Introduction To Signals and Systems MCQ Quiz - Objective Question with Answer for Introduction To Signals and Systems - 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:

The energy of a continuous-time signal x(t) is given by: \( E = \int_{-\infty}^{\infty} |x(t)|^2 dt \)

Given:

\( x(t) = \begin{cases} 5, & -2 \leq t \leq 2 \\ 0, & \text{otherwise} \end{cases} \)

Calculation:

Since \( x(t) = 5 \) from \( t = -2 \) to \( t = 2 \), we compute the energy as:

\( E = \int_{-2}^{2} |5|^2 dt = \int_{-2}^{2} 25 \, dt = 25 \times (2 - (-2)) = 25 \times 4 = 100 \)

The correct answer is: 'Option 3: i, ii, and iii only'.

Key Points

• i. E(ui) = 0
  • This assumption is correct.
  • The assumption that the expected value of the error term E(ui)=0" id="MathJax-Element-3-Frame" role="presentation" style="position: relative;" tabindex="0">E(ui)=0E(ui)=0 $E(ui) = 0$ ensures that the OLS estimators are unbiased.
  • If this condition holds, it implies that the error term does not systematically overestimate or underestimate the dependent variable, leading to unbiased parameter estimates.
  • Hence, this assumption is necessary for consistency, unbiasedness, and efficiency.
• ii. Var(ui) = σ²
  • This assumption is correct.
  • The assumption that the variance of the error term is constant Var(ui)=σ²" id="MathJax-Element-4-Frame" role="presentation" style="position: relative;" tabindex="0">Var(ui)=σ²Var(ui)=σ² $Var(ui) = σ²$ is critical for the efficiency of the OLS estimator.
  • This condition, known as homoscedasticity, ensures that the variability of the errors does not change with the level of the independent variable, which is essential for accurate hypothesis testing and confidence intervals.
  • Thus, this assumption is necessary for the OLS estimator to be efficient.
• iii. Cov(ut, ut-j) = 0 ∀ j
  • This assumption is correct.
  • The assumption that the error terms are uncorrelated over time Cov(ut,ut−j)=0∀j" id="MathJax-Element-5-Frame" role="presentation" style="position: relative;" tabindex="0">Cov(ut,ut−j)=0∀jCov(ut,ut−j)=0∀j $Cov(ut, ut-j) = 0 ∀ j$ is necessary for ensuring that the OLS estimators are consistent.
  • This condition indicates that there is no autocorrelation in the error terms, which is vital for the reliability of the OLS estimators.
  • Therefore, this assumption is necessary for consistency and efficiency of the estimator.
• iv. ut ∩ N(0, σ²)
  • This assumption is incorrect.
  • While normality of the error term is a common assumption for inference purposes (such as constructing confidence intervals), it is not strictly required for the consistency, unbiasedness, or efficiency of the OLS estimators.
  • The OLS estimators can still be consistent and efficient even if the error terms are not normally distributed, provided that the other assumptions hold.
  • Hence, this assumption is not necessary.

Additional Information

• Understanding OLS Estimator Assumptions:
  • Unbiasedness: An estimator is unbiased if its expected value equals the true parameter value. This requires the assumption E(ui)=0" id="MathJax-Element-6-Frame" role="presentation" style="position: relative;" tabindex="0">E(ui)=0E(ui)=0 $E(ui) = 0$ .
  • Efficiency: An estimator is efficient if it has the smallest variance among all unbiased estimators. This requires constant variance Var(ui)=σ²" id="MathJax-Element-7-Frame" role="presentation" style="position: relative;" tabindex="0">Var(ui)=σ²Var(ui)=σ² $Var(ui) = σ²$ .
  • Consistency: An estimator is consistent if it converges in probability to the true parameter value as the sample size increases, which is supported by the assumption of no autocorrelation in the error terms.

