EE 590 B (PMP): Numerical Examples #3
Laplace and z- Transforms.
Solutions
Wednesday, November 2, 2016
Tamara Bonaci
Department of Electrical Engineering
University of Washington, Seattle
Problem 1 – Partial Fractions I
Represent the following fraction using the partial fraction decomposition:
3x + 11
−x−6
x2
Solution:
To represent the given fraction using the partial fraction decomposition, we start as follows:
3x + 11
x2 − x − 6
=
=
3x + 11
A
B
=
+
(x − 3)(x + 2)
x−3 x+2
(A + B)x + (2A − 3B)
x2 − x − 6
(1)
Equating the corresponding terms in equation (1), we get the following system of equations:
A+B
=
3
2A − 3B
=
11
(2)
Solving the given system of equations (2) yields A = 4 and B = 1, so we can rewrite the starting
fraction as:
3x + 11
4
1
=
−
2
x −x−6
x−3 x+2
Problem 2 – Partial Fraction II
Represent the following fraction using the partial fraction decomposition:
3x3
x2 + 4
+ 4x2 − 4x
1
Solution:
To represent the given fraction using the partial fraction decomposition, we start from the following
general formula:
x2 + 4
x2 + 4
A
B
C
=
= +
+
3
2
3x + 4x − 4x
x(x + 2)(3x − 2)
x
x + 2 3x − 2
(3)
From equation (3), we can obtain the following system of equations:
x2 + 4 = A(x + 2)(3x − 2) + Bx(3x − 2) + Cx(x + 2)
(4)
We can simplify equation (4) as follows:
x2
=
(3A + 3B+ C)x2
0
=
(4A − 2B + 2C)x
4
= −4A
(5)
From the system of equations (5), we get:
A
B
C
= −1
1
=
2
5
=
2
(6)
Problem 3 – Partial Fraction III
Represent the following fraction using the partial fraction decomposition:
x2 − 29x + 5
(x − 4)2 (x2 + 3)
Solution:
To represent the given fraction using the partial fraction decomposition, we again start from the general
formula:
num
A1
A2
Ak
=
+
+ ··· +
(ax + b)k
(ax + b) (ax + b)2
(ax + b)k
(7)
Applying formula (7) to the given fraction, we can write:
x2 − 29x + 5
(x − 4)2 (x2 + 3)
A
B
Cx + D
+
+ 2
x − 4 (x − 4)2
x +3
=
(8)
Sorting out equation (8), we obtain the following system of equations:
0
=
(A + C)x3
2
=
(−4A + B − 8C + D)x2
−29x
=
(3A + 16C − 8D)x
x
5
= −12A + 3B + 16D
Problem 4 – Inverse Fourier Transform
Find the inverse Fourier Transform of:
X(jω) = ejω [sinc(ω − π) + sinc(ω + π)]
2
(9)
Solution:
From the table of Fourier Transform we know that:
t
ωT
rect
→ T sinc
T
2
Applying equation (10) to 21 rect 2t , we thus get:
t
1
rect
= sinc(ω)
2
2
Applying now the frequency shift property of the Fourier Transform, we get:
1
t
ejπt rect
→ sinc(ω − π)
2
2
1
t
e−jπt rect
→ sinc(ω + π)
2
2
Lastly, we apply the time shift property of the Fourier Transform:
t+2
t+2
1
1
+ e−jπ(t+2) rect
x(t) = ejπ(t+2) rect
2
2
2
2
t+2
= cos(π(t + 2))rect
2
(10)
(11)
(12)
(13)
Problem 5 – Frequency Response
Consider a linear time-invariant system with impulse response:
h(t) = 5sinc(5t − 10)
Find the system output if the input is x(t) =
sin( 5t )
.
t
Solution:
Based on the table of Fourier Transform, and the time shift property, we can find the Fourier transform
of h(t) to be equal to:
ω
H(jω) = πrect
e−jω2
(14)
10
The system is therefore a low-pass filter with cut-off frequency ω = 5, gain π and linear phase −2ω.
sin( t )
The Fourier transform of input signal x(t) = t 5 = 15 sinc 5t can now be found to be:
ω
X(jω) = πrect
.
2/5
The Fourier transform of the output of the system can now be found as:
Y (ω) = X(jω)H(jω) = π 2 rect(ω/0.4)e−jω2
(15)
Applying the inverse Fourier transform, we get:
y(t) = 0.2πsinc(0.2(t − 2))
3
(16)