Unit Step Signal MCQ Quiz - Objective Question with Answer for Unit Step Signal - Download Free PDF

Last updated on Apr 10, 2025

Latest Unit Step Signal MCQ Objective Questions

Unit Step Signal Question 1:

The equation for voltage waveform v(t) shown below is (where u(t) is unit step input) :

  1. u(t − 1) + u(t − 2) + u(t − 3) 
  2. u(t − 1) + 2u(t − 2) + 3u(t − 3)
  3. u(t − 1) − u(t − 2) − u(t − 3) + 3u(t − 4) 
  4. u(t − 1) +u(t − 2) +u(t − 3) − 3u(t − 4)

Answer (Detailed Solution Below)

Option 4 : u(t − 1) +u(t − 2) +u(t − 3) − 3u(t − 4)

Unit Step Signal Question 1 Detailed Solution

Concept:

The voltage waveform can be expressed using unit step functions (u(t)) to represent the changes in voltage at specific time intervals. Each step in the waveform corresponds to the addition or subtraction of a unit step function at the respective time.

Calculation:

Given:

Voltage waveform with the following characteristics:
- Starts at 0V
- Jumps to 1V at t=1
- Jumps to 2V at t=2
- Jumps to 3V at t=3
- Returns to 0V at t=4

Solution:

1. The waveform can be constructed as:

2. This represents:

  • +1V step at t=1
  • +1V step at t=2 (total 2V)
  • +1V step at t=3 (total 3V)
  • -3V step at t=4 (return to 0V)

 

Final Answer:

The correct equation is 4) u(t − 1) +u(t − 2) +u(t − 3) − 3u(t − 4)

Unit Step Signal Question 2:

For continuous time domain, select the correct property.

  1. u(t) = u(at); a < 0
  2. u(t) = u(at); a > 0
  3. u(t) = u(at + b); a > 0
  4. u(t) = u(at + b); a < 0

Answer (Detailed Solution Below)

Option 2 : u(t) = u(at); a > 0

Unit Step Signal Question 2 Detailed Solution

Concept:

  • Time advance or time delay affects the step signal
  • Time scaling will NOT affect step signal
  • Time reversal will affect step signal

 

Analysis:

Let, unit step signal; x1(t) = u(t)

Now, x2(t) = u(t + 2)

Now, x3(t) = u(2t)

Now,

x4(t) = u(-2t)

i.e.

∴ From above explanation:

We can conclude: x1(t) = x3(t)

Note: In discrete time, u[n] ≠ u[an] for a > 0 or a

Unit Step Signal Question 3:

If u(t) = 1, for t > 0 and u(t) = 0, for t

The value of the function 5 u(t) - 4 u(-t) + 0.8 u(1 - t) at t = 0.8 is:

  1. 3.8
  2. 7.8
  3. 9.8
  4. 5.8

Answer (Detailed Solution Below)

Option 4 : 5.8

Unit Step Signal Question 3 Detailed Solution

Concept:

Unit step signal:

It is defined as:

 

Calculation:

Given that, 

Function f(t) = 5 u(t) - 4 u(-t) + 0.8 u(1 - t)

The value of the function at t = 0.8,

f(0.8) = 5 u(0.8) - 4 u(-0.8) + 0.8 u(0.2)

f(0.8) = 5 - 0 + 0.8 = 5.8

Unit Step Signal Question 4:

Double integration of a unit step function would lead to

  1. an impulse
  2. a parabola
  3. a ramp
  4. a hypercube

Answer (Detailed Solution Below)

Option 2 : a parabola

Unit Step Signal Question 4 Detailed Solution

Concept:

Unit impulse signal:

Diagram

It is defined as, 

Unit step signal:

Diagram

It is defined as, 

Unit ramp signal:

Diagram

It is defined as, 

Unit parabola signal:

Diagram

It is defined as, 

The relation between these signals is given below.

Application:

From the above equations, double integration of a unit step function would lead to parabola.