CONCEPT:

Radiation Energy Comparison

• The power radiated by a source is distributed over the surface area of a sphere centered at the source. The power per unit area (intensity) decreases with the square of the distance from the source.
• The formula for intensity (I) at a distance (r) from a source with power (P) is given by:

  I = P / (4πr²)

EXPLANATION:

• For the cellphone tower:
  • Power, P1 = 1 W
  • Distance from the head, r1 = 100 m
  • Intensity at the head, I1 = P1 / (4πr1²)
  • = 1 W / (4π(100 m)²)
  • = 1 / (4π * 10,000)
  • = 1 / 125,663.7 W/m²
  • = 7.96 x 10⁻⁶ W/m²
• For the handset:
  • Power, P2 = 0.1 mW = 0.1 x 10⁻³ W
  • Distance from the head, r2 ≈ 0.01 m (assuming the handset is very close to the ear)
  • Intensity at the head, I2 = P2 / (4πr2²)
  • = 0.1 x 10⁻³ W / (4π(0.01 m)²)
  • = 0.1 x 10⁻³ / (4π * 0.0001)
  • = 0.1 x 10⁻³ / 0.001256637
  • = 0.0796 W/m²
• Comparing the intensities:
  • I2 ≈ 0.0796 W/m²
  • I1 ≈ 7.96 x 10⁻⁶ W/m²
  • I2 >> I1

Therefore, the radiation energy received by your head from the handset (E2) is much greater than that from the tower (E1).

Concept:

1. Dynamic System: System that can depend on the present, past, and future values of the input.

Ex.  y(n) = x(n+.5)+x(2n)

If any shifting or scaling property is applied to the input to get the output of the system, the system will become Dynamic.

or

any disturbance in time will lead to the system as dynamic.

2. Static System: System that depends only on present values of the input.

Ex. y(n) = x(n) + kx(n) + nx(n)

3. Causal System: System that depends only on the present and past values of the input.

Ex. y(n) = 5x(n - 2) + nx(n - 0.2)

4. Non-causal System: System that depends on future and present values of the input.

Ex. y(n) = 3x(n) + Kx(n + 10)

5. Anti-causal depends only on the future values of the input.

Ex. y(n)= Kx(n + 10)

Calculation:

Given,

\(y(n) = \frac 1 3 (x[n] + x[n - 1] + x[n + 2])\)

X[n]: Present value of the input

X[n - 1]: Past value of the input

X[n + 2]: Future value of the input

So, the given system is dynamic and non-causal.

Concept:

Shifting property of impulse function

\(\rm\displaystyle\int_{-\infty}^{\infty} x(t) \delta (t -a ) dt = x(a)\)

Scaling property of impulse function

\(\delta (at) = \frac{1}{|a|)} \delta (t)\)

Explanation:

Let:

\(\rm I =\displaystyle\int_{-\infty}^{\infty}e^{-t} \delta(2t - 2) dt\)

\( I = \rm\displaystyle\int_{-\infty}^{\infty}e^{-t} \delta[2(t - 1)] dt\)

Using scaling property of impulse function in the above equation, we'll get:

\( I =\rm\displaystyle\int_{-\infty}^{\infty}e^{-t} \frac{1}{|2|}\delta(t - 1) dt\)

Applying Shifting property of impulse function to the above equation, we'll get:

\( I =\frac{1}{2}\rm\displaystyle\int_{-\infty}^{\infty}e^{-t} \delta(t - 1) dt\)

\(I = \frac{1}{2}. \left. e^{-t} \right|_{t = 1}\)

\(\frac{1}{{2e}}\)

Given, the signal

V (t) = 30 sin 100t + 10 cos 300 t + 6 sin (500t+π/4)

So, we have

ω₁ = 100 rads

ω₂ = 300 rads

ω₃ = 500 rads

∴ The respective time periods are

\(\begin{array}{l} {T_1} = \frac{{2\pi }}{{{\omega _1}}} = \frac{{2\pi }}{{100}}sec\\ {T_1} = \frac{{2\pi }}{{{\omega _2}}} = \frac{{2\pi }}{{300}}sec\\ {T_3} = \frac{{2\pi }}{{500}}sec \end{array}\)

So, the fundamental time period of the signal is

\(LCM\left( {{T_1},{T_2},{T_3}} \right) = \frac{{LCM\left( {2\pi ,2\pi ,2\pi } \right)}}{{HCF\left( {100,\ 300,\ 500} \right)}}\)

as \({T_0} = \frac{{2\pi }}{{100}}\)

∴ The fundamental frequency, \({\omega _0} = \frac{{2\pi }}{{{T_0}}} = 100\ rad/s\)

Concept:

Any given signal x(t) can be written as the sum of its even part and odd part, i.e.

x(t) = xe(t) + xo(t)

xe(t) = Even part of g(t) calculated as:

\(x_e(t)=\frac{x(t)+x(-t)}{2}\)

xo(t) = Odd part of x(t) calculated as:

\(x_o(t)=\frac{x(t)-x(-t)}{2}\)

Calculation:

Given:

x(t) = (2 + sin t)²

x(t) = 4 + sin²t + 4 sin t

x(-t) = 4 + sin2t - 4 sin t

Even part of g(t) calculated as:

\(x_e(t)=\frac{x(t)+x(-t)}{2}\)

\(=\frac{4+sin^2⁡t+4 sin⁡t+4+sin^2⁡t-4 sin⁡t }{2}\)

= 4 + sin2t

Odd part of x(t) calculated as:

\(x_o(t)=\frac{x(t)-x(-t)}{2}\)\(\)

\(=\frac{4+sin^2⁡t+4 sin⁡t-4-sin^2⁡t+4 sin⁡t }{2}\)

= 4 sin t

Concept:

Time-shifting property: When a signal is shifted in time domain it is said to be delayed or advanced based on whether the signal is shifted to the right or left. 

For example:

Time scaling property: A signal is scaled in the time domain with the scaling factor 'a'. 

If a > 1, then the signal is contracted by a factor of 'a' along the time axis. 

if a < 1, then the signal is expanded by a factor of 'a' along the time axis. 

For example:

Analysis:

We can observe that if we scale f(t) by a factor of ½ and then shift, we will get g(t).

First scale f(t) by a factor of ½ ,

g₁(t) = f (t/2)

Shift g₁(t) by 3

\(g\left( t \right) = g_1\left( {t - 3} \right) = f\left( {\frac{{t - 3}}{2}} \right)\)

\(g\left( t \right) = f\left( {\frac{t}{2} - \frac{3}{2}} \right)\)

Concept:

Time-shifting property: When a signal is shifted in the time domain it is said to be delayed or advanced based on whether the signal is shifted to the right or left. 

For example:

Time scaling property: A signal is scaled in the time domain with the scaling factor 'a'. 

If a > 1, then the signal is contracted by a factor of 'a' along the time axis. 

if a < 1, then the signal is expanded by a factor of 'a' along the time axis. 

For example:

Analysis:

Whenever there is a combination of operations care should be taken while following the sequence of operations.

Correct approach:

x[2(t - 1.5)]

First, time scale the signal by factor 2.

Then, time shift the signal by 1.5 

Incorrect approach:

x[2(t - 1.5)]

First, time shift the signal by 1.5

Then, time scale the signal by 2.

or, 

Correct approach:

x[2t - 3]

First, time shift the signal by 3.

Then, time scale the signal by 2

Incorrect approach:

x[2t - 3]

First, time scale the signal by 2

Then, time shift the signal by 3.

Given that, y[n] = (1 + (-1)ⁿ) x [n]

y[n - n₀] = [1 + (-1)^{n - n0}] x[n - n₀]        ----(1)

y[n] = (1 + (-1)ⁿ) x[n - n₀]       ----(2)

Both the equations 1 and 2 are not equal. Hence y[n] is dependent on time. It is time variant.

For invertible systems, for each unique input x[n], there should be unique output y[n].

If x[n] = δ [n - 1]

y[n] = (1 + (-1)ⁿ) δ [n - 1]

y[1] = 0

If x[n] = k δ [n - 1]

y [n] = [1 + (-1)ⁿ] k δ [n - 1]

y[1] = 0

Unit impulse function:

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

The discrete-time version of the unit impulse is defined by

\(\delta \left[ n \right] = \left\{ {\begin{array}{*{20}{c}} {1,\;\;n = 0}\\ {0,\;\;n \ne 0} \end{array}} \right.\)

Properties:

1. \(\mathop \smallint \limits_{ - \infty }^\infty \delta \left( t \right)dt = 1\)

2. \(\delta \left( {at} \right) = \frac{1}{{\left| a \right|}}\delta \left( t \right)\)

3. x(t) δ(t – t0) = x(t0)

4. \(\mathop \smallint \limits_{ - \infty }^\infty x\left( t \right)\delta \left( {t - {t_o}} \right)dt = x\left( {{t_0}} \right)\)

5. \(\mathop \smallint \limits_{ - \infty }^\infty f\left( t \right)\delta \left( {at + b} \right)dt = \mathop \smallint \limits_{ - \infty }^\infty f\left( t \right)\frac{1}{{\left| a \right|}}\delta \left( {t + \frac{b}{a}} \right)dt\)

6. \(\mathop \smallint \limits_{ - \infty }^\infty x\left( t \right){\delta ^n}\left( {t - {t_o}} \right)dt = {\left. {\frac{{{d^n}x}}{{d{t^n}}}} \right|_{t = {t_0}}}\)

