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:

$$X(j\omega )={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}x(t)e^{-j\omega t}dt$$

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

$$x(t)=\{\begin{array}{ll}{e}^{at}& \text{for }t<0\\ e^{-at}& \text{for }t\ge 0\end{array}$$

Step 2: Split the Integral

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

$$X(j\omega )={\int }_{-\mathrm{\infty }}^0{e}^{at}{e}^{-j\omega t}dt+{\int }_0^{\mathrm{\infty }}{e}^{-at}{e}^{-j\omega t}dt$$

Simplifying the exponents inside the integrals, we get:

$$X(j\omega )={\int }_{-\mathrm{\infty }}^0{e}^{(a-j\omega )t}dt+{\int }_0^{\mathrm{\infty }}{e}^{-(a+j\omega )t}dt$$

Step 3: Evaluate the Integrals

Let's evaluate each integral separately.

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

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

Evaluating the limits:

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

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

$$\frac{1}{a-j\omega }-0=\frac{1}{a-j\omega }$$

For the second integral ${\int }_0^{\mathrm{\infty }}{e}^{-(a+j\omega )t}dt$:

$${\int }_0^{\mathrm{\infty }}{e}^{-(a+j\omega )t}dt={\left[\frac{-e^{-(a+j\omega )t}}{a+j\omega }\right]}_0^{\mathrm{\infty }}$$

Evaluating the limits:

$${\left[\frac{-e^{-(a+j\omega )t}}{a+j\omega }\right]}_0^{\mathrm{\infty }}=0-(-\frac{1}{a+j\omega })=\frac{1}{a+j\omega }$$

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:

$$X(j\omega )={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}x(t)e^{-j\omega t}dt$$

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

$$x(t)=\{\begin{array}{ll}{e}^{at}& \text{for }t<0\\ e^{-at}& \text{for }t\ge 0\end{array}$$

Step 2: Split the Integral

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

$$X(j\omega )={\int }_{-\mathrm{\infty }}^0{e}^{at}{e}^{-j\omega t}dt+{\int }_0^{\mathrm{\infty }}{e}^{-at}{e}^{-j\omega t}dt$$

Simplifying the exponents inside the integrals, we get:

$$X(j\omega )={\int }_{-\mathrm{\infty }}^0{e}^{(a-j\omega )t}dt+{\int }_0^{\mathrm{\infty }}{e}^{-(a+j\omega )t}dt$$

Step 3: Evaluate the Integrals

Let's evaluate each integral separately.

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

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

Evaluating the limits:

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

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

$$\frac{1}{a-j\omega }-0=\frac{1}{a-j\omega }$$

For the second integral ${\int }_0^{\mathrm{\infty }}{e}^{-(a+j\omega )t}dt$:

$${\int }_0^{\mathrm{\infty }}{e}^{-(a+j\omega )t}dt={\left[\frac{-e^{-(a+j\omega )t}}{a+j\omega }\right]}_0^{\mathrm{\infty }}$$

Evaluating the limits:

$${\left[\frac{-e^{-(a+j\omega )t}}{a+j\omega }\right]}_0^{\mathrm{\infty }}=0-(-\frac{1}{a+j\omega })=\frac{1}{a+j\omega }$$

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:

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

$\mathrm{y}(\mathrm{t})=\mathrm{x}(\mathrm{t})\ast \mathrm{h}(\mathrm{t})$

Taking the Fourier Transform, we get:

$\mathrm{Y}(\mathrm{f})=\mathrm{X}(\mathrm{f})\mathrm{H}(\mathrm{f})$

$Y(f)=\mathrm{X}\left(\mathrm{f}\right){\mathrm{e}}^{-\mathrm{j}4\pi \mathrm{f}}$

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

$\mathrm{y}(\mathrm{t})=\mathrm{x}(\mathrm{t}-2)$

Concept:

Time shifting property of Fourier series states that:

$Ifx\left(t\right)\stackrel{C.T.F.S}{\leftrightarrow }{c}_n$

$x\left(t-t_0\right)\stackrel{C.T.F.S}{\leftrightarrow }{c}_n{e}^{-jn{\omega }_0{t}_0}$

Since ${\omega }_0=\frac{2\pi }{T_0}$, the above expression can be written as:

$x\left(t-t_0\right)\stackrel{C.T.F.S}{\leftrightarrow }{c}_n{e}^{-jn\cdot \frac{2\pi }{T_0}\cdot t_0}$

Application:

Observing the two figures, we can write:

y(t) = x(t - 1)

$y\left(t\right)\stackrel{C.T.F.S}{\leftrightarrow }{c}_n{e}^{-jn\cdot \frac{2\pi \left(1\right)}{T_0}}$

Where c_n = Fourier series coefficient of x(t)

Since, T₀ = 2

$y\left(t\right)\stackrel{C.T.F.S}{\leftrightarrow }{c}_n{e}^{-jn\pi }=c_n^{\prime }$

$c_n^{\prime }=c_n{e}^{-j\pi n}$

Using the above expression, we get:

$c_0^{\prime }=c_0=\frac{1}{\pi }$

$c_1^{\prime }=-c_1=\frac{j}{4}$

Also for even values of n, e^-jπn = 1

$\therefore c_n^{\prime }=c_n=\frac{1}{\pi \left(1-n^2\right)}$

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

The ${\mathrm{x}}_1\left(\mathrm{t}\right)\stackrel{\mathrm{F}\mathrm{T}}{\leftrightarrow }{\mathrm{X}}_1\left(\mathrm{j}\omega \right)=\mathrm{X}\left(\mathrm{j}\omega \right){\mathrm{e}}^{-\mathrm{j}\omega }$

Now,

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

Taking Fourier transform

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

Thus,