# EE Spring Signal Processing and Linear Systems

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EE102A Spring 2008–9
Signal Processing and Linear Systems I

Problem Set #6
Due: Wednesday, 27 May 2009 at 5 PM.

1. Nonideal Sampling
This problem considers two variations on idealized sampling.
A continuous time signal g(t) is bandlimited and has 0 spectrum G(jω) for all ω with |ω| ≥
2πW .
Instead of ideal sampling with impulses, the signal is sampled using a train of very narrow
pulses, each with pulse shape p(t) and spectrum P (ω). The resulting sampled pulse train
is given by
∞
ˆ
g (t) = g(t)          p(t − nT ).
n=−∞

This is sometimes called natural sampling.

(a) Use the derivations in Lecture 13 to show that
∞
¯
p(t) =          p(t − nT ) = δT ∗ p(t),
n=−∞

∞
the convolution of p and a delta train δT (t) =         n=−∞ δ(t    − nT ). Find the Fourier
¯
transform P (jω) in terms of P (jω).
(b) What is the effect of the nonideal pulse shape on the procedure for recovering g(t)
ˆ
from g (t)? State any assumptions you make relating T , W , and p(t).
(c) Next suppose that the signal is ﬁrst multiplied by an ideal impulse train to produce a
sampled waveform
∞
y(t) = g(t)δT (t) =           g(nT )δ(t − nT ),
n=−∞

but y(t) then is passed through a ﬁlter with impulse response p(t) so that the ﬁnal
sampled signal is
∞
u(t) =          g(nT )p(t − nT ).
n=−∞

This has the form of ideal sampling, but the pulses are not impulses.
How can the original signal be recovered from this signal? Specialize your answers to
the following two examples: (assume that T = 1/2W ):
2

i. p(t) = sinc(t/2W )
ii. p(t) = rect(t/2W )

2. Find the (unilateral) Laplace transform of the following signals (including the region of
convergence):

(a) f (t) = u(t − 1)e−t
(b) f (t) = u(t)[et + e−t ]
(c) f (t) = [u(t) − u(t − 1)]e−t
(d) f (t) = te−t u(t)


      0  t<1
(e) f (t) =        1  1≤t<2
2≤t
 −2(t−2)
e


(f) One cycle of a sinusoid:

sin(2πt) 0 ≤ t < 1
f (t) =
0 1≤t

as plotted below:

1                     f (t)

0                      1             2 t
−1

Hint: Use the delay theorem, and the fact that a delayed signal is padded with zeros.

3. Find the poles and the inverse Laplace transforms h(t) of the following system functions
(Laplace transforms of causal impulse responses)

(a)
s+α
H(s) =
(s + α)2 + β 2
with real α, β
(b)
e−sT
H(s) =
s+α
(c)
1
H(s) =
(s + 2)(s − 1)
(d)
1
H(s) =
2 − e−s
3

Hint: This is a bit of a trick question as H(s) is not a ratio of polynomials as in the
cases seen in class. The poles of H(s) are the zeros of 2−e−s , the solutions to e−σ−jω =
e−σ ejω = 2, which forces separate solutions for σ and ω. To ﬁnd the inverse, H(s) can
be put into the form of a sum of simple terms by use of the geometric progression
formula, which states that if |r| < 1, then
∞
1
rn =       .
1−r
n=0

4. The waveform
y(t) = e−2t u(t)

is the output of a causal LTI with system function

s−1
H(s) =
s+1
What is x(t)?
4

Laboratory
In this lab we will look at some of the properties of sampled sinusoids, and reconstructing
sampled signals.

The sampling theorem tells us that we can reconstruct a signal perfectly provided we sample
at a rate at least twice the bandwidth of the signal. In this task we will examine what a sampled
sinusoid looks like as the sampling frequency gets closer to the Nyquist rate.
Deﬁne matlab vector x to be a 5 Hz cosine waveform, sampled at 500 Hz for an interval of 1
second. The Nyquist rate for this signal is then 10 Hz.
If we subsample x by taking every m sample we can generate the sampled signal at a sampling
rate of 500/m Hz.
For a given m,

• First plot the subsampled x as a discrete time signal, using stem(t(1:m:end),x(1:m:end)).
Try to visualize where the continuous time sinusoid would be.

