How Do You Differentiate a Triangle Wave?

A triangle wave is piecewise linear, so differentiating it means finding the slope of each rising or falling segment. The result is a square wave: a constant positive value while the triangle rises and a constant negative value while it falls.

The derivative is not defined at the triangle wave’s sharp corners. At a peak or trough, the slope changes abruptly, so the left-hand and right-hand derivatives disagree.

A standard triangle-wave definition

Suppose the wave ranges from $-A$ to $A$ and repeats every $T$ seconds. Here, $A$ is the peak amplitude, and $T$ is the period. Let $r=t\bmod T$ be the time within one period, with $0\le r<T$. One standard definition is

$$f(t)= \begin{cases} -A+\frac{4A}{T}r, & 0\le r<T/2,\\ 3A-\frac{4A}{T}r, & T/2\le r<T. \end{cases}$$

During the first half of the period, the wave rises from $-A$ to $A$. During the second half, it falls back to $-A$. Each half lasts $T/2$, and each change in height has magnitude $2A$, so the slope magnitude is

$$\frac{2A}{T/2}=\frac{4A}{T}.$$

Differentiating the two linear pieces gives

$$f'(t)= \begin{cases} \frac{4A}{T}, & 0<r<T/2,\\ -\frac{4A}{T}, & T/2<r<T. \end{cases}$$

The derivative is undefined when $r=0$ or $r=T/2$, which are the trough and peak. The endpoints are excluded because they are corners, not because the wave itself is discontinuous: the triangle wave is continuous there.

Worked example

Consider a voltage triangle wave that rises from $-2\text{ V}$ to $+2\text{ V}$ in $2\text{ ms}$, then falls back to $-2\text{ V}$ in the next $2\text{ ms}$. Its period is $T=4\text{ ms}$ and its amplitude is $A=2\text{ V}$. The rising slope is

$$\frac{4A}{T}=\frac{4(2\text{ V})}{4\text{ ms}}=2\text{ V/ms}=2000\text{ V/s}.$$

Therefore, the derivative is $+2000\text{ V/s}$ during each rising segment and $-2000\text{ V/s}$ during each falling segment. It is undefined at each maximum and minimum. The positive and negative signs indicate direction of change; the derivative’s units are volts per second.

Why the corners have no derivative

At a peak, the wave approaches along a rising segment and leaves along a falling segment. In the standard definition, the left-hand slope is $4A/T$ and the right-hand slope is $-4A/T$. A derivative exists only if these one-sided slopes agree, so there is no derivative at the peak. At a trough, the slopes switch from negative to positive, and the same issue occurs.

This is the ordinary, point-by-point derivative. In idealized signal analysis, a triangle wave’s derivative is often described as a square wave. That description refers to the constant slopes between corners; it does not make the derivative defined at the corners themselves.

Different amplitude and phase conventions

The factor in the derivative depends on what a source calls the amplitude. If a wave ranges from $0$ to $H$ instead of from $-A$ to $A$, the height change over each half-period is $H$. Its slope magnitude is then $H/(T/2)=2H/T$. In this convention, $H$ is the full peak-to-peak height, not the peak amplitude about a zero midpoint.

A time shift or a vertical offset changes where the corners occur or where the wave is centered, but it does not change the slope magnitudes. For a shifted wave, locate the shifted peaks and troughs, then use the same rising and falling slopes. If frequency $\nu$ is given instead of period, use $T=1/\nu$; for the range $-A$ to $A$, the slope magnitude is $4A\nu$.

Common mistakes

• Differentiating the wave as if it were a sine wave: A triangle wave is linear between corners, so its derivative is constant on each segment, not sinusoidal.
• Forgetting the amplitude convention: A range of $-A$ to $A$ has total height $2A$; a range of $0$ to $H$ has total height $H$. The corresponding slope magnitudes differ by a factor of two if the same letter is used for both quantities.
• Assigning a derivative at a corner: The derivative is undefined at each sharp peak and trough. A plotting or simulation program may assign a value there, but that is a chosen convention, not the classical derivative.
• Ignoring units: If time is measured in milliseconds, the direct slope calculation gives amplitude per millisecond. Convert to seconds only if the application needs those units.

Connection to Fourier series

A triangle wave can also be represented by a Fourier series. Differentiating its series produces the Fourier series of a square wave: away from the corners, it agrees with the triangle wave’s constant positive or negative slope. At a jump in that square wave, the Fourier series converges to the average of the two one-sided limits. For a symmetric square wave, that average is zero, but the triangle wave still has no classical derivative at the corner.