Calculus review for signals
Introduced| English | Meaning |
|---|---|
| derivative/dɪˈrɪvətɪv/ | The instantaneous rate of change of a function with respect to its input. |
| integral/ˈɪntɪɡrəl/ | A mathematical quantity obtained by accumulation, represented by a definite integral, or a family of antiderivatives represented by an indefinite integral. |
| rectangular pulse/rekˈtæŋɡjʊlə pʌls/ | A signal that is constant for a set time and zero at all other times. |
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.
v(t) = 3t² volts. What is the slope dv/dt at t = 2 s, in volts per second?
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$\dfrac{d}{dt}3t^{2} = 6t$, which is 12 at $t = 2$.
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.
A current of +2 mA for 1 s followed by −2 mA for 1 s delivers a net charge of 4 mC.
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Area below the axis subtracts, so the net charge is $2 - 2 = 0$ mC. 4 mC is the total without signs.
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 sensor pulse decays as 5e^(−t/0.2) mA. What total charge does it deliver?
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The area of $Ae^{-t/\tau}$ is $A\tau = 5 \times 0.2 = 1.0$ mC. The curve never touches zero, but its area is finite and exact.
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.
Charge from the area: $250\,\mu\text{A} \times 2\,\text{s} = 500\,\mu\text{C}$. Charge from the capacitor law: $C\,\Delta V = 100\,\mu\text{F} \times 5\,\text{V} = 500\,\mu\text{C}$. Two methods, one answer: that is the checking habit.
A ramp v = 2.5t volts is applied to C = 100 µF. What current flows while it rises?
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$i = C\dfrac{dv}{dt} = 100\,\mu\text{F} \times 2.5 = 250\,\mu\text{A}$, constant while the ramp rises and zero on the plateau. Constant slope, not constant value, gives constant current.
A capacitor of 220 µF is ramped steadily from 0 to 12 V in 4 s. What charging current flows, in µA?
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Slope $\dfrac{12}{4} = 3\,\dfrac{\text{V}}{\text{s}}$, so $i = 220\,\mu\text{F} \times 3 = 660\,\mu\text{A}$. Check: $660\,\mu\text{A} \times 4\,\text{s} = 2.64\,\text{mC} = 220\,\mu\text{F} \times 12\,\text{V}$.
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 $t = 0$: $\dfrac{di}{dt} = -\dfrac{5}{0.2} = -25\,\dfrac{\text{mA}}{\text{s}}$.
Check with one rectangle: height 5 mA, width $\tau = 0.2\,\text{s}$, area 1.0 mC. The match is not luck; it is what the time constant means.
A current pulse i(t) = 8e^(−t/0.05) mA flows into a capacitor. How much charge does it deliver in total, in mC?
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$A\tau = 8 \times 0.05 = 0.4\,\text{mC}$.
Three common slips
- Value for slope. A steady 5 V gives zero capacitor current; only a changing voltage drives current.
- Forgetting the sign of area. Area below the axis subtracts.
- Dropping units inside the integral. $\mu\text{A} \times \text{s} = \mu\text{C}$, every time.
Match each signal to the capacitor current it produces.
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The current follows the slope of the voltage, $i = C\dfrac{dv}{dt}$, never its height.
Recap
- Derivative = slope; integral = signed area.
- $\dfrac{d}{dt}e^{at} = a\,e^{at}$, and $\int_{0}^{\infty}Ae^{-t/\tau}\,dt = A\tau$.
- Check a result two ways, and carry the units through.