• Hold this plot with hold on,

• Then plot x as a function of t as a continuous dashed function, using plot(t,x,’--’),

to see both the continuous time cosine, and the sampled discete time cosine.
Plot the results for m equal to 10, 20, 30, and 50. What sampling rates do these correspond to?
Note that the m=30 case is still sampled well above the Nyquist rate, but the sampled signal is
very difﬁcult to recognize as a sinusoid, particularly if the dashed line isnt’t there to help. It is
remarkable that this signal can be reconstructed perfectly!
Clearly the reconstruction problem becomes more difﬁcult as the frequency gets closer to the
Nyquist rate.

In Lecture 11 we compared two different lowpass ﬁlters. The ﬁrst was an ideal lowpass ﬁlter,
that is the reconstruction ﬁlter when we are sampled exactly at the Nyquist rate. The second
requires a higher sampling rate, but is easier to implement. In this task we look at these impulse
responses. Let ωc = π for convenience, so that

h0 (t) = sinc(t)
3       3t            t
h1 (t) =   sinc         sinc
4       2             2

Both of these have unity passbands from ±π. However, h1 (t) has a linear transition band from
ωc to 2ωc .
In practice we have to truncate the ideal impulse response of the reconstruction ﬁlter, since
we can’t convolve with an inﬁnite duration impulse response. In this task we are going to assume
5

we want to truncate the impulse response at 1% of it’s peak value, and then look at how long a
segment of the impulse response we need.
First, plot h0 (t), and determine how long a segment of this signal we need to make sure the
truncated tails of the signal are below 1%. Use

>> semilogy(t,abs(h0))

to make it easier to identify the 1% level, and set the axes using

>> axis([-tmax tmax 1e-4 1]);
>> grid on

where - tmax and tmax minimum and maximum times, and the 1e-4 sets the minimum
amplitude to 0.0001, or 0.01%. Then calculate h1 (t) over this same interval. lot each of these
together on one page using subplot. Normalize by the maximum value, meaning you should
plot

>> semilogy(t,abs(h1)/max(abs(h1)).

How long a segment do we need for h1 (t) to be sure the tails of the signal are below 1% of the
peak value?

In this task we will examine the property of discrete time sinusoids with frequencies above
the Nyquist rate to appear as lower frequencies. We will use a signal that has quadratic phase,

>> t = [0:8191]/8192;
>> y = cos(2*pi*4096*(t.*t))

A continuous time signal with a quadratic phase as a function of time is known as a “chirp”, and
has a linear sweep in frequency. Listen to y using

>> sound(y,8192)

and notice that the pitch initially goes up, and then reverses direction.
Make a sketch of what the frequency is as a function of time. Since this is a cosine, you will
have both positive and negative frequencies, as well as the replicas of the spectrum centered at
plus and minus the sampling rate. Indicate where half the Nyquist rate is. Can you identify the
point that the frequency changes direction? Does it really change direction?
Hint: For a continuous time signal the instantaneous frequency f (t) is the proportional to the
derivative of the phase θ(t).
1 dθ(t)
f (t) =
2π dt
where f (t) is in Hz, and θ(t) is in radians. In this case the phase is θ(t) = 2π(4096)t2 radians.
6

You are a new Signals Analysis Ofﬁcer at an unnamed secret federal agency. You’ve captured
a signal with your Universal Software Radio. The signal has an unusual spectrum, and you
suspect that it has been encrypted. Your task is to ﬁgure out how the channel was encrypted,
and recover the message.
The signal is in the matlab ﬁle phone_number.mat on the class web site, in the matlab
variable d. The sampling rate is fs, and the sampling time is dt. Listen to the sound to see if
you can understand the message

>> sound(d, fs);

You suspect that the encryption involves shufﬂing the frequency content of the signal. The
spectrum doesn’t look the same as the AM spectra you looked at in the previous lab. As before,
you can compute the spectrum of the signal and plot it with

>>   D = fftshift(fft(fftshift(d)))
>>   ld = length(d);
>>   f = [(-ld/2):(ld/2-1)]*fs/ld;
>>   plot(f,abs(D))

Examine the signal, and compare it to what you would expect based on the previous lab. Then
see if you can ﬁgure out how to rearrange the spectrum. For example, to swap the two halves of
the spectrum, do

>> D1 = [D(ld/2+1:ld) D(1:ld/2)];

Then go back to the time domain with the inverse fft, and listen to the result

>> d1 = fftshift(ifft(fftshift(D1)));
>> sound(real(d1), fs)

We’ve speciﬁed real(), because sound() doesn’t cope well when given a complex signal. You
might want to listen to the imaginary component, or the absolute value.
Provide plots to explaining how you ﬁgured out how the data was encrypted (is there a name
for this type of spectra?). Make sure to report the message.

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