Scaling Property MCQ Quiz - Objective Question with Answer for Scaling Property - Download Free PDF
Last updated on Jul 9, 2025
Latest Scaling Property MCQ Objective Questions
Scaling Property Question 1:
Which type of property is shown by the following function.
L{K f(t)} = K F(s)
Answer (Detailed Solution Below)
Scaling Property Question 1 Detailed Solution
Concept:
The property shown by the function L{K f(t)} = K F(s) is Linearity.
Here's a brief explanation of why:
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Linearity: The linearity property of the Laplace Transform states that for constants 'a' and 'b', and functions f(t) and g(t) with Laplace Transforms F(s) and G(s) respectively, the transform of a linear combination is the linear combination of their transforms: L{a f(t) + b g(t)} = a F(s) + b G(s) The given expression L{K f(t)} = K F(s) is a specific instance of this linearity, showing that a constant multiplier can be factored out of the transform.
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Time shifting |
x(t - t0) ↔ e-jωto. X(ω) |
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Frequency shifting |
ejωt . x(t) ↔ X (ω - ω0) |
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Time scaling |
\({\bf{x}}\left( {{\bf{at}}} \right)\; \leftrightarrow \;\frac{1}{{\left| {\bf{a}} \right|}} \times \left( {\frac{{\bf{\omega }}}{{\bf{a}}}} \right)\) |
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Time Reversal |
x(-t) ↔ X (-ω) |
Scaling Property Question 2:
The magnitude of Fourier transform X(ω) of a function x(t) is shown below in figure (a). The magnitude of Fourier transform Y(ω) of another function y(t) is shown in figure (b). The phases of X(ω) and Y(ω) are zero for all ω. The magnitude and frequency units are identical in both the figures. The function y(t) can expressed in terms of x(t) as
Answer (Detailed Solution Below)
Scaling Property Question 2 Detailed Solution
We know that, expansion in frequency domain result in compression in the time domain and vice versa.
In the given question, compression is done frequency domain. So there will be expansion in time domain by same amount.
A x(t/2) ↔ 2A X(2f)
2A = 3 ⇒ A = 3/2
\(\frac{3}{2}x\left( {\frac{t}{2}} \right)\)
Top Scaling Property MCQ Objective Questions
Which type of property is shown by the following function.
L{K f(t)} = K F(s)
Answer (Detailed Solution Below)
Scaling Property Question 3 Detailed Solution
Download Solution PDFConcept:
The property shown by the function L{K f(t)} = K F(s) is Linearity.
Here's a brief explanation of why:
-
Linearity: The linearity property of the Laplace Transform states that for constants 'a' and 'b', and functions f(t) and g(t) with Laplace Transforms F(s) and G(s) respectively, the transform of a linear combination is the linear combination of their transforms: L{a f(t) + b g(t)} = a F(s) + b G(s) The given expression L{K f(t)} = K F(s) is a specific instance of this linearity, showing that a constant multiplier can be factored out of the transform.
|
Time shifting |
x(t - t0) ↔ e-jωto. X(ω) |
|
Frequency shifting |
ejωt . x(t) ↔ X (ω - ω0) |
|
Time scaling |
\({\bf{x}}\left( {{\bf{at}}} \right)\; \leftrightarrow \;\frac{1}{{\left| {\bf{a}} \right|}} \times \left( {\frac{{\bf{\omega }}}{{\bf{a}}}} \right)\) |
|
Time Reversal |
x(-t) ↔ X (-ω) |
The magnitude of Fourier transform X(ω) of a function x(t) is shown below in figure (a). The magnitude of Fourier transform Y(ω) of another function y(t) is shown in figure (b). The phases of X(ω) and Y(ω) are zero for all ω. The magnitude and frequency units are identical in both the figures. The function y(t) can expressed in terms of x(t) as
Answer (Detailed Solution Below)
Scaling Property Question 4 Detailed Solution
Download Solution PDFWe know that, expansion in frequency domain result in compression in the time domain and vice versa.
In the given question, compression is done frequency domain. So there will be expansion in time domain by same amount.
A x(t/2) ↔ 2A X(2f)
2A = 3 ⇒ A = 3/2
\(\frac{3}{2}x\left( {\frac{t}{2}} \right)\)
Scaling Property Question 5:
Which type of property is shown by the following function.
