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I am trying to determine when the two sinusoids of are equal, for the periods equal to a rational number times $2\pi$. The first signal is $\sin\left(2\pi\frac{a}{b} x\right)$ and the second signal is $\sin\left(2\pi\frac{c}{d}x\right)$ where $\frac{a}{b}$ and $\frac{c}{d}$ are rational numbers $\in Q$. When are they equal?

Obviously when x=0, we have equality, and when $x = b\cdot d$ we have $\sin\left(2 \pi \cdot ad\right) = \sin\left( 2\pi\cdot bc\right)$ which forces $ad = bc$. Are there any other values for $\frac{a}{b}$ and $\frac{c}{d}$ which will work?

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Are you wanting values $a,b,c,$ and $d$ so that $\sin(2\pi \cdot a/b\cdot x)=\sin(2\pi\cdot c/d\cdot x)$ for all $x$? –  Clayton Feb 24 '13 at 6:32

3 Answers 3

If you're looking for the functions to agree everywhere, then simply observe that their periods must match, so that $ad=bc$.

If you're looking for where sinusoids of different periods intersect (that is, if $ad\neq bc$), then note the identity

$$\sin\theta - \sin\phi = 2\sin\frac{\theta-\phi}{2}\cos\frac{\theta+\phi}{2}$$

so that

$$\sin\left(2\pi\frac{a}{b}x\right) - \sin\left(2\pi\frac{c}{d}x\right) = 0 \quad \implies \quad 2 \sin\left(\pi x\frac{ad-bc}{bd}\right) \cos\left(\pi x\frac{ad+bc}{bd}\right) = 0$$

Considering each factor, we find $$\pi x \frac{ad-bc}{bd} = n\pi \quad \text{OR} \quad \pi x \frac{ad+bc}{bd} = \frac{\pi}{2} + n\pi$$

for any integer $n$, and we can write explictly:

$$x = \frac{nbd}{ad-bc} \quad \text{OR} \quad x = \frac{(2n+1)bd}{2\left(ad+bc\right)}$$

(with the latter valid only for $ad+bc\neq 0$).

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You want values for $a,b,c$, and $d$ so that for every $x$, we have $$\sin\left(2\pi\frac{a}{b}x\right)=\sin\left(2\pi\frac{c}{d}\right).$$ Using the double angle formula for $\sin(\theta)$ and simplifying, we find that this is equivalent to $$\sin\left(\pi\frac{a}{b}x\right)\cos\left(\pi\frac{a}{b}x\right)-\sin\left(\pi\frac{c}{d}x\right)\cos\left(\pi\frac{c}{d}x\right)=0.$$But now this is easily seen to be a difference angle formula for $\sin(\theta)$, so we can reduce it further to $$\sin\left(\pi\frac{a}{b}x-\pi\frac{c}{d}x\right)=\sin\left(\pi x\left[\frac{a}{b}-\frac{c}{d}\right]\right)=0$$for all $x$. This only happens if $$\frac{a}{b}-\frac{c}{d}=0,\quad\text{that is,}\quad ad=bc.$$

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could you please elaborate the reduction from $\sin\left(\pi\frac abx\right)\cos\left(\pi\frac abx\right)-\sin\left(\pi\frac cdx\right)\cos\left(\pi\frac cdx\right)$ to $\sin\left(\pi\frac abx-\pi\frac cdx\right)$ –  lab bhattacharjee Feb 24 '13 at 8:07

$$\sin\left(\frac{2\pi a x}b\right)=\sin\left(\frac{2\pi c x}d\right)$$

So, $\frac{2\pi a x}b=n\pi+(-1)^n\frac{2\pi c x}d$ where $n$ is any integer

If $n$ is even $=2m$(say), $\frac{2\pi a x}b=2m\pi+\frac{2\pi c x}d\implies x\left(\frac ab-\frac cd\right)=m$ which is true for a certain set of values of $x$ corresponding to the values of integer $m$

If we need $\sin\left(\frac{2\pi a x}b\right)=\sin\left(\frac{2\pi c x}d\right)$ for all $x,\frac ab-\frac cd$ must be $0,$ consequently $m=0$

If $n$ is odd $=2m+1$(say), $\frac{2\pi a x}b=(2m+1)\pi-\frac{2\pi c x}d\implies x\left(\frac ab+\frac cd\right)=\frac{2m+1}2$ which is true for a certain set of values of $x$ corresponding to the values of integer $m$.

We can not make $\sin\left(\frac{2\pi a x}b\right)=\sin\left(\frac{2\pi c x}d\right)$ for all $x$ here as $2x\left(\frac ab+\frac cd\right)=2m+1$ as even if we put $\frac ab+\frac cd=0$ the RHS will be odd unlike the LHS.

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