Download Assertions - Signals and Systems - Solved Exam and more Exams Signals and Systems Theory in PDF only on Docsity!
EECS 120 Signals & Systems University of California, Berkeley: Fall 2005 Gastpar November 16, 2005
Solutions to Exam 2
Last name First name SID
- You have 1 hour and 45 minutes to complete this exam.
- The exam is closed-book and closed-notes; calculators, computing and communication devices are not permitted.
- No form of collaboration between the students is allowed. If you are caught cheating, you may fail the course and face disciplinary consequences.
- However, two handwritten and not photocopied double-sided sheet of notes is allowed.
- Additionally, you receive Tables 3.1, 3.2, 4.1, 4.2, 5.1, 5.2, 9.1, 9.2 from the class textbook.
- If we can’t read it, we can’t grade it.
- We can only give partial credit if you write out your derivations and reasoning in detail.
- You may use the back of the pages of the exam if you need more space.
*** Good Luck! ***
Problem Points earned out of
Problem 1 29
Problem 2 28
Problem 3 27
Problem 4 33
Total 117
Problem 1 (Short Questions.) 29 Points
(a) (4 Pts) For the system in Figure 1,
H(jω) =
1 , for |ω| ≤ ω 0 0 , otherwise.
Sketch the frequency response G(jω) of the overall system between x(t) and y(t).
x(t) (^) H(j !)! y(t)
Figure 1:
Solution:
y(t) = x(t) − x(t) ∗ h(t) Y (jω) = X(jω) − X(jω)H(jω) = X(jω)(1 − H(jω))
G(jω) = Y (jω) X(jω)
= 1 − H(jω)
Remark: This problem was a hint for the sampling system design in Problem 3.(c).
− ω (^00)
G(j ω)
ω ω
(b) (15 Pts) A causal LTI system is described by the following differential equation:
d^2 dt^2
y(t) + 2 d dt
y(t) + 2y(t) = d^2 dt^2
x(t) − x(t) (2)
Is this system stable? Does this system have a causal and stable inverse system?
Solution: We take the Laplace transform of both sides of the differential equation to find the transfer function H(s) of the LTI system.
s^2 Y (s) + 2sY (s) + 2Y (s) = s^2 X(s) − X(s)
H(s) = Y (s) X(s)
s^2 − 1 s^2 + 2s + 2
(s + 1)(s − 1) (s − (−1 + j))(s − (− 1 − j))
The system H(s) has poles at s = −1+j and s = − 1 −j , and zeros at s = 1 and s = −1. Since we are given that H(s) is causal, the region of convergence (ROC) of H(s) is Re{s} > −. Thus the ROC of H(s) includes the jω -axis, which implies that H(s) is stable.
sin( t)
H(j )
x(t) H(j !)
y(t)
!
sin( !ct)
cos( (^) ct) x(t)
y(t)
r(t)
!
!
c
cos( (^) ct)
!
Figure 2: Quadrature modulation.
~
H(j )
x(t) H(j !)
y(t)
!
sin( !ct)
cos( (^) ct) x(t)
y(t)
r(t)
!
!
c
cos( (^) ct)
sin( t)
!
!
G(j )
G(j )!
(^) Figure 3: “Improved” quadrature modulation.
(c) (10 Pts) As you have seen in the homework, “quadrature multiplexing” is the system shown in Figure 2, where
H(jω) =
1 , for |ω| ≤ ωM 0 , otherwise. and G(jω) =
1 , for |ω| ≥ ωc 0 , otherwise.
Both original signals are assumed to be bandlimited: X(jω) = Y (jω) = 0 , for |ω| > ωM ; and the carrier frequency is ωc > ωM. The interesting feature is that the effective bandwidth of the signal r(t) is only 2ωM , the same as for a regular AM system with only the signal x(t). Hence, y(t) can ride along for free.
Now, your colleague remembers single-sideband AM and suggests to add the filters G(jω) as shown in Figure 3. The effective bandwidth of the transmitted signal ˜r(t) is only ωM , half as much as in the original quadrature multiplexing system! Show that the “improved” system will not work. Hint: Find a pair of example spectra X(jω) and Y (jω) for which R(jω) is not zero, but R˜(jω) = 0 for all ω. Then, argue (in a few keywords) why this invalidates the “improved” quadrature modulation.
