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Which of the following is NOT one of the representations of discretetime signals?
1) Functional Representation.
Example:
2) Tabular method of representation
Example:
3) Sequence Representation
Example:
Which type/s of discretetime system do/does not exhibit the necessity of any feedback?
Recursive discrete time system:
If a system output y(n) at time n depends on any number of past output value y(n – 1), y(n – 2), …, it is called a recursive system.
This system requires two multiplication, one addition, and one memory location. This is a recursive system which means the output at time n depends on any number of a past output values.
So, a recursive system has feedback output of the system into the input. This feed back loop contains a delay element.
NonRecursive discrete time system:
If y(n) depends only on the present and past input, it is called nonrecursive. It does not exhibit the necessity of any feedback.
The convolution
y(n) = (0.5)^{n} * 2^{n} u(2n + 2) is
Concept:
Output = Input × Eigen value.
Calculation:
Let, Input x(n) = (0.5)^{n}
h(n) = 2^{n} u(2n + 2)
= 2^{n} u(n + 1)
= 2^{n+1} u(n+1) / 2
(scaling does not affect the discretetime signal)
putting 0.5 in place of Z, we get:
= 1/20
1/20 is the eigenvalue
Output
Express the following finite discretetime signal as the difference of two unit step sequences: x[n] = 1, for 0 ≤ n ≤ 5; and 0 otherwise.
Concept:
The signal u[n] is defined as:
u[n] = 1 for n ≥ 0
u[n] = 0 for n < 0
Also, the shifted version of u[n] is defined as:
u[n  n_{0}] = 1 for n ≥ n_{0}
u[n  n_{0}] = 0 for n < n_{0}
Now the combination of two discrete time signals can be analysed as:
u[n]  u[n  n_{0}] = 1 for 0 ≤ n ≤ n_{0}  1
u[n]  u[n  n0] = 0, otherwise
Application:
Since the given signal is defined as:
x[n] = 1, for 0 ≤ n ≤ 5
We can write:
x[n] = u[n]  u[n  6]
The discrete time system described by y(n) = [x(n)]^{2} is
Concept:
Linearity: Necessary and sufficient condition to prove the linearity of the system is that the linear system follows the laws of superposition i.e. the response of the system is the sum of the responses obtained from each input considered separately.
y{ax_{1}[t] + bx_{2}[t]} = a y{x_{1}[t]} + b y{x_{2}[t]}
Conditions to check whether the system is linear or not.
Causal system:
If O/P of the system is independent of the future value of input then the system is said to be causal. Causal systems are practical or physically reliable systems.
Analysis:
y(n) = [x(n)]2
Linearity check:
x_{1}(n) → x_{1}(n)^{2}
x_{2}(n) → x_{2}(n)^{2}
x_{3}(n) = [x1(n) + x_{2}(n)] → [x1(n) + x2(n)]^{2} = x_{1}(n)^{2} + x_{2}(n)^{2} + 2x_{1}(n) x_{2}(n)
≠ x1(n)^{2} + x_{2}(n)^{2}
∴ Non  linear
Causality check:
y(0) = x(0)^{2}
y(1) = x(1)^{2}
y(1) = x(1)^{2}
∴ the system is causal.
The special case of a finiteduration sequence is given as
The sequence x(n) into a sum of weighted impulse sequences will be
The arrow indicates the origin. This can be represented as:
In terms of unit impulse sequences, this can be represented as:
x(n) = 2δ(n + 1) + 4δ(n) + 3δ(n – 2)
What is the nature of the following function: y[n] = y[n1] + x[n]?
If the above recursive definition is repeated for all n, starting from 1,2.. then y[n] will be the sum of all x[n] ranging from 1 to n, making it an accumulator system.
As the value of the function depends solely on the value of the input at a time presently and/or in the past, it is a causal system.
If n tends to infinity, is the accumulator function a stable one?
The system would be unstable, as the output will grow out of bound at the maximally worst possible case.
A discretetime signal is continuous in amplitude and discrete in time. It can either be present in nature or is sampled from an analog signal. A digital signal is discrete in amplitude and time.
Determine the product of two signals: x_{1} (n) = {2, 1, 1.5, 3}; x_{2} (n) = { 1, 1.5, 0, 2}.
Product of discretetime signals is computed element by element.
⇒ x(n) = x_{1} (n) * x_{2} (n) = {2×1, 1×1.5, 1.5×0, 3×2} = {2,1.5,0,6}.
Is the function y[n] = y[n  1] + x[n] stable in nature?
It is BIBO stable in nature, i.e. bounded inputbounded output stable.
Determine the discretetime signal: x(n) = 1 for n ≥ 0 and x(n) = 0 for n < 0
Unit step is defined by: x(n) = 1 for n ≥ 0 and x(n) = 0 for n < 0.
Determine the value of the summation: ∑^{∞}_{n= ∞} δ(n + 3)(n^{2 }+ n).
∑^{∞}_{n= ∞} δ(n + 3)(n^{2 }+ n)
⇒ δ(n + 3)=1 when n= 3 otherwise 0.
Therefore, the limit reduces to n = 3
⇒ ∑^{∞}_{n = ∞} δ(n + 3)(n^{2 }+ n) = (n^{2 }+ n)_{n = 3} = (3)^{2}3 = 9 – 3 = 6.
Unit Impulse function is obtained by using the limiting process on which among the following functions?
Unit impulse function can be obtained by using a limiting process on the rectangular pulse function. Area under the rectangular pulse is equal to unity.
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