I was analyzing the circuit of an amplifier to know its response, the schematic is the following: Schematic

What I did is the following: I considered \$(v_{in}\$ as the input signal; \$v_{out}\$ as the output of the op-amp; \$v_1\$ as the value of the voltage of the positive terminal of the op-amp; \$v_2\$ as the value of the voltage of the negative terminal of the op-amp.

I named the capacitor \$C\$, the 10k resistor of the positive terminal of the op-amp as \$R_1\$, the potentiometer as a rheostat of the negative terminal as \$R_F\$, and the 220 ohm resistor as \$R_G\$.

I tried to solve it as follows:

\$i_C = i_{R_1}\$ (1)

\$C\cdot \frac d {dt}(v_{in}(t)-v_1(t)) = \frac {v_1(t) - (5V)}{R_1}\$ (2)

\$i_{R_G} = i_{R_F}\$ (3)

\$\frac {(5V) - v_2 (t)}{R_G} = \frac{v_2 (t) - v_{out}(t)}{R_F}\$ (4)

Assuming that \$v_1 (t) = v_2 (t)\$, I proceeded to expand equation (2) to solve for \$v_1(t)\$, but I got stuck:

\$C \cdot \frac {d}{dt} v_{in} (t) - C \cdot \frac {d}{dt} v_1 (t) = \frac {v_1}{R_1} - \frac {5V}{R_1}\$

I wanted to apply a Laplace transform to factor out \$v_1(t)\$ easier, to get \$V_1(s)\$, but I don't know how to manage the term \$\frac{5V}{R_1}\$, as it is a constant; I remember from school that the Laplace transform of an integral is \$\mathscr{L}\{\int_{}^{}f(t)dt\}= \frac{1}{s}F(s)\$, but making a quick search for the Laplace transform of a constant gives me \$ \mathscr {L} \{ k \} = \frac{k}{s}\$.

I know that my equations don't involve integrals so far, but I have this doubt about that if the Laplace transform of an integral is the same as the Laplace transform of a constant?

  • 1
    \$\begingroup\$ What is the ac value of the 5-V rail? It is 0 V, therefore, whether the 10-kOhm resistor is connected to 5 V or GND, it is the same. Similarly, the 200-Ohm resistor is also ac grounded. These bias points are only relevant for a dc analysis, not for an ac analysis. Your circuit is a simple differentiator (a zero at the origin) followed by a gain. \$\endgroup\$ Jun 17, 2023 at 7:13
  • \$\begingroup\$ It depends on whether the Laplace transform in question is a signal or a transfer function. if X(s) is a signal, then multiplying it by 1/s is equivalent to integrating x(t). If X(s) is a transfer function then multiplying it by 1/s produces the unit step response, so in this case, 1/s can be regarded as a unit step signal (strictly, not a 'constant'). \$\endgroup\$
    – Chu
    Jun 18, 2023 at 7:15
  • \$\begingroup\$ ... it's also important to remember that a transfer function assumes zero initial conditions. \$\endgroup\$
    – Chu
    Jun 18, 2023 at 7:36

2 Answers 2


It is true that the onesided laplace transform of a constant is the same as the one of the integral operator. The reason being that an integral from 0 to t can be thought of as the convolution of a step function with the integrand. Therefore we can write $$ \int_{0}^{t} f(\tau) \,d\tau = (\Theta * f)(t)$$ were \$\Theta\$ is the step function and \$*\$ denotes the convolution operator. Applying the onesided Laplace transform \$L_{1} \$ yields $$L_{1}(\Theta * f) = L_{1}(\Theta)\cdot L_{1}(f)$$ as the Laplace transform converts a convolution in time domain into an algebraic product in the frequency domain. And now it should be evident why the symbolic identity $$L_{1}(\int_{0}^{t}) = L_{1}(\Theta)$$ holds. Finally note that $$L_{1}(\Theta) = L_{1}(1) = \int_{0}^{\infty} 1 \cdot \exp(-t\cdot s) \,dt $$ due to the integration limits of the onesided Laplace transform. Hence the one sided laplace transform of a constant is the same as the one of a step function.


You mess with the concepts. \$ \mathscr {L} \{ k \} = \frac{k}{s}\$ is not a special case of \$\mathscr{L}\{\int_{}^{}f(t)dt\}= \frac{1}{s}F(s)\$ .

If the 2nd equation were \$\mathscr{L}\{\int_{}^{}f(t)dt\}= \frac{1}{s}f(s)\$ the 1st equation would be a special case. But it isn't.

You shouldn't let the s in the denominator fool you to mix totally different concepts. The first equation tells the L-transforms of constant functions of time. The 2nd equation connects together L-transforms of two different functions of time.


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