Time Shifting MCQ Quiz - Objective Question with Answer for Time Shifting - Download Free PDF

The correct option is 1

Concept:

If the Laplace transform of a function \( x(t) \) is known, then the Laplace transform of \( e^{-at}x(t) \) can be obtained using the time-shifting property of the Laplace transform.

This property states that:

\( \mathcal{L}\{e^{-at}x(t)\} = X(s + a) \), where \( X(s) = \mathcal{L}\{x(t)\} \)

Given:

\( \mathcal{L}\{x(t)\} = \sqrt{\frac{2}{s - 3}} \)

Calculation:

We are asked to find the Laplace transform of \( e^{-6t}x(t) \).

Using the time-shifting property:

\( \mathcal{L}\{e^{-6t}x(t)\} = \sqrt{\frac{2}{(s + 6) - 3}} = \sqrt{\frac{2}{s + 3}} \)

Explanation:

The Fourier transform of the signal \(x(t) = e^{-a|t|}\) can be calculated as follows:

Step 1: Definition of Fourier Transform

The Fourier transform of a function \(x(t)\) is defined as:

\[ X(j\omega) = \int_{-\infty}^{\infty} x(t) e^{-j\omega t} dt \]

For the given signal \(x(t) = e^{-a|t|}\), we need to consider the absolute value function. Therefore, the signal can be written as:

\[ x(t) = \begin{cases} e^{at} & \text{for } t < 0 \\ e^{-at} & \text{for } t \ge 0 \end{cases} \]

Step 2: Split the Integral

We can split the integral into two parts: one for \(t < 0\) and one for \(t \ge 0\):

\[ X(j\omega) = \int_{-\infty}^{0} e^{at} e^{-j\omega t} dt + \int_{0}^{\infty} e^{-at} e^{-j\omega t} dt \]

Simplifying the exponents inside the integrals, we get:

\[ X(j\omega) = \int_{-\infty}^{0} e^{(a - j\omega)t} dt + \int_{0}^{\infty} e^{-(a + j\omega)t} dt \]

Step 3: Evaluate the Integrals

Let's evaluate each integral separately.

For the first integral \( \int_{-\infty}^{0} e^{(a - j\omega)t} dt \):

\[ \int_{-\infty}^{0} e^{(a - j\omega)t} dt = \left[ \frac{e^{(a - j\omega)t}}{a - j\omega} \right]_{-\infty}^{0} \]

Evaluating the limits:

\[ \left[ \frac{e^{(a - j\omega)t}}{a - j\omega} \right]_{-\infty}^{0} = \frac{1}{a - j\omega} - \lim_{t \to -\infty} \frac{e^{(a - j\omega)t}}{a - j\omega} \]

Since \(a > 0\), \(e^{(a - j\omega)t}\) approaches 0 as \(t\) approaches \(-\infty\):

\[ \frac{1}{a - j\omega} - 0 = \frac{1}{a - j\omega} \]

For the second integral \( \int_{0}^{\infty} e^{-(a + j\omega)t} dt \):

\[ \int_{0}^{\infty} e^{-(a + j\omega)t} dt = \left[ \frac{-e^{-(a + j\omega)t}}{a + j\omega} \right]_{0}^{\infty} \]

Evaluating the limits:

\[ \left[ \frac{-e^{-(a + j\omega)t}}{a + j\omega} \right]_{0}^{\infty} = 0 - \left( -\frac{1}{a + j\omega} \right) = \frac{1}{a + j\omega} \]

Step 4: Add the Results

Now, combining both integrals, we get:

\[ X(j\omega) = \frac{1}{a - j\omega} + \frac{1}{a + j\omega} \]

To simplify, find a common denominator:

\[ X(j\omega) = \frac{a + j\omega + a - j\omega}{(a - j\omega)(a + j\omega)} = \frac{2a}{a^2 + \omega^2} \]

Therefore, the Fourier transform of \(x(t) = e^{-a|t|}\) is:

\[ X(j\omega) = \frac{2a}{a^2 + \omega^2} \]

Important Information:

To further understand the analysis, let’s evaluate the other options:

Option 1: \(X(j\omega) = \frac{2a}{a+\omega}\)

This option is incorrect because the denominator should be \(a^2 + \omega^2\), not \(a + \omega\).

Option 3: \(X(j\omega) = \frac{a}{a - j\omega}\)

This option is incorrect because it does not account for the full expression obtained from the Fourier transform, and it lacks the correct form of the denominator.

Option 4: \(X(j\omega) = \frac{2a}{a + j\omega}\)

This option is incorrect because it only partially matches one of the terms derived from the Fourier transform calculation and does not include the complete expression.

Conclusion:

Understanding the Fourier transform process and carefully evaluating each integral's limits and results are essential for correctly identifying the transform of a given signal. In this case, the correct Fourier transform of \(x(t) = e^{-a|t|}\) is \(X(j\omega) = \frac{2a}{a^2 + \omega^2}\), making Option 2 the correct choice.

