4.12 Exercises
4.1. Using the DTFT
of the -samples
rectangular window given in Eq. (4.7), prove that the DTFTs of the Hann and
Hamming windows, given in Eq. (4.4) and Eq. (4.3), are
and ,
respectively.
4.2. A cosine
is multiplied by a rectangular window of
samples, from
and the windowed signal
has its spectrum estimated by an FFT with
points. The FFT resolution is
rad. Using the DTFT of ,
what are the FFT values for
assuming: a)
is centered in the fourth bin and b)
is half-way the third and fourth bins?
4.3. Given the signal
in volts, what are its: a) total signal energy and b) energy within the frequency band
Hz?
4.4. What is the MS spectrum
of
volts? Using the values of ,
how can one obtain the average power of
in watts?
4.5. Assume
is the -points
FFT of a periodic signal
of period
samples. When ,
is .
Inform: a) the estimated MS spectrum
of ,
b) the signal power
and c) the new
in case
samples of
were used, together with a 16-points FFT.
4.6. Assume
is the -points
FFT of a discrete-time signal .
What are the values of its: a) DTFS, b) MS spectrum ,
c) the periodogram
(using Matlab/Octave convention of ),
d) the estimated PSD
and e) the signal power .
4.7. a) What is the PSD
of a complex exponential ,
with rad
after multiplication by a rectangular window of
samples? b) When the periodogram
of this signal is estimated with a FFT of
points, what are the values of
and ?
4.8. The periodogram of a sinusoid immersed in AWGN was calculated with an
-points
FFT as
in watts/Hz, assuming Hz.
The noise and the sinusoid are uncorrelated, such that, at the sinusoid bins, the power
is the sum of the sinusoid power and the noise power at that bin. Inform: a) the
sinusoid average power, b) the noise average power, c) the SNR in dB.
4.9. The bilateral PSD of a continuous-time white noise signal
is W/Hz.
This signal was digitized using an ideal lowpass filter and kHz,
creating a discrete-time signal .
Inform: a) the average power of ,
b) the average power of
and c) the power corresponding to a single periodogram bin of a windowed
estimated with an FFT of length
when kHz
and
are informed.
4.10. Assuming Hz,
the result of Welch’s method with 8-length FFTs for a real signal composed by a sum of two
sinusoids was S = [0,100,0,20,0] in dBW/Hz. Inform: a) the sinusoid frequencies, b) the
sinusoid powers in watts and c) their power ratio in dB. Hint: one can use Matlab/Octave to
investigate this setup with the commands:
1N=8; n=0:N-1; x1=4*cos(pi/4*n); x2=10*cos(3*pi/4*n); x=x1+x2; 2Fs=100; [S,f]=pwelch(x,rectwin(N),0,N,Fs), S*f(2), SdB=10*log10(S)
4.11. A signal has its PSD estimated via AR modeling with the result: A=[1.0, -1.8151, 0.9025] and Perror=4 (for example, with the Matlab command [A,Perror]=lpc(x,2)). What is the frequency of the peak of this PSD and its amplitude?
4.12. An autoregressive model of order one was obtained with the command [A,Perror]=lpc(x,1)), where A=[1, -0.75] and Perror=0.04 watts. a) What is the expression for assuming that incorporates the information from Perror? b) What is the expression for the PSD corresponding to this model assuming Hz (the expression must depend only on )? c) What is the value of at frequency Hz?
4.13. The unilateral PSD of a continuous-time white noise signal is W/Hz. The signal is the input to a lowpass filter with real-valued coefficients, gain and zero phase within the passband from 0 to 200 Hz (this filter is not realizable). The filter output is converted to a discrete-time signal using a C/D process with sampling frequency kHz. The average power values of and are the same and denoted as . Inform: a) the value in watts, b) the detailed graphs of the PSDs , , and corresponding to the signals , and , respectively. Note that is not a discrete-time white noise, because its PSD does not have a constant value over the range .
4.14. The periodogram of a signal with samples is estimated via Welch’s method using a rectangular window with samples. The sampling frequency is Hz. The window shift is samples, such that there is no overlapping among windows, and the samples of are organized into four blocks of samples each. After zero-padding, each block of samples is converted to frequency domain using an FFT of points. a) What is the frequency spacing in Hertz between neighboring periodogram bins? b) What is the frequency resolution in Hertz imposed by the DTFT of the windows ? Assume that is the range between the two zeros of that define its main lobe, where is the DTFT converted to continuous-time using . c) Is the overall frequency resolution limited by or ? d) Why is it that someone cannot consistently improve the overall resolution in spectral analysis by simply using zero-padding and improving by using a larger number of FFT points?