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

Concept:

Convolution in the time domain results in multiplication in the frequency domain.

To find circular convolution of two signals we can follow the following steps:

• First, make the length of the signals equal to N by adding extra zeros if needed.         
• Form two matrices, 1st matrix using the cyclic rotation of one of the signals and 2nd matrix with another signal.
• Multiply the two matrices.

Calculation:

Given:

$$\begin{gathered} x_1\left(n\right)=\left\{\begin{array}{c}1\\ \uparrow \end{array}\begin{array}{c}2\\ \end{array}\begin{array}{c}2\\ \end{array}\right\}; \\ {x}_2\left(n\right)=\left\{\begin{array}{c}1\\ \uparrow \end{array}\begin{array}{c}2\\ \end{array}\begin{array}{c}3\\ \end{array}\begin{array}{c}4\\ \end{array}\right\} \end{gathered}$$

$y\left(n\right)=x_1\left(n\right)⊛x_2\left(n\right)$

$=\left\{\begin{array}{c}1\\ \uparrow \end{array},\begin{array}{c}2\\ \end{array},\begin{array}{c}2\\ \end{array},\begin{array}{c}0\\ \end{array}\right\}⊛\left\{\begin{array}{c}1\\ \uparrow \end{array},\begin{array}{c}2\\ \end{array},\begin{array}{c}3\\ \end{array},\begin{array}{c}4\\ \end{array}\right\}$

$y\left(n\right)=\left[1220\right]\left[\begin{array}{cccc}1& 2& 3& 4\\ 4& 1& 2& 3\\ 3& 4& 1& 2\\ 2& 3& 4& 1\end{array}\right]$

$y\left(n\right)=\left\{\begin{array}{c}15\\ \uparrow \end{array},\begin{array}{c}12\\ \end{array},\begin{array}{c}9\\ \end{array},\begin{array}{c}14\\ \end{array}\right\}$

Concept:

Convolution in the time domain results in multiplication in the frequency domain.

To find circular convolution of two signals we can follow the following steps:

• First, make the length of the signals equal to N by adding extra zeros if needed.         
• Form two matrices, 1st matrix using the cyclic rotation of one of the signals and 2nd matrix with another signal.
• Multiply the two matrices.

Calculation:

Given:

$$\begin{gathered} x_1\left(n\right)=\left\{\begin{array}{c}1\\ \uparrow \end{array}\begin{array}{c}2\\ \end{array}\begin{array}{c}2\\ \end{array}\right\}; \\ {x}_2\left(n\right)=\left\{\begin{array}{c}1\\ \uparrow \end{array}\begin{array}{c}2\\ \end{array}\begin{array}{c}3\\ \end{array}\begin{array}{c}4\\ \end{array}\right\} \end{gathered}$$

$y\left(n\right)=x_1\left(n\right)⊛x_2\left(n\right)$

$=\left\{\begin{array}{c}1\\ \uparrow \end{array},\begin{array}{c}2\\ \end{array},\begin{array}{c}2\\ \end{array},\begin{array}{c}0\\ \end{array}\right\}⊛\left\{\begin{array}{c}1\\ \uparrow \end{array},\begin{array}{c}2\\ \end{array},\begin{array}{c}3\\ \end{array},\begin{array}{c}4\\ \end{array}\right\}$

$y\left(n\right)=\left[1220\right]\left[\begin{array}{cccc}1& 2& 3& 4\\ 4& 1& 2& 3\\ 3& 4& 1& 2\\ 2& 3& 4& 1\end{array}\right]$

$y\left(n\right)=\left\{\begin{array}{c}15\\ \uparrow \end{array},\begin{array}{c}12\\ \end{array},\begin{array}{c}9\\ \end{array},\begin{array}{c}14\\ \end{array}\right\}$