Electric Powertrain Torque-Speed Mapping (Electrical & Mechanical Engineering) Solution

Problem Overview

We analyze the mechanical power curve of an electric vehicle traction motor:

P(\omega) = 40 \omega \cdot e^{-0.05\omega} \cdot \left(\omega^2 + 16\right)^{-1/2}

using Logarithmic Differentiation to calculate the marginal power derivative

\frac{dP}{d\omega}

and the fractional power sensitivity

\frac{1}{P}\frac{dP}{d\omega}

\omega = 4.0\text{ (400 rad/s)}

Step-by-Step Mathematical Solution

Step 1: Apply Natural Logarithms to Decompose Products and Powers

The governing strategy is Logarithmic Differentiation:

\ln[u \cdot v \cdot w^k] = \ln(u) + \ln(v) + k\ln(w)

\frac{d}{dx}[\ln y] = \frac{1}{y}\frac{dy}{dx}

(This technique is defined in Logarithmic Differentiation.)

Take the natural logarithm of both sides of $P(\omega)$ :

\ln P(\omega) = \ln\left[40 \omega \cdot e^{-0.05\omega} \cdot \left(\omega^2 + 16\right)^{-1/2}\right]

Expand using logarithm properties:

\ln P(\omega) = \ln(40) + \ln(\omega) + \ln\left(e^{-0.05\omega}\right) + \ln\left[\left(\omega^2 + 16\right)^{-1/2}\right]

\ln P(\omega) = \ln(40) + \ln(\omega) - 0.05\omega - \frac{1}{2}\ln\left(\omega^2 + 16\right)

Step 2: Differentiate Implicitly with Respect to $\omega$

Differentiate both sides term-by-term with respect to $\omega$ :

\frac{d}{d\omega}[\ln P(\omega)] = \frac{d}{d\omega}\left[\ln(40) + \ln(\omega) - 0.05\omega - \frac{1}{2}\ln\left(\omega^2 + 16\right)\right]

\frac{1}{P}\frac{dP}{d\omega} = 0 + \frac{1}{\omega} - 0.05 - \frac{1}{2}\left(\frac{2\omega}{\omega^2 + 16}\right)

\frac{1}{P}\frac{dP}{d\omega} = \frac{1}{\omega} - 0.05 - \frac{\omega}{\omega^2 + 16}

Step 3: Evaluate Fractional Sensitivity at $\omega = 4.0$

Substitute $\omega = 4.0$ directly into the fractional sensitivity formula:

\left.\frac{1}{P}\frac{dP}{d\omega}\right|_{\omega = 4} = \frac{1}{4} - 0.05 - \frac{4}{4^2 + 16}

\left.\frac{1}{P}\frac{dP}{d\omega}\right|_{\omega = 4} = 0.25 - 0.05 - \frac{4}{16 + 16} = 0.20 - \frac{4}{32} = 0.20 - 0.125 = +0.075

Step 4: Compute the Absolute Marginal Power Rate $\frac{dP}{d\omega}$

Evaluate Baseline Power $P(4)$ : $P(4) = 40(4) \cdot e^{-0.05(4)} \cdot \left(4^2 + 16\right)^{-1/2}$ $P(4) = 160 \cdot e^{-0.2} \cdot (32)^{-1/2} = \frac{160 \cdot e^{-0.2}}{\sqrt{32}} = \frac{160 e^{-0.2}}{4\sqrt{2}} = \frac{40 e^{-0.2}}{\sqrt{2}} \approx \frac{40(0.81873)}{1.4142} \approx 23.16\text{ kW}$
Multiply by the Fractional Sensitivity: $\left.\frac{dP}{d\omega}\right|_{\omega = 4} = P(4) \cdot (+0.075) \approx 23.16 \cdot 0.075 \approx +1.737\text{ kW / (100 rad/s)}$

Engineering Conclusion

In automotive powertrain design, evaluating $\frac{1}{P}\frac{dP}{d\omega} = +0.075$ reveals that at $400\text{ rad/s}$ (approx. $3820\text{ RPM}$ ), the motor is operating on the rising side of its power envelope: every $1\%$ increase in motor angular velocity yields a $0.075\%$ increase in delivered wheel power.

Powertrain electronic control units (ECUs) continuously sample this logarithmic derivative. When $\frac{dP}{d\omega} > 0$ , the controller allows wide-open throttle acceleration. However, as rotor speed increases further, back-electromotive force (back-EMF) and rotor windage losses drive the sensitivity toward zero ( $\frac{dP}{d\omega} = 0$ , the peak power operating point) and into negative territory ( $\frac{dP}{d\omega} < 0$ ). Detecting where $\frac{dP}{d\omega}$ changes sign allows powertrain calibration engineers to schedule seamless automatic gear ratio shifts in multi-speed EV transmissions and initiate field-weakening current injection to safeguard the high-voltage battery inverter against thermal runaway.

Tony's Intertwingled Memex • Reference Manual