# How to explain IDFT of square root of a DFT?

I have this MATLAB code:

w = [1 2 3 2 1];
W=fft(w);
V=sqrt(W);
v=ifft(V);


And I need to explain how to get to $v[n]$ without using the direct formula for DFT - I need to use the properties of DFT.

I've tried solving the problem using circular shift and circular convolution but I can't seem to understand the result.

According to MATLAB, $v[n]=\textrm{[1 1 1 0 0]}$.

• Is this home work ? Jan 1 '17 at 23:53
• No, I'm studying for an exam and this is the only exercise I can't solve Jan 2 '17 at 0:03

Since $V(\omega)V(\omega)=W(\omega)$, in the time domain we expect $v[n]*v[n]=w[n]$. That is, the convolution of $v[n]$ with itself should become $w[n]$.
$w[n]$ is in the form of a triangular signal. A triangular signal of odd length $N$ can be constructed by convolving a half-rectangular signal of length $(N+1)/2$ with itself. So $v[n]$ is a rectangular pulse with $3$ elements equal to one and the remaining $2$ elements equal to zero.