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:

$$\begin{gathered} {x}_1\left(n\right)=\left\{\begin{array}{c}2\\ \uparrow \end{array}\begin{array}{c}1\\ \end{array}\begin{array}{c}2\\ \end{array}\begin{array}{c}1\\ \end{array}\right\}; \\ {x}_2\left(n\right)=\left\{\begin{array}{c}1\\ \uparrow \end{array}\begin{array}{c}2\\ \end{array}\begin{array}{c}3\\ \end{array}\begin{array}{c}4\\ \end{array}\right\} \end{gathered}$$

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

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

$y\left(n\right)=\left[2121\right]\left[\begin{array}{cccc}1& 2& 3& 4\\ 4& 1& 2& 3\\ 3& 4& 1& 2\\ 2& 3& 4& 1\end{array}\right]$

$y\left(n\right)=\left\{\begin{array}{c}14\\ \uparrow \end{array},\begin{array}{c}16\\ \end{array},\begin{array}{c}14\\ \end{array},\begin{array}{c}16\\ \end{array}\right\}$

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:

$$\begin{gathered} {x}_1\left(n\right)=\left\{\begin{array}{c}2\\ \uparrow \end{array}\begin{array}{c}1\\ \end{array}\begin{array}{c}2\\ \end{array}\begin{array}{c}1\\ \end{array}\right\}; \\ {x}_2\left(n\right)=\left\{\begin{array}{c}1\\ \uparrow \end{array}\begin{array}{c}2\\ \end{array}\begin{array}{c}3\\ \end{array}\begin{array}{c}4\\ \end{array}\right\} \end{gathered}$$

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

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

$y\left(n\right)=\left[2121\right]\left[\begin{array}{cccc}1& 2& 3& 4\\ 4& 1& 2& 3\\ 3& 4& 1& 2\\ 2& 3& 4& 1\end{array}\right]$

$y\left(n\right)=\left\{\begin{array}{c}14\\ \uparrow \end{array},\begin{array}{c}16\\ \end{array},\begin{array}{c}14\\ \end{array},\begin{array}{c}16\\ \end{array}\right\}$

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:

$$\begin{gathered} {x}_1\left(n\right)=\left\{\begin{array}{c}2\\ \uparrow \end{array}\begin{array}{c}1\\ \end{array}\begin{array}{c}2\\ \end{array}\begin{array}{c}1\\ \end{array}\right\}; \\ {x}_2\left(n\right)=\left\{\begin{array}{c}1\\ \uparrow \end{array}\begin{array}{c}2\\ \end{array}\begin{array}{c}3\\ \end{array}\begin{array}{c}4\\ \end{array}\right\} \end{gathered}$$

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

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

$y\left(n\right)=\left[2121\right]\left[\begin{array}{cccc}1& 2& 3& 4\\ 4& 1& 2& 3\\ 3& 4& 1& 2\\ 2& 3& 4& 1\end{array}\right]$

$y\left(n\right)=\left\{\begin{array}{c}14\\ \uparrow \end{array},\begin{array}{c}16\\ \end{array},\begin{array}{c}14\\ \end{array},\begin{array}{c}16\\ \end{array}\right\}$

Whenever in a discrete signal, the value of origin is not mentioned, the value of the signal is assumed to be origin value.

$\therefore x\left(n\right)=h\left(n\right)=\left\{\begin{array}{c}3\\ \uparrow \end{array},4,2,1\right\}$

$x\left(n-1\right)=\left\{\begin{array}{c}0\\ \uparrow \end{array},3,4,2,1\right\}$

$h\left(n+4\right)=\left\{3,4,2,1,\begin{array}{c}0\\ \uparrow \end{array}\right\}$

Using tabular method of convolution:

y(n) = x(n - 1) * h(n + 4)

$y\left(n\right)=\left\{0,9,24,28,\begin{array}{c}22\\ \uparrow \end{array},12,4,1,0\right\}$

y(0) = 22

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:

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

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

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

$y\left(n\right)=\left[1230\right]\left[\begin{array}{cccc}1& 2& 3& 4\\ 4& 1& 2& 3\\ 3& 4& 1& 2\\ 2& 3& 4& 1\end{array}\right]$

$y\left(n\right)=\left\{\begin{array}{c}18\\ \uparrow \end{array},\begin{array}{c}16\\ \end{array},\begin{array}{c}10\\ \end{array},\begin{array}{c}16\\ \end{array}\right\}$

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:

$$\begin{gathered} {x}_1\left(n\right)=\left\{\begin{array}{c}2\\ \uparrow \end{array}\begin{array}{c}1\\ \end{array}\begin{array}{c}2\\ \end{array}\begin{array}{c}1\\ \end{array}\right\}; \\ {x}_2\left(n\right)=\left\{\begin{array}{c}1\\ \uparrow \end{array}\begin{array}{c}2\\ \end{array}\begin{array}{c}3\\ \end{array}\begin{array}{c}4\\ \end{array}\right\} \end{gathered}$$

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

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

$y\left(n\right)=\left[2121\right]\left[\begin{array}{cccc}1& 2& 3& 4\\ 4& 1& 2& 3\\ 3& 4& 1& 2\\ 2& 3& 4& 1\end{array}\right]$

$y\left(n\right)=\left\{\begin{array}{c}14\\ \uparrow \end{array},\begin{array}{c}16\\ \end{array},\begin{array}{c}14\\ \end{array},\begin{array}{c}16\\ \end{array}\right\}$

given signals are-

$\begin{array}{rl}& x\left[n\right]=\left\{1,2,-\underset{\uparrow }{3},4\right\}\\ & h\left[n\right]=\left\{2,\underset{\uparrow }{5}\right\}\end{array}$

and let the output convolved signal is -  

$y\left[n\right]=x\left[n\right]\otimes h\left[n\right]$

To find 4 – point circular convolution make length of the input signals equal to 4

i.e

$\begin{array}{rl}& x\left[n\right]=\left\{\begin{array}{cccc}1& 2& -3& 4\\ & & \uparrow & \end{array}\right\}\mathrm{o}\mathrm{r}x\left[n\right]=\left\{\begin{array}{cccc}-3& 4& 1& 2\\ \uparrow & & & \end{array}\right\}\\ & h\left[n\right]=\left\{\begin{array}{cccc}2& 5& 0& 0\\ & \uparrow & & \end{array}\right\}\mathrm{o}\mathrm{r}h\left[n\right]=\left\{\begin{array}{cccc}5& 0& 0& 2\\ \uparrow & & & \end{array}\right\}\\ & y\left[n\right]=\left[\begin{array}{cccc}-3& 2& 1& 4\\ 4& -3& 2& 1\\ 1& 4& -3& 2\\ 2& 1& 4& -3\end{array}\right]\left[\begin{array}{c}5\\ 0\\ 0\\ 2\end{array}\right]\\ & =\left[\begin{array}{ccc}-3\times 5& +& 4\times 2\\ 4\times 5& +& 1\times 2\\ 1\times 5& +& 2\times 2\\ 2\times 5& +& -3\times 2\end{array}\right]=\left[\begin{array}{c}-7\\ 22\\ 9\\ 4\end{array}\right]\\ & \Rightarrow y\left[n\right]=\left\{\begin{array}{cccc}-7& 22& 9& 4\\ \uparrow & & & \end{array}\right\}\\ & \Rightarrow y\left[3\right]=4\end{array}$