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-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:

y(t) = x(t) + e^jtx(t)

x(t) ↔ X(jω)

e^jtx(t) ↔ X(j(ω - 1))

y(jω) at ω = 0.5 rad/sec = X(jω) + X(j(ω - 1))

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:

y(t) = x(t) + e^jtx(t)

x(t) ↔ X(jω)

e^jtx(t) ↔ X(j(ω - 1))

y(jω) at ω = 0.5 rad/sec = X(jω) + X(j(ω - 1))

The correct option is 1

Concept:

If the Laplace transform of a function \( x(t) \) is known, then the Laplace transform of \( e^{-at}x(t) \) can be obtained using the time-shifting property of the Laplace transform.

This property states that:

\( \mathcal{L}\{e^{-at}x(t)\} = X(s + a) \), where \( X(s) = \mathcal{L}\{x(t)\} \)

Given:

\( \mathcal{L}\{x(t)\} = \sqrt{\frac{2}{s - 3}} \)

Calculation:

We are asked to find the Laplace transform of \( e^{-6t}x(t) \).

Using the time-shifting property:

\( \mathcal{L}\{e^{-6t}x(t)\} = \sqrt{\frac{2}{(s + 6) - 3}} = \sqrt{\frac{2}{s + 3}} \)

Explanation:

The Fourier transform of the signal \(x(t) = e^{-a|t|}\) can be calculated as follows:

Step 1: Definition of Fourier Transform

The Fourier transform of a function \(x(t)\) is defined as:

\[ X(j\omega) = \int_{-\infty}^{\infty} x(t) e^{-j\omega t} dt \]

For the given signal \(x(t) = e^{-a|t|}\), we need to consider the absolute value function. Therefore, the signal can be written as:

\[ x(t) = \begin{cases} e^{at} & \text{for } t < 0 \\ e^{-at} & \text{for } t \ge 0 \end{cases} \]

Step 2: Split the Integral

We can split the integral into two parts: one for \(t < 0\) and one for \(t \ge 0\):

\[ X(j\omega) = \int_{-\infty}^{0} e^{at} e^{-j\omega t} dt + \int_{0}^{\infty} e^{-at} e^{-j\omega t} dt \]

Simplifying the exponents inside the integrals, we get:

\[ X(j\omega) = \int_{-\infty}^{0} e^{(a - j\omega)t} dt + \int_{0}^{\infty} e^{-(a + j\omega)t} dt \]

Step 3: Evaluate the Integrals

Let's evaluate each integral separately.

For the first integral \( \int_{-\infty}^{0} e^{(a - j\omega)t} dt \):

\[ \int_{-\infty}^{0} e^{(a - j\omega)t} dt = \left[ \frac{e^{(a - j\omega)t}}{a - j\omega} \right]_{-\infty}^{0} \]

Evaluating the limits:

\[ \left[ \frac{e^{(a - j\omega)t}}{a - j\omega} \right]_{-\infty}^{0} = \frac{1}{a - j\omega} - \lim_{t \to -\infty} \frac{e^{(a - j\omega)t}}{a - j\omega} \]

Since \(a > 0\), \(e^{(a - j\omega)t}\) approaches 0 as \(t\) approaches \(-\infty\):

\[ \frac{1}{a - j\omega} - 0 = \frac{1}{a - j\omega} \]

For the second integral \( \int_{0}^{\infty} e^{-(a + j\omega)t} dt \):

\[ \int_{0}^{\infty} e^{-(a + j\omega)t} dt = \left[ \frac{-e^{-(a + j\omega)t}}{a + j\omega} \right]_{0}^{\infty} \]

Evaluating the limits:

\[ \left[ \frac{-e^{-(a + j\omega)t}}{a + j\omega} \right]_{0}^{\infty} = 0 - \left( -\frac{1}{a + j\omega} \right) = \frac{1}{a + j\omega} \]

Step 4: Add the Results

Now, combining both integrals, we get:

\[ X(j\omega) = \frac{1}{a - j\omega} + \frac{1}{a + j\omega} \]

To simplify, find a common denominator:

\[ X(j\omega) = \frac{a + j\omega + a - j\omega}{(a - j\omega)(a + j\omega)} = \frac{2a}{a^2 + \omega^2} \]

Therefore, the Fourier transform of \(x(t) = e^{-a|t|}\) is:

\[ X(j\omega) = \frac{2a}{a^2 + \omega^2} \]

Important Information:

To further understand the analysis, let’s evaluate the other options:

Option 1: \(X(j\omega) = \frac{2a}{a+\omega}\)

This option is incorrect because the denominator should be \(a^2 + \omega^2\), not \(a + \omega\).

