Chapter 2 - Signal Properties

Periodicity - Quiz

Question 1

Which of the following signals is periodic?

Select all that apply.

• A. $\quad f_{1}(t) = \sin(2\pi t)$
• B. $\quad f_{2}(t) = \lvert\cos(2\pi t)\rvert$
• C. $\quad f_{3}(t) = \sin(2\pi \lvert t \rvert)$
• D. $\quad f_{4}(t) = \cos(2\pi \lvert t \rvert)$
• E. $\quad f_{5}(t) = \cos(2\pi t)u(t)$
• F. $\quad$ None of the above.

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Signal A is periodic with fundamental period \(T=1\):

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\[ f_{1}(t) = \sin{(2\pi t)} \]

\[ \begin{align*} f_{1}(t + 1) &= \sin{(2\pi(t + 1))} \\ &= \sin{(2\pi t + 2\pi)} \\ &= \sin{(2\pi t)} \\ &= f_{1}(t) \end{align*} \]

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Signal B is periodic with fundamental period \(T = \tfrac{1}{2}\):

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\[ f_{2}(t) = |\cos{(2\pi t)}| \]

\[ \begin{align*} f_{2}\left(t + \tfrac{1}{2}\right) &= |\sin{(2\pi(t + \tfrac{1}{2}))} \\ &= |\cos{(2\pi t + \pi)}| \\ &= |-\cos{(2\pi t)}| \\ &= |\cos{(2\pi t)}| \\ &= f_{2}(t) \end{align*} \]

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Signal C is non-periodic:

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\[ f_{3}(t) = \sin{(2\pi |t|)} \]

We can rewrite this as a piecewise function

\[ f_{3}(t) = \begin{cases}\sin{(2\pi t)},\quad &t \geq 0 \\ \sin{(-2\pi t)},\quad &t < 0\end{cases} \]

And due to the properties of the sine function, this becomes

\[ f_{3}(t) = \begin{cases}\sin{(2\pi t)},\quad &t \geq 0 \\ -\sin{(2\pi t)},\quad &t < 0\end{cases} \]

Now, if there were no absolute value sign, the signal would have a fundamental period of \(T=1\), which means that the function would also be periodic with period \(T=k,\quad k\in\mathbb{Z}\).

Let's consider these cases, as the signal definitely won't line up for any other values of \(T\).

\[ f_{3}(t - k) = \begin{cases}\sin{(2\pi t)},\quad &t - k \geq 0 \\ -\sin{(2\pi t)},\quad &t - k < 0\end{cases} \]

We can rewrite this as:

\[ f_{3}(t - k) = \begin{cases}\sin{(2\pi t)},\quad &t \geq k \\ -\sin{(2\pi t)},\quad &t < k\end{cases} \]

Now the following must hold for all \(t\), for the signal to be periodic.

\[ f_{3}(t) = f_{3}(t - k),\quad \forall \, t\in \mathbb{R} \]

But no matter what \(k\) value we have, we can always find a \(t\) value for which the statement fails to hold.

Case 1: \(k\geq 1\):

For this case we can simply test \(t=\tfrac{1}{4}\).

Since \(t=\tfrac{1}{4}>0\) we keep the sign:

\[ f_{3}\left(\tfrac{1}{4}\right) = \sin{\left(2\pi \cdot \tfrac{1}{4}\right)} = \sin{\left(\tfrac{\pi}{2}\right)} = 1 \]

Since \(t=\tfrac{1}{4}<k\) we flip the sign:

\[ f_{3}\left(\tfrac{1}{4} - k\right) = -\sin{\left(2\pi \cdot \tfrac{1}{4} - 2\pi k\right)} = -\sin{\left(\tfrac{\pi}{2}\right)} = -1 \]

\[ f_{3}\left(\tfrac{1}{4}\right) \neq f_{3}\left(\tfrac{1}{4} - k\right) \]

Case 2: \(k\leq -1\):

For the other case we can test \(t=-\tfrac{1}{4}\).

Since \(t=-\tfrac{1}{4}<0\) we flip the sign:

\[ f_{3}\left(-\tfrac{1}{4}\right) = \sin{\left(2\pi \cdot \left(-\tfrac{1}{4}\right)\right)} = \sin{\left(-\tfrac{\pi}{2}\right)} = -\sin{\left(\tfrac{\pi}{2}\right)} = -1 \]

Since \(t=-\tfrac{1}{4}>k\) we keep the sign:

\[ f_{3}\left(\tfrac{1}{4} - k\right) = -\sin{\left(2\pi \cdot \left(-\tfrac{1}{4}\right) - 2\pi k\right)} = \sin{\left(\tfrac{\pi}{2}\right)} = 1 \]

\[ f_{3}\left(-\tfrac{1}{4}\right) \neq f_{3}\left(-\tfrac{1}{4} - k\right) \]

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Signal D is periodic with fundamental period \(T=1\):

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\[ f_{4}(t) = \cos{(2\pi |t|)} \]

Unlike Signal C, the absolute value actually doesn't change the signal.

