# F06 Fun With Fourier Transforms

- The continuous signal x(t) = sin(2*pi*100*t) was sampled with sampling period T = 1/400 second to obtain a discrete-time signal x[n]. What is the resulting signal x[n]?
- The sequence x[n] = cos(pi*n/4) was obtained by sampling a continuous time signal x(t) = cos(omega * t), at a sampling rate of 1000 samples per second. What are two possible positive values of omega that could have resulted in the sequence x[n]?
- The continuous-time signal x(t) = cos(4000*pi*t) is sampled with a sampling period T to obtain a discrete-time signal x[n] = cos(pi*n/3). (a) Determine a choice for T consistent with this information. (b) Is your choice of T in part(a) unique? If so, explain why. If not, specify another choice of T consistent with the information given.
- The continuous-time signal x(t) = sin(20*pi*t) + cos(40*pi*t) is sampled with a sampling period T to obtain the discrete-time signal x[n] = sin(pi*n/5) + cos(2*pi*n/5). (a) Determine a choice for T consistent with this information. (b) Is your choice of T in part(a) unique? If so, explain why. If not, specify another choice of T consistent with the information given.
- Sketch the magnitude Fourier transform of each of the following functions. I?m not concerned about the absolute scale of the Y-axis but I am concerned about the scale of the X (frequency) axis.
- sin(2*pi*10*t)
- abs(t) < 0.1 (box of width 0.2 seconds)
- mod(t, 0.1) == 0 (a comb with 0.1 second spacing)
- sin(2*pi*10*t) * sin(2*pi*100*t)
- exp(-t^2/2) (Gaussian with standard deviation = 1)
- exp(-t^2/2) * sin(2*pi*10*t) (product of Gaussian and sine)

- The derivative with respect to time of sin(2*pi*f*t) is 2*pi*f*cos(2*pi*f*t). What does this tell you about the effect of differentiation on the Fourier transform of a signal?
- Explain why filtering after uniform sampling is, in general, not effective for anti-aliasing.
- Explain why filtering after uniform sampling is so commonly used for anti-aliasing and why it usually is effective.
- In the discrete Fourier transform of an image that is 512 by 512 pixels, what is the lowest non-zero frequency represented? What is the highest frequency?
- Download the images (oldwell1.tif, oldwell2.tif) attached to this page. Compute the 2D DFT of each (I=fft2(i)). Extract the magnitude (M=abs(I)) and phase (P=angle(I)) of each. Then reconstruct 2 new images by combining the magnitude of 1 with the phase of 2, and the magnitude of 2 with the phase of 1 (i12=real(ifft2(M1.*exp(1i*P2)))). Compare the original two images to the new pair of images. I recommend you make sure you’re doing the last step correctly by reconstructing i11 from M1 and P1 and make sure it looks like you expected.