1
$\begingroup$

it could be a donkey question but I'm a little confused.

I have this signal $w(t)=sin({\pi}0.1f)$. I have to calculate the period of this signal.

$w(t)=sin(2{\pi}\frac{1}{20}f)$

the period of this signal is: $20s$ while the frequency is $\frac{1}{20}Hz$

is it correct?

$\endgroup$
5
$\begingroup$

You are mostly correct. The equation should be $w(t) = sin(2\pi\frac{1}{20}t)$. The variable in the sin should be $t$, not $f$. The frequency is determined by the $\frac{1}{20}$, not by the variable.

Other than that you are correct. The period is 20s, which makes it a $\frac{1}{20}$ Hz sin wave.

$\endgroup$
  • $\begingroup$ I used $f$ variable because my sin function is domain of frequency and not in time domain $\endgroup$ – Mazzy Apr 14 '12 at 14:00
  • 1
    $\begingroup$ @Mazzy Okay, but then it would be w(f), not w(t). $\endgroup$ – Jim Clay Apr 15 '12 at 2:37
  • $\begingroup$ @Mazzy To add further to Jim Clay's response to your use of $f$, note that $w(f) = \sin(2\pi \frac{1}{20}f)$ is not the Fourier transform of a sinusoidal signal in the time domain. The Fourier transform of a sinusoidal time-domain signal is a pair of impulses in the frequency domain. $\endgroup$ – Dilip Sarwate Apr 15 '12 at 12:44
3
$\begingroup$

It's very important to write the equation properly. The way your equation is written originally there is no period at all. Your left side function of the variable "t". Since t doesn't show in the right side, the right side is a independent of t. Hence it's a constant and therefore the period is infinity.

Jim Clay already fixed that in his response. The other potential pitfall are units. If you write the equation as sin(2*pi*(1/20)*t), then t is a unit-less quantity since you cannot take the sin() of a quantity that has units. So the answer would be that the period is 20 (and not 20s). If you write the equation as sin(2*pi*(1/20s)*t), then the t is a time and the units of t cancel with the units in the term (1/20s). In this case the period would be 20s.

$\endgroup$
  • $\begingroup$ Why infinity and not any value? $\endgroup$ – Rojo Apr 16 '12 at 1:29
  • $\begingroup$ @Rojo: What's the frequency of the number 1? It has no frequency; it's not a sine wave. $\endgroup$ – endolith Apr 16 '12 at 16:06
  • $\begingroup$ @endolith the concept of period far exceeds sine waves $\endgroup$ – Rojo Apr 16 '12 at 17:51
  • $\begingroup$ @Rojo: What's the period of a constant, then? $\endgroup$ – endolith Apr 16 '12 at 17:52
  • $\begingroup$ @endolith, formally I am not sure, I was asking. But it would make sense to me that any nonzero value is a possible period. f(t) = f(t+P) for any P if f is a constnat. Periods are not unique. In a sin(2*pi*t), 2 is a period too. Probably the prime period in a constant is undetermined, because there's no minimum period. But I'd say "infinity" is the period of signals that never repeat themselves, not constants $\endgroup$ – Rojo Apr 16 '12 at 18:41

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.