# Modulation and filtering

When I modulate a signal $x(t)$ with $\cos(2 \pi f t)$ and the modulated signal passes through a HPF, what output do i get in the frequency domain?

• Is the cutoff frequency of the HPF higher or lower than the carrier frequency f? Commented Oct 3, 2017 at 19:32
• The cutoff freq is higher then the carrier freq
– Jona
Commented Oct 3, 2017 at 20:11

This is the core subject of linear modulation theory that's considered in analog communication systems. It's analysis is based on the modulation property of the Fourier trabsform.

Given a bandlimited baseband signal $x(t)$ such that its CTFT $X(\omega)$ is zero outside of the band $-W < \omega < W$, its multiplication with the signal $\cos(\omega_0t)$ has the following spectrum:

$$x(t) \cos(\omega_0 t) \longleftrightarrow \frac{1}{2\pi} X(\omega) \star \mathcal{F} \{ cos(\omega_0 t) \}$$ where $\star$ stands for convolution.

The Fourier transform of $\cos(\omega_0 t)$ is $$\cos(\omega_0 t) \longleftrightarrow \pi \delta(w-\omega_0) + \pi \delta(w+\omega_0)$$

Using the convolution property $f(x) \star \delta(x-a) = f(x-a)$ i.e., signal shifts to the impulse position, we deduce the spectrum of the modulated signal as:

$$x(t) \cos(\omega_0 t) \longleftrightarrow \frac{1}{2\pi} \left( \pi X(\omega-\omega_0) + \pi X(\omega+\omega_0) \right) = 0.5 X(\omega-\omega_0) + 0.5 X(\omega+\omega_0)$$

Now if you filter (convolve) this modulated signal with a (n ideal) BPF with a center frequency of $\omega_c = \omega_0$ and a bandwidth of at least $W_{BW} = 2 W$ then you would get the modulated signal spectrum alone. This can be shown by utilizing the filtering property of Fourier transform which says $$h(t) \star x(t) \longleftrightarrow H(\omega) X(\omega)$$

Now the defintion of the (ideal) BPF is such that its frequency response is: $$H(\omega) = \begin{cases} 1 &, |\omega-\omega_0| < W \\ 0 &, o.w. \\ \end{cases}$$

Then the multiplication of $H(\omega)$ with $0.5 X(\omega-\omega_0) + 0.5 X(\omega+\omega_0)$ would yield:

$$H(\omega) \left[ 0.5 X(\omega-\omega_0) + 0.5 X(\omega+\omega_0) \right] = 0.5 X(\omega-\omega_0) + 0.5 X(\omega+\omega_0)$$

Which means: $$\mathcal{F} \{ \text{BPF} \{ x(t) \cos(\omega_0 t) \} \}= 0.5 X(\omega-\omega_0) + 0.5 X(\omega+\omega_0)$$

In this example the whole spectrum is the modulated spectrum so you don't actually filter out anything, but in other situations the BPF would indeed perform some useful operations.