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Question is this. First, a ramp filter (in frequency domain) is defined by $H(Q)=|Q|$. What are the responses of a ramp filter to (1) a constant function $f(r)=c$ and (2) a sinusoid function $f(r)=\sin(wr)$? What does the response mean? Following is my work.

My work:

  1. First, take fourier transform of a function $f(r)=c$. It is $\int_{-\infty}^{\infty}f(r)e^{-2i\pi rQ}dr=c\delta(Q)$. Then multiply ramp filter and take inverse fourier transform. It is $\int_{-\infty}^{\infty}c\delta(Q)|Q|e^{2i\pi irQ}dQ=0$??

  2. Similarly, $\int_{-\infty}^{\infty}\sin(wr)e^{-2i\pi rQ}dr=\frac{\delta(Q-w/2\pi)-\delta(Q+w/2\pi)}{2i}$. So Applying the ramp filter and i.f.t gives $\frac{w(e^{iwr}-e^{-iwr})}{4i\pi}=\frac{w\sin(wr)}{2\pi}$.

It this right?

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  • $\begingroup$ Can you please clarify the math notation a little bit? Your first expression is like an integral equation not an evaluation of an integral. It is quite possible that your derivations are along the right track but with this notation it is unclear why. Also, what does your intuition say? What do you think might happen if you were to pass DC through a ramp filter? What is another name for the ramp filter? What does the ramp filter do at the end of the day? $\endgroup$ – A_A Sep 25 '16 at 17:01
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To see if your math is correct, it is useful to first understand what is, in general, the effect of a filter on a signal, and then see if you can predict what the theoretical result should look like.

If a filter has frequency response $H(Q)$, this means that its response to an input $e^{j2\pi Q_0 r}$ is the signal $H(Q_0)e^{j2\pi Q_0 r}$. In other words, a sinusoidal input of frequency $Q_0$ produces an output of the same frequency, but with amplitude $|H(Q_0)|$ and phase $\angle H(Q_0)$.

In your first question, the input has frequency $Q_0=0$. The filter's response at that frequency is $|Q_0|=0$. Then, the filter's output will be 0: frequency $Q_0=0$ is completely absorbed by the filter and it does not appear at the output.

In your second question, the input has frequency $Q_0=w/2\pi$. The filter's response at that frequency is $|Q_0|=w/2\pi$. The output, then, should be $\frac{w}{2\pi}\sin(wr)$.

As you can see, following this line of reasoning we obtain the same results as you did by just solving the equations. This corroboration should give you confidence that your results are correct, and most importantly, you can understand what the filter is doing physically.

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