Optoelectronic LED Display Panel Average Irradiance (Entertainment & Electrical Engineering) Solution

An entertainment and electrical systems engineer at the UNLV Howard R. Hughes College of Engineering is designing a modular, high-dynamic-range LED stage display panel for an interactive performance venue on the Las Vegas Strip.

The modular rectangular panel measures $0.5\text{ m}$ in width along the $x$-axis and $0.4\text{ m}$ in height along the $y$-axis, occupying the closed rectangular domain:

Because of resistive thermal heating along the internal power busbars and localized LED driver clustering, the emitted radiant surface irradiance $E(x, y)$ is non-uniform across the panel surface and is characterized by the empirical polynomial model:

Step-by-Step Solution:

1. Calculate the Surface Area of the Panel: State the Rectangular Area Formula:

   $$\text{Area}(R) = (b - a)(d - c)$$
   Substitute the domain limits $x \in [0, 0.5]$ and $y \in [0, 0.4]$:
   $$\text{Area}(R) = (0.5 - 0)(0.4 - 0) = 0.20\text{ m}^2$$

2. Evaluate Total Radiant Flux as an Iterated Double Integral: State Fubini's Theorem for Rectangular Regions (defined in Double Integrals over a Rectangular Region):

   $$\Phi = \iint_R E(x, y) \, dA = \int_{0}^{0.5} \int_{0}^{0.4} E_0 (1 + 4x^2y) \, dy \, dx$$
   Factor out the physical scalar constant $E_0$ and evaluate the inner integral with respect to $y$, holding $x$ constant:
   $$\int_{0}^{0.4} (1 + 4x^2y) \, dy = \left[ y + 2x^2y^2 \right]_{y=0}^{y=0.4}$$
   Substitute limits:
   $$= \Big(0.4 + 2x^2(0.4)^2\Big) - 0 = 0.4 + 2x^2(0.16) = 0.4 + 0.32x^2$$
   Now evaluate the outer integral with respect to $x$:
   $$\Phi = E_0 \int_{0}^{0.5} (0.4 + 0.32x^2) \, dx = E_0 \left[ 0.4x + \frac{0.32}{3}x^3 \right]_{0}^{0.5}$$
   Substitute $x = 0.5$:
   $$\Phi = E_0 \left( 0.4(0.5) + \frac{0.32}{3}(0.5)^3 \right) = E_0 \left( 0.20 + \frac{0.32}{3}(0.125) \right)$$
   $$= E_0 \left( 0.20 + \frac{0.04}{3} \right) = E_0 \left( \frac{0.60}{3} + \frac{0.04}{3} \right) = \frac{0.64}{3} E_0\text{ W}$$
   $$\Phi \approx 0.2133 E_0\text{ W}$$

3. Compute the Spatial Average Irradiance ($E_{\text{avg}}$): State the Spatial Average Value Formula (defined in Double Integrals over a Rectangular Region):

   $$E_{\text{avg}} = \frac{1}{\text{Area}(R)} \iint_R E(x, y) \, dA$$
   Substitute the total flux and surface area:
   $$E_{\text{avg}} = \frac{1}{0.20} \left( \frac{0.64}{3} E_0 \right) = 5 \times \left( \frac{0.64}{3} E_0 \right) = \frac{3.20}{3} E_0 = \frac{16}{15} E_0 \approx 1.0667 E_0\text{ W/m}^2$$

Meaning of the Answer & Real-Life Application:

• Quantitative Interpretation: The spatial average irradiance $E_{\text{avg}} = \frac{16}{15}E_0 \approx 1.067E_0$ demonstrates that despite spatial irradiance variation across the display surface, the mean radiant emission per square meter is $6.67\%$ higher than the baseline corner intensity $E_0$.
• Engineering Application: In large-scale entertainment venue video walls, human visual perception is acutely sensitive to localized brightness gradients. Entertainment and electrical engineers use spatial average irradiance calculations to balance multi-channel LED driver pulse-width modulation (PWM) look-up tables and specify anti-glare optical diffusion film thickness, ensuring that viewer-perceived luminance remains uniform across all viewing angles.