Prove the Time Shift property of the Fourier Transform (i.e., if y(t)=x(t-α) that Y(ω)=X(ω)e-jωα).
Note the similarity to the time shift property of the Laplace Transform.
This problem requires no integration.
a) For the function, p(t), what is P(ω)?
b) For the function pd(t), what is Pd(ω)?
c) For the function s(t)=sin(2π60t), what is S(ω)? You can leave the result in terms of T and ω0=2π/T.
d) Find Y(ω) for the aperiodic function y(t) shown below. You can express the result as the sum of two sinc functions. You can leave the result in terms of T and ω0=2π/T.
d) Function is product of sine and shifted pulse (pulse has width T/2 and time shift of T/4), so the result is convolution of their Fourier Transforms. Recall that the convolution of a function with a shifted impulse yields a shifted version of that function.
A function y(t) and its periodic extension yp(t) are shown below.
Find an expression for the Fourier Series coefficients of yp(t), cn. You needn't try to simplify.
Note: it can be shown (you needn't do this, but you may have done this in a previous problem set via a much more complicated procedure using direct integration) that the Fourier Series coefficients of yp(t) are also given by
For n odd, and |n|≠1, cn=0 because arguments to sinc terms are integers. This agrees with problem statement (e-jπn+1=0, for n odd).
a) Find the Fourier Transform, X(ω), for the function x(t) shown below.
This can be done without integration if you use the fact that a trapezoid is the sum of two triangular functions. A triangular function of width Tp and height 1, λ(t/Tp), is shown below.
Its Fourier Transform is given by .
b) What is the Fourier Transform, Xp(ω), of a signal, xp(t), that is a periodic extension of the one shown above. Period=T, T>2·T1.
a) We represent the function as the sum of two triangle waves of different widths, and slopes of magnitude 1/(T1-T0) (i.e., rise/run), so x(t)=f1(t)+f2(t).
Find the Fourier Transform of a rectangular pulse that is zero everywhere except between t=0 and t=T0 where it has a height of one:
a) by using a table of Fourier Transforms and Properties (this is just a rectangular pulse function of width T0 that is not centered on the origin).
b) by finding its Laplace Transform, X(s), and then using the fact that the Fourier Transform of an absolutely integrable causal function is just the Laplace Transform evaluated at s=jω. (Note: absolutely integrable means ; causal means x(t)=0 for t<0.)
c) Show that the two are equal. It may be useful to use a technique called "removing a half angle" (as well as Euler's identities for trig functions):
There is nothing complicated about this identity. If you perform the multiplication on the right you can easily see that it equals the term on the left.
a) This is just a pulse function of width T0 delayed by time T0/2.
b,c) Find Laplace Transform and replace s by jω to get Fourier Transform, and then simplify.
Engineer A is taking some data from a strain gauge and is worried about electrical interference at 60 Hz so she takes data at 150 Hz to try to avoid problems with aliasing of the 60 Hz signal. (Sampling at 150 Hz allows for representation of signals up to 75 Hz.) Apparently Engineer A is unaware that there is also a lot of noise at 120 Hz (partly due to the fact that the 60 Hz is often rectified by power supplies - remember Fourier Series?). To what frequency is the 120 Hz signal aliased? Note: the 60 Hz signal is not aliased, and need not be considered to solve this problem.
Original freqs : -120 120 (thicker lines in image)
Image freqs : -330 -270 -180 -30 30 180 270 330
Aliased freqs : -30 30
Engineer B is trying to collect bladder pressure signals, which are limited to 10 Hz and below. He decides to collect data at 25 Hz so as not to lose any information. Apparently Engineer B completely forgot about 60 Hz noise.
- To what frequency is 60 Hz aliased?
- How could this be fixed (without changing the sampling frequency)?
- In much of the world the power grid runs at 50 Hz. To what frequency is 50 Hz aliased when sampled at 25 Hz?
Original freqs : -60 60
Image freqs : -110 -90 -85 -65 -40 -35 -15 -10 10 15 35 40 65 85 90 110
Aliased freqs : -10 10
b) Filter the data to prevent aliasing (low pass with cutoff at about 10 Hz). This eliminates the signal that is aliased, while passing the signal we want.
Original freqs : -50 50
Image freqs : -100 -100 -75 -75 -50 -25 -25 0
0 25 25 50 75 75 100 100
Aliased freqs : 0 0