Calculation:

\(\mathop \smallint \limits_{ - \infty }^\infty {e^{ - t}}\delta \left( {2t - 2} \right)dt\)

\( = \frac{1}{2}\mathop \smallint \limits_{ - \infty }^\infty {e^{ - t}}\delta \left( {t - 1} \right)dt\)

\( = \frac{1}{2}{e^{ - 1}} = \frac{1}{{2e}}\)

Unit impulse function:

It is defined as

Properties:

1. \(\mathop \smallint \limits_{ - \infty }^\infty \delta \left( t \right)dt = 1\)

2. \(\delta \left( {at} \right) = \frac{1}{{\left| a \right|}}\delta \left( t \right)\)

3. x(t) δ(t – t0) = x(t0)

4. \(\mathop \smallint \limits_{ - \infty }^\infty x\left( t \right)\delta \left( {t - {t_o}} \right)dt = x\left( {{t_0}} \right)\)

5. \(\mathop \smallint \limits_{ - \infty }^\infty f\left( t \right)\delta \left( {at + b} \right)dt = \mathop \smallint \limits_{ - \infty }^\infty f\left( t \right)\frac{1}{{\left| a \right|}}\delta \left( {t + \frac{b}{a}} \right)dt\)

6. \(\mathop \smallint \limits_{ - \infty }^\infty x\left( t \right){\delta ^n}\left( {t - {t_o}} \right)dt = {\left. {\frac{{{d^n}x}}{{d{t^n}}}} \right|_{t = {t_0}}}\)

The discrete time version of the unit impulse is defined by

\(\delta \left[ n \right] = \left\{ {\begin{array}{*{20}{c}} {1,\;\;n = 0}\\ {0,\;\;n \ne 0} \end{array}} \right.\)

Properties:

1. x[n] δ[n] = x[0] δ[n]

2. \(\mathop \sum \limits_{n = - \infty }^{n = \infty } x\left[ n \right]\delta \left[ n \right] = \mathop \sum \limits_{n = - \infty }^{n = \infty } x\left[ 0 \right]\delta \left[ n \right] = x\left[ 0 \right]\)

3. x[n] δ[n – n0] = x[n0] δ[n – n0]

4. \(\mathop \sum \limits_{n = - \infty }^{n = \infty } x\left[ n \right]\delta \left[ {n - {n_0}} \right] = \mathop \sum \limits_{n = - \infty }^{n = \infty } x\left[ {{n_0}} \right]\delta \left[ {n - {n_0}} \right] = x\left[ {{n_0}} \right]\)

5. δ[an] = δ[n]

Concept:

Sampling property of impulse function is:

\(\mathop \smallint \limits_{{t_1}}^{{t_2}} x\left( t \right)\delta \left( {t - {t_0}} \right) = x\left( {{t_0}} \right)\;\;;\;\;{t_1} < {t_0} < {t_2}\) 

Unit ramp function is defined as:

\(r\left( t \right) = \left\{ {\begin{array}{*{20}{c}} {t\;\;;\;\;\;t > 0}\\ {0;\;\;\;\;t < 0} \end{array}} \right\}\) 

Calculation:

Given:

x(t) = 2r(t) – 2r (t - 2) – 4u(t - 4)

from sampling property;

\(\mathop \smallint \limits_{ - \infty }^{ + \infty } x\left( t \right)\delta \left( {t - 3} \right) = x\left( 3 \right)\) 

So,

x(3) = 2r(3) – 2r(1) – 4u(-1)

x(3) = 2 × 3 – 2 × 1 – 4 × 0

x(3) = 4

Odd Signals:

• A signal x(t) is said to be an odd signal if it satisfies the condition:

          x(-t) = - x(t) for all t (Continuous Time)

          x(-n) = - x(n) for all n (Discrete Time)

• An odd signal is anti symmetrical about the vertical axis.

Analysis:

Fro the given signal, we can write:

x[-n] = - x[n]

∴ The given signal x[n] is an odd signal.

Examples of odd signals:

Even Signals:

• Any signal x(t) is said to be an even signal if it satisfies the condition:

          x(-t) = x(t) for all t (Continuous Time)

          x(-n) = x(n) for all n (Discrete Time)

• An even signal is identical to its time-reversed signal, i.e. it is symmetrical around the vertical axis.

Examples of even signals: