I found the transfer function of the RC circuit (Figure 1). $$ G(s)=\frac{V_{out}(s)}{I_{in}(s)}=\frac{R_2}{R_2sC_1+1}=\frac{50}{50*10^{-9}s+1} $$ I can check if the transfer function is correct or not in the LTspice program. Can you tell me how to check the expression obtained by the inverse Laplace transform in the LTspice program? $$ G(t)=\frac{50}{50*10^{-9}s+1} invlaplace -> 1000000000*e^{-20000000*t} $$ I thought it was the $$ G(t)=\frac{V_{out}(t)}{I_{in}(t)} $$
signal, but it's not.


LTspice Model

  • 2
    \$\begingroup\$ Your transfer functino is wrong. You can use many methods to determine it. One hint would be to consider Z = R2 || C and then solve the circuit as if it was a voltage divider. I don't understand the LTspice part: you're asking for help on how to verify your results but, I see you know what to use. It will help you (and others reading it) to use prefixes (.e.g. 1Meg instead of 1000000, or 1n instead of 1e-9 -- easier to read, less prone to mistakes). \$\endgroup\$ Commented Dec 23, 2022 at 21:08

2 Answers 2


If you want to confirm your inverse transform with a specific circuit you need to come up with a particular solution. You can't just plug in a general solution. It seems as though you don't realize that fact.

First, let's verify your transfer function:

$$\begin{align*} \frac{v_{_\text{out}}}{R_1}+\frac{v_{_\text{out}}}{R_2}+C_1\frac{\text{d}}{\text{d}t}v_{_\text{out}}&= \frac{v_{_\text{in}}}{R_1} \\\\ \frac{v_{_\text{out}}}{R_2}+C_1\frac{\text{d}}{\text{d}t}v_{_\text{out}}&= \frac{v_{_\text{in}}-v_{_\text{out}}}{R_1} \\\\ \frac{v_{_\text{out}}}{R_2}+C_1\frac{\text{d}}{\text{d}t}v_{_\text{out}}&= i_{_\text{in}} \\\\ v_{_\text{out}}+R_2 C_1\frac{\text{d}}{\text{d}t}v_{_\text{out}}&=R_2\,i_{_\text{in}} \\\\ \left(1+R_2 C_1\frac{\text{d}}{\text{d}t}\right)v_{_\text{out}}&=R_2\,i_{_\text{in}} \end{align*}$$

So, sure. I can get \$\frac{v_{_\text{out}}}{i_{_\text{in}}}=\frac{R_2}{1+R_2C_1 s}\$. That's not the standard form, either for looking up in an inverse Laplace table or just generally. You should have re-stated it better, I think. But, that said, I can get there.

So it is confirmed.

One textbook method is to take the inverse Laplace transform, include initial conditions and take the Laplace transform, and then take the inverse Laplace transform.

I won't bother. What a pain. Since \$i_{_\text{in}}\$ is a constant, it's easier to just use annihilation.

$$\begin{align*} \left(1+R_2 C_1\frac{\text{d}}{\text{d}t}\right)v_{_\text{out}}&=R_2\,i_{_\text{in}} \\\\ \left(\frac{\text{d}}{\text{d}t}+\frac1{R_2 C_1}\right)v_{_\text{out}}&=\frac1{C_1}\,i_{_\text{in}} \\\\ \frac{\text{d}}{\text{d}t}\left(\frac{\text{d}}{\text{d}t}+\frac1{R_2 C_1}\right)v_{_\text{out}}&=\frac{\text{d}}{\text{d}t}\frac1{C_1}\,i_{_\text{in}}=0 \end{align*}$$

This is now a nice, simple homogeneous function and the zeros are at \$\frac{\text{d}}{\text{d}t}=0\$ and \$\frac{\text{d}}{\text{d}t}=-\frac1{R_2 C_1}\$.

So the general solution is:

$$\begin{align*} v_{_\text{out}}&=A_1\cdot e^{^{0\,\cdot\, t}} + A_2\cdot e^{^{\frac{-t}{R_2\,\cdot\,C_1}}} \\\\ &=A_1 + A_2\cdot e^{^{\frac{-t}{R_2\,\cdot\,C_1}}} \end{align*}$$

At \$t=\infty\$ you know that the voltage on the capacitor must be \$R_2\cdot i_{_\text{in}}\$, so \$A_1=R_2\cdot i_{_\text{in}}\$. And at \$t=0\$ you know that the voltage on the capacitor (I'm assuming you and I agree here) is \$0\:\text{V}\$. So it must be that \$A_1+A_2=0\$ and from that you can figure out that \$A_2=-A_1=-R_2\cdot i_{_\text{in}}\$.

So the particular solution is:

$$\begin{align*} v_{_\text{out}}&=R_2\cdot i_{_\text{in}}\left(1 - e^{^{\frac{-t}{R_2\,\cdot\,C_1}}}\right) \\\\ &=500\left(1 - e^{^{\frac{-t}{50\,\text{ns}}}}\right) \end{align*}$$

In LTspice, nothing more than this:

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Commonly uppercase variables are used for Constants and Laplace domain functions. Lowercase is used for time domain functions.

the expression obtained by the inverse Laplace transform [...] $$ G(t)=\frac{50}{50*10^{-9}s+1} invlaplace -> 1000000000*e^{-20000000*t} $$ I thought it was the $$ G(t)=\frac{V_{out}(t)}{I_{in}(t)} $$

You correctly write that the transfer function is \$G(s)=\frac{Vout(s)}{Iin(s)}\$. This concept of transfer function does not translate into the time domain. \$\frac{Vout(t)}{Iin(t)}\$ is not a transfer function.

$$g(t)=\mathcal{L}^{-1}G(s)$$ is called the impulse response of G(s). The impulse is also called the Dirac function or the delta function \$\delta(t)\$. \$\mathcal{L}\delta(t)=1\$.

To find the use for \$g(t)\$, consider the equation:$$Y(s)=G(s)U(s)$$ To convert to time domain, notice there is a multiplication of functions on the right-hand side. A property of Laplace transforms indicates that multiplication in the Laplace domain corresponds to convolution in the time domain. See Paul's Notes $$y(t)=\int_{0}^{t}g(\tau)u(t-\tau)d\tau\tag{1}$$

You correctly obtained that:$$G(s)=\frac{R_{2}}{R_{2}Cs+1}\Rightarrow g(t)=\frac{1}{C}e^{\frac{-t}{R_{2}C}}$$

The input step is \$u(t)=I=10A,t\ge 0\$.

Substituting into Equation (1): $$v_{out}(t)=\int_{0}^{t}\frac{I}{C}e^{\frac{-\tau}{R_{2}C}}d\tau=IR_{2}\left[1-e^{\frac{-t}{R_{2}C}}\right]$$

Jonk's (+1) answer has already dealt with LTspice, so I won't repeat it here.

The transfer function \$G(s)\$ represents the hardware in the Laplace domain. The impulse response \$g(t)\$ represents the hardware in the time domain, but is not a ratio of output to input.


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