# Circular Convolution as Cyclic Shift Operator

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.

• 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. Jan 2, 2021 at 18:41

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 $$\boldsymbol{e}_{i}^{2 N}$$.