Convolution MCQ Quiz - Objective Question with Answer for Convolution - Download Free PDF

Concept

In digital signal processing, convolution is a mathematical operation that combines two sequences to produce a third sequence. The length of the linear convolution of two sequences depends on the lengths of the individual sequences. If the lengths of the two sequences are M and N, respectively, the length of their linear convolution is given by:

Length = M + N - 1

Given signal

\(\rm x(t)=\left\{\begin{matrix}1,&-1

h(t) = δ(t + 1) + 2δ(t + 2)

According to convolution property of impulse

x(t) ⋆ δ(t - t₀) = x(t - t₀)

∴ x(t) ⋆ h(t) = x(t) ⋆ [δ(t + 1) + 2δ(t + 2)]

= x(t + 1) + 2x(t + 2)

Therefore, correct option is (3)

Concept:

Convolution is a mathematical operation on two functions that produces a third function that expresses how the shape of one is modified by the other.

The term convolution refers to both the result function. It is a mathematical way of combining two signals to form a third signal.

It is the single most important technique in Digital Signal Processing. Tabular Method is one of the methods to solve it.

x1 (n)  =  [a₁, a₂, a₃] 

x2 (n)  =  [b₁, b₂, b₃]

Calculation:

x1 (n)  =  [1, 2, 3] 

x2 (n)  =  [3, 2, 1] 

Solving by the help of Tabular Method

y(n) = x1 (n) ∗ x2 (n)

y(n) = x1 (n) ∗ x2 (n) =  [3 , 2 + 6 , 1 + 4 + 9 , 2 + 6 , 3] =  [3,8,14,8,3]

Important Points

Length Property:

The length of the y(n) signal can be calculated as:

Length of x1 (n)  +  Length of x2 (n)  -  1

3 + 3 - 1 = 5

g[n] = {…, 0, 0, 1, 0, 0, …} = δ[n]

h[n] = {1, 0, 0, 1, 0, 0, 1, 0, 0, ....}

h[n] = δ[n] + δ[n - 3] + δ[n - 6] + δ[n - 9] + …

y[n] = h[n] ⊗ g[n]

= h[n] ⊗ δ[n]

= h[n]

Any signal convoluted with the impulse signal gives the same signal.

= δ[n] + δ[n - 3] + δ[n - 6] + δ[n - 9] + …

Since δ[n - 2] term is missing in y[n] expression, y[2] = 0.

Given signal

\(\rm x(t)=\left\{\begin{matrix}1,&-1

h(t) = δ(t + 1) + 2δ(t + 2)

According to convolution property of impulse

x(t) ⋆ δ(t - t₀) = x(t - t₀)

∴ x(t) ⋆ h(t) = x(t) ⋆ [δ(t + 1) + 2δ(t + 2)]

= x(t + 1) + 2x(t + 2)

Therefore, correct option is (3)

g[n] = {…, 0, 0, 1, 0, 0, …} = δ[n]

h[n] = {1, 0, 0, 1, 0, 0, 1, 0, 0, ....}

h[n] = δ[n] + δ[n - 3] + δ[n - 6] + δ[n - 9] + …

y[n] = h[n] ⊗ g[n]

= h[n] ⊗ δ[n]

= h[n]

Any signal convoluted with the impulse signal gives the same signal.

= δ[n] + δ[n - 3] + δ[n - 6] + δ[n - 9] + …

Since δ[n - 2] term is missing in y[n] expression, y[2] = 0.

Concept:

Convolution is a mathematical operation on two functions that produces a third function that expresses how the shape of one is modified by the other.

The term convolution refers to both the result function. It is a mathematical way of combining two signals to form a third signal.

It is the single most important technique in Digital Signal Processing. Tabular Method is one of the methods to solve it.

x1 (n)  =  [a₁, a₂, a₃] 

x2 (n)  =  [b₁, b₂, b₃]

Calculation:

x1 (n)  =  [1, 2, 3] 

x2 (n)  =  [3, 2, 1] 

Solving by the help of Tabular Method

y(n) = x1 (n) ∗ x2 (n)

y(n) = x1 (n) ∗ x2 (n) =  [3 , 2 + 6 , 1 + 4 + 9 , 2 + 6 , 3] =  [3,8,14,8,3]

Important Points

Length Property:

The length of the y(n) signal can be calculated as:

Length of x1 (n)  +  Length of x2 (n)  -  1

3 + 3 - 1 = 5

$\begin{array}{rl}& x\left(t\right)=u\left(t-2\right)\ast u\left(t-3\right)\ast \delta \left(t+1\right)\\ & =r\left(t-2-3\right)\ast \delta \left(t+1\right)\\ & =r\left(t-5\right)\ast \delta \left(t+1\right)\\ & x\left(t\right)=r\left(t-5+1\right)=r\left(t-4\right)\\ & \underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}x\left(t\right).{\delta }^{\prime }\left(t-1\right)dt={[-\frac{d}{dt}x\left(t\right)|}_{t=1}\\ & ={-\frac{d}{dt}r\left(t-4\right)|}_{t=1}\\ & ={-u\left(t-4\right)|}_{t=1}\\ & =0\end{array}$

