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

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.

\(u\left( t \right)\mathop \to \limits^{t = 2t} u\left( {2t} \right)\mathop \to \limits^{t = - t} u\left( { - 2t} \right)\)

∴ From above explanation:

We can conclude: x₁(t) = x₃(t)

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

Concept:

Unit step signal:

It is defined as:

\(u\left( t \right) = \left\{ {\begin{array}{*{20}{c}} {1,\;\;t \ge 0}\\ {0,\;\;t < 0} \end{array}} \right.\)

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

Concept:

Unit impulse signal:

Diagram

It is defined as, \(\delta \left( t \right) = \left\{ {\begin{array}{*{20}{c}} {\infty ,\;\;x = 0}\\ {0,\;\;x \ne 0} \end{array}} \right.\)

Unit step signal:

Diagram

It is defined as, \(u\left( t \right) = \left\{ {\begin{array}{*{20}{c}} {1,\;\;x \ge 0}\\ {0,\;\;x < 0} \end{array}} \right.\)

Unit ramp signal:

Diagram

It is defined as, \(r\left( t \right) = \left\{ {\begin{array}{*{20}{c}} {t,\;\;x \ge 0}\\ {0,\;\;x < 0} \end{array}} \right.\)

Unit parabola signal:

Diagram

It is defined as, \(x\left( t \right) = \left\{ {\begin{array}{*{20}{c}} {\frac{{{t^2}}}{2},\;\;x \ge 0}\\ {0,\;\;x < 0} \end{array}} \right.\)

The relation between these signals is given below.

\(r\left( t \right) = \frac{d}{{dt}}\left( {x\left( t \right)} \right),\;u\left( t \right) = \frac{d}{{dt}}\left( {r\left( t \right)} \right),\;\delta \left( t \right) = \frac{d}{{dt}}\left( {u\left( t \right)} \right)\)

\(u\left( t \right) = \smallint \delta \left( t \right),\;r\left( t \right) = \smallint u\left( t \right),\;x\left( t \right) = \smallint r\left( t \right)\)

Application:

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

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

Concept:

Unit impulse signal:

Diagram

It is defined as, \(\delta \left( t \right) = \left\{ {\begin{array}{*{20}{c}} {\infty ,\;\;x = 0}\\ {0,\;\;x \ne 0} \end{array}} \right.\)

Unit step signal:

Diagram

It is defined as, \(u\left( t \right) = \left\{ {\begin{array}{*{20}{c}} {1,\;\;x \ge 0}\\ {0,\;\;x < 0} \end{array}} \right.\)

Unit ramp signal:

Diagram

It is defined as, \(r\left( t \right) = \left\{ {\begin{array}{*{20}{c}} {t,\;\;x \ge 0}\\ {0,\;\;x < 0} \end{array}} \right.\)

Unit parabola signal:

Diagram

It is defined as, \(x\left( t \right) = \left\{ {\begin{array}{*{20}{c}} {\frac{{{t^2}}}{2},\;\;x \ge 0}\\ {0,\;\;x < 0} \end{array}} \right.\)

The relation between these signals is given below.

\(r\left( t \right) = \frac{d}{{dt}}\left( {x\left( t \right)} \right),\;u\left( t \right) = \frac{d}{{dt}}\left( {r\left( t \right)} \right),\;\delta \left( t \right) = \frac{d}{{dt}}\left( {u\left( t \right)} \right)\)

\(u\left( t \right) = \smallint \delta \left( t \right),\;r\left( t \right) = \smallint u\left( t \right),\;x\left( t \right) = \smallint r\left( t \right)\)

Application:

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

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

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)

Concept:

Unit step signal:

It is defined as:

\(u\left( t \right) = \left\{ {\begin{array}{*{20}{c}} {1,\;\;t \ge 0}\\ {0,\;\;t < 0} \end{array}} \right.\)

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

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.

\(u\left( t \right)\mathop \to \limits^{t = 2t} u\left( {2t} \right)\mathop \to \limits^{t = - t} u\left( { - 2t} \right)\)

∴ From above explanation:

We can conclude: x₁(t) = x₃(t)

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

Concept:

Unit impulse signal:

Diagram

It is defined as, \(\delta \left( t \right) = \left\{ {\begin{array}{*{20}{c}} {\infty ,\;\;x = 0}\\ {0,\;\;x \ne 0} \end{array}} \right.\)

Unit step signal:

Diagram

It is defined as, \(u\left( t \right) = \left\{ {\begin{array}{*{20}{c}} {1,\;\;x \ge 0}\\ {0,\;\;x < 0} \end{array}} \right.\)

Unit ramp signal:

Diagram

It is defined as, \(r\left( t \right) = \left\{ {\begin{array}{*{20}{c}} {t,\;\;x \ge 0}\\ {0,\;\;x < 0} \end{array}} \right.\)

Unit parabola signal:

Diagram

It is defined as, \(x\left( t \right) = \left\{ {\begin{array}{*{20}{c}} {\frac{{{t^2}}}{2},\;\;x \ge 0}\\ {0,\;\;x < 0} \end{array}} \right.\)

The relation between these signals is given below.

\(r\left( t \right) = \frac{d}{{dt}}\left( {x\left( t \right)} \right),\;u\left( t \right) = \frac{d}{{dt}}\left( {r\left( t \right)} \right),\;\delta \left( t \right) = \frac{d}{{dt}}\left( {u\left( t \right)} \right)\)

\(u\left( t \right) = \smallint \delta \left( t \right),\;r\left( t \right) = \smallint u\left( t \right),\;x\left( t \right) = \smallint r\left( t \right)\)

Application:

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

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

Concept:

Unit step signal:

It is defined as:

\(u\left( t \right) = \left\{ {\begin{array}{*{20}{c}} {1,\;\;t \ge 0}\\ {0,\;\;t < 0} \end{array}} \right.\)

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

The change in slope at \(\rm{t=-2}\) is \(\rm{3}\). So, the term appearing at \(\rm{t=-2}\) is \(\rm{3r(t+2)}\). The change in slope at \(\rm{t=-1}\) is \(\rm{-3}\). The term appearing at \(\rm{t=-1}\)

is \(\rm{-3r(t+1)}\). The change in slope at \(\rm{t=1}\) is \(\rm{-3}\). The term appearing at \(\rm{t=1}\) is \(\rm{-3r(t-1)}\). The change in slope at \(\rm{t=2}\) is \(\rm{3}\) . Thus, the term appearing at \(\rm{t=2}\) is \(\rm{3r(t-2)}\). Writing all the terms together, we have 

\(\rm{x(t)=3r\left( {t + 2} \right) - 3r\left( {t + 1} \right) - 3r\left( {t - 1} \right) + 3r\left( {t - 2} \right)}\).

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)