# Is the Laplace transform of an integral the same as the Laplace transform of a constant?

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

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?

• 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. Jun 17, 2023 at 7:13
• 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').
– Chu
Jun 18, 2023 at 7:15
• ... it's also important to remember that a transfer function assumes zero initial conditions.
– Chu
Jun 18, 2023 at 7:36

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.