Solution: The basic fact to remember from the homework problem about quadrature modu- lation is that the spectra overlap and get added up. Consider the example spectra X(jω) and Y (jω) in the figure below. The spectrum X(jω) is purely real-valued. After multiplying by a cos(ωct) , it remains a purely real-valued spectrum. The trick is to select the spectrum Y (jω) as purely imaginary; that way, after multiplying by the sin(ωct), it becomes a purely real-valued spectrum, and hence, there is a chance for it to cancel out the spectrum X(jω).
To actually make this happen, we still need to pick the right shape. One example that works is given in the figure below. Note that if R(jω) looks as sketched in the figure, then R˜(jω) will be zero — the filter G(jω) removes anything below ωc.
X(j! ) Im{Y(j !)}
!!
R(j !)
!
R(j !)
!!!
!!
c
c c
c c
A few remarks: The signal with purely imaginary spectrum as given in Y (jω) is a real-valued signal (think of the spectrum of the sin function - it’s purely imaginary). This is because it is conjugate-symmetric ( Y (jω) = Y ∗(−jω) ). Also, there are many other ways to convince your colleague that the system will not work. A longer approach is to select example spectra and to show that demodulation as suggested in Figure 3 will not recover the desired data-carrying signals x(t) and y(t).
|G(jω)| =
| 1 − 13 e−jωT^ |
=
| 1 − 13 (cos(−ωT ) + j sin(−ωT )) |
=
1 − 13 cos(ωT )
3 sin(ωT^ )
1 − 23 cos(ωT ) + 19 cos^2 (ωT ) + 19 sin^2 (ωT )
=
10 9 −^
2 3 cos(ωT^ )
−π/ T π/^ T
|G(j
(b) (8 Pts) For x(t) = ejπt/(2T^ )^ , determine the corresponding output signal y(t). Your answer should not contain an integral, but apart from that, there is no need to simplify it down.
Solution: After writing x(t) = ej(π/(2T^ ))t^ we see that:
y(t) = G
j π 2 T
ej(π/(2T^ ))t^ =
1 − 13 e−jπ/^2
ejπt/(2T^ )^ =
1 + j (^13)
ejπt/(2T^ )
Problem 3 (Sampling System Design.) 27 Points
The signal x(t) has the Fourier transform shown in Figure 5.
H(j )
X(j )
Figure 5:
(a) (5 Pts) As a function of T (as in Figure 5), determine the smallest sampling frequency ωs = 2 π/Ts (where Ts is the sampling interval) for which perfect reconstruction can be guaranteed for the signal x(t). A graphical justification (sketch with labels on the frequency axis) is sufficient.
Solution: By the sampling theorem, the bandlimited signal x(t) , with X(jω) = 0 for |ω| > (^2) Tπ , can be perfectly reconstructed if we sample at frequency ωs ≥ 2
( (^2) π T
= (^4) Tπ. Graphically, we see that the shifted replica of X(jω) that is centered at ωs does not overlap with the replica at 0 if ωs − (^2) Tπ ≥ (^2) Tπ.
p (j^ ω)
−ωs −2π/T 2 π /Τ ωs − 2 π /Τ ωs
X
(b) (10 Pts) Consider the signal y(t) = h(t) ∗ x(t) , where h(t) is the impulse response of the filter H(jω) in Figure 5. Sketch the spectra of the two discrete-time signals
x[n] = x(nT ) (5) y[n] = y(nT ), (6)
where T is the same as in Figure 5. Which effect explains the difference between x[n] and y[n]?
Solution: Now we sample the signals x(t) and y(t) with frequency (^2) Tπ , and then convert the resulting impulse trains xp(t) and yp(t) to discrete-time signals x[n] and y[n]. In the frequency domain, Xp(jω) = (^) T^1
k X(j(ω^ −^2 πk/T^ )) and^ X(e jΩ) = Xp(j Ω T ) (and similarly for^ Y^ (e jΩ) ).
Note that sampling scales the vertical axis by (^) T^1 , and converting to discrete-time scales the frequency axis by T.
(c) (12 Pts) The goal is now to implement a sampler with sampling interval T 0 = T /2 , where T is as in Figure 5. Unfortunately, such a fast sampler is not available in the current technology. Instead, you have access to the following devices:
- samplers with sampling interval T , where T is the same as in Figure 5 (any number)
- anti-aliasing filters with the frequency response given in Figure 5 (any number)
- continuous-time signal adders/subtractors (any number)
- any discrete-time processing devices (ideal filters included).
