I am having great time while reading "The Scientist and Engineer's Guide to Digital Signal Processing" book.

In "Chapter-14 Introduction to Filters" author indicates we can create high pass filter from low pass filters by multiplying low pass filter's frequency kernel with a sinusodial which has 0.5 frequency. Link to the this chapter

I thought I read every chapter carefully but I can't understand why this will work? Can somebody please explain the reasoning behind of this process as easy that a computer science graduate could understand.


1 Answer 1


This can be explained by the Convolution theorem

As you state, in the time domain you have the kernel (f) multiplied by a sinusoid (g).

If we take the Fourier transform of f·g (i.e., F(f·g)), this is equivalent to the convolution between F(f) and F(g) (i.e., F(f)*F(g)). Note that the Fourier transform of the sinusoid is proportional to two delta functions located at the frequency and negative frequency of the sine wave. So when you convolve these deltas with the frequency kernel your frequency kernel "shifts" to be centered around the sinusoid frequency.

  • $\begingroup$ Thanks for your answer, what I know multiplication in time domain means convolution in F domain. So kernel is already in F domain. Because of that F(f.g) seem to me shouldn't be same as F(f)*F(g).This confuses me a little bit, Am I right. $\endgroup$ Apr 29, 2014 at 8:01
  • $\begingroup$ Oh you are right I double checked now, filter kernel was in time domain, I completely understand now. Thanks a lot . $\endgroup$ Apr 29, 2014 at 8:05
  • $\begingroup$ @KadirErdemDemir: a sinusoid at 0.5 frequency is actually like $x[n]=cos(\pi n) = (-1)^n$ whose DTFT is $X(e^{j\omega}) = \pi\delta(\omega-\pi) + \pi\delta(\omega+\pi)$ as klurie states i-since multiplicaiton in time equals convolution in frequency and ii-since convolution with impulse means means shift to impulse location, the result is that lowpass filter which is centered about zero frequency is shifted to impulse frequency at $\omega = \pi$ which is considered as the high frequency region. Hence the filter becomes a highpass filter from lowpass origin. $\endgroup$
    – Fat32
    Feb 18, 2015 at 20:28

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