# Why discrete wavelet transforms use sampling rate 2 and need signal length to be a power of 2?

I know Fourier transforms but new to wavelet transforms. I can understand Haar transform needs signal length a power of 2, since the filters have 2 taps and down-sampling and up-sampling in the pyramid algorithm form binary tree structures, so it is efficient to implement if signal length is a power of 2. But higher-order transforms (e.g. DB2) or other kinds also use sampling rate of 2 and need signal length a power of 2. Is it also for efficient implementation? Can the sampling rate be other small prime numbers 3, 5 like in FFT?

• I ran across a mention of a form of wavelet transform that uses non-$2^n$ decimation, but I can't find it now. So yes, it's out there -- but because I can't put my finger on it for you, I don't feel that this comment is really an answer. – TimWescott Dec 21 '20 at 16:09