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So I was solving a problem that resulted in a communication channel that heavily distorts high frequency but does not significantly affect low frequency and I got stuck at figuring which one of these signals would be the least distorted. I know I have to find the one with the lowest frequency but I'm not completely sure how. These are the signals:

$$x_1(t) = \delta(t);\quad x_2(t) = 5; \quad x_3(t) = 10e^{j1000t};\quad x_4(t) = \displaystyle\frac 1t$$

I just need to know how to find the frequency of one of the signals

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  • $\begingroup$ You mean nonlinear distortion? How does it distort high frequencies more than low frequencies? $\endgroup$ – endolith Nov 8 '16 at 14:59
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Normally, a signal is composed of several frequencies, which strengths are given by the Fourier Transform of the signal.

  • $x_1(t)=\delta(t) \Leftrightarrow \mathcal{F}(x_1(t))=C$, i.e. the Fourier transform is a constant (value of constant depends on the actual normalization of the Fourier Transform). Hence, it has all frequencies contained, with equal power.
  • $x_2(t)=5 \Leftrightarrow \mathcal{F}(x_2(t))=\delta(f)$, i.e. the Fourier Transform of a constant is a Dirac impulse at frequency zero. Hence, this signal has only one frequency component, the zero-frequency.
  • $x_3(t)=10e^{j1000t} \Leftrightarrow \mathcal{F}(x_3(t))=C\delta(f-1000/2\pi)$ i.e. the Fourier Transform of a complex exponential is a Dirac impulse at the position of the frequency of the exponential. Hence, it has only one component, at frequency $1000/2\pi$
  • $x_4(t)=1/t \Leftrightarrow \mathcal{F}(x_4(t))=C\text{sign}(f)$, i.e. the Fourier transform of the inverse is a sign function. Hence, it contains all frequencies.

Accordingly, the answer to your question would be signal 2, as it only contains the DC frequency ($f=0$).

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  • $\begingroup$ Thanks a lot for this, we haven't been taught this yet so I was really confused. $\endgroup$ – Drew Nov 8 '16 at 6:02
  • $\begingroup$ You can also understand the problem without the Fourier Transform: The faster/more abrupt a signal's time-domain changes, the higher frequencies it contains. Here, the function that changes slowest, is the constant (i.e. it does not change at all over time) $\endgroup$ – Maximilian Matthé Nov 8 '16 at 6:03

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