Signal and systems miscellaneous
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The power of the signal x(n) = e j2πn will be— 4
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Given, x(n) = 6e j2πn 4 or x(n) = 6 lim
s → 0
cos 2πn + j sin 2πn 
4 4
The signal is periodic with periodN = 2π m = 4 2π 4
with m = 1x(0) = 6 cos 2π · 0 + j sin 2π · 0 = 6[1 + 0] = 6 4 4 x(1) = 6 cos 2π · 1 + j sin 2π · 1 4 4 = 6 cos π + j sin π = j6 2 2 x(2) = 6 cos 2π · 2 + j sin 2π · 1 4 4
= 6 [cos π + j sin π] = 6 [– 1 + j·0] = – 6x(3) = 6 cos 2π · 3 + j sin 2π · 3 4 4 = 6 cos 3π + j sin 3π = 6[– j] = – j6. 2 2
Now, the power of the signal x(n)N - 1 P = 1/N ∑ | x(n)|2 n = 0 4 - 1 = 1/4 ∑ |x(n)|2 n = 0 = 1 [|x(0)|2 + |x(1)|2 + |x(2)|2 + |x(3)|2] 4 = 1 [|6|2 + |j6|2 + |6|2+ |– 6|2] 4 = 1 [36 + 36 + 36 + 36] 4
= 36 watt
Hence, alternative (B) is the correct choice.Correct Option: B
Given, x(n) = 6e j2πn 4 or x(n) = 6 lim
s → 0cos 2πn + j sin 2πn 4 4
The signal is periodic with periodN = 2π m = 4 2π 4
with m = 1x(0) = 6 cos 2π · 0 + j sin 2π · 0 = 6[1 + 0] = 6 4 4 x(1) = 6 cos 2π · 1 + j sin 2π · 1 4 4 = 6 cos π + j sin π = j6 2 2 x(2) = 6 cos 2π · 2 + j sin 2π · 1 4 4
= 6 [cos π + j sin π] = 6 [– 1 + j·0] = – 6x(3) = 6 cos 2π · 3 + j sin 2π · 3 4 4 = 6 cos 3π + j sin 3π = 6[– j] = – j6. 2 2
Now, the power of the signal x(n)N - 1 P = 1/N ∑ | x(n)|2 n = 0 4 - 1 = 1/4 ∑ |x(n)|2 n = 0 = 1 [|x(0)|2 + |x(1)|2 + |x(2)|2 + |x(3)|2] 4 = 1 [|6|2 + |j6|2 + |6|2+ |– 6|2] 4 = 1 [36 + 36 + 36 + 36] 4
= 36 watt
Hence, alternative (B) is the correct choice.
- Highest value of the auto correlation of a function 10 sin 10πt is equal to—
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NA
Correct Option: A
NA
- Sinusoidal function is used as basic function in electrical communication because—
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Sinusoidal function is used as basic function in electrical communication because of its convenience and response to linear sine wave.
Correct Option: C
Sinusoidal function is used as basic function in electrical communication because of its convenience and response to linear sine wave.
- A linear system is characterized by, H(jω) = e– bω3. The system is physically—
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Given that H(jω) = e– bω3
According to Paley-Wiener criterion, the given system will be physically realizable, if condition given below is satisfied.= ∫∞– ∞ |In H(jω)| · dω < ∞ 1 + ω2 = ∫∞– ∞ |In e– bω3 | dω 1 + ω2 = ∫∞– ∞ bω3 dω 1 + ω2 = b∫∞– ∞ ω2 dω 1 + ω2 = b∫∞– ∞ t , (put ω2 = t, 2ω dω = dt) 1 + t = b ∫∞-∞ 
T + 1 – 1 
dt T + 1 1 + t = b ∫∞-∞ dt - ∫∞-∞ 1 dt 1 + t
∵ ω2 is even function 1 + ω2
= 2[ω – tan– 1ω]∞ 0= 2 ∞ – π – 0 + 0 2
= 2· ∞
= ∞
Correct Option: A
Given that H(jω) = e– bω3
According to Paley-Wiener criterion, the given system will be physically realizable, if condition given below is satisfied.= ∫∞– ∞ |In H(jω)| · dω < ∞ 1 + ω2 = ∫∞– ∞ |In e– bω3 | dω 1 + ω2 = ∫∞– ∞ bω3 dω 1 + ω2 = b∫∞– ∞ ω2 dω 1 + ω2 = b∫∞– ∞ t , (put ω2 = t, 2ω dω = dt) 1 + t = b ∫∞-∞ T + 1 – 1 dt T + 1 1 + t = b ∫∞-∞ dt - ∫∞-∞ 1 dt 1 + t
∵ ω2 is even function 1 + ω2
= 2[ω – tan– 1ω]∞ 0= 2 ∞ – π – 0 + 0 2
= 2· ∞
= ∞
- Laplace transform has the following merit ever Fourier transform—
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As we know that for the existance of Laplace transform, the integral
∫∞– ∞ f(t)e– st dt must be converge.
or ∫∞– ∞ |f(t)e– st|dt < ∞
where as σ → damping coefficient + jω
∫∞– ∞ f(t) e– σt e– jωt dt
In this expression the factor e– σt is the convergence factor.
Hence, alternative (C) is the correct choice.Correct Option: C
As we know that for the existance of Laplace transform, the integral
∫∞– ∞ f(t)e– st dt must be converge.
or ∫∞– ∞ |f(t)e– st|dt < ∞
where as σ → damping coefficient + jω
∫∞– ∞ f(t) e– σt e– jωt dt
In this expression the factor e– σt is the convergence factor.
Hence, alternative (C) is the correct choice.
