Homework #6 Signals & Systems 

ECE102 Homework #6
Signals & Systems 
100 points total.
1. (32 points) Frequency Response
(a) (18 points) Consider the LTI system depicted in figure 1 whose response to an unknown
input, x(t), is
y(t) =
4e
−t − 2e
−2t


u(t)
Figure 1: System for Problem 1.
We know that for the same unknown input x(t), the intermediate signal, y1(t), is given
by:
y1(t) = 2e
−tu(t)
The overall LTI system is described by the following differential equation:
d
2
dt2
y(t) + 5 d
dty(t) + 6y(t) = 3x(t)
i. Find the frequency response, H(jω), of the overall system h(t).
ii. Find the frequency responses H1(jω) of the first LTI system and H2(jω) of the
second LTI system.
iii. Find the impulse responses h(t), h1(t) and h2(t).
(b) (6 points) Assume x(t) a real signal that is baseband, i.e., its Fourier transform X(jω)
is non-zero for |ω| ≤ ω0. We process this signal through an LTI system. Let y(t) denote
the corresponding output and let Y (jω) denote the Fourier transform of y(t). Does y(t)
have frequency components different than those of x(t)? i.e., is Y (jω) 6= 0 for some
|ω| > ω0? What if we process x(t) through a non-LTI system?
(c) (8 points) Consider the following two LTI systems with impulse responses:
h1(t) = sinc 
t
2

cos(πt)