L{K f(t)} = K F(s)
Answer (Detailed Solution Below)
Scaling Property Question 5 Detailed Solution
Concept:
The property shown by the function L{K f(t)} = K F(s) is Linearity.
Here's a brief explanation of why:
-
Linearity: The linearity property of the Laplace Transform states that for constants 'a' and 'b', and functions f(t) and g(t) with Laplace Transforms F(s) and G(s) respectively, the transform of a linear combination is the linear combination of their transforms: L{a f(t) + b g(t)} = a F(s) + b G(s) The given expression L{K f(t)} = K F(s) is a specific instance of this linearity, showing that a constant multiplier can be factored out of the transform.
|
Time shifting |
x(t - t0) ↔ e-jωto. X(ω) |
|
Frequency shifting |
ejωt . x(t) ↔ X (ω - ω0) |
|
Time scaling |
\({\bf{x}}\left( {{\bf{at}}} \right)\; \leftrightarrow \;\frac{1}{{\left| {\bf{a}} \right|}} \times \left( {\frac{{\bf{\omega }}}{{\bf{a}}}} \right)\) |
|
Time Reversal |
x(-t) ↔ X (-ω) |
Scaling Property Question 6:
Let \(\rm{X(t)}\) be Wide Sense Stationary random process with power spectral density \(\rm{S_x (f)}\). If \(\rm{Y(t)}\) is a random process defined as \(\rm{Y(t)=X(2t-1)}\), the power spectral density \(\rm{S_y (f)}\) is:
Answer (Detailed Solution Below)
Scaling Property Question 6 Detailed Solution
Power density has no effect of shifting. It is affected only by scaling
We know,
\(\rm{E\left[X\left(t\right)X\left(t+\tau\right)\right]=R_x\left(\tau\right)}\)
and \(\rm{E\left[X\left(2t+1\right)X\left(2\left(t+\tau\right)+1\right)\right]=E\left[X\left(2t+1\right)X\left(2t+2\tau+1\right)\right]=R_x\left(2\tau\right)}\)
Now, \(\rm{R_x\left(\tau\right)\mathop\to\limits^{F}S_x\left(f\right)}\)
then using the scaling property of Fourier transforms we have,
\(\rm{R_x\left(2\tau\right)\mathop\to\limits^{F}\frac{1}{2}S_x\left(\frac{f}{2}\right)}\)
Thus, the power spectral density of \(Y(t)=X(2t-1)\) is \(\rm{\frac{1}{2}S_x\left(\frac{f}{2}\right)}\).
Scaling Property Question 7:
The magnitude of Fourier transform X(ω) of a function x(t) is shown below in figure (a). The magnitude of Fourier transform Y(ω) of another function y(t) is shown in figure (b). The phases of X(ω) and Y(ω) are zero for all ω. The magnitude and frequency units are identical in both the figures. The function y(t) can expressed in terms of x(t) as
Answer (Detailed Solution Below)
Scaling Property Question 7 Detailed Solution
We know that, expansion in frequency domain result in compression in the time domain and vice versa.
In the given question, compression is done frequency domain. So there will be expansion in time domain by same amount.
A x(t/2) ↔ 2A X(2f)
2A = 3 ⇒ A = 3/2
\(\frac{3}{2}x\left( {\frac{t}{2}} \right)\)
Scaling Property Question 8:
Let x(t)<-> X(jω) be Fourier Transform pair. The Fourier Transform of the signal x(5t-3) in terms of X(jω) is given as
Answer (Detailed Solution Below)
\( \frac{1}{5}{e^{ - \frac{{j3\omega }}{5}}}X\left( {\frac{{j\omega }}{5}} \right)\)
Scaling Property Question 8 Detailed Solution
using the properties of fourier transform
Time shift x(t) --> X(jw) x(t-to) --> x(jw)
Time scaling x(t) --> X(jw) x(at) --> \( \frac{1}{a}x\left( {j\frac{w}{a}} \right)\)
We get \( x\left( {5t - 3} \right)\frac{1}{5}{e^{ - j\frac{{3w}}{5}}}x\left( {\frac{{jw}}{5}} \right)\)