Test: Discrete Time Signals - Electronics and Communication Engineering (ECE) MCQ

1) Functional Representation.

Example:

2) Tabular method of representation

Example:

3) Sequence Representation

Example:

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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.

Non-Recursive discrete time system:

If y(n) depends only on the present and past input, it is called non-recursive. It does not exhibit the necessity of any feedback.

If the two discrete signals are having the length ‘n’ and ‘m’ respectively then the resultant output signal has the length as n + m – 1

The convolution of signals in one domain is equivalent to the multiplication of signals in another domain.

Given y[n] = x[n] *h[n]

Operator * denotes the convolution of two signals.

The length of the output signal is y[n] = 3 + 5 – 1 = 7

Y(e^jω) = F(e^jω)H(e^jω)

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 - n₀] = 1 for n ≥ n₀

u[n - n₀] = 0 for n < n₀

Now the combination of two discrete time signals can be analysed as:

u[n] - u[n - n₀] = 1 for 0 ≤ n ≤  n₀ - 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]

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{ax₁[t] + bx₂[t]} = a y{x₁[t]} + b y{x₂[t]}

Conditions to check whether the system is linear or not.

• The output should be zero for zero input
• There should not be any non-linear operator present in the system.

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:

x₁(n) → x₁(n)²

x₂(n) → x₂(n)²

x₃(n) = [x1(n) + x₂(n)] →  [x1(n) + x2(n)]² = x₁(n)² + x₂(n)² + 2x₁(n) x₂(n)

≠ x1(n)² + x₂(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 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)

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.

A discrete-time 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.

Product of discrete-time signals is computed element by element.
⇒ x(n) = x₁ (n) * x₂ (n) = {2×1, 1×1.5, 1.5×0, 3×2} = {2,1.5,0,6}.

Unit step is defined by: x(n) = 1 for n ≥ 0 and x(n) = 0 for n < 0.

∑^∞_{n= -∞} δ(n + 3)(n² + n)
⇒ δ(n + 3)=1 when n= -3 otherwise 0.
Therefore, the limit reduces to n = -3
⇒ ∑^∞_{n = -∞} δ(n + 3)(n² + n) = (n² + n)|_{n = -3} = (-3)²-3 = 9 – 3 = 6.

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.

Concept:

To find the sum of two discrete-time signals, we simply add the corresponding elements at each index.

So, for n = 0, x₁(0) + x₂(0) = 2 + (-2) = 0.

For n = 1, 2 + (-1) = 1.

For n = 2, 1 + 3 = 4

For n = 3, 2 + 2 = 4

Continuing this process for n = 2 and n = 3, we get {0, 1, 4, 4}.

Given that, x(t) = cos (2πt)

Sampling frequency = 4 Hz

Time period (T) = ¼

replace t = nT_s

Discrete time sequence

x(n) = x(nT_s)