Question about delayed sampled sinusoid math expression I have been studying the digital audio processing by using the book &lt;Designing Audio Effect Plugins in C++&gt;.
For analog Sinusoid:
Complex Sinusoid = $e^{jωt}$
Delayed Sinusoid = $e^{jω(t−n)} = e^{jwt} * e^{-jwn}$, a delay of n seconds
For digital sampled version:
sampled complex sinusoid = $e^{jωnT}$, T is interval for each sample, n is the index of sample
I understand all above, but I got confused about the delayed sampled sinusoid which described as: $e^{jω ( nT −M )}$, M = samples of delay
But I think it should be described as $e^{jωT( n − M )}$, since the T is a constant for a fixed sample rate, n and M has the same unit.
At first I thought that maybe a typo, but the following computation parts of the book are all using the $e^{jω ( nT −M )}$ as basis.
Anyone can explain it for me?
Indeed a typo
$e^{j \omega ((n - k)T)}$ would be the delay for $k$ samples, the $M$ in the expression $e^{j \omega(nT - M)}$ is the delay (in seconds if $\omega$ is in $rad/s$)