BIOEN 316, Spring 2013    Name:_______________________________   

Quiz 1 practice : Fourier series and transforms 
 
1. If we amplify an ECG signal before digitizing it, which one of these we will probably 
improve?  
a.  Frequency resolution  c. Voltage resolution 
b.  Time resolution  d. Angular resolution 
Time resolution = 1/fs, frequency resolution = 1/(time over which DFT is calculated), voltage
resolution is (input voltage range) / 2b where b is number of bits in the A-to-D converter.
Amplifying increases the input voltage range. Angular resolution relates more to telescopes.

2. What should be the result if the least significant bit in our A‐to‐D converter is always zero?   
a.  It will make us sample at random times   c. It will amplify the signal 
b.  It will make our voltage resolution  d. It will make the frequency resolution 
coarser  periodic 
3. If the fastest oscillations that we want to measure are at 120 Hz, which of the following is 
the most reasonable sampling rate?  
a.  60 Hz   c. 60 kHz  
b.  anything over 0.833 Hz   d. 250 Hz 
We must sample more than twice as fast as the fastest oscillation in the measured signal.
Otherwise we will not be able to tell the difference between true low-frequency signals and their
aliases.

4. A good representation of the derivative of a step function u(t) is… 
a.  another step function  c. a ramp, mt, where m is the slope.  
b.  a complex exponential  d. a delta function δ(t)
At t=0, the step function transitions instantly from 0 to 1 so its derivative is infinity exactly at t=0
and zero everywhere else – just like a Dirac delta function.

5. It is most likely that the frequency domain representation of a unit step function would be… 
a.  All real  c. A delta function δ(t) plus an imaginary function of ω 
b.  A constant for ω > 0  d. An alias of itself
By subtracting the constant 0.5 from a step function we obtain an odd function that is –0.5 for t<0
and 0.5 for t>0. Odd functions have imaginary (and odd) Fourier transforms.

6. The administration of painkillers can be set up so that the patient causes a bolus to be 
injected by pressing a button.  We expect this to happen less often while sleeping than 
when awake.  A good mathematical description of these injections might be… 
a.  a series of hamming windows   c. an aperiodic sequence of Dirac delta functions 
b.  the discrete Fourier transform  d. (b) or (c), depending on the rate of injection 
BIOEN 316, Spring 2013    Name:_______________________________   

Originally my answer was C, but the shape of the pulse really depends on how the machine works. It
would be aperiodic if the time between injections varies during the day. This question is too vague
to put on a real quiz, but it shows how we might use delta functions as the input to a
pharmacokinetic system.

7. One advantage of using decibels to measure or display the magnitude spectrum of a signal 
is that…  
a.  It eliminates side lobes from sinc(ω)  c. It plots the frequency axis on a log scale 
b.  It helps us tell the difference between  d. It makes the magnitude periodic 
very small magnitudes  
Orders of magnitude become linear steps. It would be impossible to tell the difference between
0.01 and 0.0001 on a linear scale, but in decibels it becomes easy. There is no minimum value in
decibels so we can easily distinguish numbers as small as we want.

8. It is typical to take measurements hourly over a multiple 24‐hour periods (or at least as long 
as the subject can stay awake) to demonstrate the daily fluctuation in body temperature.   
How can we be sure that there are no fluctuations occurring faster than ½ cycle per hour?   
a.  Our hourly measurements reveal a  c. We can’t be sure about any frequencies 
fundamental frequency of one day   that are faster than half the sampling rate  
b.  We sleep only once a day  d. The combination of (a) and (b) 
We can get some certainty by sampling very fast to see what the maximum frequency actually is,
then using that information to decide how fast to sample when collecting more data later.
Sometimes we can do calculations to find theoretical limits on how fast some variable like
temperature can change.

9. If a continuous (not sampled) time‐domain signal is real (not complex), its magnitude 
spectrum should be symmetric… 
a.  Around the horizontal (frequency) axis   c. with odd symmetry 
b.  Around f = 0, f = ±fs and f = ±fs/2  d. around f = 0 only 
The magnitude spectrum cannot be odd or symmetric about the horizontal axis because it (or parts
of it) would be negative. There is no sampling frequency fs if it has not been sampled. It does have
even symmetry.

10. If a sampled (discrete) time‐domain signal is real (not complex), its magnitude spectrum 
should be symmetric… 
a.  Around the horizontal (frequency) axis   c. with odd symmetry 
b.  Around f = 0, f = ±fs and f = ±fs/2  d. around f = 0 only 
Due to periodicity and aliasing, it is symmetric about any integer multiple of fs /2.
   
