# Understanding sampling rate conversion

I am a little bit confused about the concept of sampling rate conversion. Consider the following question:

Wideband telephony signals are sampled at a sampling frequency fS1 = 16kHz. To process these signals with an automatic speech recognizer, the sampling rate is converted here to fS2 = 11.025kHz, i.e., the wideband signal is ﬁrst upsampled by a factor U, lowpassﬁltered, and ﬁnally downsampled by a factor R. Determine the lowest possible values for R and U.

I am not sure whether U / R = 16kHz / 11.025 kHz, or vice-versa?

## 1 Answer

You would upsample by 441 and downsample by 640 since $$11025 \cdot 640 = 16000 \cdot 441$$

That would be a rather expensive operation, so in practice you would either fugde it a bit: 31/45 would get you to 11022Hz and 11/16 to 11000 Hz or you would use a suitable poly-phase irrational conversion algorithm

• Thanks. I am confused, because I thought since we're going from 16k to 11.025k, it means that we will have more sample (in time-domain?!), so we should upsample by a factor of 640 and downsample by 441. – Niousha Mar 27 '19 at 21:23
• i dunno that i would associate polyphase with "irrational" ratios of sample rates. maybe polyphase with a lot of phases and linear interpolation for arbitrary conversion ratios. – robert bristow-johnson Mar 27 '19 at 21:30