Draw the block diagram of a system that takes as an input the signal x(t) (with spectrum as shown in Figure 5) as outputs the discrete-time signal x 0 [n] = x(nT 0 ). Hint: To maximize your chance of partial credit, give spectral plots of intermediate signals in your system.
Solution: The key insight is that we have to sample different parts of the signal separately. Clearly, we can low-pass filter the signal x(t) to obtain the signal y(t) in the figure below. Since y(t) is now bandlimited to π/T , we can sample it with our sampler (interval T ) with no aliasing. Separately, we want to sample the ”high-pass” part of the signal x(t). Here, we can use two standard tricks: first, to get the high-pass part, we can just take x(t) and subtract out the low-pass part, leading to the signal z(t) in the figure below. Second, in this case, the resulting signal can be sampled directly with sampling interval T , with no aliasing. This is just like in the homework problem about band-pass sampling. Alternatively, if we had access to a continuous-time mixer (multiplication by a cosine function), we could modulate it down before sampling (but as we said, for our example case, this is not necessary).
The resulting signals y[n] and z[n] now have to be combined. The main insight at this point is that we need twice the sampling rate in the end, and so, we should upsample both y[n] and z[n].
To see how to implement the combining, one really needs the spectral plots of the upsampled signals, U (ejΩ) and V (ejΩ) : It is immediately clear that we want to low-pass filter U (ejΩ) and high-pass filter V (ejΩ). Let’s define:
F (ejΩ) =
1 , for |Ω| ≤ π 2 0 , otherwise. and G(ejΩ) =
T, for |Ω| > π 2 0 , otherwise.
Recall that these are really 2π -periodic; we are merely specifying one period. The spectral plots in the figure on the next page shows that the system below works.
Grading: Most of the points were given for the two key ideas, namely, that we need two different signal paths, and that we need upsampling.
G(e )
1/2 x[n]
H(j )
Sampler
Sampler
x(t) F(e )
j
j
! "
"
y(t)
z(t)
y[n]
z[n]
u[n]
v[n]
a[n]
b[n]
B(e )
j!
#" "
U(e ) 6/T
4/T
j!
#" "
2/T
3/T
j!
#" "
Y(e ) Z(e )
2/T
j!
#" "
6/T
4/T
A(e )
j!
#" "
V(e )
4/T
j!
#" "
4/T
For the signal x 2 (t) , let’s first consider a symmetric box of width T /4 , centered at zero, and repeated at intervals of T. Call this signal v(t). Hence, again from the table, with T 1 = 1/8 , we find the FS coefficients of v(t) :
ck = sin(kπ/4) kπ
sinc(k/4). (9)
Clearly, x 2 (t) = v(t − T /8) − v(t + T /8) , and hence, using the second property (time shifting) in Table 3.1, we find the FS coefficients of the signal x 2 (t) as
bk = cke−jkπ/^4 − ckejkπ/^4 (10)
sinc(k/4)(− 2 j) sin(kπ/4) (11)
= − j 2
sinc(k/4) sin(kπ/4) (12)
Thus, b 0 = 0 , b 1 = − (^) πj and b− 1 = (^) πj.
Alternatively, you can evaluate by hand:
b 0 = T^1
∫ (^) T / 2 −T / 2
x 2 (t)dt = 0
bk = (^) T^1
∫ (^) T / 2 −T / 2
x 2 (t)e−jk(2π/T^ )tdt
= (^) T^1
∫ (^0) −T / 4
−e−jk(2π/T^ )tdt + T^1
∫ (^) T / 4 0
e−jk(2π/T^ )tdt
= (^) T^1 jk(2^1 π/T )
( 1 − ejkπ/^2
) − (^) T^1 jk(2^1 π/T )
( e−jkπ/^2 − 1
)
= (^) jk^12 π
( 2 − ejkπ/^2 − e−jkπ/^2
)
= (^) jkπ^1 (1 − cos(kπ/2))
Exercise: Show that the above two formulas for bk are, in fact, equal.
(c) (7 Pts) To actually transmit our PAM signal, we first low-pass filter it:
x˜m(t) = h(t) ∗ xm(t), where H(jω) =
1 , for |ω| ≤ (^10) Tπ 0 , otherwise.
