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Question about delayed sampled sinusoid math expression
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June 18, 2022 - 10:18 pm
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I have been studying the digital audio processing by using the book .
According to the book at page 199, for analog Sinusoid:

Complex Sinusoid = e^(jwt)

Delayed Sinusoid = e^(jw(t-n))=e^(jwt) * e^(-jwn), a delay of n seconds

For digital sampled version:

sampled complex sinusoid = e^(jwnT), 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^(jw(nT - M)) , M = samples of delay

But I think it should be described as e^(jwT(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^(jw(nT-M))as basis and drew some conclusion

Anyone can explain it for me?

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June 28, 2022 - 1:10 pm
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The book says in that section:

For the digitized signal to be delayed by M samples you multiply it by e-jωM. This is useful because neither e-jωn nor e-jωM is dependent on the time variables t or nT.

So, yeah, it can't be a typo. I can't explain this authoritatively, but I can share my take on it.

The way I see it, since M represents previous samples, it already includes in its definition the value of T.

To put it differently, the delayed sample M was at some point in the past the "current" sample. Sample M = 0 is actually the current sample and it is defined by T. When this same sample is not the current sample anymore, its relation to T is already "baked" into it.

Because of that, only the current sample needs its relation to T to be defined. The previous sample, M = 1, was already defined by T back when it was the current sample.

I don't know if I'm correct or if I make any sense... but that's how I see it. I do have the feeling though that I'm missing something...

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June 29, 2022 - 8:22 am
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I think Aro is right, there is no typo here and that your mistake may be in thinking that the units of M and n are the same. 

As you mentioned n is time in seconds. 

M on the other hand is time in samples. And the sampling rate is entirely arbitrary.

M is linked to n by the sampling rate interval (T) which itself is the ratio of a second of time to the number of samples that will fit in that 1 second.

writing nT puts that half of the equation into samples, which is the same unit as M.

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July 9, 2022 - 4:32 am
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Thank you for your replies. I think I got the point: it is a false representation.

Let's say delay a signal for M samples.

It definitely should be like this:e^(jwT(n - M)), break it apart to e^(jwnT) * e^(-jwMT)

The book wrote like this, e^(jw(nT - M)), and break it apart, it's e^(jwnT) * e^(-jwM)

What's wrong with the book is these two w in different exponents do not have the same unit, you can't use the same notation w in a single equation!!! The first one in e^(jwnT) has unit radians/sec for a specific signal, and the second w has unit radians/sample!!!

From the top equation, e^(-jwMT), we mutiply w(radians/sec) and T, wT = (2Pi*f) * (1/f_s) = 2Pi*(f/f_s) = w(raidans/sample). 

That's why in the chapter following when he evaluates the transfer function for 1st order feedforward filter:

H(w) = a_0 + a_1*e^(-jw1), the w(radians/sample)

for the Nyquist frequency, he substitutes the w by using Pi, because the Nyquist Frequency has Pi(radians/sample) speed of rotation.

 

Of course, we could stick to using w as radians/sec for the transfer function, it will be like these:

H(w) = a_0 + a_1*e^(-jw1T), the w(radians/sec)

for the Nyquist frequency, instead of substituting the w by using pi, we should use 2Pi*f_n = 2Pi*f_s/2 = Pi*f_s = Pi/T

then put the Pi/T back into the equation, the T in the equation just crossed out!!! So we could get the same result compared with using w(radians/sample).

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September 15, 2022 - 5:58 pm
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FKSky said
Thank you for your replies. I think I got the point: it is a false representation.

Let's say delay a signal for M samples.

It definitely should be like this:e^(jwT(n - M)), break it apart to e^(jwnT) * e^(-jwMT)

The book wrote like this, e^(jw(nT - M)), and break it apart, it's e^(jwnT) * e^(-jwM)

What's wrong with the book is these two w in different exponents do not have the same unit, you can't use the same notation w in a single equation!!! The first one in e^(jwnT) has unit radians/sec for a specific signal, and the second w has unit radians/sample!!!

 

I understand your point, but I don't know enough about this to agree or disagree. It would be interesting to know more...

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September 16, 2022 - 11:16 am
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The derivation here actually comes from the Steiglitz book in the references. I used this same derivation verbatim in the 1st edition of the book. It is interesting in that uses mixes analog time (t) being delayed by analog seconds (Tau) then, and then sort of shoe-horns it into a digital version. I don't want to go further into that, but definitely check out that source if you don't have it. Also, I have noticed that DSP books will use normalized sample rates and times so that fs = T = 1 and the notion of sample periods T drops out of the equations. 

Ultimately, the point is that delaying a signal by M samples means multiplying it by the second term in my version, e^-jwM -- and most importantly, this term is independent of the sample rate fs or period T and so it becomes z^-M and not z^-MT. But it comes down to semantics at that point since a delay of M samples is a delay of MT seconds. 

Though I have stopped writing books, if I ever was coerced into doing a 3rd edition at gunpoint, I would reframe this part of the discussion to align it with other books and remove this confusion. 

WP

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September 16, 2022 - 4:34 pm
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Thanks for the clarification

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September 19, 2022 - 10:38 am
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No problem - I should have stuck with t and Tau, or n and M and tried not to inject the sample period, which resulted it another hybrid equation. 

WP

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