SOLUTION MANUAL Digital Filters Using MATLAB 1st edition by Lars Wanhammar, Tapio Saramäki

, L. Wanhammar, H. Johansson and T. Saramäki: Digital Filters Using MATLAB 1

2019-05-16

∞ 0 ∞ 0 ∞
X ( jω ) = – c t e – jωt dt = ct e – jωt dt + – ct e – jωt dt = ( c – jω )t dt + – ( c + jω ) t dt
∫–∞ e ∫–∞ e ∫0 e ∫–∞ e ∫0 e
1.1 0 ∞
e ( c – jω )t e – ( c + jω ) t 1 1 2c
= ------------------ + ----------------------- = ------------ + --------------- = ------------------
c – jω –∞
– ( c + jω ) 0
c – jω c + jω ω + c2
2


Tp Tp
------ ------ ωT
∞ 2 2 sin ⎛ ---------p-⎞ ωT
Ae – jωt ⎝ 2 ⎠
1.2 X ( jω ) = ∫ x ( t )e
– jωt dt = ∫ Ae – jωt dt = ---------------- = AT p ------------------------ = AT p sinc( ---------p- ⎞
– jω ωT 2 ⎠
–∞ T T
– -----p- ---------p-
– -----p- 2 2
2

1.3 The function is periodic with period T and will therefore correspond to a discrete spectra, i.e., a Fourier
series, with components ωn = 2πn/T rad/s.
∞
1 – jω 0 nt AT nω 0 T p ⎞
X ( jω n ) = --- ∫ x ( t )e dt = ---------p- sinc( ---------------
- , n = 0, ±1, ±2, ....
T T 2 ⎠
–∞


∞
1
1.4 a) X(z) = 1 b) X ( z ) = ∑ z – n = ---------------- (geometric series which converge for z – 1 < 1
n=0
1 – z –1
The Fourier transform is obtained by replacing z with e jωT
N
1 – aN + 1
For a finite geometric series we have: ∑ a n = ----------------------- for a < 1
1–a
n=0
–1 ∞
–1
c) X ( z ) = – ∑ z –n = – ∑ z k = ----------- for z < 1
1–z
n = –∞ k=1
4
1 – z –8
d) X ( z ) = ∑ z –n = z 2 + z 2 + … + z – 3 + z – 4 = z 3 ( 1 + z – 1 + z – 2 … + z – 7 ) = z 3 ----------------
n = –3
1 – z –1
∞
1 1
e) X ( z ) = 3 – n z – n = ------------------------ for z > ---
∑ 1 – ( 3z ) –1 3
n=0
f)
∞ –1 ∞
2 –1 z 1 – 1.5z
X(z) = ∑ 2 – n z –n = ∑ 2 n z –n + ∑ 2 – n z – n = -------------------- + ------------------------ = ---------------------------------------------
1 – 2 – 1 z 1 – ( 2z ) – 1 ( 1 – 0.5z ) ( z – 0.5 )
n = –∞ n = –∞ n=0


1.5 a) x ( n ) = δ ( n ) b) x ( n ) = δ ( n + 3 ) c) x ( n ) = – 2 δ ( n – 2 ) + 2δ ( n + 1 ) + δ ( n )

N N+1 N N+1
1.6 We have X = ∑ a n and aX = ∑ a n . Hence, X – aX = ∑ a n – ∑ a n = 1 – a ( N + 1 ) and
n=0 n=1 n=0 n=1

, L. Wanhammar and H. Johansson: Digital Filters Using MATLAB 2

1 – a(N + 1) 1
X = -------------------------- . Moreover, X = ------------ for |a| <1 and N → ∞.
1–a 1–a

0.5z 0.5z 1 0.5 z –1 0.5z – 1
X ( z ) = ----------------------------------- = ------------------------------------- = ---------------- + -------------------- = --------------------- + -----------------------------
1.7 a) z 2 – 1.5z + 0.5 ( z – 1 ) ( z – 0.5 ) ( z – 1 ) ( z – 0.5 ) ( 1 – z – 1 ) ( 1 – 0.5z – 1 )
∞ ∞ ∞
= z –1 ∑ z – n – 0.5z – 1 ∑ 0.5 n z – n = ∑ ( 1 – 0.5 n )z – n , z >1
n=0 n=0 n=1
Hence, x ( nT ) = 1 – 0.5 n for n ≥ 0 and 0 otherwise.
b) We have x(n) = 0 for n < 0 and for n ≥ 0
1 1 0.5z n
x ( nT ) = -------- X ( z )z n – 1 dz = -------- ------------------------------------- dz = { Residues at z = 1 ans z = 0.5 }
j2π °C∫ j2π ( z – 1 ) ( z – 0.5 ) °C∫
0.5 0.5z n + 1
= ---------------- + -------------------- = 1 – 0.5 n
1 – 0.5 0.5 – 1

∞
1.8 y ( nT ) = h ( nT )*x ( nT ) = ∑ h ( nT – kT )x ( kT )
k = –∞
We get

k
0 y(0) = 0
1 y(1) = 0
2 y(2) = h(T)x(T) = 2
3 y(3) = h(2T)x(T) + h(T)x(2T) = 5
4 y(4) = h(3T)x(T) + h(2T)x(2T) + h(T)x(3T) = 6
5 y(5) = h(4T)x(T) + h(3T)x(2T) + h(2T)x(3T) + h(T)x(4T) = 2
6 y(6) = 0

∞
b) We have H ( z ) = ∑ h ( nT )z – n = 2z – 1 + z – 2 z ≠ 0 and
n = –∞

∞
X(z) = ∑ x ( nT )z – n = z – 1 + 2z – 2 + 2z – 3 z≠0
n = –∞

Y ( z ) = ( 2z – 1 + z – 2 ) ( z – 1 + 2z – 2 + 2z – 3 ) = 2z – 2 + 5z – 3 + 6z – 4 + 2z – 5
Hence,

2 n = 2
⎧
⎪5 n = 3
⎪
y ( nT ) = ⎨ 6 n = 4
⎪
⎪2 n = 5
⎩0 otherwise

, L. Wanhammar, H. Johansson and T. Saramäki: Digital Filters Using MATLAB 3


1.9 f1 = 100 Hz, f2 = 300 Hz, f3 = 500 Hz, and f4 = 750 Hz where A1 = 1, A2 = 0.5, A3 = 0.3, and A4 = 0.2.
a) The sampling theorem yields fsample min = 2fmax = 1500 Hz.
b) f3 appear at the frequency 1200 – 750 = 4 50 Hz

1 |X(jω)|


2πf
[kHz]
0.5 1 1.5

|X(ejωΤ)|
1/T

ωT
π 2π
[rad]

c) fsample = 1200 Hz, Amax = 0.1 dB => |H(ejωcT)| ≈ 0.99 and Amin = 40 dB => |H(ejωsT)| ≈ 0.01
The amplitude of A3 is 0.2•0.01 = 0.002, i.e., almost no folding occur.

2c 6
1.10 The analog spectrum is X ( jω ) = ------------------ = ---------------- . After sampling we get
ω2 + c2 ω2 + 9
∞ ∞
1 j2πk 1 6 6⁄T 6⁄T
X ( e jωT ) = --- ∑ X ⎛ jω – -----------⎞ = --- ∑ ------------------------------------- = … ------------------------------------- + -------------------------------------…
⎝ T ⎠
T
k = –∞
T 2πk
k = – ∞ ⎛ --------- – ω⎞ + 9
2
⎛ ---------- – ω⎞ + 9 ⎛ ---------- – ω⎞ 2 + 9
2π0 2 2π1
⎝ T ⎠ ⎝ T ⎠ ⎝ T ⎠
30 30
= … ---------------- + ------------------------------------…
ω 2 + 9 ( 20π – ω ) 2 + 9
at, for example, ω = 3 rad/s (3/2π Hz) the magnitude of the two terms are 1.6666 and 0.0083592,
respectively. Hence, the aliasing is negligible.

2f 2f L 29.4 16.6
1.11 We have from Equation(1.55) -------H- ≤ f sample ≤ -----------
- . We get ---------- ≤ f sample ≤ ------------ and for different n
n n–1 n n–1
we get
n=1 29.4 MHz ≤ fsample
n=2 14.7 MHz ≤ fsample ≤ 16.3 MHz
n=3 9.8 MHz ≤ fsample ≤ 8.3 MHz

H f
where n is an integer that satisfy: 1 ≤ n ≤ ---------------
- , i.e., 1 ≤ n ≤ 2.2968
fH – fL
Hence, we may select n = 2 and 14.7 MHz ≤ fsample ≤ 16.3 MHz. For example, fsample = 14.7 MHz and
the base band will be from 0 to 6.4 MHz.

1.12 Case 1: Passband = stopband edge. Hence, infeasible filter.
Case 2:Passband edge, fc = 30 MHz and stopband edge: fs = 60 MHz.
Case 3: Passband edge, fc = 40 MHz and stopband edge: fs = 80 MHz.