1. Given an analog signal 𝑥(𝑡) = sin(200𝜋𝑡) (𝑡:𝑠).

Determine the recovered analog signal

DIGITAL SIGNAL PROCESSING – HOMEWORK #1

when 𝑥(𝑡) is sampled at:

a. 𝑓𝑠 = 120 𝐻𝑧
Ω 200
We have: Fx = = = 100 (Hz)
2 2
With the Fx = 100 Hz, the Nyquist interval is:
−Fs Fs −120 120
[ , ]=[ , ]Hz = [−60, 60¿Hz
2 2 2 2
Signal Fx = 100Hz outside the Nyquist interval, we can’t recover this signal. The recovered
signal will be frequencies:
F0 = Fx ± m×Fs = 100 ± m120 = 100, 100 ± 120, 100 ± 240, 100 ± 360,…
Only the frequency -20Hz at m = 1 lies within the Nyquist interval, then the recovered signal

b. 𝑓𝑠 = 240 𝐻𝑧
will be -20Hz (20Hz and phase reversal).

With Fs = 240 and Fx = 100Hz, the Nyquist interval is:

−Fs Fs −240 240
[ , ]=[ , ]Hz = [−120, 120¿Hz
2 2 2 2
The sampling theorem is satisfied, then the original frequency of 100Hz will be recovered.

2. Given an analog signal 𝑥(𝑡) = 4 + 2cos(2𝜋𝑡) + 6cos(8𝜋𝑡) (𝑡:𝑠).

a. Determine the Nyquist rate (Nyquist frequency) for the signal
With the unit of time is s, the given signal has 3 frequencies:
F1 = 0 Hz, F2 = 1 Hz, F3 = 4 Hz
The highest frequency is FM = F3 = 4Hz, the Nyquist rate is 2×4Hz = 8Hz. When the signal is
sampled at rates greater than 8Hz there will be no aliasing.
b. If the signal is sampled at half the Nyquist rate, find the recovered analog signal.
When the signal is sampled at half the the Nyquist rate is Fs = 4Hz, aliasing will occur. Now
the Nyquist interval is [−2 , 2¿ Hz .The two frequencies F1 and F2 lie within this interval and,
thus, will not be aliased. The frequency F3 lie outside the Nyquist interval and, thus, will be
aliased.
F30 = F3 ± mFs = 4 ± m4 = 4, 4 ± 4, 4 ± 8, 4 ± 12,…(Hz)
Only F30 = 0 Hz at m = 1 is lie within the Nyquist interval.
The recovered signal x0(t) has the frequencies F10, F20 and F30:
x0(t) = 4cos(2𝜋F1t) + 2cos(2𝜋F2t) + 6cos(2𝜋F30t)
x0(t) = 4 + 2cos(2𝜋t) + 6cos(0t) = 4 + 2cos(2𝜋t) + 6 = 10 + 2 cos(2𝜋t).

3. Given an analog signal 𝑥(𝑡) = 4sin(2𝜋𝑡) − 2sin(6𝜋𝑡) + 4sin(12𝜋𝑡) (𝑡:𝑚𝑠) sampled at

𝑓𝑠 = 5 𝑘𝐻𝑧. Find the signal that is the alias of 𝑥(𝑡). Repeat when the signal is sampled at
𝑓𝑠 = 10 𝑘𝐻𝑧.
Convert to x(t) = Acos(Ωt):
𝑥(𝑡) = 4cos(2𝜋𝑡 − 2 ) − 2cos(6𝜋𝑡 − 2 ) + 4cos(12𝜋𝑡 − 2 ) (𝑡:𝑚𝑠)
π π π

Because the unit of time is ms, the given signal has 3 frequencies:
F1 = 1 kHz, F2 = 3 kHz, F3 = 6 kHz
The highest frequency is FM = 6kHz, the Nyquist rate is 2×6kHz = 12kHz . When the signal

With the 𝑓𝑠 = 5 𝑘𝐻𝑧, the Nyquist interval is [−2.5 , 2.5] kHz. Only the frequency F1 is lie
is sampled at rates greater than 12kHz there will be no aliasing.

within this interval and, thus, will be not aliased. The two frequencies F2 and F3 lie outside
the interval and, thus, will be aliased:
F20 = F2 ± mFs = 3 ± m5 = 3, 3 ± 5, 3 ± 10, 3 ± 15,… = 3 −¿ 5 = −¿2kHz
 F30 = F3 ± mFs = 6 ± m5 = 6, 6 ± 5, 6 ± 10, 6 ± 15,… (kHz) = 6 −¿ 5 = 1kHz
The recovered signal x0(t) has the frequencies F10, F20 and F30:
π π π
x0(t) = 4cos(2𝜋F1𝑡 − 2 ) − 2cos(2𝜋F20𝑡 − 2 ) + 4cos(2𝜋F30𝑡 − 2 )
π π π π
x0(t) = 4cos(2𝜋t − 2
) − 2cos(−4𝜋t − 2
) + 4cos(2𝜋t − 2
) = 8cos(2𝜋t − 2
) −
π
2cos(−4𝜋t − 2 ) (ms:t)
With the 𝑓𝑠 = 10 𝑘𝐻𝑧, the Nyquist interval is [−5 , 5] kHz. The two frequency F1 and F2 is
lie within this interval and, thus, will be not aliased. The frequency F3 lie outside the interval
and, thus, will be aliased:
F30 = F3 ± mFs = 6 ± m5 = 6, 6 ± 5, 6 ± 10, 6 ± 15,… (kHz) = 6 −¿ 10 =−¿4kHz
The recovered signal x0(t) has the frequencies F10, F20 and F30:
π π π
x0(t) = 4cos(2𝜋F1𝑡 − 2 ) − 2cos(2𝜋F2𝑡 − 2 ) + 4cos(2𝜋F30𝑡 − 2 )
π π π
x0(t) = 4cos(2𝜋t − 2 ) − 2cos(6𝜋t − 2 ) + 4cos(−8𝜋t − 2 ) (ms:t)

DIGITAL SIGNAL PROCESSING 2025 – HOMEWORK #2

1. Sketch the following sequences.
a. 𝑥(𝑛) = 3𝛿(𝑛 + 2) − 0.5𝛿(𝑛) + 5𝛿(𝑛 − 1) − 4𝛿(𝑛 − 5)
𝑥(𝑛) = {3; 0; −0,5; 5; 0; 0; 0; 0; −4}.
b. 𝑥(𝑛) = 𝛿(𝑛 + 1) − 2𝛿(𝑛 − 1) + 5𝛿(𝑛 − 4)
𝑥(𝑛) = {1; 0; −2; 0; 0; 5}.
c. 𝑥(𝑛) = 2𝛿(𝑛 + 3) − 0.5𝛿(𝑛 + 1) − 5𝛿(𝑛 − 2) − 4𝛿(𝑛 − 5)
𝑥(𝑛) = {2; 0; −0.5; 0; 0; −5;4 0; 0; −4}.
2. Write an expression for each digital signal using the unit sample sequence and its
shifted sequences.

a. x(n) = δ(n – 1) – δ(n – 2) + δ(n – 4) – δ(n – 5)

x(n) = {0; 0; 1; –1; 0; 1; –1}.
b. x(n) = 3δ(n) + δ(n – 1) + 2δ(n – 2) + δ(n – 3) + δ(n – 5)
x(n) = {0; 3; 1; 2; 1; 0; 1}.

3. Sketch the block diagram of the discrete-time system (𝑦(𝑛) = 0, 𝑥(𝑛) = 0, 𝑛 < 0).
a. 𝑦(𝑛) − 𝑦(𝑛 − 1) = 0.5𝑥(𝑛) − 2𝑥(𝑛 − 1)
b. 𝑦(𝑛) = 𝑥(𝑛) − 0.7𝑥(𝑛 − 1) − 0.2𝑥(𝑛 − 3)

c. 𝑦(𝑛) = 0.5𝑦(𝑛 − 1) + 2𝑥(𝑛)

4. Determine which of the following systems is a linear system.

a. 𝑦(𝑛) = 5𝑥(𝑛) + 2𝑥2(𝑛)
𝑦1(𝑛) = 5𝑥1(𝑛) + 2𝑥12(𝑛); 𝑦2(𝑛) = 5𝑥2(𝑛) + 2𝑥22(𝑛)
x3 (n) = a1𝑥1(n) + a2𝑥2(n)
y3(n) = T[𝑥3(n)] = 5𝑥3(n) + 2𝑥32(n)

𝑦3(𝑛) = T[a1𝑥1(n) + a2x2(n)] =5[a1x1(n) + a2x2(n)] + 2[a1𝑥1(n) + a2𝑥2(n)]3

A linear combination of the two input sequences:

a1𝑦1(n) + a2𝑦2 (n) = a15𝑥1(n) + 2 a1𝑥12(n) + a25𝑥2(n) + 2a2𝑥22(n)

A linear combination of the two output:

 The system is nonclear.

b. 𝑦(𝑛) = 𝑥(𝑛 − 1) + 4𝑥(𝑛)
𝑦1(𝑛) = 𝑥1(𝑛 − 1) + 4𝑥1(𝑛); 𝑦2(𝑛) = 𝑥2(𝑛 − 1) + 4𝑥2(𝑛)
𝑥 3 (n) = a1𝑥1(n) + a2𝑥2(n)
𝑦3(n) = T[𝑥3(n)] = 𝑥3(𝑛 − 1) + 4𝑥3(𝑛)

𝑦3(𝑛) = T[a1𝑥1(n) + a2𝑥2(n)] = a1𝑥1(𝑛 − 1) + 4a1𝑥1(𝑛) + a2𝑥2(𝑛 − 1) + 4a2𝑥2(𝑛)

A linear combination of the two input sequences:

a1𝑦1(n) + a2𝑦2(n) = a1[𝑥1(𝑛 − 1) + 4𝑥1(𝑛)] + a2[𝑥2(𝑛 − 1) + 4𝑥2(𝑛)]

A linear combination of the two output:

 The system is linear.

c. 𝑦(𝑛) = 4𝑥3(𝑛 − 1) − 2𝑥(𝑛)
 𝑦1(𝑛) = 4𝑥13(𝑛 − 1) − 2𝑥1(𝑛); 𝑦2(𝑛) = 4𝑥23(𝑛 − 1) − 2𝑥2(𝑛)
𝑥 3 (n) = a1𝑥1(n) + a2𝑥2(n)
𝑦3(n) = T[𝑥3(n)] = 4𝑥33(𝑛 − 1) − 2𝑥3(𝑛)

A linear combination of the two input sequences:

𝑦3(𝑛) = T[a1x1(n) + a2x2(n)] = 4[a1𝑥1(n−1) + a2𝑥2(n−1)]3 – 2[a1𝑥1(n) + a2𝑥2(n)]

A linear combination of the two output:

a1𝑦1(n) + a2𝑦2(n) = 4a1𝑥13(𝑛 − 1) – 2a1𝑥1(𝑛) + 4a2𝑥23(𝑛 − 1) – 2a2𝑥2(𝑛)

=>The system is nonlinear

5. Given the linear systems, find which one is time-invariant.
a. 𝑦(𝑛) = −5𝑥(𝑛 − 10)
The response of this system to (n−k) is 𝑦’(𝑛) = −5𝑥(𝑛 – k – 10)
Now if we delay 𝑦(𝑛) by k units in time 𝑦(n−k) = −5𝑥(𝑛 – k – 10).
 𝑦’(𝑛) = 𝑦(n−k), Time-invariant
b. 𝑦(𝑛) = 𝑥(−𝑛)
The response of this system to (n−k) is 𝑦’(𝑛) = 𝑥(–n – k) = 𝑥(–n – k)
Now if we delay 𝑦(𝑛) by k units in time 𝑦(n−k) = 𝑥(–(n –k)) = 𝑥(–n + k).
 𝑦’(𝑛) != 𝑦(n−k), Time-variant
c. 𝑦(𝑛) = 4𝑥(𝑛2)
The response of this system to (n−k) is 𝑦’(𝑛) = 4𝑥(n2 – k) = 4𝑥(n2 – k)
Now if we delay 𝑦(𝑛) by k units in time 𝑦(n−k) = 4𝑥((n – k)2) = 4𝑥(n2 + k2 – 2nk).
 𝑦’(𝑛) != 𝑦(n−k), Time-variant
6. Determine which of the following linear systems is causal.
Causal system: if the output of the system at any time n depends only on present and past
Noncausal system: The requirement above do no match
a. 𝑦(𝑛) = 0.5𝑥(𝑛) + 100𝑥(𝑛 − 2) − 20𝑥(𝑛 − 10)
0.5𝑥(𝑛)  present value  casual
100𝑥(𝑛 − 2)  past value  casual
20𝑥(𝑛 − 10)  past value  casual
 Because the output of the system at any time n depends only on present and past, the system
is causal.
b. 𝑦(𝑛) = 𝑥(𝑛 + 4) + 0.5𝑥(𝑛) − 2𝑥(𝑛 − 2)
𝑥(n+4)  future value  non-casual
𝑥(n)  present value  casual
𝑥(n – 2)  past value  casual
 Because the output depends on x(n+4) (a future input), the system is non-causal
systems.
c. 𝑦(𝑛) = −0.4𝑦(𝑛 − 1) + 𝑥(𝑛 + 2)
−0.4𝑦(𝑛 − 1)  past value  casual
𝑥(𝑛 + 2) future value  non-casual
Because the output depends on 𝑥(𝑛 + 2) (a future input), the system

DIGITAL SIGNAL PROCESSING 2025 – HOMEWORK #3

1. Find the unit-impulse response for the following linear systems:
a. 𝑦(𝑛) = 0.5𝑥(𝑛) − 0.5𝑥(𝑛 − 2), 𝑛 ≥ 0, 𝑥(−2) = 𝑥(−1) = 0
h(𝑛) = 0.5 δ (𝑛) − 0.5δ(𝑛 − 2)
h(0) = y(0) = 0.5δ(0) − 0.5δ(0 − 2) = 0.5
h(1) = y(1) = 0.5δ(1) − 0.5δ(0 − 1) = 0
h(2) = y(2) = 0.5δ(2) − 0.5δ(0 − 2) = −0.5
 h(n) = {0.5, 0, −0.5}
b. 𝑦(𝑛) = 0.75𝑦(𝑛 − 1) + 𝑥(𝑛), 𝑛 ≥ 0, 𝑦(−1) = 0 (Vinh)
h(𝑛) = 0.75h(𝑛 − 1) + δ(𝑛)
h(0) = 0.75h(− 1) + δ (0) = 1 = 0.750
h(1) = 0.75h(0) + δ (1) = 0.751
h(2) = 0.75h(1) + δ (2) = 0.752 = 0.5625
h(3) = 0.75h(2) + δ (3) = 0.753 = 0.421875
 y(n) = h(n) = 0.75n × u(n)
c. 𝑦(𝑛) = −0.8𝑦(𝑛 − 1) + 𝑥(𝑛 − 1), 𝑛 ≥ 0, 𝑥 (−1) = 0, 𝑦(−1) = 0
h(n) = −0.8h(𝑛 − 1) + δ (𝑛 − 1)
h(0) = −0.8h(− 1) + δ (− 1) = 0
h(1) = −0.8h(0) + δ (0) = 1 = (−0.8)1-1
h(2) = −0.8h(1) + δ (1) = −0.8 = (−0.8)2-1
h(3) = −0.8h(2) + δ (2) = 0.64 = (−0.8)3-1
h(n) = −0.8h(3) + δ (3) = −0.512 = (−0.8)4-1
 y(n) = h(n) = (−0.8)n-1 × u(n−1)
d. 𝑦(𝑛) = 0.5𝑦(𝑛 − 1) + 0.5𝑥(𝑛), 𝑛 ≥ 0, 𝑦(−1) = 0 (Nam)
h(n) = 0.5h(n-1) + δ(n)
h(0) = 0.5h(-1) + δ(0) = 1 = 0.50
h(1) = 0.5h(0) + δ(1) = 0.5 = 0.51
h(2) = 0.5h(1) + δ(2) = 0.25 = 0.52
h(3) = 0.5h(2) + δ(3) = 0.125 = 0.53

2. Given the sequence ℎ(𝑘). Determine ℎ(−𝑘), ℎ(2 − 𝑘), ℎ(−𝑘 − 3).
 y(n) = h(n) = 0.5n× u(n).

a. ℎ(𝑘) = {0, 𝟐, 2, 2, 3, 3, 0},

 ℎ(−𝑘) = {0, 3, 3, 2, 2, 2, 0}, n start form 0
h(2 − 𝑘) = {0, 3, 3, 2, 2, 2, 0}, n start form 0
ℎ(−𝑘− 3) = {0, 3, 3, 2, 2, 2, 0, 0, 0}, n start form 0
b. ℎ(𝑘) = {0, −𝟏, −1, 2, 2, −2, 0}
ℎ(− 𝑘) = {0, −2, 2, 2, −1, −1, 0} , n start form 0
ℎ(2 − 𝑘) = {0, −2, 2, 2, −1, −1, 0}, n start form 0
ℎ(− 𝑘 − 3) = {0, −2, 2, 2, −1, −1, 0, 0, 0}, n start form 0
3. Evaluate the convolution sum 𝑦(𝑛) = ∑∞ 𝑘=−∞ 𝑥(𝑘)ℎ(𝑛 − 𝑘).

a. ℎ(𝑛) = {0, 𝟐, 2, 2, 1, 1, 0} and 𝑥(𝑛) = {0, 𝟐, 1, 1, 0}

h(n)\x(n) 0 2 1 1 0
0 0 0 0 0 0
2 0 4 2 2 0
2 0 4 2 2 0
2 0 4 2 2 0
1 0 2 1 1 0
1 0 2 1 1 0
0 0 0 0 0 0

 y(n) = (0, 0 , 4, 6 ,8 ,6, 5, 2, 1, 0, 0).

b. ℎ(𝑛) = {0, 𝟏, 2, 2, 0} and 𝑥(𝑛) = {1, 𝟐, 3, 4, 5}

h(n)\x(n) 1 2 3 4 5

0 0 0 0 0 0

1 1 2 3 4 5

2 2 4 6 8 10

2 2 4 6 8 10

0 0 0 0 0 0

 y(n) = (0, 1, 4, 9, 14, 19, 18, 10, 0).

c. ℎ(𝑛) = {0, −1, −𝟐, −2, 1, 0} and 𝑥(𝑛) = {0, 2, 1, 𝟐, 1, 0}

h(n)\x(n) 0 2 1 2 1 0

0 0 0 0 0 0 0

−1 0 −2 −1 −2 −1 0

−2 0 −4 −2 −4 −2 0
 −2 0 −4 −2 −4 −2 0

1 0 2 1 2 1 0

0 0 0 0 0 0 0
 y(n) = {0, 0, −2, −5, −8, −5, −5, 0, 1, 0}
d. ℎ(𝑛) = 𝑢(𝑛) and 𝑥(𝑛) = 𝑎𝑛𝑢(𝑛), where |𝑎| < 1
∞
y ( n )= ∑ x ( k ) ⋅h ( n−k )

With 𝑥(k) = 𝑎k𝑢(k), |a| < 1:

k=−∞

𝑥(k)
1.2

0.8

0.6

0.4

0.2

0
-1 0 1 2 3

And h ( n−k ) = u(n−k ), in case n = 0:

ℎ(𝑛−𝑘)
1.2

0.8

0.6

0.4

0.2

0
-4 -3 -2 -1 0 1 2 3

Series 1
For n < 0, The plot of h(n−k) is shifted to the left by n units, so we have:
∞
y ( n )= ∑ x ( k ) ×h ( n−k ) = 0
k=−∞
For n ≥ 0, The plot of h(n−k) is shifted to the right by n units, so we have:
 ∞ +∞ n+1
a∨¿
y ( n )= ∑ x ( k ) ×h ( n−k ) = ∑ a × 1 = 1−¿
k
¿
k=−∞ k=−∞ 1−¿ a∨¿ ¿

4. Determine the stability for the linear systems:

a. 𝑦(𝑛) = 0.5𝑥(𝑛) + 100𝑥(𝑛 − 2) − 20𝑥(𝑛 – 10)
∞

∑ ¿ h( k)∨¿ ¿ = |0.5| + |100|+ |– 20| = 120.5

k=−∞

 Is stable

b. 𝑦(𝑛) = ∑ 0.75 𝑥(𝑛−𝑘)

∞
k

k=0

h(k) = 0.75k
because 0.75 < 1:
∞ ∞
1
∑ ¿ h( k)∨¿=∑ ¿ 0.75∨¿ k =
1−¿ 0.75∨¿=4 ¿
¿¿
k=−∞ k=0

 Is stable

c. 𝑦(𝑛) =
∞

∑ 2k x (n−k )
k=0

h(k) = 2k
Because | a |= 2 > 1, therefore:
∞ ∞
a∨¿ n+1
∑ ¿ h( k)∨¿=∑ ¿ 0.75∨¿ =1−¿ k

2∨¿n +1
¿¿¿
=> ∞ (Not stable)
k=−∞ k=0
1−¿ a∨¿=1−¿ ¿¿
1−¿ 2∨¿ ¿

d. 𝑦(𝑛) =
∞

∑ (−1.5)k x(n−k)
k=0

h(k) =(−1.5)k
Because | a |= 1.5 > 1, therefore
∞ ∞
a∨¿n +1
∑ ¿ h( k)∨¿=∑ ¿−1.5∨¿k =1−¿
−1.5∨¿ n+1
¿¿¿
=> ∞ (Not stable)
k=−∞ k=0
1−¿ a∨¿=1−¿ ¿¿
1−¿−1.5∨¿ ¿

e. 𝑦(𝑛) =
∞

∑ (−1.5)k x(n−k)
k=0

h(k) =(−0.5)k
Because | a |= 0.5 < 1, therefore:
∞ ∞
1
∑ ¿ h( k)∨¿=∑ ¿−0.5∨¿ k =
1−¿−0.5∨¿=2 ¿
¿¿
k=−∞ k=0

 Is stable
1. Find the DTFT 𝑋(𝜔) of the sequences 𝑥(𝑛). Find the magnitude and the phase of
DIGITAL SIGNAL PROCESSING 2025 – HOMEWORK #4

the 𝑋(𝜔).
a. 𝑥(𝑛) = {1, 2, 3, 4, 𝟎, 4, 3, 2, 1}

𝑋 (𝜔 ) =
+∞

∑ x ( n ) e− jωn = x (−4 ) e j 4 ω + x (−3 ) e j 3 ω+ x (−2 ) e j2 ω+ x (−1 ) e jω + x ( 0 ) e j 0 + x (1 ) e− jω

n=∞
− j 2ω − j3ω
+ x (2 ) e +x (3 )e + x ( 4 ) e− j 4 ω
𝑋(𝜔) =2cos(4𝜔)+ 4cos(2𝜔) + 6cos(3𝜔) + 8cos(𝜔)
Magnitude: ∣X(ω)∣=∣2cos(4ω)+4cos(3ω)+6cos(2ω)+8cos(ω)∣

Phase: arg[𝑋(𝜔)] = { 0 , X ( ω ) >0

, X ( ω ) <0

b. 𝑥(𝑛) = 𝛿(𝑛) + 2𝛿(𝑛 − 1) − 3𝛿(𝑛 − 2)

 𝑥(𝑛) = {0, 2, − 3}

𝑋(𝜔) =
+∞

∑ x ( n ) e− jωn = x ( 0 ) e− j 0+ x (1 ) e− jω+ x (2 ) e− j 2ω
𝑋(𝜔) =2e −jω −3e −j2ω = 2[cos(𝜔) −¿ jsin(𝜔)] − 3[cos(2𝜔) −jsin(2𝜔)]
n=∞

= 2cos(𝜔) −3 cos (ω)+ j ¿ 2sin(ω ¿ ¿

Magnitude: | 𝑋(𝜔)| = √ ( 2 cos ( ω )−3 cos ( ω ) )2 + ( 3 sin ( 2 ω )−2sin ( ω ) )2

Phase: arg[𝑋(𝜔)] =arg[

3 sin (2 ω )−2 sin ( ω )
]
2 cos ( ω )−3 cos ( ω )

c. 𝑥(𝑛) = 𝑢(𝑛 + 2) − 𝑢(𝑛 − 3) = 0 , otherwise {1 ,−2 ≤ n≤ 2

𝑋(𝜔) =
+∞ 2

∑ x (n) e − jωn
= ∑ x ( n) e
− jωn
= e2jω+ejω+1+e−jω+e−2jω

𝑋(𝜔) = 2cos(2ω)+2cos(ω)+1
n=∞ n=−2

Magnitude: | 𝑋(𝜔)| = | 2cos(2ω)+2cos(ω)+1|

𝑋(𝜔) always ≥ 0 so:
Phase: arg(𝑋(𝜔)) = 0.
d. 𝑥(𝑛) = 0.5𝑛𝑢(𝑛)

𝑋(𝜔) =
+∞ +∞
1 1
∑ x (n )e
− jωn
=∑ (0.5 e ¿¿− jω)n ¿ =
1−0.5 e
− jω
=
1−0.5 cos ( ω ) + j 0.5 sin ( ω )
n=−∞ n=0
𝑋(𝜔) =
1−0.5 cos ( ω )− j 0.5 sin ( ω ) 1−0.5 cos ( ω )− j0.5 sin ( ω )
2 2 =
[ 1−0.5 cos ( ω ) ] + [ 0.5 sin ( ω ) ] 1.25−cos ( ω )

Magnitude: |𝑋(𝜔)| =
√( )( )
2 2
1−0.5 cos ( ω ) −0.5 sin ( ω )
+
1.25−cos ( ω ) 1.25−cos ( ω )
− j 0.5 sin ( ω )

Phase: arg[X(𝜔)] = arg[

1.25−cos ( ω )
]
1−0.5 cos ( ω )
1.25−cos ( ω )
e. 𝑥(𝑛) = 𝑛(𝑎𝑛)𝑢(𝑛), |𝑎| < 1

𝑋(𝜔) =
− jω
+∞ +∞
ae acos ( ω )−ajsin ( ω )
∑ x (n) e − jωn
= ∑ n(ae ¿¿− jω) ¿ =
n

( 1−ae − jω 2
)
=
1−2 acos ( ω ) + a
2
n=∞ n=0

XR(𝜔) = 2 ; XI(𝜔) =
acos ( ω ) −ajsin ( ω )
;
1−2 acos ( ω ) +a 1−2 acos ( ω ) +a2
Magnitude: |𝑋(𝜔)| =√ X R2 + X I 2

Phase: arg[𝑋(𝜔)] = arg[

X I ( ω)
]
X R ( ω)

f. 𝑥(𝑛) = 𝑢(𝑛) − 𝑢(𝑛 − 𝑀) = 0 , otherwise {1, 0 ≤ n ≤ M −1

𝑋(𝜔) =
+∞ M −1

∑ ∑ ( e− jω )
− jωn n
x (n )e =
n=−∞ n=0

This is a geometric series with common ratio q=e−jω and M terms.

We use the geometric sum formula:

1 × ( 1−q M ) 1−e
M −1 − jω M
X ( ω )= ∑ q =
n
= − jω
n=0 1−q 1−e

M
Multiply numerator and denominator by e−jω (to symmetrize):
2

X ( ω )=e
− jω
( M −1)
2
×
sin( M2ω )
sin ( )
ω
2
Magnitude: |𝑋(𝜔)| = |
( 2 )
Mω
sin
|
sin ( )
ω
2

Phase: arg[𝑋(𝜔)] = −ω ×
M −1

2. Find the IDTFT 𝑥(𝑛) of the spectrum 𝑋(𝜔).

2

a. 𝑋(𝜔) = 1 − 2𝑒−𝑗2𝜔 + 3𝑒−𝑗3𝜔

𝑥(𝑛)= ∫ X ( ω ) e
π
1 jωn

𝑋(𝜔) = 1 𝑒 0 − 2𝑒−𝑗2𝜔 + 3𝑒−𝑗3𝜔

2 −π

 𝑥(𝑛)=δ(n)−2δ(n−2)+3δ(n−3)

( { )
ω0

b. 𝑋(𝜔) =
, n=0
{
1, ω ∈ [−ω 0 , ω 0 ] ;
0 , ω ∈ [ −π , π ] [ −ω 0 , ω 0 ] .
x ( n ) = π
sin ( ω 0 n )
,n≠0
πn

n≠ 0 :

𝑥(𝑛)= ∫ X ( ω ) e = ∫ e d ω=
π ω0
1 1 1 1 j ω n j −ω n 2 jsin ( ω ) sin ( ω )
jωn jωn
( e −e )= =0 0

2 −π 2 −ω 2 jn 0
2 jn n

n = 0:

𝑥(0)= ∫ X ( ω ) e = ∫ e d ω= ( ω 0 +ω 0 )= 0 = ❑0
π ω0
1 jω0 1 0 1 2ω ω
2 −π 2 −ω 2 0
2

( { )
ω0

c. 𝑋(𝜔) =
1−
, n=0
{
1 ,ω ∈ [ −π , π ] [− ω0 , ω0 ] ;
0 ,ω ∈ [ −ω 0 ,ω 0 ] .
x ( n ) = π
−sin ( ω 0 n )
,n≠0
πn

n≠ 0 :

𝑥(𝑛)= ∫ X ( ω ) e =
π −ω0 π
1 1 1
2 −π
jωn
2 ∫e jωn
d ω+¿ ∫ e d ω ¿
2ω
jωn

−π 0
 1 1 j−ω n j−π n j π n j ω n
x ( n )= [ e −e + e −e ]= 1 [ 2 jsin ( πn )−2 jsin ( ω0 n ) ]
0 0

2 jn 2 jn

sin ( πn )−sin ( ω 0 n ) −sin ( ω 0 n )

x ( n )= = ; sin ( πn ) = 0 for all n.
πn πn

n=0:

𝑥(0)= ∫ X ( ω ) e = ∫ 1 d ω+¿ ∫ 1 d ω= ( −ω0 + π + π−ω0 ) ¿

π −ω0 π
1 j0 1 1 1
2 −π 2 −π 2ω 2 0

𝑥(0)= 1
−ω0
❑

({ )
ω2−ω1

d.𝑋(𝜔)=
, n=0
{
1, ω ∈ [−ω 2 ,−ω1 ] ∪ [ ω1 , ω2 ] ;
0 , ω ∈ [−,−ω2 ] ∪ [−ω1 , ω1 ] ∪ [ ω2 , π ] .
x ( n ) π
sin ( ω 2 n ) −sin ( ω 1 n )
,n≠0
πn

n≠ 0 :

𝑥(𝑛) = ∫ X ( ω ) e = ∫ e d ω+¿ ∫ e d ω ¿
π −ω1 ω2
1 jωn 1 jωn 1 jωn
2 −π 2 −ω 2ω 2 1

𝑥(𝑛) =
1 −jω n −jω n jω n jω n 1
(e −e + e −e )=
1 2
(2 jsin ( ω2 n ) −2 jsin ( ω 1 n ) )
2 1

2 jn 2 jn

𝑥(𝑛) =
sin ( ω2 n ) −sin ( ω 1 n )
n

n = 0:

𝑥(0)= ∫ X ( ω ) e = ∫ 1 d ω+¿ ∫ 1 d ω ¿
π −ω1 ω2
1 0 1 1
2 −π 2 −ω 2ω 2 1

𝑥(0)= (−ω 1 +ω 2 +ω2 −ω1 ) =

1 ω2−ω1
2 π
 ({ )
ω 2−ω1

e. 𝑋(𝜔)=
1− ,n=0
{
0 , ω ∈ [−ω 2 ,−ω1 ] ∪ [ ω1 , ω2 ] ;
1 ,ω ∈ [−,−ω 2 ] ∪ [−ω 1 , ω 1 ] ∪ [ ω 2 , π ] .
x ( n ) π
sin ( ω1 n )−sin ( ω2 n )
, n≠ 0
πn

n≠ 0 :

𝑥(𝑛) = ∫ X ( ω ) e = ¿ ¿
π
1 jωn 1
2 −π 2

𝑥(𝑛) =
1
( e j−ω n−e j−π n + e jn−e j−ω n +e j π n−e j ω n )
2 1 2

2 jn

𝑥(𝑛) =
1
2 jn
( j 2 sin ( πn )− j 2 sin ( ω2 n ) + j2 sin ( ω 1 n ) ); j 2 sin ¿)¿ 0 for all n

𝑥(𝑛) = 1 (− j2 sin ( ω 2 n ) + j2 sin ( ω 1 n ) ) =

sin ( ω1 n )−sin ( ω2 n )
2 jn n

n = 0:

𝑥(𝑛) = ∫ X ( ω ) e = ¿ ¿
π
1 jωn 1
2 −π 2

𝑥(𝑛) = 2 (−ω 2+ π + ω1 + ω1 + π−ω2 ) = 1

1 −ω2 ω1
+
π π
3. Find the frequency response, magnitude response and phase response of the systems.
a. 𝑦(𝑛) = 𝑥(𝑛) + 2𝑥(𝑛 − 1) + 3𝑥(𝑛 − 3)
h(n) = (n) + 2(n−1) + 3(n−3)
frequency response:
∞
H ( ω )= ∑ h ( n ) e− jωn = 1e− j 0 + 2e− jω+3 e− j 3 ω=¿1 + 2e− j2 ω +3 e− j 3 ω
n=−∞

magnitude response:
| H ( ω )| = |1+2 cos ( ω )−2 jsin ( ω ) +3 cos ( 3 ω )−3 jsin ( 3 ω )|
∣H(ω)∣=(√ ¿ ¿ ¿
phase response:
arg( H ( ω )) = arg ¿
b. 𝑦(𝑛) = 0.8𝑦(𝑛 − 1) + 𝑥(𝑛)
c.
DTFT:
∞ ∞ ∞

∑ y ( n) e− jωn
=0.8 ∑ y (k )e − jω ( k+1)
+ ∑ x ( n ) e− jωn ; k=n−1
n=−∞ k=−∞ n=−∞
− jω
Y(ω) = 0.8e Y(ω) + X(ω)
frequency response:
Y (ω) 1 1 1−0.8 cos ( ω )−0.8 jsin ( ω )
H ( ω )= = − jω =
=
X ( ω ) 1−0.8 e 1−0.8 cos ( ω ) +0.8 jsin ( ω ) ( 1−0.8 cos ( ω ) )2 + ( 0.8 sin ( ω ) )2

magnitude response:

∣H(ω)∣=(
√( ) (( )
2 2
1−0.8 cos ( ω ) −0.8 jsin ( ω )
2 2
+ 2 2
( 1−0.8 cos ( ω ) ) + ( 0.8 sin ( ω ) ) 1−0.8 cos ( ω ) ) + ( 0.8 sin ( ω ) )

phase response:

( )
−0.8 sin ( ω )
2 2

arg( H ( ω )) =arg
( 1−0.8 cos ( ω ) ) + ( 0.8 sin ( ω ) )
1−0.8 cos ( ω )
2 2
= arg ( −0.8 sin ( ω )
1−0.8 cos ( ω ) )
( 1−0.8 cos ( ω ) ) + ( 0.8 sin ( ω ) )
d. 𝑦(𝑛) = 0.5𝑦(𝑛 − 1) + 𝑥(𝑛) + 0.5𝑥(𝑛 − 1)(Thành)
DTFT:
∞ ∞ ∞ ∞

∑ y ( n )=0.5 ∑ y ( k ) e− jω (k+ 1)+ ∑ x ( n ) +0.5 ∑ x ( k ) e− jω ( k+1 )

n=−∞ k=−∞ n=−∞ k=−∞

k = n −1 ;
Y(ω) = 0.5e− jωY(ω) + X(ω) + 0.5e− jωX(ω)
frequency response:
Y (ω) 1+0.5 e
− jω
H ( ω )= = =
X ( ω ) 1−0.5 e− jω
1+ 0.5 cosω− j 0.5 sinω ( 1+0.5 cosω− j 0.5 sinω ) ( 1−0.5 cosω+ j0.5 sinω )
=
1−0.5 cosω+ j 0.5 sinω ( 1−0.5 cosω )2 + ( 0.5 sinω )2
1−0.25 ( cos 2 ( ω )−sin 2 ( ω ) ) + j 0.5 sin ( ω ) cos ( ω )
H ( ω )=
( 1−0.5 cosω )2+ ( 0.5 sinω )2
magnitude response:
 ∣H(ω)∣=
|1+ 0.5 e− jω| √( 1+ 0.5 cos ( ω ) )2+ ( sin ( ω ) )2
=
|1−0.5 e− jω| √( 1−0.5 cos ( ω ) )2+ ( sin ( ω ) )2
phase response:

( )
0.5 sin ( ω ) cos ( ω )

( )
2 2
( 1−0.5 cosω ) + ( 0.5 sinω ) 0.5 sin ( ω ) cos ( ω )
arg( H ( ω )) =arg = arg
1−0.25 ( cos ( ω ) −sin ( ω ) )
2 2
1−0.25 ( cos2 ( ω ) −sin2 ( ω ) )
( 1−0.5 cosω )2+ ( 0.5 sinω )2
4. Find the output 𝑦(𝑛) of the system (applying DTFT properties).
a. 𝑥(𝑛) = ℎ(𝑛) = {1, 𝟎, 1}
y(n) = h(n) * x(n) (convolution propertie)
∞
y ( n )= ∑ x ( k ) h ( n−k )
k=−∞

h(n)\x(n) 1 0 1
1 1 0 1
0 0 0 0
1 1 0 1
y(n) = {1, 0, 2, 0, 1}
b. 𝑥(𝑛) = 𝑢(𝑛 + 2) − 𝑢(𝑛 − 3), ℎ(𝑛) = 2𝛿(𝑛 − 1) − 3𝛿(𝑛 − 2)
𝑥(𝑛) ={1, 1, 1, 1, 1}
h(n) ={0, 2, − 3}
y(n) = 𝑥(𝑛)* ℎ(𝑛) (convolution)
∞
y ( n )= ∑ x ( k ) h ( n−k )
k=−∞

h(n)\x(n) 1 1 1 1 1
0 0 0 0 0 0
2 2 2 2 2 2
−3 −3 −3 −3 −3 −3
 y(n) = (0, 2, −1, −1 ,−1 ,−1 ,−3)
c. ℎ(𝑛) = 0.5𝑛𝑢(𝑛), 𝑥(𝑛) = 0.25𝑛𝑢(𝑛)
y(n) = 𝑥(𝑛)* ℎ(𝑛) (convolution properties)
DTFT:
Y(ω) = H(ω ¿ × X (ω)
∞ ∞
1
H(ω ¿= ∑ h ( n ) e − jωn
=∑ 0.5 u ( n ) e− jωn =
n=−∞ n=1 1−0.5 e− jω
∞ ∞
1
X(ω) ¿ ∑ x ( n ) e− jωn=∑ 0.25u ( n ) e− jωn=
1−0.25 e− jω
n=−∞ n=1

1 −1 2
Y(ω) = ↔ Y ( ω )= +
( 1−0.5 e − jω
) ( 1−0.25 e − jω
) 1−0.25 e
− jω
1−0.5 e
− jω

[ ]
−1
−1 1
y(n) = DTFT( Y ( ω ) ) =DTFT =an u ( n )
1−a e− jω

y(n) = −0.25 n u ( n ) +2 [ 0.5n u ( n ) ] = ( 2 ×0.5 n−0.25n ) u ( n )

3 7 15
, , , … , ( 2 ×0.5 −0.25 ) }
n n
y(n) = {1,
4 16 64
jnπ
d. h(n) = 0.8 n u ( n ) , x ( n )=e 2

y(n) = 𝑥(𝑛)* ℎ(𝑛) (convolution properties)

{10,,nn<k≥ k
∞
y ( n )= ∑ x ( k ) h ( n−k ); u ( n−k )=
k=−∞

∞ jkπ n jkπ
y ( n )= ∑ e 2
0.8 ( n−k )
u ( n−k ) = ∑ e 2
0.8( n−k )
k=−∞ k=−∞

set m = n −k → k =n−m, when k ≤ n , m≥ 0

j ( n− k ) π
x(n−k ¿=e 2
is infinite support for all n∈ Z
h(m) = 0 when m<0

∑ (0.8 e )
∞ j ( n −m ) π jnπ ∞ − jmπ jnπ n − jπ m
y ( n )= ∑ 0.8 e
m 2
=e 2
∑ 0.8 m e 2
=e 2 2

m=0 m=0 m=0

jnπ jnπ
jnπ
1 2 1 1
y ( n )=e 2
× e × e 2
×
=
( ( π2 )− j 0.8 sin ( π2 )) = ( π2 )
− jπ
1−0.8 e 2 1− 0.8 cos 1+ j 0.8 sin

jnπ
1
y(n) =e 2
×
1+ j 0.8
e. 𝑥(𝑛) = 2 − 5 sin(𝑛𝜋) , ℎ(𝑛) = 0.25𝑛𝑢(𝑛)
y(n) = 𝑥(𝑛)* ℎ(𝑛) (convolution properties)
∞ n
y ( n )= ∑ x ( k ) h ( n−k )=¿ ∑ 2 ×0.25( n−k ) u ( n−k ) ¿
k=−∞ k=−∞

we have: u ( n−k )= {10,,nn<k≥ k

set m = n−k , when n ≥ k , m≥ 0:
h(m) = 0 when m<0
x(n −m) = 2 for all m
so:
n
y ( n )=2 ∑ 0.25m
m=0

This is a geometric series with common ratio q=0.25 and n+1 terms.

We use the geometric sum formula:

n
1× ( 1−qn +1) 1−0.25
n+ 1
1−0.25
n +1
8
y ( n )=2 ∑ q =2 = ( 1−0.25 )
m n+1
=2 =2
m=0 1−q 1−0.25 1−0.25 3

1. Find the 4-point DFT 𝑋(𝑘) of the sequences, then compute its IDFT 𝑥(𝑛).
DIGITAL SIGNAL PROCESSING 2025 – HOMEWORK #5

a. 𝑥(𝑛) = {𝟏, 1, −1, 0}

N=4
3 3 − jnπk
DFT: X(k) = ∑ x ( n ) W kn
4 =∑ x ( n ) e
2

n=0 n=0

X(0) = x ( 0 ) W 04 + x ( 1 ) W 04 + x ( 2 ) W 04 + x ( 3 ) W 04=1+1−1+ 0=1

X(1) = x ( 0 ) W 14 ×0+ x (1 ) W 14× 1+ x ( 2 ) W 14 ×2 + x ( 3 ) W 14×3 =1− j+1+ 0=2− j
X(2) = x ( 0 ) W 24 ×0+ x (1 ) W 24× 1+ x ( 2 ) W 24 ×2 + x ( 3 ) W 24× 3=1−1−1+ 0=−1
X(3) = x ( 0 ) W 34 ×0+ x (1 ) W 34× 1+ x ( 2 ) W 34 ×2 + x ( 3 ) W 44 ×3 =1+ j+1+0 j=2+ j

The 4-point DFT X(k) is {1, 2− j , −1, 2+ j}

 N −1 3
1 1
IDFT: x(n) =
N
∑ X (k ) W −kn
N = ∑ X ( k ) W −kn
4 k=0 4
k=0

1
x (0) =
4
( X ( 0 ) W 4 + X ( 1) W 4 + X ( 2)W 4 + X ( 3 ) W 4 )
−0 ×0 −1× 0 −2 ×0 −3 ×0

1
x ( 0 )=
4
[ 1+ ( 2− j )−1+ ( 2+ j ) ]=1
1
x (1) =
4
( X ( 0 ) W 4 + X ( 1) W 4 + X ( 2) W 4 + X ( 3) W 4 )
−0 ×1 −1 ×1 −2× 1 −3 ×1

1
x (1 )=
4
[ 1+ ( 2− j ) j−1 (−1 ) −( 2+ j ) j ]=1

x (2) =( X ( 0 ) W −0
4
×2
+ X ( 1 ) W −1×
4
2
+ X ( 2 ) W −2
4
×2
+ X ( 3 ) W −3
4
×2
)
1
x (2 )=
4
[ 1+ ( 2− j )(−1 )−1 ( 1 ) + ( 2+ j )(−1 ) ]=−1

x (3) =( X ( 0 ) W −0
4
×3
+ X (1 ) W −1×
4
3
+ X ( 2 ) W −2
4
×3
+ X ( 3 ) W −3×
4
3
)
1
x (3 )=
4
[ 1+ ( 2− j )(− j )−1 (−1 )+ ( 2+ j ) j ]=0
b. 𝑥(𝑛) = {𝟒, −3, 2, 1}
N=4
DFT:
3 3 − jnπk
X(k) = ∑ x ( n ) W kn
4 =∑ x ( n ) e
2

n=0 n=0

For k = 0:

X(0) = x(0)e0 + x(1)e0 + x(2)e0 + x(3)e0 = 4 + (−¿3) + 2 + 1 = 4

For k = 1:

X(1) = x(0)e0 + x(1)e−j2π + x(2)e−jπ + x(3)e−j3π/2 = 4(1) + (-3)(-j) + 2(-1) + 1(j)

X(1) = 2 + 4j

For k = 2:

X(2) = x(0)e0 + x(1)e−jπ + x(2)e−j2π + x(3)e−j3π = 4(1) + (−¿3)(−1) + 2(1) + 1(−1)=8

For k = 3:
X(3) = x(0)e0 + x(1)e−j3π/2 + x(2)e−j3π + x(3)e−j9π/2

X(3) = 4(1) + (−3)(j) + 2(−1) + 1(−j) = 2 − 4j

The 4-point DFT X(k) is {4, 2 + 4j, 8, 2 – 4j}

N −1 3
1 1
IDFT: x(n) =
N
∑ X ( k ) W −kn
N = ∑ X ( k )W 4
4
−kn

k=0 k=0

For n = 0:

1 1
x(0) =
4
[ X ( 0 ) e + X ( 1 ) e + X ( 2 ) e + X ( 3 ) e ]= [ 4 + ( 2+4 j ) +8+ ( 2−4 j ) ]
0 0 0 0
4

1
x(0) = [4 + 2 + 8 + 2] = 4
4

For n = 1:
1
x(1) =
4
[ 0 jπ / 2 jπ
X ( 0 ) e + X ( 1 ) e + X ( 2 ) e + X (3 ) e
j3 π / 2
]
1
x(1) = [4(1) + (2 + 4j)(j) + 8(−1) + (2 – 4j)(−j)] = −3
4
For n = 2:
1
x(2) = [X(0)e0 + X(1)ejπ + X(2)ejπ/2 + X(3)ej3π]
4
1
x(2) = [4(1) + (2 + 4j)(−¿1) + 8(1) + (2 −¿ 4j)(−¿1)] = 2
4
For n = 3:
1
x(3) = [X(0)e0 + X(1)ej3π/2 + X(2)ej3π + X(3)ej9π/2]
4
1
x(3) = [4(1) + (2 + 4j)(−¿j) + 8(−¿1) + (2 −¿ 4j)(j)]
4

c) 𝑥(𝑛) = {𝟎. 𝟖, 0.6} ={ 𝟎. 𝟖, 0.6, 0, 0 }

x(3) = 1

N=4
3 3 − jnπk
DFT: X(k) = ∑ x ( n ) W kn
4 =∑ x ( n ) e
2

n=0 n=0
 3 3 − jnπk
X(k) = ∑ x ( n ) W =∑ x ( n ) e
kn
4
2

n=0 n=0

For k = 0:

X(0) = x(0)e0 + x(1)e0 + x(2)e0 + x(3)e0 = 0.8 + 0.6 + 0 + 0= 1.4

For k = 1:

X(1) = x(0)e0+x(1)e−j2π+x(2)e−jπ+x(3)e−j3π/2 = 0.8(1) + 0.6(−j) + 0(−1) + 0(j)

X(1) = 0.8 − 0.6j

For k = 2:

X(2) = x(0)e0+x(1)e−jπ+x(2)e−j2π+x(3)e−j3π = 0.8(1) + 0.6(−1) + 0(1) + 0(−1)

X(2) = 8

For k = 3:

X(3) = x(0)e0 + x(1)e−j3π/2 + x(2)e−j3π + x(3)e−j9π/2

X(3) = 0.8(1) + 0.6(j) + 0(−1) + 0(−j) = 0.8 + 0.6j

The 4-point DFT X(k) is {1.4,0.8−0.6j,0.2,0.8+0.6j}

N −1 3
1 1
IDFT: x(n) =
N
∑ X ( k ) W −kn
N = ∑ X ( k )W 4
4
−kn

k=0 k=0

For n = 0:

1
x(0) =
4
[ X ( 0) e + X ( 1) e + X (2) e + X ( 3) e ]
0 0 0 0

1
x(0) =
4
[ 1.4+ ( 0.8−0.6 j )+ 0.2+ ( 0.8+0.6 j ) ]

1
x(0) = [1.4 + 0.8 + 0.2 + 0.8 − 0.6j + 0.6j] = 0.8
4

For n = 1:
1
x(1) =
4
[ 0 jπ / 2 jπ
X ( 0 ) e + X ( 1 ) e + X ( 2 ) e + X (3 ) e
j3 π / 2
]
 1 1
x(1) = [1.4+(0.8j−0.6 j 2 )−0.2−(0.8j+0.6 j 2 )] = ¿1.4 + 0.8j + 0.6 − 0.2 − 0.8j + 0.6)
4 4
=0.6
For n = 2:
1
x(2) = [X(0)e0 + X(1)ejπ + X(2)ejπ/2 + X(3)ej3π]
4
1
x(2) = [1.4(1) + (0.8−0.6j)(−1) + 0.2(1) + (0.8+0.6j)(−1)]
4
1
x(2) = (1.4 − 0.8 + 0.2 − 0.8 + 0.6j − 0.6j) = 0
4
For n = 3:
1
x(3) = [X(0)e0 + X(1)ej3π/2 + X(2)ej3π + X(3)ej9π/2]
4
1
x(3) = [1.4(1) + (0.8 − 0.6j)(−j) + 0.2(−1) + (0.8 + 0.6j)(j)]
4

1
x(3) = (1.4 − 0.8j − 0.6 − 0.2 + 0.8j − 0.6) = 0
2. Consider a digtal sequence sampled at the rate of 20kHz. If we use 𝑁 = 8000 point
4

DFT to compute the spectrum 𝑋(𝑘), determine the frequency resolution and the
frequency when 𝑘 = 2, 𝑁/2, 𝑁 − 2, 𝑁. the frequency resolution is:
fs 20000
∆f= = =2.5 Hz
N 8000

3. Compute the circular convolution of 𝑥(𝑛) and ℎ(𝑛). Compare to the linear
convolution.
a. 𝑥(𝑛) = ℎ(𝑛) = {𝟏,1,1}
circular convolution:
x(n) 1 1 1 y(n)
1 1 1 3
h(n) 1 1 1 3
1 1 1 3
y(n) = {3, 3, 3}
Linear convolution:
x(n)\h(n) 1 1 1
1 1 1 1
1 1 1 1
1 1 1 1
y(n) = {1, 2, 3, 2, 1}
The linear have length is 3 + 3−1=5
b. x(𝑛) = {𝟎,1,2,3}, ℎ(𝑛) = {𝟏,2,3,4}
circular convolution:
x(n) 0 1 2 3 y(n)
1 4 3 2 16
2 1 4 3 15
h(n)
3 2 1 4 16
4 3 2 1 10
y(n) = {16,15,16,10}

linear convolution:

h(n)\x(n) 0 1 2 3
1 0 1 2 3
2 0 2 4 6
3 0 3 6 9
4 0 4 8 12
y(n) ={0, 1, 4, 10, 16, 17, 12}
The linear length is 4 + 4−1=7

c. 𝑥(𝑛) = {𝟎,1,2,3,0,0,0}, ℎ(𝑛) = {𝟏,2,3,4,0,0,0}

The circular length is 4

circular convolution:
x(n) 0 1 2 3 0 0 0 y(n)
h(n) 1 0 0 0 4 3 2 0
 2 1 0 0 0 4 3 1
3 2 1 0 0 0 4 4
4 3 2 1 0 0 0 10
0 4 3 2 1 0 0 16
0 0 4 3 2 1 0 17
0 0 0 4 3 2 1 12
y(n) = {0, 1, 4, 10, 16, 17, 12}

linear convolution:

h(n)\x(n) 0 1 2 3 0 0 0
1 0 1 2 3 0 0 0
2 0 2 4 6 0 0 0
3 0 3 6 9 0 0 0
4 0 4 8 12 0 0 0
0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0
y(n)={0, 1, 4, 10, 16, 17, 12, 0, 0, 0, 0, 0, 0}
The linear have length is 7 + 7−1=13
The circular length is 7
4. Evaluate 𝑋(𝑘) using the decimation-in-time FFT method. Determine the number
of complex multiplications
a. x(𝑛) = {𝟏,1}
N= 2
f1(n) = x(2n) = x(0)
f2(n) = x(2n+1) = x(1)
N
X(k) = F 1 ( k )+W kN F 2 ( k ), k = −1=0
2
k=0: X ( 0 )=F1 ( 0 ) +W 02 F 2 ( 0 )
 N N
X(k+ ¿ = F 1 ( k )−W kN F2 ( k ) , k = −1=0
2 2
N=2 → X ( k +1 )=F 1 ( k )−W kN F 2 ( k )
k=0: X(1) = F 1 ( 0 )−W 02 F2 ( 0 )

X(0)=x(0)+ W 02 x (1 )=1+1=2
X(1)=x(0)−W 20 x (1 )=1−1=0
N
number of complex multiplications: log 2 ( N )=1 log 2 ( 2 )=1
2
b. x(𝑛) = {𝟎.𝟖,0.4,−0.4,−0.2}
N= 4
f1(n) = x(2n) = x(0), x(2)
f2(n) = x(2n+1) = x(1), x(3)
N
X(k) = F 1 ( k )+W kN F 2 ( k ), k =0, , −1=1
2
k=0: X ( 0 )=F1 ( 0 ) +W 04 F 2 ( 0 )
k=1: X(1) = F 1 ( 1 ) +W 14 F 2 (1 )
N N
X(k+ ¿ = F 1 ( k )−W kN F2 ( k ) , k =0,1,… , −1=1
2 2
N=2 → X ( k +2 )=F 1 ( k )−W k4 F2 ( k )
k=0: X(2) = F 1 ( 0 )−W 04 F 2 ( 0 )
k=1: X(3) = F 1 ( 1 )−W 14 F2 ( 1 )
X ( 0 )=[ x ( 0 )+ x ( 2 ) ] +W 4 [ x ( 1 )+ x ( 3 ) ] =0.4+1 ×0.2=0.6
0

X ( 1 )= [ x ( 0 ) + x ( 2 ) ] + W 4 [ x ( 1 ) −x ( 3 ) ] =0.4− j ( 0.4+ 0.2 )=0.4−0.6 j

1

X ( 2 )= [ x ( 0 ) + x ( 2 ) ] −W 4 [ x ( 1 ) + x ( 3 ) ] =0.4−0.2=0.2
1

X ( 3 ) =[ x ( 0 )−x ( 2 ) ]− W 4 [ x ( 1 ) −x ( 3 ) ] =1.2− j 0.6

1

N
number of complex multiplications: log 2 ( N )=2 log 2 ( 4 )=4
2

c. 𝑥(𝑛) = {𝟒,3,2,1,1,2,3,4}
N= 8
f1(n) = x(2n) = x(0), x(2), x(4), x(6)
f2(n) = x(2n+1) = x(1), x(3), x(5), x(7)
8
X(k) = F 1 ( k )+W kN F 2 ( k ), k =0,1,…, −1=3
2
k=0: X ( 0 )=F1 ( 0 ) +W 08 F 2 ( 0 )
k=1: X(1) = F 1 ( 1 ) +W 18 F 2 ( 1 )
k=2: X(2) = F 1 ( 2 ) +W 28 F 2 ( 2 )
k=3: X(3) = F 1 ( 3 ) +W 38 F 2 ( 3 )
N 8
X(k+ ¿ = F 1 ( k )−W kN F2 ( k ) , k =0,1,… , −1=3
2 2
N=8 → X ( k + 4 )=F 1 ( k )−W k4 F2 ( k )
k=0: X ( 4 )=F 1 ( 0 )−W 08 F 2 ( 0 )
k=1: X(5) = F 1 ( 1 )−W 18 F 2 (1 )
k=2: X(6) = F 1 ( 2 )−W 28 F 2 2
k=3: X(7) = F 1 ( 3 ) −W 38 F 2 ( 3 )
For N = 8, indices are 3-bit binary:
Decimal Binary Bit-reversed New index
0 000 000 0
1 001 100 4
2 010 010 2
3 011 110 6
4 100 001 1
5 101 101 5
6 110 011 3
7 111 111 7

0
X ( 0 )=F1 ( 0 ) +W 8 F 2 ( 0 )=10+10=20
1 2
(2 2
2 )
X(1) = F 1 ( 1 ) +W 8 F 2 ( 1 )=3+ (− j ) (−1 ) + √ − j √ ( 1+ (− j )(−3 ))

X(2) = F 1 ( 2 ) +W 28 F 2 ( 2 )=0+W 28 ⋅0=0

3
( −2√ 2 − j √22 ) ( 1−3 j )
X(3) = F 1 ( 3 ) +W 8 F 2 ( 3 )=( 3− j ) +
0
X ( 4 )=F 1 ( 0 )−W 8 F 2 ( 0 )=10−1⋅10=0

1
X(5) = F 1 ( 1 )−W 8 F 2 (1 )=( 3+ j ) − ( −2√ 2 − j √22 ) ( 1+3 j )
X(6) = F 1 ( 2 )−W 28 F 2 2=0−(− j ) ⋅0=0
3
X(7) = F 1 ( 3 ) −W 8 F 2 ( 3 )=( 3− j )− ( −2√ 2 − j √22 ) ( 1−3 j )
 N
number of complex multiplications: log 2 ( N )=4 log 2 ( 8 ) =12
2

1. Determine the 𝑧-transforms of the following signals. Find the ROC.

DIGITAL SIGNAL PROCESSING 2025 – HOMEWORK #6

a. 𝑥(𝑛) = {𝟏, 2, 3, 4}
∞ 3
X ( z )= ∑ x ( n ) z−n = ∑ x ( n ) z−n=z 0 +2 z−1+ 3 z −2 + 4 z−3
n=−∞ n=0

ROC: all except z =0

b. 𝑥(𝑛) = {1, 2, 3, 𝟒}
∞ 0
X ( z )= ∑ −n
x (n ) z = ∑ x ( n ) z−n=4 z 0+ 3 z 1 +2 z 2+1 z 3
n=−∞ n=−3

ROC: all except ∞

c. 𝑥(𝑛) = 10𝑢(𝑛)
∞ ∞
X(z)= ∑ 10u ( n ) z −n = 10 ∑ ¿ ¿
n=−∞ n=0

ROC consists of |z|¿ 1

d. 𝑥(𝑛) = 0.5𝑛𝑢(𝑛)
∞ ∞
X(z)= ∑ n
0.5 u ( n ) z −n
= ∑ ¿¿
n=−∞ n=0

ROC consists of |z|¿|0.5|

e. 𝑥(𝑛) = 10sin(0.25𝜋𝑛)𝑢(𝑛)
∞ ∞
X(z) = ∑ 10 sin ( 0.25 πn ) u ( n ) z =10 ∑ sin ( 0.25 πn ) u ( n ) z −n
−n

n=−∞ n=0
−1
z sin ( 0.25 π )
X(z) =
1−2 z−1 cos ( 0.25 π ) + z −2

ROC consists of |z|¿|1|

 f. 𝑥(𝑛) = 3𝑢(𝑛 − 4)
∞ ∞ ∞
X(z)= ∑ 3u ( n−4 ) z−n=¿ 3 ∑ u ( n−4 ) z−n=¿ ¿ ¿ 3 ∑ u ( k ) z−k−4
n=−∞ n=4 k=0

∞
1
X(z)= 3 z
−4
∑ u ( k ) z−k =3 z −4 × 1−z −1
k=0

ROC consists of |z|¿|1|

g. 𝑥(𝑛) =2(−0.5 )n u ( n )

∞ ∞
X(z)= ∑ 2¿ ¿ = 2 ∑ ¿ ¿
n=−∞ n=0

2z
X(z)=
1−(−0 ,5 z ¿¿−1)¿

ROC consists of |z|¿|−0.5|

h. 𝑥(𝑛) = −2𝑢(𝑛) − 0.75𝑛𝑢(𝑛)
∞ ∞
X1(z)= ∑ −2u ( n ) z −n
= −2 ∑ ¿ ¿
n=−∞ n=0

∞ ∞ −1
X2(z)= ∑ −0.75 u ¿ ¿ = −∑ ¿ ¿ = X(z)=
n
1−(0 ,75 z ¿¿−1)=
−z
¿
n=−∞ n=0
z −0.75
2
−z 2 z −3 z +2.5 z
X(z)= − =
z−1 z−0.75 ( z−1 ) ( z−0.75 )

ROC consists of |z|¿|1|

2. Find the inverse 𝑧-transform for the following 𝑋(𝑧) (ROC: |𝑧|>1).
−10 z z
a. X(z) = 4 −
z−1 z+ 0.5
−1
Z {4} = 4δ(n)
−1 −10 z
Z {
z−1
} = −10 × Z
−1 z
z−1 { }
=−10 u ( n )
 −1
Z {
−z
z+ 0.5
}=Z
−1 −z
{
z− (−0.5 ) }
=−(−0.5 )n u ( n )

 x(n) = 4δ(n) −10 u ( n )− (−0.5 )n u ( n )

−5 z 10 z 2z
b. X(z) = z−1 + +
( z −1 )2 ( z−0.8 )2
−5 z z
−1
Z { } =−5 × Z−1{ }=−5 ×u ( n )
z−1 z−1

−1
10 z −1
z n
Z { 2 } =10 × Z { 2 }=10 ×n ×1 ×u ( n )
( z−1 ) ( z−1 )
2z −1 2.5 ×0.8 z −1 0.8 z
−1
Z { 2} =
Z 2
=2.5 × Z { n
2 }=2.5 ×n × ( 0.8 ) ×u ( n )
( z−0.8 ) ( z−0.8 ) ( z−0.8 )

 x ( n )=−5 u ( n ) +10 n ( 1 )n u ( n ) +2.5 n ( 0.8 )n u ( n )

−4 −1 −5
4z z −8 z
c. X ( z )= + 2
+z +
z−1 ( z−1 ) z−0.5
Z { z X ( z )}= x(n−n 0)
−1 −n 0

−4
4z −1 −5 z
−1
Z { }=4 × Z z =4 u ( n−5 )
z−1 z−1

{ }
−1
z −1 −2 z ( n−2 )
−1
Z { 2 }=
Z z 2
= ( n−2 ) 1 u ( n−2 )=( n−2 ) u ( n−2 )
( z−1 ) ( z−1 )
Z { z ×1} =δ ( n−8 )
−1 −8

 x ( n )=4 u ( n−5 )+ ( n−2 ) u ( n−2 )+ δ ( n−8 )

1
d. X ( z )=
z 2−0.3 z−0.04
X ( z) 1 −25 5 20
= = + +
z z ( z −0.4 ) ( z +0.1 ) z ( z−0.4 ) ( z +0.1 )
5z 20 z
X(z)= −25+ +
( z−0.4 ) ( z +0.1 )
−1
Z {−25}¿−25 δ (n)
5z −1 z
−1
Z { }= 5 × Z−1{ z }= 5 ( 0.4 )n u ( n )
( z−0.4 ) ( z−0.4 )
20 z z
−1
Z { }= 20 × Z−1{ }= 20 (−0.1 )n u ( n )
( z+ 0.1 ) ( z+ 0.1 )
 x ( n )=¿ 5 ( 0.4 )n u ( n )+20 ( 0.1 )n u ( n ) −25 δ (n)
z
e. X(z)=
( z−0.2 ) ( z+ 0.4 )
 X ( z) 1 5 5
= = −
z ( z−0.2 ) ( z+ 0.4 ) 3 ( z −0.2 ) 3 ( z +0.4 )
5z 5z
X(z)= −
3 ( z −0.2 ) 3 ( z +0.4 )
5z 5 n
−1
Z { }= ( 0.2 ) u ( n )
3 ( z −0.2 ) 3
−5 −5 n
−1
Z { }= (−0.4 ) u ( n ) x ( n )
3 ( z +0.4 ) 3
5 n 5 n
x(n)= ( 0.2 ) u ( n )− (−0.4 ) u ( n )
3 3
z ( z+ 0.5 )
f. X(z) =
( z−0.1 ) 2 ( z−0.6 )
X ( z) ( z +0.5 ) −83.33 7.33 76 12
= = z + z−0.6 + z−0.1 −
z 2
z ( z−0.1 ) ( z−0.6 ) ( z−0.1 )2

−83.33 7.33 z 76 z 12 z
X(z) = + + −
1 z−0.6 z−0.1 ( z−0.1 )2
Z { −83.33 }= −83.33 δ ( n )
−1

7.33 z
−1
Z { }= 7.33 ( 0.6 )n u ( n )
z−0.6

76 z
−1
Z { }= 76 ( 0.1 )n u ( n )
z−0.1

−12 z
−1
Z { }= −12 n ( 0.1 )n u ( n )
( z−0.1 )2
x(n)= −83.33 δ ( n )−7.33 ( 0.6 )n u ( n ) +76 ( 0.1 )n u ( n ) +12 n ( 0.1 )n u ( n )
3. Find transfer functions 𝐻(𝑧) of the following systems.
a. 𝑦(𝑛) = 0.5𝑥(𝑛) − 0.5𝑥(𝑛 − 1)
Z{y(n)}=Z{0.5x(n)}−Z{0.5x(n−1)}
Y(z)=0.5X(z)−0.5Z{x(n−1)}
We have Z{x(n−n0)}=z−n0X(z)
For a delay of one sample (n0=1), this becomes Z{x(n−1)}=z−1X(z)
Y(z)=0.5X(z)−0.5z−1X(z)
Y(z)=X(z)(0.5−0.5z−1)
 Y (z)
H(z)= =0.5−0.5z−1
X ( z)

b. 𝑦(𝑛) = 𝑥(𝑛) − 0.5𝑦(𝑛 − 1)

Z{y(n)}=Z{x(n)}−Z{0.5y(n−1)}
We have Z{x(n−n0)}=z−n0X(z)
For a delay of one sample (n0=1), this becomes Z{y(n−1)}=z−1Y(z)
Y(z) = X(z) − 0.5 z−1 Y ( z )
X(z)=Y(z)(1+ 0.5 z−1 ¿
Y (z)
H(z)= =1+ 0.5 z−1
X ( z)
c. 𝑦(𝑛) = 𝑥(𝑛) − 0.25𝑥(𝑛 − 2) − 1.1𝑦(𝑛 − 1) − 0.28𝑦(𝑛 − 2)
Z{y(n)}=Z{x(n)}−Z{0.25𝑥(𝑛 − 2)} − Z{1.1𝑦(𝑛 − 1) }− Z{0.28𝑦(𝑛 − 2)}
We have Z{x(n−n0)}=z−n0X(z)
Y(z)=X(z) − 0.25 z−2 X ( z ) −1.1 z −1 Y ( z ) −0.28 z−2 Y ( z )
 Y(z)(1+1.1 z−1+ 0.28 z −2)=X(z)(1− 0.25 z−2)
Y (z) 1−0.25 z
−2
 H(z)= =
X ( z ) 1+1.1 z −1+ 0.28 z −2
d. 𝑦(𝑛) = 0.25𝑥(𝑛 − 2) + 0.5𝑦(𝑛 − 1) − 0.2𝑦(𝑛 − 2)

Y(z) = 0.25z-2X(z) + 0.5z-1Y(z) − 0.2z-2Y(z)

Y(z) − 0.5z-1Y(z) + 0.2z-2Y(z) = 0.25z-2X(z)

Y(z)[1 − 0.5z-1 + 0.2z-2] = 0.25z-2X(z)

Y (z) 0.25 z
−2
0.25
 H(z)= = = 2
( ) −1
X z 1−0.5 z +0.2 z−2
z −0.5 z +0.2
4. Convert the following transfer functions into the different equations.
1
a. H ( z )= −1
1−0.3 z
Y ( z) 1
H ( z )= =
X ( z ) 1−0 , 3 z−1

We have Z{x(n−n0)}=z−n0X(z)
Y(z)−0 , 3 z−1 Y ( z )=X ( z )
 y(n) −0 , 3 y ( n−1 )=x ( n )
  y(n)=x(n)+ 0 , 3 y ( n−1 )
b. H(z)=0.1+0.2 z−1 +0.3 z−2
Y ( z)
H ( z )= = 0.1 + 0.2z⁻¹ + 0.3z⁻²
X (z )
We have Z{x(n−n0)}=z−n0X(z)
Y(z) = X(z) · (0.1 + 0.2z⁻¹ + 0.3z⁻²)
Y(z) = 0.1X(z) + 0.2z⁻¹X(z) + 0.3z⁻²X(z)
 y(n) = 0.1x(n) + 0.2x(n−1) + 0.3x(n−2)
2 −2
z −0.25 1−0.25 z
c. H(z)= 2
= −1 −2
z +1.1 z+ 0.18 1+1.1 z +0.18 z
Y (z) 1−0.25 z
−2
H(z)= =
X ( z ) 1+1.1 z −1+ 0.18 z −2
Y(z)+Y(z) 1.1 z−1+Y ( z ) 0.18 z−2=X ( z )−X ( z ) 0.25 z −2
We have Z{x(n−n0)}=z−n0X(z)
 y(n)+1.1y(n−1 ¿+0.18 y (n−2)=x (n)−0.25 x (n−2)
 y(n)= x (n)−0.25 x (n−2)¿−1.1 y (n−1)−0.18 y (n−2)
2
z −0.1 z +0.3
d. H(z)= 3
z
Y ( z ) −1
H(z)= =z −0.1 z −2 +0.3 z−3
X ( z)
Y(z)=X(z) z−1− X ( z ) 0.1 z −2+ X ( z ) 0.3 z−3
We have Z{x(n−n0)}=z−n0X(z)
 y(n)=u(n−1 ¿−0.1u(n−2)+0.3 u(n−3)
5. Sketch the z-plane pole-zero plot and determine the stability of the system
2
z +0.25
a. H ( z )=
( z −0.5 ) ( z 2+3 z +2.5 )
Numerator (Zeros): z2+0.25=0
 z=± j0.5
So the zeros are at:
z=+j0.5z
z=−j0.5z
Denominator (Poles):(z−0.5)( z 2+3z+2.5)=0

{
z=0.5
 z=1.5 ± j
2
Pole at:

{
z=0.5 ,|z|<1
j
z=1.5 ± ,|z|=1.58>1 ( outsideunitcircle )
2
 System is not stable
z−0.5
b. H(z)=
( z+ 0.25 ) ( z 2+ z+ 0.5 )
Numerator (Zeros): z−0.5 =0
 z=0.5
Denominator (Poles): ( z +0.25 ) ( z 2+ z +0.5 )=0

{
z =−0.5
 j
z=−0.5 ±
2

Pole at:

{
z =−0.5 ,|z|<1
j
z=−0.5 ± ,| z|=0.707< 1
2
  System is stable
2
z + 2 z +1
c. H(z)= 2
z +5 z +6

Numerator (Zeros): z 2+ 2 z +1= 0

 z=−1(double)
Denominator (Poles): z 2+ 5 z +6 =0
 {z=−2
z=−3
Pole at:

{ z=−2 ,|z|>1
z=−3 ,|z|>1
 System is unstable

2
z + 4 z+5
d. H(z) = 3 2
z +2 z + 6 z

Numerator (Zeros): z 2+ 4+ 5= 0
 z=−2 ± j
Denominator (Poles): z 3 +2 z 2 +6 z =0
 {z=−1z=0± j √ 5
Pole at:

{ z=0 ,|z|<1
z=−1 ± j √ 5 ,|z|=2.449>1
 System is unstable

DIGITAL SIGNAL PROCESSING 2025 – HOMEWORK #7

1. Given the following FIR filter y ( n )=0.1 x (n)+0.25 x (n−1)+0.2 x (n−2). Determine
the transfer function, filter length, nonzero coefficients, and impulse response.
Solution:
Transfer function:
y ( n )=0.1 x ( n ) +0.25 x ( n−1 ) +0.2 x (n−2)
Y (z)
¿> H ( z )= =0.1+0.25 z−1+ 0.2 z −2
X ( z)
Filter length: N = 3
Nonzero coefficients:
a={ 1 } (Coefficients of y(n))
b={ 0.1 , 0.25 , 0.2 } (Coefficients of x(n))
Impulse response:
Replace h(n) for y(n) andδ (n) for x(n)
=> h ( n )=0.1 δ (n)+ 0.25 δ(n−1)+ 0.2 δ (n−2)

{
δ ( n )= 1 ,n=0
0,n≠0
We take k from 0 to 2 so that we calculate h(n) from 0 to 2 with result different from 0
and
n from 3 to infinite the result will be 0
n=0=¿ h ( 0 )=0.1n=1=¿ h ( 1 )=0.25n=2=¿ h ( 2 ) =0.2
h ( n )={ 0.1 , 0.25 ,0.2 }

2. Design a three-tap lowpass FIR filter with a cutoff frequency of 800Hz, and a
sampling rate of 8000Hz. Determine the transfer function and difference equation of
the system.
Using Rectangular window
Solution:
f c =800 Hz
f s=8000 Hz
f c ×2 π 800 ×2 π
ωc= = =0.2 π
fs 8000

{
ω0

{
, n=0 0.2, n=0
h ( n ) = π =¿ h ( n ) =
Lowpass filter: sin ( 0.2 πn )
sin ( ω 0 n ) ,n≠0
,n≠0 πn
πn
Choose the filter length N = 3
n=0=¿ h ( 0 )=0.2n=1=¿ h ( 1 )=0.187n=−1=¿ h (−1 ) =0.187
¿> h ( n )=[0.187 ,0.2 , 0.187]
To achieve causality
'
(
h ( n )=hd n−
N−1
2 ) =hd ( n−1 )

From n=−1 ,0 ,1 we calculated h' ( n ) at n = 0, 1, 2

=> h ' ( n )=[0.187 , 0.2 , 0.187]

Case using rectangular window, so we have ω R = {10,,nOtherwise

≤ 0≤ N −1

n=0=¿ ω R (0)=1n=1=¿ ω R ( 1 )=1n=2=¿ ω R ( 2 )=1

=>ω R =[1 ,1 , 1] (2)
h ' ( n )=[0.187 , 0.2 , 0.187] (1)
From (1) and (2) we will calculate the h1 ( n )=h' ( n ) × ω R ( n )=[0.187 , 0.2, 0.187 ]
−1 −2
¿> H eff 1 ( z )=0.187+0.2 z + 0.187 z
¿> y eff 1 (n)=0.187 x ( n ) +0.2 x ( n−1 ) +0.187 x (n−2)

3. Design a three-tap highpass FIR filter with a cutoff frequency of 1000Hz, and a
sampling rate of 8000Hz, using Hamming window. Determine the transfer function
and difference equation of the system.
Solution:
f c =1000 Hz
f s=8000 Hz
f c ×2 π 1000 ×2 π
ωc= = =0.25 π
fs 8000

{
ω0

{
1− , n=0 0.75 , n=0
h ( n ) = π =¿ h ( n ) =
Highpass filter: −sin ( 0.25 πn )
−sin ( ω 0 n ) , n≠ 0
,n ≠ 0 πn
πn
Choose the filter length N = 3
n=0=¿ h ( 0 )=0.75n=1=¿ h ( 1 )=−0.225n=−1=¿ h (−1 ) =−0.225
=>h ( n )=[−0.225 , 0.75 ,−0.225]
To achieve causality
h' ( n )=hd n−(N−1
2 )
=hd ( n−1 )

From n=−1 ,0 ,1 we calculated h' ( n ) at n = 0, 1, 2

=> h ' ( n )=[−0.225 , 0.75 ,−0.225]
Case using Hann (Hanning) window, so we have

{
2 πn
0.5−0.5 cos ⁡( ), 0 ≤n ≤ N−1
ω hann= N −1
0 , Otherwise
n=0=¿ ωhann (0)=0n=1=¿ ω hann ( 1 )=1n=2=¿ ω hann ( 2 )=2−2=0
=>ω hann=[0 ,1 , 0] (3)
h ' ( n )=[−0.225 , 0.75 ,−0.225] (1)
From (1) and (3) we will calculate the h2 ( n ) =h' ( n ) × ωhann ( n )=[0 , 0.75 , 0]
¿> heff 2 ( n )=[0 ,0.75 ]
−1
¿> H eff 2 ( z )=0.75 z
¿> y eff 2 (n)=0.75 x ( n−1 )

4. Design a five-tap bandpass FIR filter with a lower cutoff frequency of 2000Hz, an
upper cutoff frequency of 2400Hz, and a sampling rate of 8000Hz, using Hann
window. Determine the transfer function and difference equation of the system.
Solution:
f cl=2000 Hz
f cu=2400 Hz
f s=8000 Hz
f cl ×2 π 2000 ×2 π
ω cl= = =0.5 π
fs 8000
f ch ×2 π 2400 ×2 π
ω cu= = =0.6 π
fs 8000
 {
ω cu−ω cl

{
,n=0 0.1 , n=0
π
Bandpass filter: h ( n )= =¿ h ( n )= sin ( 0.6 πn ) −sin ( 0.5 πn )
sin ( ω cu n ) −sin ( ω cl n ) , n≠ 0
,n≠0 πn
πn
Choose the filter length N = 5
n=−2=¿ h (−2 )=−0.0935
n=−1=¿ h (−1 ) =−0.0156
n=0=¿ h ( 0 )=0.1
n=1=¿ h ( 1 )=−0.0156
n=2=¿ h ( 2 ) =−0.0935
=>h ( n )=[−0.0935 ,−0.0156 , 0.1 ,−0.0156 ,−0.0935]
To achieve causality
h' ( n )=hd n−( N−1
2 )
=hd ( n−1 )

From n = -2, -1, 0, 1, 2we calculated h' ( n ) at n = 0, 1, 2, 3, 4

=> h ' ( n )=[−0.0935 ,−0.0156 , 0.1 ,−0.0156 ,−0.0935]
Case using Hann (Hanning) window, so we have

{
2 πn
0.5−0.5 cos ⁡( ), 0 ≤n ≤ N−1
ω hann= N −1
0 , Otherwise
n=0=¿ ωhann ( 0 )=0n=1=¿ ω hann ( 1 )=0.5n=2=¿ ω hann ( 2 )=1n=3=¿ ωhann ( 3 )=0.5
n=4=¿ ω hann ( 4 )=0
=>ω hann=[0 , 0.5 ,1 , 0.5 , 0] (3)
h ' ( n )=[−0.0935 ,−0.0156 , 0.1 ,−0.0156 ,−0.0935](1)
From (1) and (3) we will calculate the h2 ( n ) =h' ( n ) × ωhann ( n )
h2 ( n ) =[0 ,−0.0078 ,0.1 ,−0.0078 , 0]
¿> heff 2 ( n )=[0 ,−0.0078 , 0.1 ,−0.0078 , 0]
−1 −2 −3
¿> H eff 2 ( z )=−0.0078 z +0.1 z −0.0078 z
=> y eff 2=−0.0078 x ( n−1 ) +0.1 x ( n−2 ) +0.0078 x (n−3)

5. Design a five-tap bandstop FIR filter with a lower cutoff frequency of 3000Hz, an
upper cutoff frequency of 3600Hz, and a sampling rate of 8000Hz, using Blackman
window. Determine the transfer function and difference equation of the system.
Solution:
f cl=3000 Hz
f cu=3600 Hz
f s=8000 Hz
f cl ×2 π 3000 ×2 π
ω cl= = =0.75 π
fs 8000
f cu ×2 π 3600 ×2 π
ω cu= = =0.9 π
fs 8000
 {
ωcu −ω cl

{
1− , n=0 0.85 , n=0
π
Bandpass filter: h ( n )= =¿ h ( n )= sin ( 0.75 πn )−sin ( 0.9 πn )
sin ( ω cl n ) −sin ( ω cu n ) ,n ≠ 0
,n≠0 πn
πn
Choose the filter length N=5
n=−2=¿ h (−2 )=−0.0656
n=−1=¿ h (−1 ) =0.127
n=0=¿ h ( 0 )=0.85
n=1=¿ h ( 1 )=0.127
n=2=¿ h ( 2 ) =−0.0656
¿> h ( n )=[−0.0656 , 0.127 , 0.85 ,0.127 ,−0.0656 ]
To achieve causality

(
h' ( n )=hd n−
N−1
2 )
=hd ( n−1 )

From n = -2, -1, 0, 1, 2we calculated h' ( n ) at n = 0, 1, 2, 3, 4

h ' ( n )=[−0.0656 , 0.127 , 0.85 , 0.127 ,−0.0656]
Case using blackman window, so we have

{ ( N2 −1 )+0.08 cos ( N4−1 ) , n ≤ 0≤ N−1

πn πn
0.42−0.5 cos
ω B=
0 ,Otherwise
n=0=¿ ωB ( 0 )=0n=1=¿ ω B ( 1 )=0.34n=2=¿ ω B ( 2 )=1n=3=¿ ωB ( 3 ) =0.34n=4=¿ ω B ( 4 )=0
ω B= [ 0 ,0.34 ,1 , 0.34 , 0 ] (3)
h ' ( n )=[−0.0656 , 0.127 , 0.85 , 0.127 ,−0.0656] (1)
From (1) and (3) we will calculate the h2 ( n ) =h' ( n ) × ω B ( n )
h2 ( n ) =[0 , 0.04318 , 0.85 ,0.04318 ,0 ]
¿> heff 2 ( n )=[0 ,0.04318 ,0.85 , 0.04318 , 0]
−1 −2 −3
¿> H eff 2 ( z )=0.04318 z +0.85 z −0.04318 z
=> y eff 2 ( n )=0.04318 x ( n−1 )+ 0.85 x ( n−2 )−0.04318 x (n−3)

DIGITAL SIGNAL PROCESSING 2025 – HOMEWORK #8

1. Given the following IIR filter y (n)=0.2 x (n)+0.5 y (n−1). Determine the
transfer function, nonzero coefficients, and impulse response.
Solution:
=> y ( n )−0.5 y ( n−1 )=0.2 x ( n )
=>Y ( z )−0.5 z−1 Y ( z )=0.2 X ( z )
Transfer function:
Y (z) 0.2
¿> =H ( z )=
X ( z) 1−0.5 z−1
Impulse response: ¿> h ( n )=0.2 × ( 0.5 )n × u(n)
Nonzero coefficients: a={ 1 ,−0.5 } ; b={ 0.2 }
 Ωp
2. Transform the single pole lowpass Butterworth filter H ( s )= into:
s +Ω p

Solution:
a. Highpass filter with a cutoff frequency Ωc
Ωp s
H 1 ( s )= =
Ωp Ωc Ωc + s
+Ω p
s

b. Bandpass filter with upper and lower band edge frequencies Ωu and Ωl respectively
Ωp 1 S(Ω u−Ωl )
H 2 ( s )= 2
= 2 = 2
S +Ω l Ω u S +Ωl Ωu S + S ( Ω u−Ωl ) +Ω l Ωu
Ωp +Ω p +1
S(Ω u−Ωl ) S (Ωu −Ωl )

c. Bandstop filter with upper and lower band edge frequencies Ωu and Ωl respectively
Ωp
H 3 ( s )=
Ωp S ¿ ¿ ¿
2
S +Ω u Ωl
¿ 2
S + S ( Ω u−Ωc ) +Ω c Ωu

100
3. Given an analog filter H ( s )= . Convert it to the digital transfer function and
s +100
difference equation, using a sampling period T = 0.01s and bilinear transformation
design

Solution:
100
Lowpass : H ( s )= ; T =0 , 01(s)
s +100

Transfer function:

100 100 (z+ 1)

¿> H (z)= =
( )
2 z−1
T z+ 1
+100
20 0 ( z−1 ) +100 (z+ 1)

1 1 −1
+ z
−1
1 00+100 z 3 3
¿> H (z)= −1 =
3 00−100 z 1
1− z −1
3

Difference equation:
 1 1 1
¿> y (n)= y (n−1)+ x (n)+ x (n−1)
3 3 3

1
4. Use the normalized lowpass filter with a cutoff frequency of 1 rad/s H ( s )= ,
s +1
bilinear
transformation and a sampling rate of 90Hz to design digital IIR filters:
a. Lowpass filter with a cutoff frequency of 15Hz
b. Highpass filter with a cutoff frequency of 15Hz
c. Bandpass filter with a lower cutoff frequency of 10Hz, an upper cutoff frequency
of 30Hz
d. Bandstop filter with a lower cutoff frequency of 10Hz, an upper cutoff frequency
of 30Hz
Solution:
a. Lowpass
f c =15 Hz
fc 15 π
¿> ωc =2 π =2 π =
fs 90 3
2
( ) ω
Ωc = tan c =180 tan
T 2 ( ) π
3 ×2
≈ 103,923

1 Ωc 103,923
H ( s )= = =
s s +Ωc s+103,923
+1
Ωc
−1
103,923 103,923 z +103,923
¿> H ( z )= =
( )
−1
z−1 283 , 923−76,077 z
180 × + 103,923
z +1
−1
0,366+ 0,366 z
¿> H ( z )= −1
1−0,268 z
¿> y (n)=0,366 x (n)+0,366 x (n−1)+0,268 y ( n−1)

b. Highpass
f c =15 Hz
fc 15 π
¿> ωc =2 π =2 π =
fs 90 3
2
( )ω
Ωc = tan c =180 tan
T 2 ( )
π
3 ×2
≈ 103,923
1 s
H ( s )= =
Ωc Ωc + s
+1
s
 ¿> H ( z )=
180 × ( z +1 )
z−1

103 , 923+180 × (
z+1 )
z−1

180 z −180
¿> H ( z )=
103 , 923 z+103,923+ 180 z−180
180 z−180
¿> H ( z )=
283 , 923 z−76,077
−1
( ) 0,634−0,634 z
¿> H z = −1
1−0,268 z
¿> y (n)=0,634 x (n)−0,634 x (n−1)+0,268 y (n−1)

c. Bandpass
ωu × f s fu 15 2π
f u=30 Hz=¿ f u= ¿> ωu= ×2 π= × 2 π =
2π fs 90 3
ωl × f s fl 10 2π
f l=10 Hz=¿ f l= ¿> ωl= × 2 π= ×2 π =
2π fs 90 9

{
¿> u
Ω ≈ 311, 77
Ωl ≈ 65 ,51
1 246 , 26 s
H ( s )= 2 = 2
s +20424 s +246 , 26 s +20424
+1
s × 246 ,26

¿> H ( z )=
( )
246 , 26 ×180
z −1
z+1

( )
2
( z−1 ) z−1
1802 +246 , 26 ×180 +20424
( z +1 )
2
z +1
−2
0 , 46−0 , 46 z
¿> H ( z )= −1 −2
1−0 , 25 z −0 , 09 z
¿> y (n)=0 , 46 x (n)−0 , 46 x (n−2)+0 , 25 y (n−1)−0 , 09 y (n−2)

d. Bandstop
ωu × f s fu 15 2π
f u=30 Hz=¿ f u= ¿> ωu= ×2 π= × 2 π =
2π f s 90 3
ωl × f s fl 10 2π
f l=10 Hz=¿ f l= ¿> ωl= × 2 π= ×2 π =
2π fs 90 9

{
¿> u
Ω ≈ 311, 77
Ωl ≈ 65 ,51
 2
1 s +20424
¿> H ( s ) = = 2
s ×246 , 26
2
+1 s +20424 +246 , 26 s
s +20424
( z−1 )2
1802 × 2
+20424
( z +1 )
¿> H ( z )= 2
( z −1 )
1802 ×
( z+1 )
2
+246 , 26 ×180
z−1
( )
z +1
+20424
−2
0 , 54−0 ,25 z +0 , 54 z
( )
⇒H z = −2
1−0 , 25 z−0 , 09 z
¿> y (n)=0 , 54 x (n)−0 , 25 x (n−1)+0 , 54 x (n−2)+ 0 ,25 y (n−1)−0 , 09 y (n−2)