LTI systems - Networks, Signals and Systems - Electronics & Communication Engineering Previous Year Questions

1

2012 · Electronics & Communication Engineering · Networks, Signals and Systems · LTI systems

Electronics & Communication Engineering (EC) 2012

The input \( x(t) \) and output \( y(t) \) of a system are related as \( y(t) = \int_{-\infty}^{t} x(\tau) \cos(3\tau) d\tau \). The system is

A

time-invariant and stable

B

stable and not time-invariant

C

time-invariant and not stable

D

not time-invariant and not stable

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2

2013 · Electronics & Communication Engineering · Networks, Signals and Systems · LTI systems

Electronics & Communication Engineering (EC) 2013 [Session 1]

Two systems with impulse responses \(h_1(t)\) and \(h_2(t)\) are connected in cascade. Then the overall impulse response of the cascaded system is given by

A

product of \(h_1(t)\) and \(h_2(t)\)

B

sum of \(h_1(t)\) and \(h_2(t)\)

C

convolution of \(h_1(t)\) and \(h_2(t)\)

D

subtraction of \(h_2(t)\) from \(h_1(t)\)

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3

2013 · Electronics & Communication Engineering · Networks, Signals and Systems · LTI systems

Electronics & Communication Engineering (EC) 2013 [Session 1]

The impulse response of a system is \( h(t) = t u(t) \). For an input \( u(t-1) \), the output is

A

\(\frac{t^2}{2} u(t) \)

B

\(\frac{t(t-1)}{2} u(t-1) \)

C

\(\frac{(t-1)^2}{2} u(t-1) \)

D

\(\frac{t^2-1}{2} u(t-1) \)

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4

2013 · Electronics & Communication Engineering · Networks, Signals and Systems · LTI systems

Electronics & Communication Engineering (EC) 2013 [Session 1]

Which one of the following statements is NOT TRUE for a continuous time causal and stable LTI system?

A

All the poles of the system must lie on the left side of the \(j\omega \) axis.

B

Zeros of the system can lie anywhere in the \(s \) -plane.

C

All the poles must lie within \(|s| = 1 \).

D

All the roots of the characteristic equation must be located on the left side of the \(j\omega \) axis.

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5

2013 · Electronics & Communication Engineering · Networks, Signals and Systems · LTI systems

Electronics & Communication Engineering (EC) 2013 [Session 1]

Assuming zero initial condition, the response \( y(t) \) of the system given below to a unit step input \( u(t) \) is

A

\(u(t) \)

B

\(t u(t) \)

C

\(\frac{t^2}{2} u(t) \)

D

\(e^{-t} u(t) \)

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6

2013 · Electronics & Communication Engineering · Networks, Signals and Systems · LTI systems

Electronics & Communication Engineering (EC) 2013 [Session 1]

Let \( g(t) = e^{-\pi t^2} \), and \( h(t) \) is a filter matched to \( g(t) \). If \( g(t) \) is applied as input to \( h(t) \), then the Fourier transform of the output is

A

\(e^{-\pi f^2} \)

B

\(e^{-\pi f^2 / 2} \)

C

\(e^{-\pi |f|} \)

D

\(e^{-2\pi f^2} \)

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7

2013 · Electronics & Communication Engineering · Networks, Signals and Systems · LTI systems

Electronics & Communication Engineering (EC) 2013 [Session 1]

The impulse response of a continuous time system is given by \( h(t) = \delta(t-1) + \delta(t-3) \). The value of the step response at \( t = 2 \) is

A

0

B

1

C

2

D

3

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8

2013 · Electronics & Communication Engineering · Networks, Signals and Systems · LTI systems

Electronics & Communication Engineering (EC) 2013 [Session 1]

A system described by a linear, constant coefficient, ordinary, first order differential equation has an exact solution given by y(t) for t > 0, when the forcing function is x(t) and the initial condition is y(0). If one wishes to modify the system so that the solution becomes -2y(t) for t > 0, we need to

A

change the initial condition to -y(0) and the forcing function to 2x(t)

B

change the initial condition to 2y(0) and the forcing function to -x(t)

C

change the initial condition to \(j\sqrt{2}y(0)\) and the forcing function to \(j\sqrt{2}x(t)\)

D

change the initial condition to -2y(0) and the forcing function to -2x(t)

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9

2013 · Electronics & Communication Engineering · Networks, Signals and Systems · LTI systems

Electronics & Communication Engineering (EC) 2013 [Session 2]

The impulse response of a system is $h(t) = t u(t)$. For an input $u(t-1)$, the output is

A

$\frac{t^2}{2} u(t)$

B

$\frac{t(t-1)}{2} u(t-1)$

C

$\frac{(t-1)^2}{2} u(t-1)$

D

$\frac{t^2-1}{2} u(t-1)$

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10

2013 · Electronics & Communication Engineering · Networks, Signals and Systems · LTI systems

Electronics & Communication Engineering (EC) 2013 [Session 2]

A system described by a linear, constant coefficient, ordinary, first order differential equation has an exact solution given by \(y(t)\) for \(t > 0\), when the forcing function is \(x(t)\) and the initial condition is \(y(0)\). If one wishes to modify the system so that the solution becomes \(-2y(t)\) for \(t > 0\), we need to

A

change the initial condition to \(-y(0)\) and the forcing function to \(2x(t)\)

B

change the initial condition to \(2y(0)\) and the forcing function to \(-x(t)\)

C

change the initial condition to \(j\sqrt{2}y(0)\) and the forcing function to \(j\sqrt{2}x(t)\)

