1}{13}*\begin{bmatrix}
93 & 68 & 80\\
68 & 61 & 58\\
80 & 58 & 69\\
\end{bmatrix}\), |A|=13.
9. Given the equations are 4x+2y+z=8, x+ y+ z=3, 3x+y+3z=9. Find the values of x, y and z.
a) 5/3, 0, 2/3
b) 1, 2, 3
c) 4/3, 1/3, 5/3
d) 2, 3, 4
Explanation: The matrix from the equations is represented as M=\(\begin{bmatrix}
4 & 2 & 1\\
1 & 1 & 1\\
3 & 1 & 3\\
\end{bmatrix}\)
The another matrix is X = \(\begin{bmatrix}
8\\
3\\
9\\
\end{bmatrix}\)
Then |M| = 6
For x=\(\begin{bmatrix}
8 & 2 & 1\\
3 & 1 & 1\\
9 & 1 & 3\\
\end{bmatrix}\) = 5/3
Similarly, y=0, z=-2/3.
10. Find the adjacent A as A=\(\begin{bmatrix}
1 & 7 & -3\\
5 & 4 & -2\\
6 & 8 & -6\\
\end{bmatrix}\).
a) \(\begin{bmatrix}
1 & 1 & 1\\
1 & 2 & 3\\
2 & 3 & 4\\
\end{bmatrix}\)
b) \(\begin{bmatrix}
31 & 39 & 80\\
39 & 45 & 74\\
80 & 74 & 136\\
\end{bmatrix}\)
c) \(\begin{bmatrix}
1 & 0 & 0\\
0 & 1 & 0\\
0 & 0 & 1\\
\end{bmatrix}\)
d) \(\begin{bmatrix}
35 & 34 & 32\\
56 & 67 & 48\\
98 & 74 & 52\\
\end{bmatrix}\)
Explanation: The adjacency of A is given by AAT
AT = \(\begin{bmatrix}
1 & 5 & 6\\
7 & 4 & 8\\
-3 & -2 & -6\\
\end{bmatrix}\),
AAT = \(\begin{bmatrix}
1 & 7 & -3\\
5 & 4 & -2\\
6 & 8 & -6\\
\end{bmatrix} × \begin{bmatrix}
1 & 5 & 6\\
7 & 4 & 8\\
-3 & -2 & -6\\
\end{bmatrix}\)
adjA=\(\begin{bmatrix}
31 & 39 & 80\\
39 & 45 & 74\\
80 & 74 & 136\\
\end{bmatrix}\).