Problem 1: (The following is taken from the Signal Processing PhD Quals written exam for January 2007. Note: this is not the complete exam.) Assume we have a signal x(t) with a Fourier transform X(jΩ) given by where X0 is some real valued number, W is a real valued positive number, and Ω is specified in units of radians/second.
(a) What is the value of x(t) at t = 0?
(b) For an arbitrary t, what is the relationship between x(t) and x(−t)? (c) What is the value of ʃ ∞ −∞ x(t)dt?
(d) What is the value of ʃ ∞ −∞ |x(t)| 2 dt?
Answer: These solutions are all based on the elementary properties of the Fourier transform (see the class handout).
(b) Using the symmetry properties, we note that X ( j!) is real, therefore x(!t) = x(t), that is they are complex conjugates.
(c) This one is a little tricky! We use the property that
BUT note that there is a singularity at ! = 0. The question is: what is the value of X ( j0) ? The problem statement specifies that 0 X ( j!) = X for 0 " !
On the other hand if you approximate the step discontinuity with a smooth function (say erf()) around ! = 0, you can argue that the value of X ( j0) = 0.75X0 , or
So the answer is dependent on your assumption about the discontinuity! (d) From Parseval’s theorem
Problem 2: An impulse δ(t) is passed through an ideal low-pass filter with frequency response H(jΩ) = 1 for |Ω| < Ωc and H(jΩ) = 0 otherwise. Find and sketch y(t), the output of the filter. Is this a causal filter?
Answer:
If and impulse is passed through the filter, we obtain the impulse response h(t) F {H( j!}
The filter is acausal.
Problem 3: After measurement and curve-fitting it is determined that a causal
signal processing filter has an impulse response h(t) = 5e−3t for t > 0.. What is the filter’s (a) transfer function, and (b) its frequency response function? Determine the -6 dB cutoff frequency, that is the frequency at which the output amplitude is one half of the low frequency response
Answer: h (t) = 5e- 3t Let’s compute the Fourier transform of h(t)
Note: lower limit in integral is 0 because a real filter is a causal system.
a) The transfer function can be found by taking the Laplace transform, which can be viewed as a Fourier transform where jω is replaced by s=σ+jω .
b) The frequency response is given by H(jω) computed previously.
c) We find the cut-off frequency by solving: