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

Concept:

The inverse Fourier Transform is defined as:

$x\left(t\right)=\frac{1}{2\pi }\underset{-\mathrm{\infty }}{\overset{+\mathrm{\infty }}{\int }}X\left(\omega \right)e^{j\omega t}d\omega$

Properties of delta function:

1) $\underset{-\mathrm{\infty }}{\overset{+\mathrm{\infty }}{\int }}\delta \left(t\right)dt=1$

2) $\underset{-\mathrm{\infty }}{\overset{+\mathrm{\infty }}{\int }}\delta \left(t-t_o\right)x\left(t\right)dt=x\left(t_o\right)$

3) $\delta \left(at\right)=\frac{\delta \left(t\right)}{\left|a\right|}$

4) $\delta \left(t-t_o\right).x\left(t\right)=x\left(t_0\right).\delta \left(t-t_o\right)$

5) $\delta \left(-t\right)=\delta \left(t\right)$

Calculation:

Given:

$X(\omega )=2\pi \delta \left(\omega -{\omega }_o\right)$

The inverse Fourier Transform will be:

$x(t)=\frac{1}{2\pi }\underset{-\mathrm{\infty }}{\overset{+\mathrm{\infty }}{\int }}2\pi \delta \left(\omega -{\omega }_o\right)e^{j\omega t}d\omega$
Using Property (2) (Frequency shifting property) of delta signal, we get:

$\underset{-\mathrm{\infty }}{\overset{+\mathrm{\infty }}{\int }}\delta \left(\omega -{\omega }_o\right)e^{j\omega t}d\omega =e^{j{\omega }_{o}t}$

$x(t)=e^{j{\omega }_{o}t}$

Concept:

The inverse Fourier Transform is defined as:

$x\left(t\right)=\frac{1}{2\pi }\underset{-\mathrm{\infty }}{\overset{+\mathrm{\infty }}{\int }}X\left(\omega \right)e^{j\omega t}d\omega$

Properties of delta function:

1) $\underset{-\mathrm{\infty }}{\overset{+\mathrm{\infty }}{\int }}\delta \left(t\right)dt=1$

2) $\underset{-\mathrm{\infty }}{\overset{+\mathrm{\infty }}{\int }}\delta \left(t-t_o\right)x\left(t\right)dt=x\left(t_o\right)$

3) $\delta \left(at\right)=\frac{\delta \left(t\right)}{\left|a\right|}$

4) $\delta \left(t-t_o\right).x\left(t\right)=x\left(t_0\right).\delta \left(t-t_o\right)$

5) $\delta \left(-t\right)=\delta \left(t\right)$

Calculation:

Given:

$X(\omega )=2\pi \delta \left(\omega -{\omega }_o\right)$

The inverse Fourier Transform will be:

$x(t)=\frac{1}{2\pi }\underset{-\mathrm{\infty }}{\overset{+\mathrm{\infty }}{\int }}2\pi \delta \left(\omega -{\omega }_o\right)e^{j\omega t}d\omega$
Using Property (2) (Frequency shifting property) of delta signal, we get:

$\underset{-\mathrm{\infty }}{\overset{+\mathrm{\infty }}{\int }}\delta \left(\omega -{\omega }_o\right)e^{j\omega t}d\omega =e^{j{\omega }_{o}t}$

$x(t)=e^{j{\omega }_{o}t}$

Analysis:

v(t) = cos (2πf_ct), w(t) = cos [2π (f_c + A)t]

From above block diagram, we can write:

x(t) = m(t) × v(t)

So. The frequency spectrum of x(t) will be:

Now, this signal passes through HPF having frequency f_c. The resultant spectrum will be:

The output of the second multiplier will be:

z(t) = y(t) × w(t)

A time-domain multiplication is the frequency domain convolution, i.e.

