Signal and systems miscellaneous
- If a unit step current is passed through a capacitor what will be the voltage across the capacitor?
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Voltage across capacitor = 1 ∫idt C
= Integration of step input
= Ramp function
Hence, alternative (C) is the correct choice.Correct Option: C
Voltage across capacitor = 1 ∫idt C
= Integration of step input
= Ramp function
Hence, alternative (C) is the correct choice.
- System represented by equations—
(i) y[n] = anx[n]
(ii) y[n] = ax[n – 1] + bx[n – 3]
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(i) y[n] = an x[n]
F[x(n)] = an x[n]
The response to a delayed excitation.
F[x(n – n0)] = an[x(n – n0)] …(A)
The delayed response is
y[(n – n0)] = a[(n – n0) x(n – n0] …(B)
Since F[x(n – n0)] ≠ y(n – n0)
Hence, the system is not time invariant i.e., the system is time dependent.
(ii) y[n] = ax[n – 1] + bx[n – 2]
F[x(n)] = ax[n – 1] + bx[n – 2]
The response to a delayed excitation
F[x(n – n0)] = ax[n – n0 – 1] + bx[n – n0 – 2]
The delayed response is
y(n – n0) = ax[n – n0 – 1] + bx[n – n0 – 2]
Since, F[n – n0] = y[n – n0]
Therefore, the given system is time invariant.
Hence, alternative (A) is the correct choice.Correct Option: A
(i) y[n] = an x[n]
F[x(n)] = an x[n]
The response to a delayed excitation.
F[x(n – n0)] = an[x(n – n0)] …(A)
The delayed response is
y[(n – n0)] = a[(n – n0) x(n – n0] …(B)
Since F[x(n – n0)] ≠ y(n – n0)
Hence, the system is not time invariant i.e., the system is time dependent.
(ii) y[n] = ax[n – 1] + bx[n – 2]
F[x(n)] = ax[n – 1] + bx[n – 2]
The response to a delayed excitation
F[x(n – n0)] = ax[n – n0 – 1] + bx[n – n0 – 2]
The delayed response is
y(n – n0) = ax[n – n0 – 1] + bx[n – n0 – 2]
Since, F[n – n0] = y[n – n0]
Therefore, the given system is time invariant.
Hence, alternative (A) is the correct choice.
- System represented by equation— y[n] = n[ x(n)]2 is—
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Given that y[n] = n[x(n)]2
First check for linearity
F[x1(n)] = n[x1(n)]2
and F[x2(n)] = n[x2(n)]2
Therefore,
F[x1(n)] + F[x2(n)] = n[{x2(n)}2 + {x2(n)}2]
F[x1(n) + x2(n)] = n[x1(n) + x2(n)]2
= n[{x2(n)}2 + {x2(n)}2 + 2x1(n) x2(n)]
∴ F[x1(n)] + F[x2(n)] ≠ F[x1(n) + x2(n)]
These for the system is non-linear. Now, check for time variance.
y[n] = n[x(n)]2
The response of delayed excitation is
F[x(n – n0)] = n[x(n – n0)2
and the delayed response is
y(n – n0)= (n – n0) [x(n – n0)]2
Since, y(n – n0) ≠ F[x(n – n0)]
Therefore, the system is time variant.
Hence, alternative (D) is the correct choice.Correct Option: D
Given that y[n] = n[x(n)]2
First check for linearity
F[x1(n)] = n[x1(n)]2
and F[x2(n)] = n[x2(n)]2
Therefore,
F[x1(n)] + F[x2(n)] = n[{x2(n)}2 + {x2(n)}2]
F[x1(n) + x2(n)] = n[x1(n) + x2(n)]2
= n[{x2(n)}2 + {x2(n)}2 + 2x1(n) x2(n)]
∴ F[x1(n)] + F[x2(n)] ≠ F[x1(n) + x2(n)]
These for the system is non-linear. Now, check for time variance.
y[n] = n[x(n)]2
The response of delayed excitation is
F[x(n – n0)] = n[x(n – n0)2
and the delayed response is
y(n – n0)= (n – n0) [x(n – n0)]2
Since, y(n – n0) ≠ F[x(n – n0)]
Therefore, the system is time variant.
