Periodic Functions

If we graph $f(\theta)=sin(\theta)$, the codomain varies from $-1$ to $1$, repeating with a period of exactly $2\pi$.

Definition Period. The number of repeats per second which occur for a periodic function. The most common symbol for a period is “T”.

If the equation is represented as $f(x)=A\cdot sin(B\cdot x)$, then the period, $T=2\pi/B$. Try to remember that the oscillating graph for $\sin(\theta)$ has the $x$ axis in units of angles, probably rads.

SineWave_1.png
Figure 1: Figure shows the $\sin(x)$ with amplitude one. This “no frills” sine wave has a natural period, $T$, of $2\pi$.
Definition Frequency: Frequency, f, is the number of peaks that occur in one second. Frequency is the inverse of period. Frequency is $1/T$. $$f=cycles/sec$$ $$Period=T=\frac{seconds}{cycle}$$

Unfortunately, the two words, “cycles” and “rads”, used to talk about period and frequency, are not units. Or to put it another way, they are unit-less.

If we subtract a constant from the $x$ inside the function, it has the effect of moving the function to the right. To move the function up or down, we add or subtract a constant outside the function. $f(x)=cos(x)+1$ will have a codomain from $0$ to $2$.

Definition Phase Shift. The phase shift, $\phi$, is a value on the $x$-axis where the sine or cosine wave begins. Normally, if we write the standard equation as $f(x)=A\cdot sin(Bx+C)$, then $C$ represents the amount that the wave is shifted. However, it isn't direct. To obtain the phase shift, we set $$B\phi+C=0$$ $$\phi=\frac{-C}{B}$$

phase shift.png
Figure 2: Phase shift.

If a constant is multiplied times the sine or cosine function, it changes the codomain. The constant is called an amplitude. If we had $f(x)=2\cdot sin(x)$, then the codomain would go from $-2$ to $2$. So, if the function is expressed as $f(x)=A\cdot \sin(Bx)$, then the amplitude is $|A|$. Nothing inside of the function can change the codomain.

Example: Design a sine function with an amplitude of 2, a period of 3, a vertical offset of +2, and a right shift of 0.5.
Answer: We will use the form $A\cdot sin(Bx+C)+D$. Amplitude is direct, so $A=2$. The period is $P=2\pi/B=3$, so we solve and $B=2\pi/3$. The vertical shift is again direct, so $D=2$. Finally, we want the value of the sine function to be zero when $\phi=1/2$. $$B\cdot\phi+C=0$$ $$\frac{2\pi}{3}\cdot\frac{1}{2}+C=0$$ $$C=-\frac{2\pi}{6}=\frac{-\pi}{3}$$ Therefore, the answer is $f(x)=2\cdot sin(\frac{2\pi}{3}x-\frac{\pi}{3})+2$