Question about ramp filter used in filtered backprojection

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?

• 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? – A_A Sep 25 '16 at 17:01

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 $\frac{w}{2\pi}\sin(wr)$.