# How can I adjust the size of an integrated signal?

I want to test whether I have solved my differential equation correctly. I have this block wiring diagram.

But after the signal passed through the integrator, the amplitude is much smaller than I would have expected. If the block wiring diagram represents the mathematical equation, I would have expected the amplitude to be as shown in the picture. Does somebody maybe know how to adjust the amplitude of the signal after it went through the integrator, or why it is reduced at all? I am very grateful for any help.

## 1 Answer

The output has the correct amplitude, since the integral's amplitude is frequency-dependent:

$$\int \sin(2\pi f t){\,\rm d}t = -\frac{1}{2\pi f} \cos(2\pi f t) + c$$

Note the integral's amplitude $$\1/(2\pi f)\$$: that's the scaling you observed.

To get a unity output, the frequency must be $$\ f=1/2\pi \$$:

$$\int \sin \left(2\pi \frac{1}{2\pi} t \right){\,\rm d}t = -\cos(t) + c$$

Also note that the integrator is free to add an arbitrary offset voltage to the output.

We can simulate this using CircuitLab simulator built into this site (!):

simulate this circuit – Schematic created using CircuitLab

We can observe that the resulting $$\ c=1 \$$:

How can I adjust the size of an integrated signal?

If you want to normalize the output amplitude to be equal to the input amplitude, must multiply the integrator's output by the $$\ 2\pi f \$$ factor:

$$H(s) = \frac{2 \pi f}{s}.$$

We may as well remove the DC offset, which is equal to the amplitude of the sine wave:

simulate this circuit

I highly suggest playing with CircuitLab - for such simple exploration, it's entirely adequate and you don't need Simulink.

It is also very flexible, since most of its functions support variables, so e.g. you can set up the frequency and amplitude as variables, and use them in all kinds of sources, Laplace block polynomials, passive component parameters, active component model parameters, etc.

The variables $$\f\$$ and $$\A\$$ are interactively adjustable, and the transient response time domain simulation is already set up, so it's very easy to play with those circuits.

• +1, through I'd say it's not that the integrator is free to add its own C, but rather integrating a sine will give you just that: a raised cosine. Because the initial conditions are zero and the first quarter period is a rising slope, therefore the output is idt(sin())=-cos(), starting from zero. Apr 15 at 17:19
• @aconcernedcitizen "Starting from zero" is nice of course :) Apr 20 at 12:31