# Amplitude spectrum (transfer function) of signal?

I have one question related to finding amplitude spectrum (transfer function) of signal knowing that output signal is time derivative of input signal. I have the answer graph but I don't understand the reason.

HINT:

If $$X(j\omega)$$ is the Fourier transform of a signal $$x(t)$$, try to remember (or look up) how $$X(j\omega)$$ is related to the Fourier transform of the derivative $$x'(t)$$.

Some people are more familiar with this in the Laplace transform domain: the time derivative corresponds to which operation in the Laplace transform domain? Replace $$s$$ with $$j\omega$$, and you'll understand the given figure.

• Yep! I got it, thank you very much! Mar 14 at 15:50

Based on what you explained, the output signal $$y(t)$$ is given by: $$y(t) = \frac{dx(t)}{dt}$$ where x(t) is the input signal.

By transforming this problem in the Laplace domain, we can re-express the above equation as: $$Y(s) = sX(s) - x(0^{-})$$ If we know that $$x(0^{-})=0$$, then this can be simplified as $$Y(s) = sX(s)$$. Using this expression we can now obtain the transfer function of our system in the Laplace domain $$H(s)$$ as: $$H(s) = \frac{Y(s)}{X(s)} = s$$ In the Fourier domain, we can replace $$s=j\omega$$, such that: $$H(j\omega) = \frac{Y(j\omega)}{X(j\omega)} = j\omega$$ Then we can find the amplitude of $$H(j\omega)$$, which is equal to: $$|H(j\omega)| = |j\omega| = |\omega|$$ You can also express this in terms of linear frequency $$f$$ (in terms of Hz), where the angular frequency $$\omega$$ is related to linear frequency by $$\omega = 2\pi f$$. This gives us: $$|H(f)| = 2\pi |f|$$

If you simulate this, you will get the following plot, which is the same as the one you posted.