Given the following signal vectors: $$ γ=[ψ_0,0,ψ_1,0,ψ_2,0,…,ψ_{N-1},0]^T\in \mathbb{R}^{2N}, ϕ=[1,\frac{1}{2},0,…,0,\frac{1}{2}]^T \in \mathbb{R}^{2N}$$

I want to show that the convolution of $γ$ and $ϕ$ is actually a cyclical-shift operator. i.e.: $$γ*ϕ = γ + \frac{1}{2}T_1(γ) + \frac{1}{2}T_{2N-1}(γ) $$ Where $T_{t_0}$ is cyclical-shift operator with offset of $t_0$ places.

I tried to develop it according to the definition of (circular) convolution: $$h_l=∑_{n=0}^{2N-1} γ_n ϕ_{l-n (mod \space 2N) } $$

However, I couldn't come up with something helpful...

How can I prove this identity? Thanks.

  • 5
    $\begingroup$ Convolution is a linear operation. Express $\phi$ as the sum of three simpler vectors, convolve $\gamma$ by the three simpler vectors individually, then add up the three individual convolutions into the final result. $\endgroup$ – Andy Walls Jan 2 at 18:41

You may solve it by 3 steps:

  1. Show yourself that a Cyclic Convolution with a vector $ \boldsymbol{e}_{i}^{N} $ is a Cyclic Shift Operator $ {T}_{i - 1} \left( \cdot \right) $. Where $ \boldsymbol{e}_{i}^{N} $ is defined as a vector of length $ N $ which all its elements is zero but the $ i $- th element which is 1.
  2. Decompose $ \phi $ into $ 3 $ vectors of type $ \boldsymbol{e}_{i}^{2 N} $.
  3. Use linearity of the Cyclic Convolution to prove your assertion.

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