Hence, alternative (D) is the correct choice.
- The system represented by equation—
y[n] = a[x(n)]2 + bx[n]
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Given y(n) = a[x(n)]2 + bx(n)
First check for linearity
Since here,
F[x1(n)] = a[x1(n)]2 + bx1(n)
and F[x2(n)] = a[x2(n)]2 + bx2(n)
∴ F[x1(n)] = F[x2(n)]
= a[{x2(n)}2 + {x2(n)2}] + b[x1(n) + x2(n)]
Also, F[x1(n) + x2(n)]
= a[x1(n) + x2(n)]2 + b[x1(n) + x2(n)]
= a[{x1(n)}2 + {x2(n)}2 + 2x1(n) x2(n)] + bx1(n) + bx2(n)
Since here,
F[x1(n) + x2(n)] ≠ F[x1(n)] + F[x2(n)]
Therefore, the system is non-linear.
Check now for the time-in-variance.
y(n) = a[x(n)]2 + bx(n)
The response of delayed excitation is
F[x(n – n0)] = a[x(n – n0)]2 + b[x(n – n0)]
and the delayed response is
y[(n – n0)] = a[x(n – n0)]2 + b[x(n – n0)]
Since, y[(n – n0)] = F[x(n – n0)]
Therefore, the system is time-invariant.
Hence, alternative (C) is the correct choice.Correct Option: C
Given y(n) = a[x(n)]2 + bx(n)
First check for linearity
Since here,
F[x1(n)] = a[x1(n)]2 + bx1(n)
and F[x2(n)] = a[x2(n)]2 + bx2(n)
∴ F[x1(n)] = F[x2(n)]
= a[{x2(n)}2 + {x2(n)2}] + b[x1(n) + x2(n)]
Also, F[x1(n) + x2(n)]
= a[x1(n) + x2(n)]2 + b[x1(n) + x2(n)]
= a[{x1(n)}2 + {x2(n)}2 + 2x1(n) x2(n)] + bx1(n) + bx2(n)
Since here,
F[x1(n) + x2(n)] ≠ F[x1(n)] + F[x2(n)]
Therefore, the system is non-linear.
Check now for the time-in-variance.
y(n) = a[x(n)]2 + bx(n)
The response of delayed excitation is
F[x(n – n0)] = a[x(n – n0)]2 + b[x(n – n0)]
and the delayed response is
y[(n – n0)] = a[x(n – n0)]2 + b[x(n – n0)]
Since, y[(n – n0)] = F[x(n – n0)]
Therefore, the system is time-invariant.
Hence, alternative (C) is the correct choice.
- The following statements are given below—
(i) y[n] = 3x[n – 2] + 8x[n + 2]
(ii) y[n] = x[n – 1] + ax[n – 2]
(iii) y[n] = x[– n] is non-causal
The correct statements is/are—
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(i) y(n)= 3x(n – 2) + 8x(n + 2)
Here, y(n) is determined the past input samples value 3x(n – 2) and future input value 8x(n – 2) and hence the system is non-causal.
(ii) y(n) = x(n – 1) + ax(n – 2)
Here, y(n) is determined the past input samples value both for term x(n – 1) and ax(n – 2). Hence, the system is causal.
(iii) y(n) = x(– n)
Here, the input sample value is located on the negative time axis and for any negative value output depends on the future value. Hence, the system is non-causal.
Hence, alternative (B) is the correct choice.Correct Option: B
(i) y(n)= 3x(n – 2) + 8x(n + 2)
Here, y(n) is determined the past input samples value 3x(n – 2) and future input value 8x(n – 2) and hence the system is non-causal.
(ii) y(n) = x(n – 1) + ax(n – 2)
Here, y(n) is determined the past input samples value both for term x(n – 1) and ax(n – 2). Hence, the system is causal.
(iii) y(n) = x(– n)
Here, the input sample value is located on the negative time axis and for any negative value output depends on the future value. Hence, the system is non-causal.
Hence, alternative (B) is the correct choice.
