Problem 9.2. This problem uses the Modulation Property of the Fourier Transform and is similar to HW 8.3 in Fall 2017 except part (c) uses fc = 10 Hz instead of fc = 5 Hz. Please see homework #8 solutions in Fall 2017.

Problem 9.2(b). Please see the above note for problem 9.2. Here is a key Fourier transform property for this part:

 
                   1
F { x(t) h(t) } = ---- X(w) * H(w)
                  2 pi

That is, multiplication in the time domain is convolution in the frequency domain. The scaling factor of 1/(2 pi) is due to the fact that w = 2 pi f. The derivation of the multiplication-in-time property is available in Handout T Multiplication-in-Time Fourier Transform Property

Another property that you might need for this problem is that

              / oo
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x(t) * d(t) = | x(v) d(t - v) dv = x(t)
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             / -oo

where d(t) is the Dirac delta. This is due to the sifting property of the Dirac delta (see lecture slides 12-7 through 12-9). Likewise,

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x(t) * d(t - t0) = | x(v) d(t - t0 - v) dv = x(t - t0)
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                  / -oo

Problem 9.2(d). Bandwidth is defined as the non-zero extent of the magnitude of the frequency response in positive frequencies. The difficulty is in quantifying non-zero extent. The simplest method is to eyeball the bandwidth based on the plot of the magnitude of the Fourier transform given in part (c).