# 1.5 Sinusoidal signals

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## Sinusoidal signals

Sinusoidal signals are perhaps the most important type of signal that we will encounter in signal processing. There are two basic types of signals, the cosine :

$x\left(t\right)=Acos\left(\Omega t\right)$

and the sine :

$x\left(t\right)=Asin\left(\Omega t\right)$

where $A$ is a real constant. Plots of the sine and cosine signals are shown in [link] . Sinusoidal signals are periodic signals. The period of the cosine and sine signals shown above is given by $T=2\pi /\Omega$ . The frequency of the signals is $\Omega =2\pi /T$ which has units of rad/sec . Equivalently, the frequency can be expressed as $1/T$ , which has units of $se{c}^{-1}$ , cycles/sec , or Hz . The quantity $\Omega t$ has units of radians and is often called the phase of the sinusoid. Recalling the effect of a time shift on the appearance of a signal, we can observe from [link] that the sine signal is obtained by shifting the cosine signal by $T/4$ seconds, i.e.

$sin\left(\Omega t\right)=cos\left(\Omega \left(t-T/4\right)\right)$

and since $T=2\pi /\Omega$ , we have

$sin\left(\Omega t\right)=cos\left(\Omega t-\pi /2\right)\right)$

Similarly, we have

$cos\left(\Omega t\right)=sin\left(\Omega t+\pi /2\right)\right)$

Using Euler's Identity, we can also write:

$Acos\left(\Omega t\right)=\frac{A}{2}\left({e}^{j\Omega t},+,{e}^{-j\Omega t}\right)$

and

$Asin\left(\Omega t\right)=\frac{A}{2j}\left({e}^{j\Omega t},-,{e}^{-j\Omega t}\right)$

The quantity ${e}^{j\Omega t}$ is called a complex sinusoid and can be expressed as

${e}^{±j\Omega t}=cos\left(\Omega ,t\right)±jsin\left(\Omega ,t\right)$

There are a number of trigonometric identities which are sometimes useful. These are shown in [link] . [link] shows some basic calculus operations on sine and cosine signals.

 $sin\left(\theta \right)=cos\left(\theta -\pi /2\right)$ $cos\left(\theta \right)=sin\left(\theta +\pi /2\right)$ $sin\left({\theta }_{1}\right)sin\left({\theta }_{2}\right)=\frac{1}{2}\left[cos,\left({\theta }_{1}-{\theta }_{2}\right),-,cos,\left({\theta }_{1}+{\theta }_{2}\right)\right]$ $sin\left({\theta }_{1}\right)cos\left({\theta }_{2}\right)=\frac{1}{2}\left[sin,\left({\theta }_{1}-{\theta }_{2}\right),-,sin,\left({\theta }_{1}+{\theta }_{2}\right)\right]$ $cos\left({\theta }_{1}\right)cos\left({\theta }_{2}\right)=\frac{1}{2}\left[cos,\left({\theta }_{1}-{\theta }_{2}\right),+,cos,\left({\theta }_{1}+{\theta }_{2}\right)\right]$ $acos\left(\theta \right)+bsin\left(\theta \right)=\sqrt{{a}^{2}+{b}^{2}}cos\left(\theta ,-,{tan}^{-1},\left(\frac{b}{a}\right)\right)$ $cos\left({\theta }_{1}±{\theta }_{2}\right)=cos\left({\theta }_{1}\right)cos\left({\theta }_{2}\right)\mp sin\left({\theta }_{1}\right)sin\left({\theta }_{2}\right)$ $sin\left({\theta }_{1}±{\theta }_{2}\right)=sin\left({\theta }_{1}\right)cos\left({\theta }_{2}\right)±sin\left({\theta }_{1}\right)cos\left({\theta }_{2}\right)$
 $\frac{d}{dt}cos\left(\Omega t\right)=-\Omega sin\left(\Omega t\right)$ $\frac{d}{dt}sin\left(\Omega t\right)=\Omega cos\left(\Omega t\right)$ $\int cos\left(\Omega t\right)dt=\frac{1}{\Omega }sin\left(\Omega t\right)$ $\int sin\left(\Omega t\right)dt=-\frac{1}{\Omega }cos\left(\Omega t\right)$ ${\int }_{0}^{T}sin\left(k{\Omega }_{o}t\right)cos\left(n{\Omega }_{o}t\right)dt=0$ ${\int }_{0}^{T}sin\left(k{\Omega }_{o}t\right)sin\left(n{\Omega }_{o}t\right)dt=0,k\ne n$ ${\int }_{0}^{T}cos\left(k{\Omega }_{o}t\right)cos\left(n{\Omega }_{o}t\right)dt=0,k\ne n$ ${\int }_{0}^{T}{sin}^{2}\left(n{\Omega }_{o}t\right)dt=T/2$ ${\int }_{0}^{T}{cos}^{2}\left(n{\Omega }_{o}t\right)dt=T/2$

Now suppose that we have a sum of two sinusoids, say

$x\left(t\right)=cos\left({\Omega }_{1}t\right)+cos\left({\Omega }_{2}t\right)$

It is of interest to know what the period $T$ of the sum of 2 sinusoids is. We must have

$\begin{array}{ccc}\hfill x\left(t-T\right)& =& cos\left({\Omega }_{1}\left(t-T\right)\right)+cos\left({\Omega }_{2}\left(t-T\right)\right)\hfill \\ \hfill & =& cos\left({\Omega }_{1}t-{\Omega }_{1}T\right)+cos\left({\Omega }_{2}t-{\Omega }_{2}T\right)\hfill \\ \hfill \end{array}$

It follows that ${\Omega }_{1}T=2\pi k$ and ${\Omega }_{2}T=2\pi l$ , where $k$ and $l$ are integers. Solving these two equations for $T$ gives $T=2\pi k/{\Omega }_{1}=2\pi l/{\Omega }_{2}$ . We wish to select the shortest possible period, since any integer multiple of the period is also a period. To do this we note that since $2\pi k/{\Omega }_{1}=2\pi l/{\Omega }_{2}$ , we can write

$\frac{{\Omega }_{1}}{{\Omega }_{2}}=\frac{k}{l}$

so we seek the smallest integers $k$ and $l$ that satisfy [link] . This can be done by finding the greatest common divisor between $k$ and $l$ . For example if ${\Omega }_{1}=10\pi$ and ${\Omega }_{2}=15\pi$ , we have $k=2$ and $l=3$ , after dividing out 5, the greatest common divisor between 10 and 15. So the period is $T=2\pi k/{\Omega }_{1}=0.4$ sec. On the other hand, if ${\Omega }_{1}=10\pi$ and ${\Omega }_{2}=10.1\pi$ , we find that $k=100$ and $l=101$ and the period increases to $T=2\pi k/{\Omega }_{1}=20$ sec. Notice also that if the ratio of ${\Omega }_{1}$ and ${\Omega }_{2}$ is not a rational number, then $x\left(t\right)$ is not periodic!

If there are more than two sinusoids, it is probably easiest to find the period of one pair of sinusoids at a time, using the two lowest frequencies (which will have a longer period). Once the frequency of the first two sinusoids has been found, replace them with a single sinusoid at the composite frequency corresponding to the first two sinusoids and compare it with the third sinusoid, and so on.

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