EE3414
Homework #1 Solution
1. What are the magnitude, frequency, and phase of the following sinusoidal signals?
a) 10sin (20πt+π/2)
b) 10cos (10πt+π/4)
Solution:
Magnitude (A)
a)
b)
10
10
Frequency ( f 0 )
10
5
Phase (  )
0
π/4
Notes:
a) 10sin( 20πt+π/2) = 10cos(20πt+0)
2. What are fundamental frequency and Fourier series coefficients of the following signal (Hint:
You don’t need to calculate any integral!). Illustrate the line spectrum of the signal.
s(t )  10sin(20 t )  2 cos(40 t   / 3)  0.5sin(80 t   / 4)
Solution:




s (t )  10 cos(20 t  )  2 cos(40 t  )  0.5cos(80 t   )
2
3
4 2



 0  10 cos(2 10  t  )  2 cos(2  20  t  )  0.5cos(2  40  t  )
2
3
4
The fundamental frequency: f 0 =10
Fourier series coefficients (note this solution is in terms of single sided Fourier series
representation):
A0  0, A1  10,1   / 2, A2  2,2   / 3, A3  0, A4  0.5,4   / 4, S k  0, k  4.
To obtain the two-sided Fourier series representation, we can write, using Euler formula
1
1
S(t)  0  5e j / 2e j 2 (10)t  5e j / 2e j 2 (10)t  e j /3e j 2 (20)t  e  j /3e  j 2 (20)t  e  j / 4e j 2 (40)t  e j / 4e  j 2 (40)t
4
4
Thus the two-sided Fourier series coefficients are:
1
1
S0  0, S1  5e  j / 2 , S 1  5e j / 2 , S2  e j / 3 , S 2  e  j / 3 , S 4  e  j / 4 , S 4  e j / 4
4
4
The two-sided line spectrum of the signal is as follows (line height only indicates the
magnitude):
5e j / 2
e j / 3
e  j / 3
1 j / 4
e
4
-40
5e  j / 2
-30
-20
-10
0
10
20
1  j / 4
e
4
30
40
f
3. Determine the Fourier series expansion (i.e. all the expansion coefficients) for the following
periodic triangular wave, and illustrate the line spectrum. Compared to the line spectrum of the
square wave signal derived in the lecture, which signal has a wider bandwidth?
Solution:
1
2
Using the double-sided Fourier series transform formula, we have
Fundamental period T0  2 , and fundamental frequency f 0 
T /2
S0 
0
1
1 0
1
1
1
s (t )dt   (1  t )dt   (1  t )dt 

T0 T0 / 2
2 1
20
2
T /2
1 0
Sk 
s (t ) exp(  j 2kf0t )dt
T0 T0 / 2
 1/ 2
k 0
0
1
 2
1
1
  (1  t ) exp(  j 2kf0t )dt   (1  t ) exp(  j 2kf0t )dt   2 2 k  odd
2 1
20
 k  k  even
 0
The two-sided line spectrum is as follows
Line Spectrum of a Triangular Wave with T0=2
0.7
0.6
0.5
k
|s |
0.4
0.3
0.2
0.1
0
-5
-4
-3
-2
-1
0
f
1
2
3
4
5
On the other hand, the square wave example in the class has the following double-sided Fourier
series coefficients:
S0  0
 2

S k   jk
 0
k  odd
k  even
Its line spectrum is shown below
Line Spectrum of a Square Wave with T0=1
0.7
0.6
0.5
k
|s |
0.4
0.3
0.2
0.1
0
-5
-4
-3
-2
-1
0
f
1
2
3
4
5
Comparing the two figures, we can see that the square wave has a wider bandwidth. For
example, if we use a magnitude threshold of 0.1
For square wave:
f min  1, f max  5, B  4
For above triangular wave:
f min  0, f max  0.5, B  0.5
The square wave has much sharper transition than the triangular wave. This sharp transition
requires many high frequency components to synthesize. Consequently, the square wave has a
wider bandwidth.
4. Determine the Fourier transform of the following signal and illustrate the magnitude
spectrum of the signal. Compare to the spectrum of the pulse signal derived in the lecture,
which signal has a wider bandwidth?
S(t)
1
-1
1
Solution:
Recall Fourier analysis, we have

0
1

1
0
S  f    s  t  exp   j 2 ft  dt   1  t  exp   j 2 ft  dt   1  t  exp   j 2 ft  dt
After calculation, we get
2  e j 2 f  e j 2 f 1  cos 2 f sin 2  f
S f 


 sin c 2  f 
2
2 2
2 2
4 f
2 f
 f 
The magnitude spectrum of this signal is:
magnitude of spectrum
1
0.9
0.8
0.7
|S(f)|
0.6
0.5
0.4
0.3
0.2
0.1
0
-10
-8
-6
-4
-2
0
f
2
4
6
8
10
The comparison of the two signal spectrums is shown as following:
1
spectrum of pulse function
spectrum of triangular wave
threshold
0.8
0.6
0.4
0.2
0
-0.2
-0.4
-10
-8
-6
-4
-2
0
2
4
6
8
10
We can see that the pulse signal derived in the lecture has a wider bandwidth. For a magnitude
threshold of 0.1, the triangular wave has a bandwidth less than 1 (indicated by the green line),
whereas the pulse wave has a bandwidth slightly less than 3 (indicated by the orange line). The
reason is the same as their periodic counter part. The pulse signal has a very sharp transition,
and correspondingly higher frequency.
5.
The figure below is part of a speech waveform. As can be seen, it has a nearly periodic
structure, and within each period, there exists many fine ripples. The frequency
associated with the basic period is called the pitch, the shortest fine ripple determines the
highest frequency in the signal. Estimate its fundamental period by determining the
length of a basic cycle. Also, estimate the maximum frequency by determining the
length of a shortest ripple.
Note that this signal is a discrete time signal obtained by sampling a continuous time signal.
The sampling frequency used was f_s=22050Hz. The horizontal axis of the plot indicates
the sample index k. Hence the actual time interval between two indices is the sampling
interval =1/f_s=1/22050 sec.
Blown-Up of the Signal
A Short Section of a Speech Signal
0.04
0.04
0.03
0.03
0.02
0.02
0.01
0.01
0
-0.01
0
-0.02
-0.01
-0.03
-0.04
2000
2100
2200
2300 2400 2500 2600 2700
k, t=k  ,  =1/fs ,fs =22050Hz
2800
2900
3000
-0.02
2450
2460
2470
2480
2490
2500
2510
2520
2530
2540
2550
Solution:
Looking at the waveform plot (left), the first 4 cycle occupy 300 samples. Therefore, each cycle
takes about 75 samples. The fundamental period is the time corresponding to 1 cycle,
therefore T0  75*(1/22050)=3.40e-3 sec. The pitch frequency (i.e. fundamental frequency) is
f0  1  1
 294 Hz
T0
3.40e  3
There are several very narrow ripples (see the blown-up plot on the right). For example, between
2490 and 2500, there are two narrow ripples spanning approximately 6 samples. So the shortest
ripple takes about 3 samples, with time Tmin  3 / 22050  1.33E  4 sec. The corresponding
maximum frequency is f max  1 / Tmin  7518 Hz.