- Fourier transforms videos
- Signals & Systems slides and handouts
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- Problem 9.1(d).
An impulse train in the time domain models instantaneous sampling in time.
The Fourier transform of an impulse train is an impulse train.
Please see lecture slides 16-3 and 16-4 on Sampling & Reconstruction and
homework #8 solutions in Fall 2017 for the solution to HW 8.4.
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- 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:
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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.
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See the handout on
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The derivation of the multiplication-in-time property is available in
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Handout T Multiplication-in-Time Fourier Transform Property
Another property that you might need for this problem is that
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x(t) * d(t) = | x(v) d(t - v) dv = x(t)
|
/ -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,
/ oo
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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 9.1(c).
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The Fourier transform will be of the form
C(w) = F(w) + G(w)
To sketch the magnitude response by hand,
you can use the following inequality:
| C(w) | = | F(w) + G(w) | ≤ | F(w) | + | G(w) |
The inequality decouples the plotting of | F(w) | and | G(w) |.
For the hand sketch, you could draw the spectrum without using
a specific numeric value for fc in mind, although fc would have
to be larger than the bandwidth of the lowpass signal prior to
modulation to get a bandpass spectrum.
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For the MATLAB plot of the magnitude response, you could modify the
code from Lecture Slide 14-5 to match the range of frequencies in
this problem:
fmaxplot = 20;
wmaxplot = 2*pi*fmaxplot;
w = -wmaxplot : 0.01 : wmaxplot;
H = 2 ./ (2 + j*w);
Hmag = abs(H);
Hphase = phase(H);
figure;
plot(w, Hmag);
title('Magnitude Response');
figure;
plot(w, Hphase);
title('Phase Response');
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You'll need to put in your formulas for H.
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You'll need to put in your formula for H.
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- 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).
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Problem 9.2. Additional question on frequency selectivity asked in the second part of the problem.
Approach #1. We are treating the signal as the impulse response of an LTI system.
In order to have a frequency response that is bounded in magnitude over all frequenices,
the LTI system must be bounded-input bounded-output (BIBO) stable.
In continuous time, a BIBO stable LTI system must have its transfer function in
the Laplace domain have a region of convergence that includes the imaginary axis s = j w.
For a causal system, this means that the poles must be in the left-half of the Laplace plane.
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(The equivalent condition for a causal discrete-time LTI system is that all poles must be include
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(The equivalent condition for a causal discrete-time LTI system is that all poles must be included
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the unit circle.)
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For this second part, two of the three impulse responses represent BIBO unstable systems
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For this second part, the impulse responses in parts (a) and (c) represent BIBO unstable systems
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because the region of convergence does not include the imaginay (j w) axis,
and hence the frequency selectivity cannot be determined.
And that would sufficient for an answer in those two cases.
Approach #2. On the other hand, we could analyze the frequency content of the signal.
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To determine the frequency content in each signal, one can either
- convert the Laplace transform to the frequency domain by substituting s = j w if the
the substitution is valid, i.e. the region of convergence for the Laplace transform
includes the imaginary axis, or
- compute the continuous-time Fourier transform of the signal
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The Fourier transform of the unit step u(t) is
pi delta(w) + 1 / (j w)
That transform pair is the second entry in the
visual dictionary of Fourier transform pairs.
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For part (a), we can compute the Fourier transform of cos(w0 t) u(t) using the Fourier transform
property that multiplication in the time domain is convolution in the frequency domain:
X(j w) = (1/(2 pi)) F{ cos(w0 t) } * F{ u(t) }
X(j w) = (1/(2 pi)) (pi delta(w + w0) + pi delta(w - w0)) * ( pi delta(w) + 1 / (j w) )
Recall that
delta(w) * G(w) = G(w)
delta(w - w0) * G(w) = G(w - w0)
X(j w) = (1/2) ( pi delta(w + w0) + pi delta(w - w0) ) + 1 / ( j (w + w0) ) + 1 / ( j (w - w0) )
A plot of |X(j w)| is below:
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Note that |X(j w)| is unbounded at w = w0 due to the 1 / ( j (w - w0) ) term and
unbounded at w = -w0 due to the 1 / ( j (w + w0) ) term.
The frequency content has a bandpass shape.
Problem 9.2. Notch filters.
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A notch filter eliminates a specific frequency w0:
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A notch filter eliminates a specific frequency w0.
Recall that a cosine of frequency w0 also has a frequency component at -w0:
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cos( w0 t ) = (1/2) e-j w0 t + (1/2) ej w0 t
The notch filter would remove frequencies at w0 and -w0.
Here's an example notch filter magnitude response:
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Here's an example pole-zero configuration in the Laplace domain for the above filter:
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Send comments to
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bevans@ece.utexas.edu
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Videos:
Brian Douglas (13:03) -
Darryl Morrell, Part 1 (9:02) -
Darryl Morrell, Part 2 (11:07)