Then, we transmit the signals y 1 (t) = ˜x 1 (t) cos( (^40) Tπ t) and y 2 (t) = ˜x 2 (t) cos( (^40) Tπ t). Sketch the Fourier transforms of these two signals in the plots provided below, carefully labeling the frequency axis. In the magnitude plots (i.e., |Y 1 (jω)| and |Y 2 (jω)| , respectively), the amplitudes need not be exact. Remark: The current labels on the frequency axis in the plots are for your convenience only. If you prefer, you can cross them out and start from scratch.
Solution: Since x 1 (t) is periodic, its spectrum X 1 (jω) is a “line spectrum”, i.e., composed of delta functions. From Table 4.2, first entry, the delta functions are spaced 2π/T apart (the fundamental frequency of x 1 (t) ), and their amplitudes are 2πak , where ak are the Fourier series coefficients of x 1 (t) :
X 1 (jω) = 2 π
∑^ ∞
k=−∞
akδ(ω − k 2 π T
The low-pass filtering throws away anything at frequencies higher than 10π/T , i.e., only the center and 5 delta functions on each side of the origin survive, but otherwise, the spectrum X˜ 1 (jω) looks exactly like the spectrum of X 1 (ω) :
X˜ 1 (jω) = 2 π
∑^5
k=− 5
akδ(ω − k 2 π T
Multiplying by cos( (^40) Tπ t) simply places two copies of the spectrum of X˜ 1 (jω) , one centered at 40 π/T , the other at − 40 π/T. To get the amplitudes right (but no points were taken off for minor errors in the amplitudes), remember that the spectrum of the cosine consists of two delta functions of amplitude π each, and the multiplication property has a factor of 1/(2π) , hence,
Y 1 (jω) =
2 π · π 2 π
k=− 5
akδ(ω − k
2 π T
40 π T
∑^5
k=− 5
akδ(ω − k
2 π T
40 π T
For the signal y 2 (t) , the argument is exactly the same, leading to
Y 2 (jω) =
2 π · π 2 π
k=− 5
bkδ(ω − k
2 π T
40 π T
∑^5
k=− 5
bkδ(ω − k
2 π T
40 π T
(d) (10 Pts) The communication channel’s effect on the signal can be described by the following band-pass filter:
Hchannel(jω) =
sin^2 (ωT / 4 − π) for 36π/T < |ω| < 38 π/T 1 , for 38π/T ≤ |ω| ≤ 42 π/T sin^2 (ωT / 4 − π) for 42π/T < |ω| < 44 π/T 0 , otherwise.
The channel output signal is then z 1 (t) = y 1 (t) ∗ hchannel(t) and z 2 (t) = y 2 (t) ∗ hchannel(t) , respectively. Assuming that s[n] = 1 , for all n , find the power of z 1 (t) and z 2 (t). These are the received powers. Which pulse is more efficient for transmission across this channel?
Solution: The key insight is that z 1 (t) is still a periodic signal. Hence, its power is calculated as
Pr 1 =
T
T
|z 1 (t)|^2 dt. (19)
From Parseval, this is the same as
Pr 1 =
∑^ ∞
n=−∞
|ck|^2 , (20)
where ck are the Fourier series coefficients of the signal z 1 (t). From Part (c), we know the Fourier Transform Z 1 (jω) : Passing Y 1 (jω) through the bandpass filter, only 6 delta pulses survive, namely the three centered around 40π/T and the three centered around − 40 π/T. To find the Fourier series coefficients, we have to divide the amplitude of the delta function by 2π (see Table 4.2, first entry), and so we can read directly our of the figure:
|c 20 | = |c− 20 | = 1/ 4 , |c 19 | = |c− 19 | = c 21 = c− 21 = 1/(2π), (21)
and all other Fourier series coefficients are zero, from which we find
Pr 1 =
∑^ ∞
n=−∞
|ck|^2 = 2 · 1 /16 + 4 · 1 /(2π)^2 =
π^2
By the same token, we can read out the Fourier series coefficients for the signal z 2 (t) , and find
Pr 2 =
∑^ ∞
n=−∞
|dk|^2 = 4 · 1 /(2π)^2 =
π^2
Hence, the received power for the pulse x 1 (t) is higher, and we conclude that this pulse is more efficient.
Note: There is also a subtlety that we swept under the carpet. Really, it is not clear whether both pulses have the same transmitted power — We only calculated the powers of x 1 (t) and x 2 (t) , respectively, but the transmitted signals, really, are y 1 (t) and y 2 (t). These powers are most easily found along the lines of the above calculation, but we would need some more of the Fourier series coefficients.