Unit Step Signal Question 5:

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

  1. 0
  2. 1
  3. 2
  4. 3

Answer (Detailed Solution Below)

Option 2 : 1

Unit Step Signal Question 5 Detailed Solution

h (t) = δ (t – 1) + δ (t – 3)

From convolution property, we get

x(t) * δ (t – t0) = x (t – t0)

Now, for x(t) = u(t)

y(t) = u (t) * h (t)

= u (t) * [ δ (t – 1) + δ (t – 3)]

= u (t -1) + u (t - 3)

At t = 2,

y(2) = u (2 – 1) + u (2 – 3)

= u (1) + u (-1)

= 1 + 0

= 1

Top Unit Step Signal MCQ Objective Questions

Double integration of a unit step function would lead to

  1. an impulse
  2. a parabola
  3. a ramp
  4. a hypercube

Answer (Detailed Solution Below)

Option 2 : a parabola

Unit Step Signal Question 6 Detailed Solution

Download Solution PDF

Concept:

Unit impulse signal:

Diagram

It is defined as, 

Unit step signal:

Diagram

It is defined as, 

Unit ramp signal:

Diagram

It is defined as, 

Unit parabola signal:

Diagram

It is defined as, 

The relation between these signals is given below.

Application:

From the above equations, double integration of a unit step function would lead to parabola.

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

  1. 0
  2. 1
  3. 2
  4. 3

Answer (Detailed Solution Below)

Option 2 : 1

Unit Step Signal Question 7 Detailed Solution

Download Solution PDF

h (t) = δ (t – 1) + δ (t – 3)

From convolution property, we get

x(t) * δ (t – t0) = x (t – t0)

Now, for x(t) = u(t)

y(t) = u (t) * h (t)

= u (t) * [ δ (t – 1) + δ (t – 3)]

= u (t -1) + u (t - 3)

At t = 2,

y(2) = u (2 – 1) + u (2 – 3)

= u (1) + u (-1)

= 1 + 0

= 1

The equation for voltage waveform v(t) shown below is (where u(t) is unit step input) :

  1. u(t − 1) + u(t − 2) + u(t − 3) 
  2. u(t − 1) + 2u(t − 2) + 3u(t − 3)
  3. u(t − 1) − u(t − 2) − u(t − 3) + 3u(t − 4) 
  4. u(t − 1) +u(t − 2) +u(t − 3) − 3u(t − 4)

Answer (Detailed Solution Below)

Option 4 : u(t − 1) +u(t − 2) +u(t − 3) − 3u(t − 4)

Unit Step Signal Question 8 Detailed Solution

Download Solution PDF

Concept:

The voltage waveform can be expressed using unit step functions (u(t)) to represent the changes in voltage at specific time intervals. Each step in the waveform corresponds to the addition or subtraction of a unit step function at the respective time.

Calculation:

Given:

Voltage waveform with the following characteristics:
- Starts at 0V
- Jumps to 1V at t=1
- Jumps to 2V at t=2
- Jumps to 3V at t=3
- Returns to 0V at t=4

Solution:

1. The waveform can be constructed as:

2. This represents:

  • +1V step at t=1
  • +1V step at t=2 (total 2V)
  • +1V step at t=3 (total 3V)
  • -3V step at t=4 (return to 0V)

 

Final Answer:

The correct equation is 4) u(t − 1) +u(t − 2) +u(t − 3) − 3u(t − 4)

Unit Step Signal Question 9:

If u(t) = 1, for t > 0 and u(t) = 0, for t

The value of the function 5 u(t) - 4 u(-t) + 0.8 u(1 - t) at t = 0.8 is:

  1. 3.8
  2. 7.8
  3. 9.8
  4. 5.8

Answer (Detailed Solution Below)

Option 4 : 5.8

Unit Step Signal Question 9 Detailed Solution

Concept:

Unit step signal:

It is defined as:

 

Calculation:

Given that, 

Function f(t) = 5 u(t) - 4 u(-t) + 0.8 u(1 - t)

The value of the function at t = 0.8,

f(0.8) = 5 u(0.8) - 4 u(-0.8) + 0.8 u(0.2)

f(0.8) = 5 - 0 + 0.8 = 5.8

Unit Step Signal Question 10:

For continuous time domain, select the correct property.

  1. u(t) = u(at); a < 0
  2. u(t) = u(at); a > 0
  3. u(t) = u(at + b); a > 0
  4. u(t) = u(at + b); a < 0

Answer (Detailed Solution Below)

Option 2 : u(t) = u(at); a > 0

Unit Step Signal Question 10 Detailed Solution

Concept:

  • Time advance or time delay affects the step signal
  • Time scaling will NOT affect step signal
  • Time reversal will affect step signal

 

Analysis:

Let, unit step signal; x1(t) = u(t)

Now, x2(t) = u(t + 2)

Now, x3(t) = u(2t)

Now,

x4(t) = u(-2t)

i.e.

