1. **State the problem:** We want to analyze the function $$y(t) = \int_0^t 62.5 \sin(2\pi \cdot 60 \tau) e^{-(t-\tau)} d\tau$$ which represents a convolution integral involving a sinusoidal input and an exponential decay. 2. **Formula and explanation:** This integral is a convolution of the function $$f(\tau) = 62.5 \sin(2\pi \cdot 60 \tau)$$ with the exponential decay kernel $$g(t-\tau) = e^{-(t-\tau)}$$ over the interval from 0 to $$t$$. 3. **Key properties:** - The sinusoidal term oscillates with frequency 60 Hz. - The exponential term causes decay depending on the difference $$t-\tau$$. - The integral accumulates the weighted sinusoidal input over time with decay. 4. **Intermediate work:** Rewrite the integral as a convolution: $$y(t) = (f * g)(t) = \int_0^t f(\tau) g(t-\tau) d\tau$$ Using Laplace transforms, the convolution in time domain corresponds to multiplication in Laplace domain: $$Y(s) = F(s) G(s)$$ Where: $$F(s) = \mathcal{L}\{62.5 \sin(2\pi \cdot 60 t)\} = \frac{62.5 \cdot 2\pi \cdot 60}{s^2 + (2\pi \cdot 60)^2}$$ $$G(s) = \mathcal{L}\{e^{-t}\} = \frac{1}{s+1}$$ Therefore: $$Y(s) = \frac{62.5 \cdot 2\pi \cdot 60}{(s+1)(s^2 + (2\pi \cdot 60)^2)}$$ 5. **Inverse Laplace transform:** Using partial fraction decomposition and known inverse transforms, the solution in time domain is: $$y(t) = \frac{62.5 \cdot 2\pi \cdot 60}{(1)^2 + (2\pi \cdot 60)^2} \left( e^{-t} - \cos(2\pi \cdot 60 t) + \frac{1}{2\pi \cdot 60} \sin(2\pi \cdot 60 t) \right)$$ 6. **Interpretation:** - The output $$y(t)$$ is a combination of a decaying exponential and sinusoidal oscillations. - The amplitude is scaled by the factor $$\frac{62.5 \cdot 2\pi \cdot 60}{1 + (2\pi \cdot 60)^2}$$ which is small due to the high frequency. This describes the shape and behavior of the graph: a sinusoidal wave modulated by exponential decay.