# 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?

• i was studying the same book and I agree with your question, i think the author is wrong, 𝑛𝑇 is in "seconds" but M is in unit of samples – cs guy May 1 at 12:56

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 $$rad/s$$)

It seems like an incorrect notation.

When a function of time is sampled, $$t = nT_s$$ is considered where $$n$$ is the sample index and $$T_s$$ is the sampling period.

Generally, if a function of time is given as e.g. $$x(t)$$, the sampled function is written as $$x[n]$$ rather than $$x(nT_s)$$ but this is an acceptable notation too.

In the case of the given exponential function, let’s say $$y(t) = e^{j{\omega}t}$$, the sampled function will be $$y[n] = e^{j{\omega}n}$$.

If the sample function $$y[n]$$ is shifted $$M$$ samples towards the right-hand side of the sample axis for $$M>0$$, the resultant shifted function becomes $$y[n-M] = e^{j{\omega}[n-M]}$$.

Alternatively, if both the time and samples are taken into account at the same time, the shifted function can be written down as $$y((n-M)T_s) = e^{j{\omega}(n-M)T_s}$$.

However, the shifting notation $$(nT_s-{M})$$ is wrong as in this case, the shifting is done in the time domain as $$nT_s$$ is still representing a moment in time because $$n$$ is a constant value and the sampling period is multiplied by a constant. This process results in a moment in the units of seconds, milliseconds, etc.

This incorrect notation is analogous to subtracting a current value from a voltage value.

I prefer to use the sample function notation e.g. $$x[n]$$ rather than $$x(nT_s)$$ to minimise confusion.