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I have a series RL circuit (with zero initial conditions) and I want to find the voltage across the inductor. The formula I got is:

$$\text{V}_\text{L}\left(t\right)=\int_0^t\text{V}_\text{in}\left(\tau\right)\cdot\mathscr{L}_\text{s}^{-1}\left[\frac{\text{sL}}{\text{sL}+\text{R}}\right]_{\left(t-\tau\right)}\space\text{d}\tau\tag1$$

Question: is this formula correct?

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1 Answer 1

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Yes, the equation you gave is correct.

Please note that the uppercase letters are usually reserved for frequency domain and/or DC values. For time-varying signals such as an inductor voltage, the lowercase notation \$v_L(t)\$ is far more common than the uppercase notation \$V_L(t)\$.


The impedance of the circuit is

$$Z(s) = R + sL$$

The voltage on the inductor is:

$$V_L(s) = V_{in}(s) \frac{Z_L(s)}{Z(s)} = V_{in}(s) \frac{sL}{R + sL}$$

The impulse response is:

$$G(s) = \frac{sL}{R + sL}, \quad V_{in}(s) = \mathcal{L}\left\{\delta(t)\right\} = 1$$

where \$\delta(t)\$ is the impulse (Dirac) function. The \$G(s)\$ is also known as the transfer function.

In the time domain the impulse response is

$$g(t) = \mathcal{L}^{-1} \left\{\frac{sL}{R + sL}\right\}$$

The response to any input in the time domain can be found by convolution:

$$v_L(t) = v_{in}(t) * g(t) = \int_{0}^{t}{v_{in}(\tau)g(t-\tau)d\tau}$$

The only catch is that the system must be linear and time-invariant (LTI). The convolution integral will not work if you have nonlinear elements in your circuit, eg a diode etc.

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  • \$\begingroup\$ limit of convolution integral 0 to t is valid only when both Vin(t) and h(t) are zero when t<0 , although h(t) is zero for t<0 but there are no such restrictions on Vin(t) as given by OP , so isn't it would be better to take limit from 0 to infinity? \$\endgroup\$
    – user215805
    Commented Nov 22, 2021 at 20:22
  • \$\begingroup\$ Check definition section of convolution integral in Wikipedia, good article on this \$\endgroup\$
    – user215805
    Commented Nov 22, 2021 at 20:35
  • \$\begingroup\$ The upper limit of the convolution integral is always \$t\$ and never \$\infty\$. As for the lower limit, it normally goes from \$-\infty\$, but if ANY of the two signals is \$0\$ for \$t<0\$ then the lower limit becomes \$0\$. Any of the two signals, not both! I have never seen someone define signal for \$t < 0\$ in engineering, but feel free to provide an example where it would be useful. P.S. SE did not allow me to edit my previous comment after 5 mins, so I deleted the old one and wrote new one. \$\endgroup\$ Commented Nov 22, 2021 at 20:41

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