Writing this piecewise, we get:

\[ f_{4}(t) = \begin{cases}\cos{(2\pi t)},\quad &t \geq 0 \\ \cos{(-2\pi t)},\quad &t < 0\end{cases} \]

But due to the properties of cosine, this becomes:

\[ f_{4}(t) = \begin{cases}\cos{(2\pi t)},\quad &t \geq 0 \\ \cos{(2\pi t)},\quad &t < 0\end{cases} \]

So we simply have:

\[ f_{4}(t) = \cos{(2\pi t)} \]

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Signal E is non-periodic:

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\[ f_{5}(t) = \cos{(2\pi t)}u(t) \]

We can write this as a piecewise function:

\[ f_{5}(t) = \begin{cases} \cos{(2\pi t)},\quad &t\geq 0 \\ 0,\quad &t<0 \\ \end{cases} \]

Now, if there were no unit step function, the signal would have a fundamental period of \(T=1\), which means that the function would also be periodic with period \(T=k,\quad k\in\mathbb{Z}\).

Let's consider these cases, as the signal definitely won't line up for any other values of \(T\).

\[ f_{5}(t - k) = \begin{cases}\cos{(2\pi t)},\quad &t - k \geq 0 \\ 0,\quad &t - k < 0\end{cases} \]

We can rewrite this as:

\[ f_{5}(t - k) = \begin{cases}\cos{(2\pi t)},\quad &t \geq k \\ 0,\quad &t < k\end{cases} \]

Now the following must hold for all \(t\), for the signal to be periodic.

\[ f_{5}(t) = f_{5}(t - k),\quad \forall \, t\in \mathbb{R} \]

But no matter what \(k\) value we have, we can always find a \(t\) value for which the statement fails to hold.

Case 1: \(k\geq 1\):

For this case we can simply test \(t=\tfrac{1}{2}\).

Since \(t=\tfrac{1}{2}>0\) the function is non-zero:

\[ f_{5}\left(\tfrac{1}{2}\right) = \cos{\left(2\pi \cdot \tfrac{1}{2}\right)} = \cos{\left(\pi\right)} = -1 \]

Since \(t=\tfrac{1}{2}<k\) the function goes to zero:

\[ f_{5}\left(\tfrac{1}{2} - k\right) = 0 \]

\[ f_{5}\left(\tfrac{1}{2}\right) \neq f_{5}\left(\tfrac{1}{2} - k\right) \]

Case 2: \(k\leq -1\):

For the other case we can test \(t=-\tfrac{1}{2}\).

Since \(t=-\tfrac{1}{2}<0\) the function goes to zero:

\[ f_{5}\left(-\tfrac{1}{2}\right) = 0 \]

Since \(t=-\tfrac{1}{2}>k\) the function is non-zero:

\[ f_{5}\left(\tfrac{1}{2} - k\right) = \cos{\left(2\pi \cdot \left(-\tfrac{1}{2}\right) - 2\pi k\right)} = \cos{\left(\pi\right)} = -1 \]

\[ f_{5}\left(-\tfrac{1}{2}\right) \neq f_{5}\left(-\tfrac{1}{2} - k\right) \]

Question 2

Which of the following signals is non-periodic?

Select all that apply.

• A. $\quad g_{1}(t) = \sin{(2\pi t)} + \sin{(4\pi t)}$
• B. $\quad g_{2}(t) = \sin{(t)} + \sin{(\sqrt{2}t)}$
• C. $\quad g_{3}(t) = \sin{\left(2t\right)} + \cos{(3.14t)} + \sin{\left(1.\overline{81}t\right)}$ (repeating)
• D. $\quad g_{4}(t) = \cos{(2\pi t)} + 2\cos{(12\pi t)}\sin{(25\pi t)}$
• E. $\quad g_{5}(t) = \sin{(2\pi t)} + \sin{(t)}$
• F. $\quad$ None of the above.

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Question 3

Which of the following discrete-time signals is periodic?

Select all that apply.

• A. $\quad h_{1}[n] = (-1)^{n}$
• B. $\quad h_{2}[n] = \sin{\left(\frac{\pi n}{2}\right)}$
• C. $\quad h_{3}[n] = \cos{\left(\frac{2\pi n}{10}\right)} + \cos{\left(\frac{4\pi n}{10}\right)}$
• D. $\quad h_{4}[n] = \cos{(n)} + \cos{(2n)}$
• E. $\quad h_{5}[n] = n\ \mathrm{mod}\ 4$ (i.e., $\{0,1,2,3,0,1,\dots\}$)
• F. $\quad$ None of the above.

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