After passing through the LPF centered at f_c, the Fourier spectrum of output s(t) will be:

Now, the required Bandwidth (BW) will be:

BW = (A - B) – 0

= 100 – 40

= 60 Hz

Calculation:

Analyzing the circuit by taking each Option into consideration we proceed as follows:

First considering option (1)

i.e. if the Frequency Spectrum of A is as shown:

The output when the signal x(t) passes through A will simply be the multiplication of j in the frequency spectrum of x(t). The output signal spectrum from Block A will be, therefore:

When the above spectrum is multiplied with j (as seen in the circuit given), the output becomes;

When the above signal is then passed through a summer, the resultant spectrum will be as shown:

But given y(t) is a shifted spectrum of the above, and we know that;

If x(t) « X(f)

$x\left(t\right).e^{-j{\omega }_{o}t}\text{ X}(\text{f}+{\text{f}}_{\text{o}})$

i.e. when multiplied with a single negative frequency exponent, the spectrum is shifted to the left by f₀.

If B = 

The output spectrum will be:

So Option (1) shows the Correct representation of A and B.

$x\left(t\right)\stackrel{F.T}{\leftrightarrow }\frac{5}{\omega }\cos \left(\pi \omega \right)$

By frequency shifting properties.

$\begin{array}{l}x\left(t\right)\stackrel{F.T}{\leftrightarrow }x\left(\omega \right)\\ x\left(t\right).e^{j{\omega }_{0}t}\stackrel{F.T}{\leftrightarrow }x\left(\omega -{\omega }_0\right)\\ e^{j6t}x\left(t\right)\stackrel{F.T}{\leftrightarrow }\frac{5}{\omega -6}\cos \left\{\pi \left(\omega -6\right)\right\}\end{array}$  

Given,

x(t) = cos(6πt) + sin(8πt)

The bandwidth of the signal will be:

$\frac{8\pi }{2\pi }=4\mathrm{H}\mathrm{z}$

Now, y(t) = x(2t + 5) can be written as:

$y(t)=\mathrm{x}\left(2\left(\mathrm{t}+\frac{5}{2}\right)\right)$

Taking the Fourier transform, we get:

$\mathrm{Y}\left(\mathrm{j}\omega \right)=\frac{1}{\left|2\right|}\mathrm{X}\left(\frac{\mathrm{j}\omega }{2}\right){\mathrm{e}}^{\frac{\mathrm{j}5\omega }{2}}$

$\mathrm{Y}\left(\mathrm{j}\omega \right)=\frac{1}{2}\mathrm{X}\left(\frac{\mathrm{j}\omega }{2}\right){\mathrm{e}}^{\frac{\mathrm{j}5\omega }{2}}$

Thus, the frequency spectrum of X(jω)  expands by 2.

This will make the highest frequency component of Y(jω) as:

2× 4 = 8 Hz 

Hence, the Nyquist rate will be:

f_s = 2 × 8 = 16 samples/sec

Analysis:

v(t) = cos (2πf_ct), w(t) = cos [2π (f_c + A)t]

From above block diagram, we can write:

x(t) = m(t) × v(t)

So. The frequency spectrum of x(t) will be:

Now, this signal passes through HPF having frequency f_c. The resultant spectrum will be:

The output of the second multiplier will be:

z(t) = y(t) × w(t)

A time-domain multiplication is the frequency domain convolution, i.e.

After passing through the LPF centered at f_c, the Fourier spectrum of output s(t) will be:

Now, the required Bandwidth (BW) will be:

BW = (A - B) – 0

= 100 – 40

= 60 Hz

Calculation:

Analyzing the circuit by taking each Option into consideration we proceed as follows:

First considering option (1)

i.e. if the Frequency Spectrum of A is as shown:

The output when the signal x(t) passes through A will simply be the multiplication of j in the frequency spectrum of x(t). The output signal spectrum from Block A will be, therefore:

When the above spectrum is multiplied with j (as seen in the circuit given), the output becomes;

When the above signal is then passed through a summer, the resultant spectrum will be as shown:

But given y(t) is a shifted spectrum of the above, and we know that;

If x(t) « X(f)

$x\left(t\right).e^{-j{\omega }_{o}t}\text{ X}(\text{f}+{\text{f}}_{\text{o}})$

i.e. when multiplied with a single negative frequency exponent, the spectrum is shifted to the left by f₀.

If B = 

The output spectrum will be:

So Option (1) shows the Correct representation of A and B.

Given,

x(t) = cos(6πt) + sin(8πt)

The bandwidth of the signal will be:

$\frac{8\pi }{2\pi }=4\mathrm{H}\mathrm{z}$

Now, y(t) = x(2t + 5) can be written as:

$y(t)=\mathrm{x}\left(2\left(\mathrm{t}+\frac{5}{2}\right)\right)$

Taking the Fourier transform, we get:

$\mathrm{Y}\left(\mathrm{j}\omega \right)=\frac{1}{\left|2\right|}\mathrm{X}\left(\frac{\mathrm{j}\omega }{2}\right){\mathrm{e}}^{\frac{\mathrm{j}5\omega }{2}}$

$\mathrm{Y}\left(\mathrm{j}\omega \right)=\frac{1}{2}\mathrm{X}\left(\frac{\mathrm{j}\omega }{2}\right){\mathrm{e}}^{\frac{\mathrm{j}5\omega }{2}}$

Thus, the frequency spectrum of X(jω)  expands by 2.

This will make the highest frequency component of Y(jω) as:

2× 4 = 8 Hz 

Hence, the Nyquist rate will be:

f_s = 2 × 8 = 16 samples/sec

Concept:

The inverse Fourier Transform is defined as:

$x\left(t\right)=\frac{1}{2\pi }\underset{-\mathrm{\infty }}{\overset{+\mathrm{\infty }}{\int }}X\left(\omega \right)e^{j\omega t}d\omega$

Properties of delta function:

1) $\underset{-\mathrm{\infty }}{\overset{+\mathrm{\infty }}{\int }}\delta \left(t\right)dt=1$

2) $\underset{-\mathrm{\infty }}{\overset{+\mathrm{\infty }}{\int }}\delta \left(t-t_o\right)x\left(t\right)dt=x\left(t_o\right)$

3) $\delta \left(at\right)=\frac{\delta \left(t\right)}{\left|a\right|}$

4) $\delta \left(t-t_o\right).x\left(t\right)=x\left(t_0\right).\delta \left(t-t_o\right)$

5) $\delta \left(-t\right)=\delta \left(t\right)$

Calculation:

Given:

$X(\omega )=2\pi \delta \left(\omega -{\omega }_o\right)$

The inverse Fourier Transform will be:

$x(t)=\frac{1}{2\pi }\underset{-\mathrm{\infty }}{\overset{+\mathrm{\infty }}{\int }}2\pi \delta \left(\omega -{\omega }_o\right)e^{j\omega t}d\omega$
Using Property (2) (Frequency shifting property) of delta signal, we get:

$\underset{-\mathrm{\infty }}{\overset{+\mathrm{\infty }}{\int }}\delta \left(\omega -{\omega }_o\right)e^{j\omega t}d\omega =e^{j{\omega }_{o}t}$

$x(t)=e^{j{\omega }_{o}t}$

Concept:

The inverse Fourier Transform is defined as:

$x\left(t\right)=\frac{1}{2\pi }\underset{-\mathrm{\infty }}{\overset{+\mathrm{\infty }}{\int }}X\left(\omega \right)e^{j\omega t}d\omega$

Properties of delta function:

1) $\underset{-\mathrm{\infty }}{\overset{+\mathrm{\infty }}{\int }}\delta \left(t\right)dt=1$

2) $\underset{-\mathrm{\infty }}{\overset{+\mathrm{\infty }}{\int }}\delta \left(t-t_o\right)x\left(t\right)dt=x\left(t_o\right)$

3) $\delta \left(at\right)=\frac{\delta \left(t\right)}{\left|a\right|}$

4) $\delta \left(t-t_o\right).x\left(t\right)=x\left(t_0\right).\delta \left(t-t_o\right)$

5) $\delta \left(-t\right)=\delta \left(t\right)$

Calculation:

Given:

$X(\omega )=2\pi \delta \left(\omega -{\omega }_o\right)$

The inverse Fourier Transform will be:

$x(t)=\frac{1}{2\pi }\underset{-\mathrm{\infty }}{\overset{+\mathrm{\infty }}{\int }}2\pi \delta \left(\omega -{\omega }_o\right)e^{j\omega t}d\omega$
Using Property (2) (Frequency shifting property) of delta signal, we get:

$\underset{-\mathrm{\infty }}{\overset{+\mathrm{\infty }}{\int }}\delta \left(\omega -{\omega }_o\right)e^{j\omega t}d\omega =e^{j{\omega }_{o}t}$

$x(t)=e^{j{\omega }_{o}t}$

Analysis:

v(t) = cos (2πf_ct), w(t) = cos [2π (f_c + A)t]

From above block diagram, we can write:

x(t) = m(t) × v(t)

So. The frequency spectrum of x(t) will be:

Now, this signal passes through HPF having frequency f_c. The resultant spectrum will be:

The output of the second multiplier will be:

z(t) = y(t) × w(t)

A time-domain multiplication is the frequency domain convolution, i.e.

After passing through the LPF centered at f_c, the Fourier spectrum of output s(t) will be:

Now, the required Bandwidth (BW) will be:

BW = (A - B) – 0

= 100 – 40

= 60 Hz

Calculation:

Analyzing the circuit by taking each Option into consideration we proceed as follows:

First considering option (1)

i.e. if the Frequency Spectrum of A is as shown:

The output when the signal x(t) passes through A will simply be the multiplication of j in the frequency spectrum of x(t). The output signal spectrum from Block A will be, therefore:

When the above spectrum is multiplied with j (as seen in the circuit given), the output becomes;

When the above signal is then passed through a summer, the resultant spectrum will be as shown:

But given y(t) is a shifted spectrum of the above, and we know that;

If x(t) « X(f)

$x\left(t\right).e^{-j{\omega }_{o}t}\text{ X}(\text{f}+{\text{f}}_{\text{o}})$

i.e. when multiplied with a single negative frequency exponent, the spectrum is shifted to the left by f₀.

If B = 

The output spectrum will be:

So Option (1) shows the Correct representation of A and B.

$x\left(t\right)=\frac{1}{a^2+t^2}$ 

Given, $x\left(\omega \right)=\frac{\pi }{a}{e}^{-a\left|\omega \right|}$

We know that $\cos bt=\frac{e^{jbt}+e^{-jbt}}{2}$

then, $x\left(t\right)\cos bt=\frac{1}{2}\left[x\left(t\right)e^{jbt}+x\left(t\right)e^{-jbt}\right]$

$x\left(\omega \right)=\frac{\pi }{a}{e}^{-a\left|\omega \right|}$ 

By using frequency shifting

$x\left(t\right)e^{j{\omega }_{0}t}\leftrightarrow x\left(\omega -{\omega }_0\right)$ 

$x\left(t\right)\cos bt\leftrightarrow \frac{1}{2}\frac{\pi }{a}\left[e^{-a\left|\omega -b\right|}+e^{-a\left|\omega +b\right|}\right]$ 

Fourier transform of $x\left(t\right)\cos bt=\frac{\pi }{2a}\left[e^{-a\left|\omega -b\right|}+e^{-a\left|\omega +b\right|}\right]$

x(t) = 4 + cos (4πt) - sin (8πt)

$f_1=0,f_2=\frac{4\pi }{2\pi }=2Hz,f_3=\frac{8\pi }{2\pi }=4Hz$ 

Sinc²(t) ↔ Tri(f)

Using frequency shifting property,

$H\left(f\right)=\frac{Tri\left(f-8\right)+Tri\left(f+8\right)}{2}$