∴ From above explanation:

We can conclude: x1(t) = x3(t)

Note: In discrete time, u[n] ≠ u[an] for a > 0 or a

Unit Step Signal Question 11:

Double integration of a unit step function would lead to

  1. an impulse
  2. a parabola
  3. a ramp
  4. a hypercube

Answer (Detailed Solution Below)

Option 2 : a parabola

Unit Step Signal Question 11 Detailed Solution

Concept:

Unit impulse signal:

Diagram

It is defined as, 

Unit step signal:

Diagram

It is defined as, 

Unit ramp signal:

Diagram

It is defined as, 

Unit parabola signal:

Diagram

It is defined as, 

The relation between these signals is given below.

Application:

From the above equations, double integration of a unit step function would lead to parabola.

Unit Step Signal Question 12:

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

  1. 0
  2. 1
  3. 2
  4. 3

Answer (Detailed Solution Below)

Option 2 : 1

Unit Step Signal Question 12 Detailed Solution

h (t) = δ (t – 1) + δ (t – 3)

From convolution property, we get

x(t) * δ (t – t0) = x (t – t0)

Now, for x(t) = u(t)

y(t) = u (t) * h (t)

= u (t) * [ δ (t – 1) + δ (t – 3)]

= u (t -1) + u (t - 3)

At t = 2,

y(2) = u (2 – 1) + u (2 – 3)

= u (1) + u (-1)

= 1 + 0

= 1

Unit Step Signal Question 13:

If

u(t) = 1, for t > 0

u(t) = 0, for t

The value of the function 5 u(t) - 4 u(-t) + 0.8 u(1 - t) at t = 0.8 is:

  1. 3.8
  2. 7.8
  3. 9.8
  4. 5.8
  5. 4.2

Answer (Detailed Solution Below)

Option 4 : 5.8

Unit Step Signal Question 13 Detailed Solution

Concept:

Unit step signal:

It is defined as:

Calculation:

given that 

Function f(t) = 5 u(t) - 4 u(-t) + 0.8 u(1 - t)

value of the function at t = 0.8

f(0.8) = 5 u(0.8) - 4 u(-0.8) + 0.8 u(0.2)

f(0.8) = 5 - 0 + 0.8 = 5.8

Unit Step Signal Question 14:

The signal in terms of elementary signal is

Answer (Detailed Solution Below)

Option 4 :

Unit Step Signal Question 14 Detailed Solution

The change in slope at  is . So, the term appearing at  is . The change in slope at  is . The term appearing at 

is . The change in slope at  is . The term appearing at  is . The change in slope at  is  . Thus, the term appearing at  is . Writing all the terms together, we have 

.

Unit Step Signal Question 15:

The equation for voltage waveform v(t) shown below is (where u(t) is unit step input) :

  1. u(t − 1) + u(t − 2) + u(t − 3) 
  2. u(t − 1) + 2u(t − 2) + 3u(t − 3)
  3. u(t − 1) − u(t − 2) − u(t − 3) + 3u(t − 4) 
  4. u(t − 1) +u(t − 2) +u(t − 3) − 3u(t − 4)

Answer (Detailed Solution Below)

Option 4 : u(t − 1) +u(t − 2) +u(t − 3) − 3u(t − 4)

Unit Step Signal Question 15 Detailed Solution

Concept:

The voltage waveform can be expressed using unit step functions (u(t)) to represent the changes in voltage at specific time intervals. Each step in the waveform corresponds to the addition or subtraction of a unit step function at the respective time.

Calculation:

Given:

Voltage waveform with the following characteristics:
- Starts at 0V
- Jumps to 1V at t=1
- Jumps to 2V at t=2
- Jumps to 3V at t=3
- Returns to 0V at t=4

Solution:

1. The waveform can be constructed as:

2. This represents:

  • +1V step at t=1
  • +1V step at t=2 (total 2V)
  • +1V step at t=3 (total 3V)
  • -3V step at t=4 (return to 0V)

 

Final Answer:

The correct equation is 4) u(t − 1) +u(t − 2) +u(t − 3) − 3u(t − 4)

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