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I am trying to figure out the bandwidth of \$f_1f_2\$, where \$f_1 = sinc^2(3t)\$ and \$f_2 = sin(100t)\$. So when I take the Fourier Transform, I can rewrite the equation as such: \$F(\omega) \leftrightarrow F_1 * F_2\$. Easy so far.

Moving on, \$F_1 = 3\pi\Delta(\omega/12)\$, and \$F_2 = j\pi\delta(\omega+100) - j\pi\delta(\omega-100)\$. This is where I get stuck.

When you convolve anything with \$\delta(t+\tau)\$, it is merely placing the function you are convolving \$\delta(t+\tau)\$ with at time \$\tau\$. When taking bandwidths of frequencies, I know you only look past time \$t=0\$.

At this point I need to find the bandwidth of the function \$-3j\pi\Delta(\frac{w-100}{12})\$. Without this being in the imaginary frequency domain, for \$\omega \geq 0\$ there would be no bandwidth (everything is zero or has a negative amplitude for that frequency). However, we are in the imaginary frequency domain, so what would the bandwidth of this filter be?

The graph of the transform is
fourier transform [sinc^2(3t)sin(100t)]
enter image description here
(also on wolfram-alpha)

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However, we are in the imaginary frequency domain, so what would the bandwidth of this filter be?

I'm not sure why you're having difficulty with the fact that the frequency domain signal is, in this case, imaginary. The symmetries of the Fourier transform are usually taught early on in signal processing courses:

  • If the time domain signal is real and odd, e.g., \$\sin\omega t\$, the associated frequency domain signal is imaginary and odd.
  • If the time domain signal is real and even, e.g., \$\cos\omega t\$, the associated frequency domain signal is real and even.

But, for bandwidth calculation, you're interested in the magnitude in the frequency domain (think Bode magnitude plot). Since your frequency domain signal is pure imaginary, it couldn't be any easier; simply remove the j factor.

If, however, your frequency domain signal were complex, you would need to multiply the function by its conjugate and take the square root in order to find the magnitude.

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  • \$\begingroup\$ This is what I was looking for. Our professor didn't explain this very well at the beginning of the course, and this makes it much easier to understand. Thank you! \$\endgroup\$ – nathpilland Dec 2 '13 at 17:21
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There seems to be a slight confusion in your understanding of amplitude, phase and frequency:

Without this being in the imaginary frequency domain, for \$\omega\ge0\$ there would be no bandwidth (everything is zero or has a negative amplitude for that frequency).

Firstly, your frequencies are not imaginary here. Your frequency-domain function \$F1∗F2\$ has imaginary values, which we can see because of the \$j\$ in its formula.

Secondly, amplitude cannot be negative (by definition). Amplitude is the magnitude of the complex number representing the signal at a given frequency. (The phase is then defined as the angle of the complex number.)

Keeping those things in mind, it should be easier to find the bandwidth now. Take the amplitude of the complex function, which will just be the absolute value of its imaginary part in this case. Then you can find the bandwidth by looking at the lowest and highest frequencies that have non-zero amplitude. As far as I can tell, this is 12 rad/s.

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  • \$\begingroup\$ Sorry for the confusing comment, I was merely stating that there is no bandwidth in the real frequency domain. So if I am looking in the positive\$\omega\$ direction in the imaginary frequency domain, my bandwidth would be 12? And this imaginary frequency bandwidth can be counted as well and isn't discounted just because it's imaginary? \$\endgroup\$ – nathpilland Nov 25 '13 at 7:10
  • \$\begingroup\$ One thing I am trying to explain is that there is no imaginary frequency domain. All the frequencies are real. The other thing I am saying is that the bandwidth is determined by the magnitude of the signal, not its real value. \$\endgroup\$ – richarddonkin Nov 25 '13 at 7:17
  • \$\begingroup\$ Oh I think I see what you're saying. So everything is plotted on one graph? I think I'm just confused why the i is even involved in the frequency domain. I also had another question about Nyquist frequencies and these involved \$i*\delta(t)\$ impulses, and I don't understand the difference between these and regular \$\delta(t)\$ impulses \$\endgroup\$ – nathpilland Nov 25 '13 at 7:34
  • \$\begingroup\$ A signal with the form \$a+jb\$ means that the signal component in the frequency domain is not real, it is complex. That means the signal component at a given frequency has a phase offset from the \$\cos\$ basis function. \$\endgroup\$ – richarddonkin Nov 25 '13 at 12:01

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