**bitmedi**
BIMU3009
Signal Processing
Bütünleme Sınavı
Istanbul University - Cerrahpaşa
Computer Engineering Department - Fall 2023
January 30th , 2024 09:50-11:00
Soru
S1 (35p)
S2 (35p)
S3 (30p)
ÖÇ
1,2
1,2
1
PÇ
1
2
7
Toplam
Puan
LÜTFEN OKUYUN:
ˆ Sınava sizin için belirlenen sınıfta giriniz.
ˆ Bu sınavın süresi 70 dakikadır. Süre bittiğinde cevap kağıdını doldurmaya devam edenler kopya çekmiş sayılır.
Bu sınavda 1 hesap makinası, ön ve arkasına kendi notlarınızı yazdığınız soru çözümleri içermeyen elle yazılmış ya
da fotokopi bir A4 kağıdı ve kitabın arkasında bulunan, ve üzerine başka birşey yazılı olmamak kaydıyla
Appendix A’nın fotokopisini kullanabilirsiniz. Bunların dışında her türlü defter, kitap, notlar, sözlük ve
elektronik sözlük yasaktır.
ˆ Lütfen soruları kurşun kalemle, İNGİLİZCE, kısa ve anlaşılır olarak CEVAP KAĞIDINA cevaplayınız.
Anlaşılmayan, muğlak ifadeler kullanmak, kötü yazı yazmak notunuza negatif olarak etki edecektir. Çözüm
adımlarınız anlaşılır olmalıdır. Soru kağıdına cevaplarınızı yazmayınız, değerlendirmeye alınmayacaktır.
ˆ Soruları çözerken gidiş yolunu göstermeniz gerekmektedir.
ˆ Materyalin paylaşılması yasaktır. Hesap makinası ve silgi paylaşmak kopya sayılacaktır!
ˆ Bilgisayar, PDA, cep telefonu türünden elektronik cihazlar kullanmak yasaktır.
ˆ Soruları çözmeye başlamadan lütfen okuyun.
ˆ Soru kağıtlarına isim ve numaranızı yazınız.
ˆ Soru kağıtlarınızı, çıkarken cevap kağıdınızla beraber teslim ediniz. A4 kağıdınız ve Appendix A’nın fotokopisi
sizde kalabilir.
ˆ Bu sınavda toplam 100 puanlık soru vardır.
ˆ SINAVDA KOPYA ÇEKENLER, KOPYA VERENLER VE BUNLARA TEŞEBBÜS EDENLER
SINAVDAN ”0” ALACAKTIR VE DEKANLIĞA ŞİKAYET EDİLECEKLERDİR!.
Başarılar. (Mustafa Dağtekin)
Lütfen buraya ”okudum” yazınız.: ................
Some useful equations
Assuming α ∈ R, α ̸= 0
m
X
k=i
αk =
αm+1 − αi
α−1
m
X
k=0
αk =
Assuming −1 < α < 1, α ̸= 0
1 − αm+1
1−α
∞
X
k=0
αk =
1
1−α
∞
X
k=i
αk =
αi
1−α
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QUESTIONS
Q1:
Consider an DT LTI system, H1 , with the following impulse response:
h1 [n] = (−2)n u[−n]
Answer the following questions:
(a) (8 pts) Is H1 stable? Why? Please show your work.
Solution 1a :
For a Discrete Time LTI system to be stable, the impulse response must be absolutely
summable. In other words, the following must be true:
∞
X
|h1 [n]| < ∞
n=−∞
Let’s check this condition:
∞
X
|h1 [n]| =
n=−∞
0
X
|(−2)n u[−n]|
n=−∞
=
0
X
|(−2)n |
n=−∞
=
=
0
X
2n
n=−∞
∞
X
−n
2
n=0
=
1
1−
1
2
=2<∞
Therefore, H1 is stable.
(b) (1 pts) Is H1 causal? Just write CAUSAL or NON-CAUSAL. You don’t need to explain.
Solution 1b :
H1 is NOT-CAUSAL because the impulse function is not zero for all n < 0.
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(c) (1 pts) Is H1 memoryless? Just write MEMORYLESS or NON-MEMORYLESS. You
don’t need to explain.
Solution 1c :
H1 is NOT-MEMORYLESS because the impulse function is not zero for all n ̸= 0.
(d) (10 pts) Find the step response of H1 . Please show your work.
Solution 1d :
Step response is the output of the system when the input is the unit step function. In other
words, it is the output of the system when the input is x[n] = u[n]. We can find the step
response by convolving the impulse response with the unit step function, as follows:
s[n] =
n
X
h1 [k]
k=−∞
Let’s find the step response for n ≤ 0:
n
X
s[n] =
=
=
k=−∞
n
X
k=−∞
n
X
h1 [k]
(−2)k u[−k]
(−2)k
k=−∞
n+1
=
(−2)
(−2) − 1
=−
(−2)n+1
3
for n ≤ 0
0
X
n
X
Let’s find the step response for n > 0:
s[n] =
k=−∞
=
0
X
h1 [k] +
k=1
(−2)k + 0
k=−∞
=
2
3
for n > 0
h1 [k]
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Therefore, the step response of H1 is:
s[n] =