BIOEN 316, Spring 2013    Name:_______________________________   

11. To change from a frequency resolution of 1 Hz to a resolution of 0.2 Hz, what could we do?   
a.  Sample at 200 Hz instead of 1 kHz   c. Use a DFT operation that covers 5 seconds 
of the signal instead of 1 second 
b.  Average every 5 samples  d. Sample every 5 seconds instead of every 1 
second 
The term “frequency resolution” applies best to the output of the discrete Fourier transform. It is
the difference between the frequencies of two adjacent points along the frequency axis, i.e. Δf =
mfs/N – (m–1) fs/N = fs/N = 1/NΔt = 1/T where T is the amount of time over which the DFT is
calculated. We can get more time in the DFT by sampling longer, or we can add zeros at the end of
the signal samples (zero padding). The result of zero padding is not as good as sampling longer, but
sometimes we can’t go back and collect a longer signal.

12. The body temperature is known to vary over each 24‐hour day, higher in the afternoon and 
lower in the morning.  What is the absolute maximum time between measurements that 
would allow us to see that there is a daily oscillation in temperature?   
a.  Two hours  c. Twenty‐four hours 
b.  Twelve hours  d. Forty‐eight hours 
Actually, 12 hours isn’t really good enough because it might just hit all of the average values and
miss all the oscillations. It’s the best out of these options, however.

13. The ECG is almost a periodic signal, but not quite.  Suppose we sampled an actual heartbeat 
over 20 seconds when the heart was beating at 60 beats per minute.  How could we modify 
our stored signal so that each heart beat appears identical to the others?   
a.  Resample the signal in time, slower than  c. Apply DFT, sample the signal in frequency, 
before, e.g. discard 19 out of every 20  (i.e. set 19 out of each 20 points equal to 
samples.    zero), then use inverse DFT  
b.  Resample the signal faster in time, by  d. No need; the implied periodicity of the DFT 
interpolating between each two original  forces all of the beats to appear the same. 
samples. 
Deleting 19 out of every 20 points destroys the information that tells how each heartbeat differs
from the others. As for (d), the implied periodicity means that an infinite signal would be composed
of identical 20-second periods, but it says nothing about periodicity within the captured signal.

14. The following expression represents a periodic function as the complex form of a Fourier 
series.  Find the numerical values of the complex coefficients C1 and C2.  
1 1
v(t )   e j3nt , n  0  
2
3  n  2

I get C1 = 1/(3pi^2) = 0.034 and C2 = 1/(12pi^2) = 0.00844.

15.    
   
BIOEN 316, Spring 2013    Name:_______________________________   

16. Convert the Fourier series from the previous problem into the trigonometric form, i.e. a 
sum of an cos(nω0) and bn sin(nω0) terms.  Find the numerical value of a0, a2 and b2.   
There is no constant term, so a0 = 0. The coefficients are all real, so the signal is all even. Even
functions have no sine component, so bn = 0. Then, two Cn’s (each pair of positive and negative n)
add to make each cosine, so an = 2 Cn; a2 = 0.0169.

 
The following plot shows one part of a function v1(t), which is periodic with a period of 6 
seconds, shown by the arrows.  Each segment has the function v(t) = t2, and the entire plot is 
centered at t = 0.   
10
8
6
4
2
0
 
Suppose that this function is represented by the first Fourier series below.  Answer each of 
the questions below about the second Fourier series, v2(t).  
1 
9 9
a.   v(t )  3   cos(n )e jnt / 3   cos(n )e jnt / 3  
2 2 2 2
 n  1 n 
1 
9j  jnt / 3 9j  jnt / 3
b.   v2(t )  3   cos(n )e  cos(n )e  
2 2 2 2
 n  1 n 
 
17. The negative sign in the exponent of v2(t) will cause…  
a.  No effect    c. The phase to be shifted by ±π 
b.  the signal to be inverted, such that each  d. v2 to be imaginary 
period makes an inverted parabola 
The n’s cover all of the positive and negative integers, so either way, half of the exponents will be
positive and half negative.

18. The shape of v2 will be…  
a.  Basically the same as v1 but shifted   b. Quite different 
Explain your answer. 
This is an interesting case in which the magnitude spectrum has even symmetry, but the time-
domain signal is complex. Multiplying all of the terms except C0 by j is the same as multiplying the
periodic parabola by j, so v2(t) is a real constant plus a purely imaginary function. I do not plan to
put this question on the quiz.