I'm unsure how to sketch a tri signal with horizontal shift.

Eg , let's say we want to sketch tri((t-4)/4). Can I reduce it to tri( (t/4) - 1 )? This would mean sketching a tri(t/4) signal and shifting it to the right by 1. Or should it be a tri(t/4) signal shifted to the right by 4?

I've found out on wikipedia that the rect signal is defined as rect((X-Y)/T) where you shift the center of the signal by varying Y. But I'm not sure if sketching tri applies the same concept.

  • $\begingroup$ Using tri(t) as a reference, plot tri( (t-4)/4 ) by plugging in points for t and you have your answer. $\endgroup$ – CMDoolittle Oct 5 '15 at 16:07
  • $\begingroup$ You stretch by a factor of 4 and then shift right by 4. $\endgroup$ – CMDoolittle Oct 5 '15 at 16:09

The peak of the tri signal is at the point where the argument of tri (that thingy inside the parentheses following tri) equals $0$. The base of the tri signal extends from the point where the argument equals $-\frac 12$ to the point where the argument equals $+\frac 12$ (or $-1$ to $+1$ depending on what you are calling tri). Thus, tri$\left(\frac{t-t_0}{T}\right)$ is an isoceles triangle whose apex is at the point $(t_0,1)$ and whose base extends from $t_0-\frac{T}{2}$ to $t_0+\frac{T}{2}$ (or from $t_0-T$ to $t_0+T$ for the other interpretation of tri).


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