Are Trackers 'geeky'?

I got told by a mate (in real life) that trackers are the music sequencer choice for the geeks. now I have no idea what that is about!? is it the hex? 00 to FF isn’t rocket science! and I’m not a pale, spotty glasses wearing maths nerd! heh

infact, trackers rock and you know it! :drummer:

Pretty much anything that involves a computer is labeled geeky in most people’s books.

But seriously? How is tracking any more or less “geeky” than using a paino roll? It’s all technical as hell.
Now if your mate is a traditional musician, :guitar: he’s just saying this out of fear that the kudos will one day turn the opposite direction in which he stands. :o

People need to deal, computers are not going anywhere any time soon.

From one closet geek to another, track on dude! :walkman:

You must have missed that the minimum system requirements for Renoise included “neckbeard”. RTFM

sure its for geeks.
I attended a LAN party on my birthday, and I’m a renoise user…
see?
:D

people use renoise because it’s a great sequencer, it has little or nothing to do with involuntary abstinence and pocket protectors

In LTI system theory, control theory, and in digital or analog signal processing, the relationship between the input signal, \displaystyle x(t), to output signal, \displaystyle y(t), of an LTI system is governed by:

y(t) = h(t) * x(t) \ \stackrel{\mathrm{def}}{=}\ \int_{-\infty}^{\infty} x(u) h(t-u) , du

Or, in the frequency domain,

Y(s) = H(s) X(s) ,

where

X(s) = \mathcal{L}\left { x(t) \right } \ \stackrel{\mathrm{def}}{=}\ \int_{-\infty}^{\infty} x(t) e^{-st}, dt

Y(s) = \mathcal{L}\left { y(t) \right } \ \stackrel{\mathrm{def}}{=}\ \int_{-\infty}^{\infty} y(t) e^{-st}, dt

and

H(s) = \mathcal{L}\left { h(t) \right } \ \stackrel{\mathrm{def}}{=}\ \int_{-\infty}^{\infty} h(t) e^{-st}, dt .

Here \displaystyle h(t) is the time domain impulse response of the LTI system and \displaystyle X(s), \displaystyle Y(s), \displaystyle H(s), are the Laplace transforms of \displaystyle x(t), \displaystyle y(t), and \displaystyle h(t), respectively. \displaystyle H(s) is called the transfer function of the LTI system and, as does the impulse response, \displaystyle h(t), fully defines the input-output characteristics of the LTI system.

When such a system is driven by a quasi-sinusoidal signal, (a sinusoid with a slowly changing amplitude envelope \displaystyle A(t), relative to the change of phase, \displaystyle\omega, of the sinusoid),

x(t) = A(t) \cos(\omega t + \theta) \

the output of such an LTI system is very well approximated as

y(t) = |H(i \omega)| A(t-\tau_g) \cos\left(\omega (t-\tau_{\phi}) + \theta\right) \

if

\frac{d \log \left( A(t) \right)}{dt} \ll \omega \

and \displaystyle\tau_g and \displaystyle\tau_\phi, the group delay and phase delay respectively, are as shown below and potentially functions of ω. In a linear phase system (with non-inverting gain), both \displaystyle\tau_g and \displaystyle\tau_\phi are equal to the same constant delay of the system and the phase shift of the system increases linearly with frequency ω.

It can be shown that for an LTI system with transfer function H(s) that if such is driven by a complex sinusoid of unit amplitude,

x(t) = e^{i \omega t} \

the output is

\begin{align} y(t) & = H(i \omega) e^{i \omega t} \ \ & = \left( |H(i \omega)| e^{i \phi(\omega)} \right) e^{i \omega t} \ \ & = |H(i \omega)| e^{i \left(\omega t + \phi(\omega) \right)} \ \ \end{align} \

where the phase shift \displaystyle\phi is

\phi(\omega) \ \stackrel{\mathrm{def}}{=}\ \arg \left{ H(i \omega) \right} \

Additionally, it can be shown that the group delay, \displaystyle\tau_g, and phase delay, \displaystyle\tau_\phi, are related to the phase shift \displaystyle\phi as

\tau_g = - \frac{d \phi(\omega)}{d \omega} \

\tau_{\phi} = - \frac{\phi(\omega)}{\omega} \ .

In physics, and in particular in optics, the term group delay has the following meanings:

  1. The rate of change of the total phase shift with respect to angular frequency,

\tau_g = -\frac{d\phi}{d\omega}

through a device or transmission medium, where \phi \ is the total phase shift in radians, and \omega \ is the angular frequency in radians per unit time, equal to 2 \pi f \ , where f \ is the frequency (hertz if group delay is measured in seconds).

