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The inverse fourier transform of sgn(f)
- a)-jπt
- b)j/πt
- c)1/jπt
- d)None of these
Correct answer is option 'B'. Can you explain this answer?
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The inverse fourier transform of sgn(f)a)-jπtb)j/πtc)1/jπtd)N...
then
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The inverse fourier transform of sgn(f)a)-jπtb)j/πtc)1/jπtd)N...
The inverse Fourier transform of sgn(f)a)-j is given by:
F^(-1)[sgn(f)a)-j] = (1/2π) ∫[from -∞ to +∞] sgn(f)a)-j * e^(j2πft) df
To simplify this expression, we can consider the sign function sgn(f) separately. The sign function sgn(f) is defined as:
sgn(f) =
-1, if f < />
0, if f = 0,
1, if f > 0.
Since the sign function is only non-zero for positive values of f, the integral can be simplified to:
F^(-1)[sgn(f)a)-j] = (1/2π) ∫[from 0 to +∞] a)-j * e^(j2πft) df
Now, we can further simplify the expression by substituting a = |a|e^(jθ), where |a| is the magnitude of a and θ is the phase angle of a.
F^(-1)[sgn(f)a)-j] = (1/2π) ∫[from 0 to +∞] |a|e^(jθ) * e^(j2πft) df
Simplifying further, we get:
F^(-1)[sgn(f)a)-j] = (1/2π) |a| e^(jθ) ∫[from 0 to +∞] e^(j(2πf + θ)t) df
Using the property ∫[from -∞ to +∞] e^(jωt) dt = 2πδ(ω), where δ(ω) is the Dirac delta function, we can evaluate the integral:
F^(-1)[sgn(f)a)-j] = (1/2π) |a| e^(jθ) [2πδ(2πf + θ)]
Finally, simplifying the expression, we obtain:
F^(-1)[sgn(f)a)-j] = |a| e^(jθ) δ(2πf + θ)
F^(-1)[sgn(f)a)-j] = (1/2π) ∫[from -∞ to +∞] sgn(f)a)-j * e^(j2πft) df
To simplify this expression, we can consider the sign function sgn(f) separately. The sign function sgn(f) is defined as:
sgn(f) =
-1, if f < />
0, if f = 0,
1, if f > 0.
Since the sign function is only non-zero for positive values of f, the integral can be simplified to:
F^(-1)[sgn(f)a)-j] = (1/2π) ∫[from 0 to +∞] a)-j * e^(j2πft) df
Now, we can further simplify the expression by substituting a = |a|e^(jθ), where |a| is the magnitude of a and θ is the phase angle of a.
F^(-1)[sgn(f)a)-j] = (1/2π) ∫[from 0 to +∞] |a|e^(jθ) * e^(j2πft) df
Simplifying further, we get:
F^(-1)[sgn(f)a)-j] = (1/2π) |a| e^(jθ) ∫[from 0 to +∞] e^(j(2πf + θ)t) df
Using the property ∫[from -∞ to +∞] e^(jωt) dt = 2πδ(ω), where δ(ω) is the Dirac delta function, we can evaluate the integral:
F^(-1)[sgn(f)a)-j] = (1/2π) |a| e^(jθ) [2πδ(2πf + θ)]
Finally, simplifying the expression, we obtain:
F^(-1)[sgn(f)a)-j] = |a| e^(jθ) δ(2πf + θ)
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The inverse fourier transform of sgn(f)a)-jπtb)j/πtc)1/jπtd)None of theseCorrect answer is option 'B'. Can you explain this answer? for Electronics and Communication Engineering (ECE) 2025 is part of Electronics and Communication Engineering (ECE) preparation. The Question and answers have been prepared according to the Electronics and Communication Engineering (ECE) exam syllabus. Information about The inverse fourier transform of sgn(f)a)-jπtb)j/πtc)1/jπtd)None of theseCorrect answer is option 'B'. Can you explain this answer? covers all topics & solutions for Electronics and Communication Engineering (ECE) 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The inverse fourier transform of sgn(f)a)-jπtb)j/πtc)1/jπtd)None of theseCorrect answer is option 'B'. Can you explain this answer?.
The inverse fourier transform of sgn(f)a)-jπtb)j/πtc)1/jπtd)None of theseCorrect answer is option 'B'. Can you explain this answer? for Electronics and Communication Engineering (ECE) 2025 is part of Electronics and Communication Engineering (ECE) preparation. The Question and answers have been prepared according to the Electronics and Communication Engineering (ECE) exam syllabus. Information about The inverse fourier transform of sgn(f)a)-jπtb)j/πtc)1/jπtd)None of theseCorrect answer is option 'B'. Can you explain this answer? covers all topics & solutions for Electronics and Communication Engineering (ECE) 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The inverse fourier transform of sgn(f)a)-jπtb)j/πtc)1/jπtd)None of theseCorrect answer is option 'B'. Can you explain this answer?.
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