Given signal

\(\rm x(t)=\left\{\begin{matrix}1,&-1

h(t) = δ(t + 1) + 2δ(t + 2)

According to convolution property of impulse

x(t) ⋆ δ(t - t₀) = x(t - t₀)

∴ x(t) ⋆ h(t) = x(t) ⋆ [δ(t + 1) + 2δ(t + 2)]

= x(t + 1) + 2x(t + 2)

Therefore, correct option is (3)

$y\left(t\right)=\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}x\left(\tau \right)h\left(t-\tau \right)d\tau$ 

$=\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}{e}^{-\tau }u\left(\tau \right).e^{-2\left(t-\tau \right)}u\left(t-\tau \right)d\tau$ 

$=f_0^t{e}^{-\tau }{e}^{-2\left(t-\tau \right)}.d\tau ;t\ge 0=e^{-2t}\underset{0}{\overset{t}{\int }}{e}^{\tau }d\tau$ 

$=e^{-2t}\left(e^t-1\right);t\ge 0$ 

Since, y(t) = 0 for t < 0

Concept

In digital signal processing, convolution is a mathematical operation that combines two sequences to produce a third sequence. The length of the linear convolution of two sequences depends on the lengths of the individual sequences. If the lengths of the two sequences are M and N, respectively, the length of their linear convolution is given by:

Length = M + N - 1

g[n] = {…, 0, 0, 1, 0, 0, …} = δ[n]

h[n] = {1, 0, 0, 1, 0, 0, 1, 0, 0, ....}

h[n] = δ[n] + δ[n - 3] + δ[n - 6] + δ[n - 9] + …

y[n] = h[n] ⊗ g[n]

= h[n] ⊗ δ[n]

= h[n]

Any signal convoluted with the impulse signal gives the same signal.

= δ[n] + δ[n - 3] + δ[n - 6] + δ[n - 9] + …

Since δ[n - 2] term is missing in y[n] expression, y[2] = 0.

T = 2, ω₀ = π

$x\left(t\right)\ast z\left(t\right)\stackrel{FS,\pi }{\leftrightarrow }TX\left[k\right]Z\left[k\right]$

$x\left(t-1\right)\stackrel{FS,\pi }{\leftrightarrow }{e}^{-jk\pi }X\left[k\right]$

$\mathrm{x}\left(\mathrm{t}\right)\leftrightarrow \frac{{\mathrm{e}}^{\mathrm{j}\frac{\pi }{4}\mathrm{t}}-{\mathrm{e}}^{-\mathrm{j}\frac{\pi }{4}\mathrm{t}}}{2\mathrm{j}}$

$\mathrm{y}\left(\mathrm{t}\right)\leftrightarrow \frac{{\mathrm{e}}^{\mathrm{j}\frac{\pi }{4}\mathrm{t}}+{\mathrm{e}}^{-\mathrm{j}\frac{\pi }{4}\mathrm{t}}}{2}$

Now, $\mathrm{z}\left(\mathrm{t}\right)=\mathrm{x}\left(\mathrm{t}\right)\star \mathrm{y}\left(\mathrm{t}\right)\leftrightarrow {\mathrm{c}}_{\mathrm{k}}=\mathrm{T}.{\mathrm{a}}_{\mathrm{k}}.{\mathrm{b}}_{\mathrm{k}}$

Where

$\mathrm{z}\left(\mathrm{t}\right)\leftrightarrow {\mathrm{c}}_{\mathrm{k}}$

$\mathrm{x}\left(\mathrm{t}\right)\leftrightarrow {\mathrm{a}}_{\mathrm{k}}$

$\mathrm{y}\left(\mathrm{t}\right)\leftrightarrow {\mathrm{b}}_{\mathrm{k}}$

Thus,

Now, 

${\mathrm{C}}_{\mathrm{k}}=\mathrm{T}.{\mathrm{C}}_{\mathrm{k}}^1=\left(\frac{2\pi }{\frac{\pi }{4}}\right)\left({\mathrm{e}}^{\mathrm{j}\frac{\pi }{4}\mathrm{t}}-{\mathrm{e}}^{-\mathrm{j}\frac{\pi }{4}\mathrm{t}}\right)\frac{1}{4\mathrm{j}}$

$\Rightarrow \mathrm{z}\left(\mathrm{t}\right)=8\left(\frac{1}{4\mathrm{j}}\right)\left(2\mathrm{j}\sin \left(\frac{\pi }{4}\mathrm{t}\right)\right)\Rightarrow \mathrm{z}\left(\mathrm{t}\right)=4\sin \left(\frac{\pi }{4}\mathrm{t}\right)$