  1. In an optical fiber, the transit time required for optical power, traveling at a given mode’s group velocity, to travel a given distance.

Note: For optical fiber dispersion measurement purposes, the quantity of interest is group delay per unit length, which is the reciprocal of the group velocity of a particular mode. The measured group delay of a signal through an optical fiber exhibits a wavelength dependence due to the various dispersion mechanisms present in the fiber.

Source: from Federal Standard 1037C

It is often desirable for the group delay to be constant across all frequencies; otherwise there is temporal smearing of the signal. Because group delay is \tau_g(\omega) = -\frac{d\phi}{d\omega}, as defined in (1), it therefore follows that a constant group delay can be achieved if the transfer function of the device or medium has a linear phase response (i.e., \phi(\omega) = \phi(0) - \tau_g \omega \ where the group delay \tau_g \ is a constant). The degree of nonlinearity of the phase indicates the deviation of the group delay from a constant.

all digital music making is geeky, except renoise.

I have a beautiful girlfriend, Renoise can’t possibly be geeky ;)

Buddhists shave their heads. Not everyone who shaves his head is buddhist.

You’re getting geeks and nerds mixed up… geeks are nerds with social skills. ;)

Being geeky when you re 14-18 is tragic. After that it just means you ve found something to obsess over and spend your time on whilst everyone else is sinking into boredom.

Its not like cubase and logic exactly impress girls…

as a girl whose father was a composer (so she was by no means foreign to music) once said (when observing FT2 play a tune): “wow, you can really see the music!”…

funny how you guys go from geeks to girls…
I wonder what Captain Freud would say on the matter… ;)

in real life i am developer and because i love trackers)

freud would surely have something to say about that.

however, I actually don’t care about the question if renoise is “geeky” or not (or computers in general). I sit alot of time infront of my PC doing things I enjoy (no, not searching for porn), I like to browse around wiki, I am long out of school but still enjoy reading things about current breakthroughs in science and math, I spend time on this forum talking with people I like, I actually have a shirt saying “I am currently away from the computer” which is the biggest “Hi, I am a geek”-hint I can give to anyone and I am wearing that quite often.

But I also have a working social life, other interests, I have non-geeky friends who actually value me for the fact that I do my own music, do fresh mixsets from time to time, that I actually know alot of things I have read in the past and that I am a valuable source in discussions about alot of things etc. etc.

The point of all this is that I would not give that up for anyone (especially the music part). If I would met a woman who would somehow be disgusted by these facts and would demand me to stop doing music (or cutting it down massively) and all that other “stuff I do infront of a computer” I would have to wave goodbye to her. Wouldn’t work out, because that’s what I basically am. And this doesn’t only apply to women.

i think the way renoise looks and works is much more badass than most other software.
i consider us the bad boys of electronic music! :)

hmmm, I never took the topic as serious as I should have?

I think anyone who is deeply into ‘something’ can be considered
a ‘geek’ by anyone else who is not into that certain ‘something’.
Some people consider D&D folk ‘geeky’, while others put the tag
solely on computer programmers’ heads. With all the ones and
zeros going on in a tracker, I guess it has a highly potential ‘geekfactor’.

But dedication has often been taken the wrong way throughout time.
In the end, it led to some of the biggest breakthroughs in human history.

Hence why geeks are cool. Period. Ask a ninja: :ph34r:

Interesting responses.

I think geeks and girls are associated together because girls, females, women in this pop culture don’t find this quality in a man very attractive and that makes many men feel insecure.

I could go with the bad boy theory, but we all know that’s a weak argument in the end as most women don’t actually enjoy being played, beaten and lied to.

Plain and simple, anything, any activity whether it’s sitting at a computer or riding a Harely with no shirt that steals attention away from a female is a one way kick in the arse to the curb.

Women love attention, they love excitement, romance, and a great many of them love shopping. They don’t like when they are not your definite drive and ambition in life. Yet, oddly, they find it rather annoying when they get it. (another topic in itself) And they will trample over anything to get it. Trust that to be the truth friends.

You on the other hand, love your tracker to a fault. Which is another word for craft, art, skill or focus. And if you like, geek.

If you’re into your art form and she doesn’t have any hobbies, all I got to say is get ready for the wicked storm. :panic:

Most people don’t understand the meaning of love anyway so I say if you’re an artist you’re better off by yourself, unless you have one of those females that have actually evolved.

yes. trackers are geeky… and if you are geek enough, its the best piece of music software there is :)

cheers!