RL SERIES CIRCUIT

Transfer Function

The transfer function associated to an ODE is the relation between the Laplace Transform of the output and the Laplace Transform of the input when the input is applied and the initial conditions are zero.

H(s) = \frac{Y(s)}{U(s)} = \frac{\sum_{j=0}^{n}b_{j}s^{j}}{\sum_{i=0}^{n}a_{i}s^{i}}

In general, the transfer function is represented by H(s).

Convolution of Two Continuos Time Functions

Consider two continuous time functions x_{1}(t) and x_{2}(t). The Convolution of the two functions, which generates a new function y(t), is given by the integral:
y(t) = x_{1}(t) * x_{2}(t) = \int_{0}^{t} x_{1}(t - \tau) x_{2} (\tau) d \tau = \int_{0}^{t} x_{1}(\tau) x_{2} (t - \tau) d \tau

Laplace Transform

The Laplace Transform pair of a continuos time signal x(t); ~ t > 0; ~ t ~ \epsilon ~\mathfrak{R} is:

X(s) = \mathcal{L}[x(t)] = \int_{0}^{\infty} x(t) e^{-st} dt

x(t) = \mathcal{L}^{-1}[X(s)] = \frac{1}{2j \pi} \int_{0-j \infty}^{0+ \infty} X(s) e^{st} ds

SIMULAÇÕES EM ENGENHARIA ELÉTRICA

RL SERIES CIRCUIT

THE MATHEMATICAL MODEL OF THE RL SERIES CIRCUIT

The impulse response of the two voltages are:

h_{VL}(t) = \delta(t) - \frac{R}{L} e^{- \frac{R}{L}t} u_{-1}(t)

h_{VR}(t) = \frac{R}{L} e^{- \frac{R}{L}t} -u_{-1}(t)

Where -u_{-1}(t) and \delta(t) are, respectively, the unit step and the unit impulse functions.

When the input is not a unit pulse but a generic voltage v(t), the corresponding outputs can be computed using the Convolution of the impulse responses with the input. If you want to remember the concept of the Convolution of two continuous time functions, click on the magnifying glass.

The two voltage responses due to a generic input, v(t), are:

v_{L}(t) = h_{VL}(t) * v(t) = \int_{0}^{t}h_{VL} (t - \tau) v(\tau) d \tau = \int_{0}^{t} h_{VL}(\tau) v(t- \tau) d \tau

v_{R}(t) = h_{VR}(t) * v(t) = \int_{0}^{t}h_{VR} (t - \tau) v(\tau) d \tau = \int_{0}^{t} h_{VR}(\tau) v(t - \tau) d \tau

This object addresses the impulse responses, the responses to the unit step and the responses to a sinusoid of variable frequency and amplitude.

The second part of this object focuses on the frequency responses of the voltage variables. In order to study the frequency responses, it is necessary to determine the transfer functions associated to each one. If you want to remember the concept of the Transfer Function, click on the magnifying glass.

In order to determine the transfer funcion, the Laplace Transform was used. If you want to remember the concept of the definition of the Laplace Transform, click on the magnifying glass.

The transfer functions for the voltages in each element are:

H(s)_{VL} = \frac{s}{s + \frac{R}{L}} = \frac{V_{L}(s)}{V(s)}

H(s)_{VR} = \frac{\frac{R}{L}}{s + \frac{R}{L}} = \frac{V_{R}(s)}{V(s)}

The outputs in the frequency domain are computed from the transfer functions expressions and the Laplace Transforms of the input functions. The Bode Diagramas for the outputs are part of this object.