D

change the initial condition to \(-2y(0)\) and the forcing function to \(-2x(t)\)

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11

2013 · Electronics & Communication Engineering · Networks, Signals and Systems · LTI systems

Electronics & Communication Engineering (EC) 2013 [Session 2]

The impulse response of a continuous time system is given by \(h(t) = \delta(t-1) + \delta(t-3)\). The value of the step response at \(t = 2\) is

A

0

B

1

C

2

D

3

Open complete paper

12

2013 · Electronics & Communication Engineering · Networks, Signals and Systems · LTI systems

Electronics & Communication Engineering (EC) 2013 [Session 3]

Two systems with impulse responses \( h_1(t) \) and \( h_2(t) \) are connected in cascade. Then the overall impulse response of the cascaded system is given by

A

product of \(h_1(t) \) and \(h_2(t) \)

B

sum of \(h_1(t) \) and \(h_2(t) \)

C

convolution of \(h_1(t) \) and \(h_2(t) \)

D

subtraction of \(h_2(t) \) from \(h_1(t) \)

Open complete paper

13

2013 · Electronics & Communication Engineering · Networks, Signals and Systems · LTI systems

Electronics & Communication Engineering (EC) 2013 [Session 3]

Assuming zero initial condition, the response \(y(t)\) of the system given below to a unit step input \(u(t)\) is

A

\(u(t)\)

B

\(t u(t)\)

C

\(\frac{t^2}{2} u(t)\)

D

\(e^{-t} u(t)\)

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14

2013 · Electronics & Communication Engineering · Networks, Signals and Systems · LTI systems

Electronics & Communication Engineering (EC) 2013 [Session 3]

The impulse response of a continuous time system is given by \(h(t)=\delta(t-1)+\delta(t-3)\). The value of the step response at \(t=2\) is

A

0

B

1

C

2

D

3

Open complete paper

15

2013 · Electronics & Communication Engineering · Networks, Signals and Systems · LTI systems

Electronics & Communication Engineering (EC) 2013 [Session 4]

The impulse response of a system is \(h(t) = t u(t)\). For an input \(u(t-1)\), the output is

A

\(\frac{t^2}{2} u(t)\)

B

\(\frac{t(t-1)}{2} u(t-1)\)

C

\(\frac{(t-1)^2}{2} u(t-1)\)

D

\(\frac{t^2-1}{2} u(t-1)\)

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16

2013 · Electronics & Communication Engineering · Networks, Signals and Systems · LTI systems

Electronics & Communication Engineering (EC) 2013 [Session 4]

A system described by a linear, constant coefficient, ordinary, first order differential equation has an exact solution given by y(t) for t > 0, when the forcing function is x(t) and the initial condition is y(0). If one wishes to modify the system so that the solution becomes −2y(t) for t > 0, we need to

A

change the initial condition to −y(0) and the forcing function to 2x(t)

B

change the initial condition to 2y(0) and the forcing function to −x(t)

C

change the initial condition to j√2⋅y(0) and the forcing function to j√2⋅x(t)

D

change the initial condition to −2y(0) and the forcing function to −2x(t)

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17

2013 · Electronics & Communication Engineering · Networks, Signals and Systems · LTI systems

Electronics & Communication Engineering (EC) 2013 [Session 4]

The impulse response of a continuous time system is given by \(h(t) = \delta(t-1) + \delta(t-3)\). The value of the step response at \(t=2\) is

A

0

B

1

C

2

D

3

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18

2014 · Electronics & Communication Engineering · Networks, Signals and Systems · LTI systems

Electronics & Communication Engineering (EC) 2014 [Session 2]

The input-output relationship of a causal stable LTI system is given as $y[n] = \alpha y[n-1] + \beta x[n]$. If the impulse response $h[n]$ of this system satisfies the condition $\sum_{n=0}^{\infty} h[n] = 2$, the relationship between $\alpha$ and $\beta$ is

A

$\alpha = 1 - \beta/2$

B

$\alpha = 1 + \beta/2$

C

$\alpha = 2\beta$

D

$\alpha = -2\beta$

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19

2014 · Electronics & Communication Engineering · Networks, Signals and Systems · LTI systems

Electronics & Communication Engineering (EC) 2014 [Session 2]

An unforced linear time invariant (LTI) system is represented by \[\begin{bmatrix} \dot{x}_1 \\ \dot{x}_2 \end{bmatrix} = \begin{bmatrix} -1 & 0 \\ 0 & -2 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \end{bmatrix}\]. If the initial conditions are \(x_1(0) = 1\) and \(x_2(0) = -1\), the solution of the state equation is

A

\(x_1(t) = -1, x_2(t) = 2\)

B

\(x_1(t) = -e^{-t}, x_2(t) = 2e^{-t}\)

C

\(x_1(t) = e^{-t}, x_2(t) = -e^{-2t}\)

D

\(x_1(t) = -e^{-t}, x_2(t) = -2e^{-t}\)

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20

2014 · Electronics & Communication Engineering · Networks, Signals and Systems · LTI systems

Electronics & Communication Engineering (EC) 2014 [Session 3]

Let \(h(t)\) denote the impulse response of a causal system with transfer function \(\frac{1}{s+1}\). Consider the following three statements: S1: The system is stable. S2: \(\frac{h(t+1)}{h(t)}\) is independent of \(t\) for \(t > 0\). S3: A non-causal system with the same transfer function is stable. For the above system,

A

only S1 and S2 are true

B

only S2 and S3 are true

C

only S1 and S3 are true

D

S1, S2 and S3 are true

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