Is $x(t) = \cos t + \sin\left(\frac{1}{2}t\right)$ a periodic signal?

Is $x(t) = \cos t + \sin\left(\frac{1}{2}t\right)$ a periodic signal?

The answer provided by the book is different from my answer. Book says its not a periodic signal. Can you guys tell me why is it not a periodic signal?

$\cos(t)$ is periodic as $2\pi f_1 = 1 \Rightarrow f_1=\frac{1}{2\pi} \Rightarrow T_1=2\pi$

$\sin\left(\frac{1}{2}t\right)$ is also periodic as $2\pi f_2 = \frac{1}{2} \Rightarrow f_2=\frac{1}{4\pi} \Rightarrow T_2=4\pi$

Therefore $\frac{T_1}{T_2} = \frac{2\pi}{4\pi} = \frac{1}{2}$ is rational number

Therefore the given $x(t)$ is a periodic signal.

• Which book says so? – Matt L. Dec 11 '15 at 12:44
• Pranav: Welcome to DSP.SE! You've asked a homework / self-study question in precisely the right way: asked the question, made it clear that it's a textbook question you're trying to answer, and shown what you think the answer is (or shown as much as you understand). Well done! – Peter K. Dec 11 '15 at 13:06
• Typographical errors, not to mention outright blunders, are not unknown in the Solutions Manual or the "answers to odd-numbered problems" that are included in textbooks because while a professor might have written the text and carefully proof-read it, the Solutions Manual and the answers to exercises are worked out by harried graduate student assistants and not as carefully checked by the professor. This is not to say that the professor is absolved of responsibility for the errors, but just an explanation of why the "book answer" is not to be trusted as the gospel truth in all cases. – Dilip Sarwate Dec 11 '15 at 14:06
• Thank you friends for clearing my doubt. You guys are the best! Thanks alot :) It was from the coaching institute book, one of the practice problem. :) – Pranav Peethambaran Dec 13 '15 at 8:16
• SE.DSP wishes you a happy new year 2017, with a kind reminding signal that your question or its answers may require some action (update, votes, acceptance, etc.) – Laurent Duval Jan 2 '17 at 22:57

As for every $t_0\in\mathbb{R}$ and $k\in\mathbb{Z}$ $$\begin{eqnarray} &x(t_0+4k\pi) &=\cos(t_0+4k\pi)+\sin(t_0/2+2k\pi)\\ & &=\cos(t_0)+\sin(t_0/2)\\ & &=x(t_0) \end{eqnarray}$$ you answer is correct: $x(t)$ is periodic.

To add a contrarian answer: If your time index, $t$, is an integer, then your signal is not periodic.

The definition of periodic is: $x[t]$, $t\in\mathbb{Z}$ is periodic with period $P\in \mathbb{Z}$ iff $$x[t] = x[t+P]$$

So we need $$\cos(t) = \cos(t+P)$$ so for periodicity we require $$P = 2\pi k$$ with $k\in \mathbb{Z}$.

Since $\pi$ is irrational, that cannot be the case.

Hence the first component of your signal cannot be periodic, so the entire signal cannot be periodic.

• There is no need for P to be integer, so cos (t) = cos (t+P). For any t cos (t) = cos(t +2π), because cos(t+2π) = sin(t)*sin(2π) + cos(t)*cos(2π) =sin(t)*0+cos(t)*1 = cos(t) – Stoleg Dec 11 '15 at 13:58
• Nice! but if the book is following the convention that $x(t)$ denotes a continuous-time signal and $x[t]$ denotes a discrete-time signal with $t$ taking on integer values only, then it does not apply.... – Dilip Sarwate Dec 11 '15 at 14:00
• @Stoleg : No, $P$ must be an integer in this case. Because the index of $x$ must be an integer for discrete time signals. Otherwise $x[t+P]$ is undefined. – Peter K. Dec 11 '15 at 14:13
• @LaurentDuval : Thanks! Yes, the continuous-time signal is definitely periodic $x(t) = x(t+P)$, for $t\in\mathbb{R}$ and $\forall P \in \mathbb{R}$. And, as you say, the sampled version (discrete time version) is not periodic, but the reconstructed version is (provided we've sampled quickly enough... though perhaps an aliased+reconstructed version may also be periodic...hmmm.). – Peter K. Dec 11 '15 at 16:32
• $\mathbf{F}_{un}$!!! – Peter K. Dec 13 '15 at 14:18

The double-angle formulae for trigonometric identities tell you that $\cos \left(\frac{2t}{t}\right) = 1 - 2\sin^2(\frac{t}{2})$.

You thus ave $x(t) =1 +\sin(\frac{t}{2}) - 2\sin^2(\frac{t}{2})$. Hence, your signal is composed of functions (as adds and multiplies) that all admit $4\pi$ as a period (yes, the constant function $x\to 1$ is $4\pi$ periodic as well).

Thus your function looks very periodic.