Calculus review for signals

A trace that rises and flattens

• A scope trace rises smoothly from 0 to 5 V across a capacitor, then flattens. At every instant the capacitor's current depends on how fast the trace moves.
• The charge delivered over an interval is the area under the current trace. Signals are functions of time, and calculus reads them in exactly these two ways.

The derivative is the slope

• The derivative 导数 $\dfrac{dv}{dt}$ is the instantaneous rate of change: the slope of the trace. For powers of $t$: $\dfrac{d}{dt}t^{n} = n\,t^{n-1}$.
• A straight ramp has one slope everywhere; a flat line has slope zero. For a capacitor, $i = C\dfrac{dv}{dt}$, so current follows the slope, not the height.

The integral is signed area

• The integral 积分 $\int i\,dt$ adds up area under the curve: $\int_{0}^{T} t^{n}\,dt = \dfrac{T^{n+1}}{n+1}$. Area below the axis counts as negative.
• A rectangular pulse 矩形脉冲 of height $A$ and width $\tau$ has area $A\tau$; a triangle has half its enclosing rectangle.

Exponentials are their own slopes

• $\dfrac{d}{dt}e^{at} = a\,e^{at}$: the exponential factors out of its own difference quotient. That is why circuits that vary as $e^{at}$ keep that shape when differentiated.
• The area under a decaying exponential is finite: $\int_{0}^{\infty} A e^{-t/\tau}\,dt = A\tau$. Its "effective width" is the time constant.

A linear ramp, checked two ways

• A capacitor charges linearly from 0 to 5 V in 2 s, so $v = 2.5t$ and the slope is $2.5\,\dfrac{\text{V}}{\text{s}}$.
• With $C = 100\,\mu\text{F}$: $i = C\dfrac{dv}{dt} = 100\,\mu\text{F} \times 2.5 = 250\,\mu\text{A}$, constant while the ramp rises and zero once it flattens.

An exponential pulse, by its area

• A sensor pulse decays as $i(t) = 5e^{-t/0.2}$ mA: peak 5 mA, time constant 0.2 s.
• Total charge: $5 \times 0.2 = 1.0\,\text{mC}$. The steepest slope is at $t = 0$: $\dfrac{di}{dt} = -\dfrac{5}{0.2} = -25\,\dfrac{\text{mA}}{\text{s}}$.

Three common slips

Recap