# Question about delayed sampled sinusoid math expression

I have been studying the digital audio processing by using the book <Designing Audio Effect Plugins in C++>.

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?

$$e^{j \omega ((n - k)T)}$$ 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 $$rad/s$$)