What is the bandwidth of an imaginary convolution?

Question

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)]

(also on wolfram-alpha)

Answer 1

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.

Answer 2

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.