n+1

 −(−2)
3
for n ≤ 0

2
for n > 0
3
If we plot the step response, it will look like this:
0.66 s [n]
1
0.33
−3
−8
−7
−6
−5
−1
−4
n
1
2
3
4
5
6
7
−0.33
−0.66
(e) (15 pts) Find the output of H1 when the input is x1 [n] =
work.
1 n
2
u[n + 1]. Please show your
Solution 1e :
The output of H1 when the input is x1 [n] can be found by convolving the input with the
impulse response, as follows:
y1 [n] = x1 [n] ∗ h1 [n]
∞
X
=
x1 [k] h1 [n − k]
k=−∞
Let’s plot x1 [k] and h1 [n − k] to see the range of k and n values that will contribute to the
output:
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2
x1 [k]
1.5
1
0.5
k
−8
−7
−6
−5
−4
−3
−2
−1
1
1
2
3
4
5
6
7
8
h1 [n − k]
0.5
+
1
n
n
n
−
1
k
−0.5
So, for n < −1, the output can be found as follows:
y1 [n] =
=
=
∞
X
x1 [k] h1 [n − k]
k=−∞
∞ X
k=−1
∞ X
k=−1
1
2
k
1
2
k
(−2)n−k
(−2)n (−2)−k
∞ k
X
1
= (−2)
(−2)−k
2
k=−1
k
∞
X
1 k
1
n
−
= (−2)
2
2
k=−1
∞ X
1 k
= (−2)n
−
4
k=−1
−1
− 14
n
= (−2)
1 − − 41
16
n
= − (−2)
5
n
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=−
16
(−2)n
5
for n < −1
For n ≥ −1, the output can be found as follows:
y1 [n] =
=
∞ k
X
1
k=n
∞ X
k=n
2
1
2
n
= (−2)
= (−2)n
k
(−2)n−k
(−2)n (−2)−k
∞ k
X
1
k=n
∞ X
k=n
∞ X
2
1
2
(−2)−k
k −
1
2
k
1 k
−
= (−2)
4
k=n
1 n
n − −4
= (−2)
− 14 − 1
n
4
1
n
= (−2)
−
5
4
n
4 1
=
5 2
n
Therefore, the output of H1 when the input is x1 [n] is:
y1 [n] =
 16
n
− 5 (−2)
4
5
Q2:
1 n
2
for n < −1
for n ≥ −1
Consider the following CT LTI system, H2 , with the following impulse response:
h2 (t) = e−t [u(t + 1) − u(t − 1)]
Answer the following questions:
(a) (8 pts) Is H2 stable? Why? Please show your work.
Solution 2a :
For a Continuous Time LTI system to be stable, the impulse response must be absolutely
integrable. In other words, the following must be true:
Z
∞
|h2 (t)| dt < ∞
−∞
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Let’s check this condition:
Z
∞
Z
∞
e−t [u(t + 1) − u(t − 1)] dt
|h2 (t)| dt =
−∞
−∞
Z
∞
=
e−t [u(t + 1) − u(t − 1)] dt
−∞
Z
1
=
e−t [u(t + 1) − u(t − 1)] dt
−1
Z
1
=
e−t dt
−1
= −e−t
1
−1
= −e−t
1
−1
= e − e−1
=e−
1
<∞
e
Therefore, H2 is stable.
(b) (1 pts) Is H2 causal? Just write CAUSAL or NON-CAUSAL. You don’t need to explain.
Solution 2b :
H2 is NON-CAUSAL because the impulse function is NOT zero for all t < 0.
(c) (1 pts) Is H2 memoryless? Just write MEMORYLESS or NON-MEMORYLESS. You
don’t need to explain.
Solution 2c :
H2 is NON-MEMORYLESS because the impulse function is NOT zero for all t ̸= 0.
(d) (10 pts) Find the step response of H2 . Please show your work.
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Solution 2d :
Step response is the output of the system when the input is the unit step function. In other
words, it is the output of the system when the input is x(t) = u(t). We can find the step
response by convolving the impulse response with the unit step function, as follows:
t
Z
s(t) =
h2 (τ ) dτ
−∞
Let’s find the step response for t ≤ −1:
Z
t
s(t) =
h2 (τ ) dτ
−∞
Z t
=
0 dτ
−∞
for t ≤ −1
=0
Let’s find the step response for −1 < t ≤ 1:
Z
t
s(t) =
h2 (τ ) dτ
−∞
Z −1
0 dτ +
=
=
t
Z
= −e
=e−e
e−τ dτ
−1
−∞
t
−e−τ −1
−t
+e
−t
for − 1 < t ≤ 1
Let’s find the step response for t > 1:
Z
t
s(t) =
h2 (τ ) dτ
−∞
Z −1
=
=
Z
1
0 dτ +
−∞
1
−e−τ −1
−1
=e−e
−τ
e
Z
dτ +
−1
t
0 dτ
1
+0
for t > 1
Therefore, the step response of H2 is:


0





s(t) = e − e−t





e − e−1
for t ≤ −1
for − 1 < t ≤ 1
for t > 1
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If we plot the step response, it will look like this:
s2 (t)
2.35
t
−2
−1.5
−1
−0.5
0.5
1
1.5
2
2.5
3
(e) (15 pts) Find the output of H2 when the input is x(t) = e−t u(t). Please show your work.
Solution 2e :
The output of H2 when the input is x(t) can be found by convolving the input with the
impulse response, as follows:
y(t) = x(t) ∗ h2 (t) = h2 (t) ∗ x(t)
Z ∞
=
x(τ ) h2 (t − τ ) dτ
−∞
Z ∞
=
h2 (τ ) x(t − τ ) dτ
−∞
Let’s plot x(t − τ ) and h2 (τ ) to see the range of τ and t values that will contribute to the
output:
3
e h2 (τ )
2
1
1
e
−4
−3
−2
−1
τ
1
2
3
4
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1 x(t − τ )
0.5
τ
t−2
t−1
t
−0.5
So, for t < −1, the output will be zero. For −1 < t < 1, the output will be:
t
Z
e−τ eτ −t dτ
y(t) =
−1
t
Z
e−t 1 dτ
=
−1
= e−t τ |t−1
= e−t (t + 1)
for − 1 < t < 1
For t > 1, the output will be:
Z
1
y(t) =
e−τ eτ −t dτ +
−1
Z
1
=
Z
t
e−τ 0 dτ
1
e−t 1 dτ + 0
−1
= e−t τ |1−1
= e−t (1 + 1)
=2e
−t
for t > 1
for t > 1
Therefore, the output of H2 when the input is x(t) = e−t u(t) is:


0
for t ≤ −1





y(t) = e−t (t + 1) for − 1 < t < 1





2 e−t
for t > 1
If we plot the output, it will look like this:
**bitmedi**
1
y2 (t)
0.8
0.6
0.4
0.2
t
−1.5
−0.5
0.5
1
1.5
2
2.5
3
3.5
4
4.5
Consider the following DT signal, a[n]:
2
1
a[n]
Q3:
−1
0
−1
−2
−3
−2
−1
0
n
1
2
Answer the following questions:
(a) (15 pts) Find and sketch the following signal: b[n] = 2a[−n] − a[n − 1]. Please show your
work.
Solution 3a :
Let’s use tables to find b[n]:
n
−2
−1
0
1
2
3
···
a[n]
0
−2
−3
2
0
0
···
2a[−n]
0
4
−6
−4
0
0
···
a[n − 1]
0
0
−2
−3
2
0
···
b[n]
0
4
−4
−1
−2
0
···
If we plot b[n], it will look like this:
5
b[n]
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4
3
2
1
0
−1
−2
−3
−4
−2
−1
0
n
1
2
P
(b) (15 pts) A periodic signal, c[n], is created via the following operation: c[n] = ∞
k=−∞ a[n −
3k]. Find the fundamental period of c[n] and sketch the signal such that it includes at
least three periods. Please show your work. (Sinyalin temel periyodunu bulunuz ve en az 3
periyodunu dahil edecek şekilde çiziniz.)
Solution 3b :
So if we expand c[n]:
c[n] =
∞
X
a[n − 3k]
k=−∞
= · · · + a[n + 6] + a[n + 3] + a[n] + a[n − 3] + a[n − 6] + · · ·
If we plot c[n] between (-7,7), it will look like this:
2
c[n]
1
0
−1
−2
−3
−7
−6
−5
−4
−3
−2
−1
0
n
1
2
3
As can be seen from the plot, the fundamental period of c[n] is 3.
4
5
6
7