and
h2(t) = 2 sinc (2t)
Find the output of each system to the following input x(t) = cos(3πt) cos(4πt). If we
are given an input-output pair of an unknown LTI system, can we always identify this
system?
Hint: Recall the input output relationship that we demonstrated in discussion. If h(t)
is real,
x(t) = cos(ω0t) → y(t) = |H(jω0)| cos(ω0t + ∠H(jω0))
2. (18 points) Filters
(a) (6 points) Consider an ideal low-pass filter hLP,1(t) with frequency response HLP,1(jω)
depicted below in figure 2.
Figure 2: An ideal low pass filter
Using this filter, we construct the following new system:
Figure 3: New system
We are given two choices for α: 1 or -1. Which value should we choose so that the
new system is a high-pass filter? Does the new filter have any phase in its frequency
response?
(b) (3 points) Why are the ideal filters non-realizable systems?
(c) (5 points) We want to design a causal non-ideal low-pass filter hLP,2(t), using the following frequency response:
HLP,2(jω) = k
β + jω
Find k and β so that HLP,2(jω) is unity for ω = 0 and its cutoff frequency is ω0 = 2π
rad/s, (i.e., the magnitude of HLP,2(jω) is 1/
√
2 for ω = 2π rad/s).
(d) (4 points) We again consider the system of part (a) where instead of the ideal low-pass
filter, we are going to use the non-ideal low-pass filter of part (c).
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Figure 4: The system of part (a) with the non-ideal low pass filter
For the same value of α you found in part (a), find the frequency response of the
equivalent system. Explain if the new system behaves as a high-pass filter and why.
3. (13 points) Moving average filters
We now consider the moving average filter, also known as a “boxcar” filter, and is one of the
most primitive filters in practice. Its impulse response is shown below:
(a) (8 points) What is the frequency response H(jω) of this filter?
(b) (5 points) Sketch the amplitude response |H(jω)| of the filter. What happens to |H(jω)|
as ω → ∞?
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4. (25 points) Modulation and Demodulation
(a) (15 points) Consider the communication system shown below:
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3. Single Sideband (SSB) Communication.
The double sideband (DSB) systems require a transmission bandwidth that is twice the signal bandwidth. In this question we’ll look at a system where the transmission bandwidth
and signal bandwidth are the same.
The system is plotted below:
× m(t)
cos(ωct)
×
cos(ωct)
m(t)
Propagation
Transmitter Receiver 2
Ideal Bandpass Ideal Lowpass
x(t) y(t) z(t)
H(jω)
The signal m(t) is first modulated by cos(ωct), and then passed through an ideal bandpass
filter. The spectrum of the input M(jω) and the frequency response of the ideal bandpass
filter H(jω) are
−2πB 2πB ω
2πB
−ωc ωc
−ω ω c ωc
2πB
(b)
(c)
1
M(jω)
H(jω)
The modulated signal is x(t), and the output of the ideal bandpass is y(t). This signal is
transmitted, and processed by a synchronous receiver.
Sketch the signal spectrum at
(a) the output of the modulator X(jω),
(b) the output of the ideal bandpass Y (jω), and
(c) the output of the demodulator, Z(jω).
Label the bands in Y (jω), and show where they end up in Z(jω).
Does this system recover m(t)?
y(t)
The signal m(t) is first modulated by cos(ωct), and then passed through an ideal bandpass filter. The spectrum of the input M(jω) and the frequency response of the ideal
bandpass filter H(jω) are:
2
3. Single Sideband (SSB) Communication.
The double sideband (DSB) systems require a transmission bandwidth that is twice the signal bandwidth. In this question we’ll look at a system where the transmission bandwidth
and signal bandwidth are the same.
The system is plotted below:
× m(t)
cos(ωct)
×
cos(ωct)
m(t)
Propagation
Transmitter Receiver 2
Ideal Bandpass Ideal Lowpass
x(t) y(t) z(t)
(a)
H(jω)
The signal m(t) is first modulated by cos(ωct), and then passed through an ideal bandpass
filter. The spectrum of the input M(jω) and the frequency response of the ideal bandpass
filter H(jω) are
−2πB 2πB ω
2πB
−ωc ωc
−ω ω c ωc
2πB
1
M(jω)
H(jω)
The modulated signal is x(t), and the output of the ideal bandpass is y(t). This signal is
transmitted, and processed by a synchronous receiver.
Sketch the signal spectrum at
(a) the output of the modulator X(jω),
(b) the output of the ideal bandpass Y (jω), and
(c) the output of the demodulator, Z(jω).
Label the bands in Y (jω), and show where they end up in Z(jω).
Does this system recover m(t)?
The modulated signal is x(t), and the output of the ideal bandpass is y(t). This signal
is transmitted through a channel. We assume that this channel does not introduce
distortion into y(t). The received signal y(t) is then processed by a receiver. Sketch the
signal spectrum at
i. the output of the modulator, i.e., X(jω),
ii. the output of the ideal bandpass, Y (jω), and
iii. the output of the demodulator, Z(jω)
Does this system recover m(t)?
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(b) (10 points) In the subquestion (a) of this problem, you have seen that to demodulate
the received signal y(t) to get the signal m(t), we multiply y(t) by cos(ωct) first and then
we low-pass filter the result. In this subquestion, you are asked to show that you can
achieve the same effect with an ideal sampler. In other words, we assume the following
block diagram of the receiver instead: where the ideal sampler is drawn as a switch
2. Demodulation with an Ideal Sampler
Multiplying a signal f(t) with cos(ωct) produces a modulated signal
z(t) = f(t) cos(ωct)
where ωc is the carrier frequency. One way to demodulate this signal and recover f(t) is to
multiply z(t) by cos(ωct), and lowpass filter the result.
In this problem you will show that you can achieve the same effect with an ideal sampler.
The block diagram of the receiver is
Lowpass
t = nT
Sampler
−2πB 2πB
z(t) m(t) y(t) H(jω)
where the ideal sampler is drawn as a switch that closes instantaneously every T seconds
to acquire a new sample. Assume that the signal f(t) is bandlimited, with F(jω) = 0 for
|ω| > 2πB, and that the carrier frequency ωc  2πB. Other than that F(jω) is unspecified
(plot it as a bandlimited triangle in the plots below).
(a) Show that we can recover f(t) if the ideal sampler operates at a frequency ωc (i.e.
samples at a rate of ωc/2π samples/s). Draw the spectrum of the input signal Z(jω),
and the spectrum of the signal right before the lowpass filter R(jω).
(b) What is the lowest frequency at which the sampler can operate, and still recover f(t)?
Draw the spectrum of R(jω).
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that closes instantaneously every T seconds to acquire a new sample. Show that we can
recover m(t) if the ideal sampler operates at a frequency ωc (i.e. samples at a rate of
ωc/2π samples/s). Draw the spectrum of the signal (Z(jω)) right before the lowpass
filter.
5. (12 points) Sampling
(a) Assume x(t) a real bandlimited signal where X(jω) is non-zero for |ω| ≤ 2πB rad/s. If
Fs Hz is the Nyquist rate of x(t), determine the Nyquist rate of the following signals in
terms of B:
i. x(t + 1)
ii. cos(2πBt)x(t)
iii. x(t) + x(2t)
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