Option 3: \(X(j\omega) = \frac{a}{a - j\omega}\)

This option is incorrect because it does not account for the full expression obtained from the Fourier transform, and it lacks the correct form of the denominator.

Option 4: \(X(j\omega) = \frac{2a}{a + j\omega}\)

This option is incorrect because it only partially matches one of the terms derived from the Fourier transform calculation and does not include the complete expression.

Conclusion:

Understanding the Fourier transform process and carefully evaluating each integral's limits and results are essential for correctly identifying the transform of a given signal. In this case, the correct Fourier transform of \(x(t) = e^{-a|t|}\) is \(X(j\omega) = \frac{2a}{a^2 + \omega^2}\), making Option 2 the correct choice.

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-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:

From the given pictures of x(t) and y(t)

We get,

y(t) = - x(2t + 2)

y(t) is time scaled and time shifted version of x(t)

Step 1:

If x(t) ↔ X(ω)

Then, x(t + 2) ↔ e^j.2.ω X(ω)

Step 2:

Using time shifting property

x(2t + 2) ↔ ½ e^jω X(ω/2)

Step 3:

Using time scaling property:

\( \;Y\left( \omega \right) = - \frac{1}{2}{e^{j\omega }}X\left( {\frac{\omega }{2}} \right)\)

\(\rm y(t)=x(t)*h(t)\)

Taking the Fourier Transform, we get:

\(\rm Y(f)=X(f)H(f)\)

\(Y(f)\rm = X\left( f \right){e^{ - j4\pi f}} \)

Taking the inverse Fourier Transform of Y(f), we get:

\(\rm y(t)= x(t-2)\)

Concept:

Time shifting property of Fourier series states that:

\(If\;x\left( t \right)\mathop \leftrightarrow \limits^{C.T.F.S} {c_n}\)

\(x\left( {t - {t_0}} \right)\mathop \leftrightarrow \limits^{C.T.F.S} {c_n}{e^{ - jn{\omega _0}{t_0}}}\)

Since \({\omega _0} = \frac{{2π }}{{{T_0}}}\), the above expression can be written as:

\(x\left( {t - {t_0}} \right)\mathop \leftrightarrow \limits^{C.T.F.S} {c_n}{e^{ - jn \cdot \frac{{2π }}{{{T_0}}} \cdot {t_0}}}\)

Application:

Observing the two figures, we can write:

y(t) = x(t - 1)

\(y\left( t \right)\mathop \leftrightarrow \limits^{C.T.F.S} {c_n}{e^{ - jn \cdot \frac{{2π \left( 1 \right)}}{{{T_0}}}}}\)

Where c_n = Fourier series coefficient of x(t)

Since, T₀ = 2

\(y\left( t \right)\mathop \leftrightarrow \limits^{C.T.F.S} {c_n}{e^{ - jnπ }} = c_n'\)

\(c_n' = {c_n}{e^{ - jπ n}}\)

Using the above expression, we get:

\(c_0' = {c_0} = \frac{1}{π }\)

\(c_1' = - {c_1} = \frac{j}{4}\)

Also for even values of n, e^-jπn = 1

\(\therefore c_n' = {c_n} = \frac{1}{{π \left( {1 - {n^2}} \right)}}\)

Let \({{\rm{x}}_1}\left( {\rm{t}} \right){\rm{}} = {\rm{x}}\left( {{\rm{t}} - {\rm{}}1} \right)\)

The \({{\rm{x}}_1}\left( {\rm{t}} \right)\mathop \leftrightarrow \limits^{{\rm{FT}}} {{\rm{X}}_1}\left( {{\rm{j\omega }}} \right) = {\rm{X}}\left( {{\rm{j\omega }}} \right){{\rm{e}}^{ - {\rm{j\omega }}}}\)

Now,

Let \({{\rm{x}}_2}\left( {\rm{t}} \right) = \frac{{{{\rm{d}}^2}{{\rm{x}}_1}\left( {\rm{t}} \right)}}{{{\rm{d}}{{\rm{t}}^2}}} = \frac{{{{\rm{d}}^2}{\rm{x}}\left( {{\rm{t}} - 1} \right)}}{{{\rm{d}}{{\rm{t}}^2}}}\)

Taking Fourier transform

\(\begin{array}{l} {{\rm{X}}_2}\left( {{\rm{j\omega }}} \right){\rm{}} = {\rm{}}{\left( {{\rm{j\omega }}} \right)^2}{\rm{}}{{\rm{X}}_1}\left( {{\rm{j\omega }}} \right)\\ \Rightarrow {{\rm{X}}_2}\left( {{\rm{j\omega }}} \right) = - {{\rm{\omega }}^2}{\rm{X}}\left( {{\rm{j\omega }}} \right){{\rm{e}}^{ - {\rm{j\omega }}}} \end{array}\)

Thus,