Advanced Signals and Systems - Linear and Cyclic Convolution

 

17. Linear convolution of sequences.

Task

Find the convolution sum \(v(n) = v_1(n) \ast v_2(n)\) of the following sequences \(v_1(n)\) and \(v_2(n)\)

\begin{align} v_1(n) =& \rho _1^n \cdot \gamma_{-1}(n) \nonumber \\ v_2(n) =& \rho _2^n \cdot \gamma_{-1}(n)\nonumber \end{align}

where \(0 < \rho_1\) and \( \rho_2 < 1\).

Amount and difficulty

  • Working time: approx. xx minutes
  • Difficulty: xx

Additional material

Solution

\begin{align*} v(n) &= \sum \limits_{k=-\infty}^{\infty} v_1(k) \cdot v_2(n-k) \\ &= \sum \limits_{k=-\infty}^{\infty} \rho_1^k \cdot \gamma_{-1}(k) \cdot \rho_2^{n-k} \cdot \gamma_{-1}(n-k) \\ &= \cdots\\ &= \begin{cases} \frac{\rho_1^{n+1}-\rho_2^{n+1}}{\rho_1-\rho_2}\cdot \gamma_{-1}(n) &, \rho_1\neq \rho_2\\ \rho_2^n\cdot(n+1)\cdot \gamma_{-1}(n) &, \rho_1 = \rho_2 \end{cases}\end{align*}

 

18. Linear and cyclic convolution.

Task

Given two sequences \(v_1(n)\) and \(v_2(n)\) of length \(M=5\):

\begin{align} v_1(n) =& [5,4,3,2,1] \nonumber \\ v_2(n) =& [1,2,3,4,5]\nonumber \end{align}

Determine the linear convolution \(v_3(n)\) and the cyclic convolution \(v_4(n)\) of the sequences. Give a method to calculate the linear convolution.

Amount and difficulty

  • Working time: approx. xx minutes
  • Difficulty: xx

Additional material

Linear convolution

Cyclic convolution

Calculating the linear convolution by cyclic convolution:

Solution

  • Linear convolution

    \begin{align*} v_3(n) &= v_1(n) * v_2(n) = \sum \limits_{k=-\infty}^{\infty} v_1(k) \cdot v_2(n-k)\\ &\cdots\\ v_3(n) &= [5, 14, 26, 40, 55, 40, 26, 14, 5] \ \ \ \text{ for } \ \ 0\leq n \leq M-1 \end{align*}

  • Cyclic convolution

    \begin{align*} v_4(n) &= \sum \limits_{k=-\infty}^{\infty} v_1(k) \cdot v_2(\text{mod}(n-k,M))\\ &\cdots\\ v_4(n) &= [45, 40, 40, 45, 55] \end{align*}

  • Modified cyclic convolution to get correct result

    The cyclic convolution of \(v'_1(n)\) and \(v'_2(n)\) is calculated, where \(v'_1(n)\) and \(v'_2(n)\) denote the sequences \(v_1(n)\) and \(v_2(n)\) padded with \(M-M_2\) and \(M-M_1\) zeros, where \(M\) is the length of the resulting sequence \(v_5(n)\) and \(M=M_1+M_2+1\).

    \begin{align*} v_5(n) &= \sum \limits_{k=-\infty}^{\infty} v'_1(k) \cdot v'_2(\text{mod}(n-k,M))\\ &\cdots\\ v_5(n) &= [5, 14, 26, 40, 55, 40, 26, 14, 5] \ \ \ \text{ for } \ \ 0\leq n \leq M-1 \end{align*}