Signal and systems miscellaneous
- The value of sgn (f)
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We know that
x(t) = sgn (t) ↔ 2 jω
By using duality property if
x(t) → x(jω)
then x(t) → 2π x(– jω)2 = – 2π sgn (ω) jt 2 = – 2π sgn (f) jt or sgn (f) = – 1 × j = j πjt j πt
Hence, alternative (B) is the correct choice.Correct Option: B
We know that
x(t) = sgn (t) ↔ 2 jω
By using duality property if
x(t) → x(jω)
then x(t) → 2π x(– jω)2 = – 2π sgn (ω) jt 2 = – 2π sgn (f) jt or sgn (f) = – 1 × j = j πjt j πt
Hence, alternative (B) is the correct choice.
- If g(t) = 2 1 + t 2, then G(jω)—
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We know that
x(t) = e– a|t| ↔ 2a a2 + ω2
If a = 1,then e– a|t| ↔ 2 1 + ω2
By using duality property,2 ↔ 2πe– |– ω| 1 + t2 or 2 ↔ 2πe– |ω| 1 + t2
Hence, alternative (A) is the correct choice.Correct Option: A
We know that
x(t) = e– a|t| ↔ 2a a2 + ω2
If a = 1,then e– a|t| ↔ 2 1 + ω2
By using duality property,2 ↔ 2πe– |– ω| 1 + t2 or 2 ↔ 2πe– |ω| 1 + t2
Hence, alternative (A) is the correct choice.
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then its Fourier transform X(jω) will be—If x(t) = 1000 sin C(1000t) π
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We know that
x(t) = 
A |t| ≤ T/2 0 otherwise then x(jω) = AT.sin C 
ω T 
2
However, given that
x(t) = 100 sin C (100t) π
By using duality propertyx(jω)=2πA rect ω t
The value of a and ω can be calculated astT = 1000 t 2
or T = 2000AT = 1000 π or A = 1000 = 1000 = 1 π·T π·2000 2π Thus, x(jω)=2π· 1 rect 
ω 
2π 2000 or x(jω) = rect ω 2000
Hence, alternative (D) is the correct choice.
Correct Option: D
We know that
x(t) = A |t| ≤ T/2 0 otherwise then x(jω) = AT.sin C ω T 2
However, given thatx(t) = 100 sin C (100t) π
By using duality propertyx(jω)=2πA rect ω t
The value of a and ω can be calculated astT = 1000 t 2
or T = 2000AT = 1000 π or A = 1000 = 1000 = 1 π·T π·2000 2π Thus, x(jω)=2π· 1 rect ω 2π 2000 or x(jω) = rect ω 2000
Hence, alternative (D) is the correct choice.
- For the signal x(n) = b|n|, b > 0 The z-transform is given by—
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x(n) = b|n|, b > 0
or x(n) = b|n| u(n) + b– n u[– n – 1]
The sequence x(n) = b|n| for |b| < 1,
function is converges while for sequence x(n) = bn for |b| > 1, function is divergesbn u[n] ←z→ 1 , |z| > b 1 – bz– 1 or bn u[n] ←z→ z , |z| > b b z – b b– 1 u[– n – 1] ←z→ – 1 |z| < 1 1 – b– 1 z– 1 b or b– 1 u[– n – 1] ←z→ – bz |z| < 1 bz– 1 b
Thus, the z-transform for the composite function isx(z) = z – bz , b < |z| < 1 z – b bz – 1 b
Hence, alternative (A) is the correct choice.Correct Option: A
x(n) = b|n|, b > 0
or x(n) = b|n| u(n) + b– n u[– n – 1]
The sequence x(n) = b|n| for |b| < 1,
function is converges while for sequence x(n) = bn for |b| > 1, function is divergesbn u[n] ←z→ 1 , |z| > b 1 – bz– 1 or bn u[n] ←z→ z , |z| > b b z – b b– 1 u[– n – 1] ←z→ – 1 |z| < 1 1 – b– 1 z– 1 b or b– 1 u[– n – 1] ←z→ – bz |z| < 1 bz– 1 b
Thus, the z-transform for the composite function isx(z) = z – bz , b < |z| < 1 z – b bz – 1 b
Hence, alternative (A) is the correct choice.
- If x(z) = log (1 + az– 1), |z| > |a| then x(n) is given by—
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If x(n) ←z→ X(z)
then nx(n) ←z→ – z d X(z) az– 1 dz 1 + az– 1
and if X (Z) = log (1 + az– 1)then nx(n) ←z→ – z d X(z) az– 1 dZ 1 + az– 1
As we know that– an u(n) ←z→ 1 1 + az– 1
Similarly, by using time-shifting property.(– a)n– 1 u(n – 1) ←z→ 1 · z– 1 1 + az– 1 or a(– a)n– 1 u(n – 1) = az– 1 1 + az– 1 or – (– a)n u(n – 1) = az– 1 1 + az– 1
or – (– a)n u(n – 1) = nx(n)or x(n) = – (– a)n u(n – 1) n
Hence, alternative (B) is the correct choice.Correct Option: B
If x(n) ←z→ X(z)
then nx(n) ←z→ – z d X(z) az– 1 dz 1 + az– 1
and if X (Z) = log (1 + az– 1)then nx(n) ←z→ – z d X(z) az– 1 dZ 1 + az– 1
As we know that– an u(n) ←z→ 1 1 + az– 1
Similarly, by using time-shifting property.(– a)n– 1 u(n – 1) ←z→ 1 · z– 1 1 + az– 1 or a(– a)n– 1 u(n – 1) = az– 1 1 + az– 1 or – (– a)n u(n – 1) = az– 1 1 + az– 1
or – (– a)n u(n – 1) = nx(n)or x(n) = – (– a)n u(n – 1) n
Hence, alternative (B) is the correct choice.
