Test: Continuous Time Convolution - Electrical Engineering (EE) MCQ

Convolution of a function with a delta function shifts accordingly.

Convolution of a function with a delta function shifts accordingly.

Use the convolution formula.

Divide it into 2 sectors and apply the convolution formula.

Convolution of a function with a delta function shifts accordingly.

Divide it into 2 sectors and apply the convolution formula.

Convolution of a function with a delta function shifts accordingly.

Apply the convolution formula. The above corollary exists for any x(t) [not impulsive].

The resultant impulse response will be the convolution of all the subsequent impulse responses.

The resultant impulse response will be the convolution of all the subsequent impulse responses.

The resultant impulse response will be the convolution of all the subsequent impulse responses.

The delta function convolved with another function results in the shifted function.

The resultant impulse response will be the convolution of all the subsequent impulse responses.

The resultant impulse response will be the convolution of all the subsequent impulse responses.

By using the impulse response of a system, convolution can be used to calculate a system's zero state response (i.e., its response when it has zero initial conditions) to an arbitrary input. Linearity and superposition are used. 
The convolution can be defined as:

u(t)*u(t) = r(t)

Calculation:

Given signals are x(t) = u(t - 1) - u(t - 3) and h(t) = u(t) - u(t - 2) 

x(t) and h(t) is graphically represented by:

c(t) = x(t) * h(t)

c(t) = [u(t -1) - u(t - 3)]*[u(t) - u(t -2)]

c(t) = u(t - 1)*u(t) - u(t -1)*u(t - 2) - u(t - 3)*u(t) + u(t - 3)*u(t - 2)

c(t) = r(t - 1) - r(t - 3) - r(t - 3) + r(t + 5)

c(t) = r(t - 1) -2r(t - 3) + r(t + 5)
The output signal is represented as:

Discrete Time System:

The convolution of two a signal with a system with impulse response h(n) is represented as:
y[n] = x[n] ∗ h[n]
Continuous Time System:

The convolution of two a signal with a system with impulse response h(t) is represented as:
y(t) = x(t) ∗ h(t)

x(n) = {2, 3, -1, -4}
h(n) = {2, 4, 1}


Final Sample = -4
Sum = 4 + (-4) = 0

Concept:

By using the impulse response of a system, convolution can be used to calculate a system's zero state response (i.e., its response when it has zero initial conditions) to an arbitrary input.  Linearity and superposition are used. 
The convolution can be defined as:

u(t)*u(t) = r(t)

Calculation:

Given signals are x(t) = u(t - 1) - u(t - 3) and h(t) = u(t) - u(t - 2) 

x(t) and h(t) is graphically represented by:

c(t) = x(t) * h(t)

c(t) = [u(t -1) - u(t - 3)]*[u(t) - u(t -2)]

c(t) = u(t - 1)*u(t) - u(t -1)*u(t - 2) - u(t - 3)*u(t) + u(t - 3)*u(t - 2)

c(t) = r(t - 1) - r(t - 3) - r(t - 3) + r(t + 5)

c(t) = r(t - 1) -2r(t - 3) + r(t + 5)

The output signal is represented as:

Discrete Time System:

The convolution of two a signal with a system with impulse response h(n) is represented as:

y[n] = x[n] ∗ h[n]

Continuous Time System:

The convolution of two a signal with a system with impulse response h(t) is represented as:

y(t) = x(t) ∗ h(t)

x(t) = δ(t + 1) – δ(t – 1)

x(t) * h(t) = y(t)

= h(t + 1) – h(t – 1)



y(t) = h(t + 1) – h(t – 1)