For an LTI system, why does the Fourier transform of the impulse response give the frequency response?

Question

I know that for a given system, the Fourier transform of its impulse response gives its frequency response. I want to find where this property comes from, but haven't been able to find if it's a definition or if there's a mathematical proof available, for both continuous and discrete-time systems.

Answer 1

Let $h(t)$ denote the impulse response of an LTI system. Then, for any input $x(t)$, the output is $$y(t) = \int_{-\infty}^\infty h(\tau)x(t-\tau)\,\mathrm d\tau.$$ In particular, the response to the input $x(t) = \exp(j2\pi ft)$ is $$\begin{align} y(t) &= \int_{-\infty}^\infty h(\tau)\exp(j2\pi f(t-\tau))\,\mathrm d\tau\\ &= \exp(j2\pi ft)\int_{-\infty}^\infty h(\tau)\exp(-j2\pi f\tau)\,\mathrm d\tau\\ &= H(f)\exp(j2\pi ft),\tag{1} \end{align}$$ where $H(f)$ is the Fourier transform of the impulse response $h(t)$. In words, for an LTI system with impulse response $h(t)$, the input $\exp(j2\pi ft)$ produces output $H(f)\exp(j2\pi ft)$. This is precisely what we have as the definition of the frequency response of an LTI system (call this $FR(f)$ for now):

for every frequency $f$, the response to $\exp(j2\pi ft)$ is $FR(f)\exp(j2\pi ft)$ which is just a (complex) constant times the input complex exponential.

But $(1)$ shows that $FR(f)$ is just $H(f)$, the Fourier transform of the impulse response $h(t)$, which is what you wanted to prove.

Answer 2

The intuitive answer is that an impulse in time at t=0 contains all frequencies of equal magnitude, so applying an impulse to an LTI system is the same as applying all frequencies at once, thus the result is the response of the system to all frequencies, i.e., the frequency response. For a real world example, you can find the total frequency response of a mechanical system by instrumenting it with accelerometers and hitting it with a hammer, which approximates an impulse. When I worked on the Space Shuttle in the early eighties, this is exactly how we determined the frequency response of the Ku-band integrated communications and radar system, a complex mechanical LTI system. Once knowing the frequency response, we filtered input control signals to avoid excitation of mechanical resonant modes that could make the system unstable.

Answer 3

The key ingredient is that the base functions of the Fourier transform $\exp(i \omega t)$ are eigenfunctions of LTI systems. That means the LTI system can be represented as a diagonal linear operator in the Fourier basis. Or in other words: To apply an LTI system in frequency domain, you just multiply their frequency responses. And applying an LTI system to a signal means multiplying the system frequency response with the Fourier transform of the signal.

Now to measure the frequency response of a system, we can let the system act on a signal with a known nowhere-vanishing Fourier transform, ideally even one with constant unit Fourier transform $F\{s(t)\}(\omega)=1$, and that is the unit impulse function in time domain. The resulting Fourier transform is then 1 multiplied with the LTI frequency response, so it's the frequency response of the system.

This is not precisely what you asked for, because the calculation happened in frequency domain. But we can switch between time and frequency domain at any point during the calculation. So you can run your unit impulse signal through the system in time domain to get the impulse response of the system, and only then take the Fourier transform in the end. That's why the Fourier transform of the impulse response is the frequency response.

All this can be spelled out mathematically and more rigorously, but I think you should be able to see the connection even with this slightly superficial description. If you need more details, let me know.

Answer 4

Frequency response is the eigenvalue of the system when the input is an eigenfunction of the form $e$^{$j\omega n$}.

Solve it by yourself to get $\displaystyle \sum_{k=-\infty}^\infty h[k]$e^{$-j\omega k$}, which is the Fourier transform of $h[n]$.