Frequency domain analysis of a convolved signal

When two signals are convolved in the time domain, what frequency components will be present in the frequency domain? Is it similar to $\text{(frequency 1 + frequency 2)}$ and $\text{(frequency 1 - frequency 2)}$? No, it isn’t. That formula applies to the time-domain multiplication of two sinusoidal signals.

According to the Discrete Convolution Theorem, convolution of two discrete signals in the time domain is equivalent to multiplication of their DFTs in the frequency domain:

F{x[n] ∗ h[n]} = X[k] ⋅ H[k]

where X[k] and H[k] are the DFTs of x[n] and h[n], respectively.

Thus, the convolution y[n] in the time domain can be computed by taking the inverse DFT of the product:

y[n] = F^-1{X[k] ⋅ H[k]}

In general, the frequency components present in X[k]⋅H[k] correspond to the frequencies where both X[k] and H[k] have significant values. Frequencies in X[k] that align with the passband of  H[k] will be prominent in X[k]⋅H[k], while those outside of it will be suppressed. Thus, X[k]⋅H[k] in the frequency domain represents a signal that has been shaped by the frequency response of h[n], retaining only the frequencies that h[n] allows to pass or emphasizing the frequency range defined by H[k].

For example, if the frequency domain representation of a signal is $X(k)$, and is multiplied by a factor (e.g., 5), then the frequency components of the resulting signal will remain the same, but the amplitude and phase of the signal will change. The amplitude of the new signal in the time domain will be scaled by a factor of 5.

Example

Suppose we are convolving two signals, $S1$ and $S2$, in the time domain, where $S1=\sin (2\pi \cdot 100\cdot t)$ and $S2 = \cos(2 \pi \cdot 1000 \cdot t)$. In the frequency domain, you will see that the frequency components are centered at 100 Hz and 1000 Hz, with significant peak frequencies at 100 Hz and 1000 Hz.

Further Reading

1. What is convolution (full convolution)?
2. Power Spectral Density Calculation Using FFT in MATLAB